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A First Course in Mathematical Analysis
Mathematical Analysis (often called Advanced Calculus) is generally found by students
to be one of their hardest courses in Mathematics. This text uses the so-called sequential
approach to continuity, differentiability and integration to make it easier to understand
the subject.
Topics that are generally glossed over in the standard Calculus courses are given
careful study here. For example, what exactly is a ‘continuous’ function? And how
exactly can one give a careful definition of ‘integral’? This latter is often one of the
mysterious points in a Calculus course – and it is quite tricky to give a rigorous
treatment of integration!
The text has a large number of diagrams and helpful margin notes, and uses many
graded examples and exercises, often with complete solutions, to guide students
through the tricky points. It is suitable for self study or use in parallel with a standard
university course on the subject.
A First Course inMathematical
Analysis
D A V I D A L E X A N D E R B R A N N A N
Published in association with The Open University
CAMBRIDGE UNIVERSITY PRESS
Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo
Cambridge University PressThe Edinburgh Building, Cambridge CB2 8RU, UK
First published in print format
ISBN-13 978-0-521-86439-8
ISBN-13 978-0-511-34857-0
© The Open University 2006
2006
Information on this title: www.cambridge.org/9780521864398
This publication is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.
ISBN-10 0-511-34857-6
ISBN-10 0-521-86439-9
Cambridge University Press has no responsibility for the persistence or accuracy of urls for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.
Published in the United States of America by Cambridge University Press, New York
www.cambridge.org
hardback
eBook (EBL)
eBook (EBL)
hardback
To my wife Margaret
and my sons David, Joseph and Michael
Contents
Preface page ix
1 Numbers 1
1.1 Real numbers 2
1.2 Inequalities 9
1.3 Proving inequalities 14
1.4 Least upper bounds and greatest lower bounds 22
1.5 Manipulating real numbers 30
1.6 Exercises 35
2 Sequences 37
2.1 Introducing sequences 38
2.2 Null sequences 43
2.3 Convergent sequences 52
2.4 Divergent sequences 61
2.5 The Monotone Convergence Theorem 68
2.6 Exercises 79
3 Series 83
3.1 Introducing series 84
3.2 Series with non-negative terms 92
3.3 Series with positive and negative terms 103
3.4 The exponential function x 7! ex 122
3.5 Exercises 127
4 Continuity 130
4.1 Continuous functions 131
4.2 Properties of continuous functions 143
4.3 Inverse functions 151
4.4 Defining exponential functions 161
4.5 Exercises 164
5 Limits and continuity 167
5.1 Limits of functions 168
5.2 Asymptotic behaviour of functions 176
5.3 Limits of functions – using " and � 181
5.4 Continuity – using " and � 193
5.5 Uniform continuity 200
5.6 Exercises 203
vii
6 Differentiation 205
6.1 Differentiable functions 206
6.2 Rules for differentiation 216
6.3 Rolle’s Theorem 228
6.4 The Mean Value Theorem 232
6.5 L’Hopital’s Rule 238
6.6 The Blancmange function 244
6.7 Exercises 252
7 Integration 255
7.1 The Riemann integral 256
7.2 Properties of integrals 272
7.3 Fundamental Theorem of Calculus 282
7.4 Inequalities for integrals and their applications 288
7.5 Stirling’s Formula for n! 303
7.6 Exercises 309
8 Power series 313
8.1 Taylor polynomials 314
8.2 Taylor’s Theorem 320
8.3 Convergence of power series 329
8.4 Manipulating power series 338
8.5 Numerical estimates for p 346
8.6 Exercises 350
Appendix 1 Sets, functions and proofs 354
Appendix 2 Standard derivatives and primitives 359
Appendix 3 The first 1000 decimal places offfiffiffi
2p
, e and p 361
Appendix 4 Solutions to the problems 363
Chapter 1 363
Chapter 2 371
Chapter 3 382
Chapter 4 393
Chapter 5 402
Chapter 6 413
Chapter 7 426
Chapter 8 443
Index 457
viii Contents
Preface
Analysis is a central topic in Mathematics, many of whose branches use
key analytic tools. Analysis also has important applications in Applied
Mathematics, Physics and Engineering, where a good appreciation of the
underlying ideas of Analysis is necessary for a modern graduate.
Changes in the school curriculum over the last few decades have resulted
in many students finding Analysis very difficult. The author believes that
Analysis nowadays has an unjustified reputation for being hard, caused by
the traditional university approach of providing students with a highly polished
exposition in lectures and associated textbooks that make it impossible for
the average learner to grasp the core ideas. Many students end up agreeing with
the German poet and philosopher Goethe who wrote that ‘Mathematicians are
like Frenchmen: whatever you say to them, they translate into their own
language, and forthwith it is something entirely different!’
Since 1971, the Open University in United Kingdom has taught Mathematics
to students in their own homes via specially written correspondence texts, and
has traditionally given Analysis a central position in its curriculum. Its philo-
sophy is to provide clear and complete explanations of topics, and to teach these
in a way that students can understand without much external help. As a result,
students should be able to learn, and to enjoy learning, the key concepts of the
subject in an uncluttered way. This book arises from correspondence texts for
its course Introduction to Pure Mathematics, that has now been studied suc-
cessfully by over ten thousand students.
This book is therefore different from most Mathematics textbooks! It adopts
a student-friendly approach, being designed for study by a student on their own
OR in parallel with a course that uses as set text either this text or another text.
But this is the text that the student is likely to use to learn the subject from. The
author hopes that readers will gain enormous pleasure from the subject’s
beauty and that this will encourage them to undertake further study of
Mathematics!
Once a student has grasped the principal notions of limit and continuous
function in terms of inequalities involving the three symbols ", X and �, they
will quickly understand the unity of areas of Analysis such as limits, continu-
ity, differentiability and integrability. Then they will thoroughly enjoy the
beauty of some of the arguments used to prove key theorems – whether their
proofs are short or long.
Calculus is the initial study of limits, continuity, differentiation and integra-
tion, where functions are assumed to be well-behaved. Thus all functions
continuous on an interval are assumed to be differentiable at most points in
the interval, and so on. However, Mathematics is not that simple! For example,
there exist functions that are continuous everywhere on R , but differentiable
Johann Wolfgang von Goethe(1749–1832) is said to havestudied all areas of science ofhis day except mathematics –for which he had no aptitude.
ix
nowhere on R ; this discovery by Karl Weierstrass in 1872 caused a sensation in
the mathematical community. In Analysis (sometimes called Advanced
Calculus) we make no assumptions about the behaviour of functions – and
the result is that we sometimes come across real surprises!
The book has two principal features in its approach that make it stand out
from among other Analysis texts.
Firstly, this book uses the ‘sequential approach’ to Analysis. All too often
students starting on the subject find that they cannot grasp the significance of
both " and � simultaneously. This means that the whole underlying idea about
what is happening is lost, and the student takes a very long time to master the
topic – or, in many cases in fact, never masters the topic and acquires a strong
dislike of it. In the sequential approach they proceed at a more leisurely pace to
understand the notion of limit using " and X – to handle convergent sequences –
before coming across the other symbol �, used in conjunction with " to handle
continuous functions. This approach avoids the conventional student horror at
the perceived ‘difficulty of Analysis’. Also, it avoids the necessity to re-prove
broadly similar results in a range of settings – for example, results on the sum
of two sequences, of two series, of two continuous functions and of two
differentiable functions.
Secondly, this book makes great efforts to teach the "–� approach too. After
students have had a first pass at convergence of sequences and series and at
continuity using ‘the sequential approach’, they then meet ‘the "–� approach’,
explained carefully and motivated by a clear ‘"–� game’ discussion. This
makes the new approach seem very natural, and this is motivated by using
each approach in later work in the appropriate situation. By the end of the book,
students should have a good facility at using both the sequential approach and
the "–� approach to proofs in Analysis, and should be better prepared for later
study of Analysis than students who have acquired only a weak understanding
of the conventional approach.
Outline of the content of the book
In Chapter 1, we define real numbers to be decimals. Rather than give a heavy
discussion of least upper bound and greatest lower bound, we give an intro-
duction to these matters sufficient for our purposes, and the full discussion is
postponed to Chapter 7, where it is more timely. We also study inequalities,
and their properties and proofs.
In Chapter 2, we define convergent sequences and examine their proper-
ties, basing the discussion on the notion of null sequence, which simplifies
matters considerably. We also look at divergent sequences, sequences
defined by recurrence formulas and particular sequences which converge
to p and e.
In Chapter 3, we define convergent infinite series, and establish a number of
tests for determining whether a given series is convergent or divergent. We
demonstrate the equivalence of the two definitions of the exponential function
x 7! ex, and prove that the number e is irrational.
In Chapter 4, we define carefully what we mean by a continuous function, in
terms of sequences, and establish the key properties of continuous functions.
We also give a rigorous definition of the exponential function x 7! ax.
We define ex as limn!1
1þ xn
� �n
and asP
1
n¼0
xn
n!.
x Preface
In Chapter 5, we define the limit of a function as x tends to c or as x tends
to 1 in terms of the convergence of sequences. Then we introduce the "–�definitions of limit and continuity, and check that these are equivalent to the
earlier definitions in terms of sequences. We also look briefly at uniform
continuity.
In Chapter 6, we define what we mean by a differentiable function, using
difference quotients Q(h); this enables us to use our earlier results on limits to
prove corresponding results for differentiable functions. We establish some
interesting properties of differentiable functions. Finally, we construct the
Blancmange function that is continuous everywhere on R , but differentiable
nowhere on R .
In Chapter 7, we give a careful definition of what we mean by an integrable
function, and establish a number of related criteria for establishing whether
a given function is integrable or not. Our integral is the so-called Riemann
integral, defined in terms of upper and lower Riemann sums. We check the
standard properties of integrals and verify a number of standard approaches for
calculating definite integrals. Then we give a number of applications of
integrals to limits of certain sequences and series and prove Stirling’s Formula.
Finally, in Chapter 8, we study the convergence and properties of power
series. The chapter ends with a marvellous proof of the irrationality of the
number p that uses a whole range of the techniques that have been met in the
previous chapters.
For completeness and for students’ convenience, we give a brief guide to our
notation for sets and functions, together with a brief indication of the logic
involved in proofs in Mathematics (in particular, the Principle of Mathematical
Induction) in Appendix 1. Appendix 2 contains a list of standard derivatives
and primitives and Appendix 3 the first 1000 decimal places in the values of the
numbersffiffiffi
2p
, p and e. Appendix 4 contains full solutions to all the problems set
during each chapter.
Solutions are not given to the exercises at the end of each chapter, however.
Lecturers/instructors may wish to use these exercises in homework assignments.
Study guide
This book assumes that students have a fair understanding of Calculus. The
assumptions on technical background are deliberately kept slight, however, so
that students can concentrate on the newer aspects of the subject ‘Analysis’.
Most students will have met some of the material in the early chapters
previously. Although this means that they can therefore proceed fairly quickly
through some sections, it does NOT mean that those sections can be ignored –
each section contains important ideas that are used later on and most include
something new or have a different emphasis.
Most chapters are divided into five or six sections (each often further divided
into sub-sections); sections are numbered using two digits (such as ‘Section
3.2 ’) and sub-sect ions using thr ee digits (such as ‘Sub- section 3.2.4 ’).
Generally a section is considered to be about one evening’s work for an
average student.
Chapter 7 on Integration is arguably the highlight of the book. However,
it contains some rather complicated mathematical arguments and proofs.
Stirling’s Formula says that,for large n, n! is ‘roughly’ffiffiffiffiffiffiffiffi
2pnp
ne
� �n, in a sense that we
explain.
Preface xi
Therefore, when reading Chapter 7, it is important not to get bogged down in
details, but to keep progressing through the key ideas, and to return later on to
reading the things that were left out at the earlier reading. Most students will
require three or four passes at this chapter before having a good idea of most of it.
We use wide pages with a large number of margin notes in which we place
teaching comments and some diagrams to aid in the understanding of particular
points in arguments. We also provide advice on which proofs to omit on a first
study of the topic; it is important for the student NOT to get bogged down in
a technical discussion or a proof until they have a good idea of the message
contained in the result and the situations in which it can be used. Therefore
clear encouragement is given on which portions of the text to leave till later, or
to simply skim on a first reading.
The end of the proof of a Theorem is indicated by a solid symbol ‘&’ and the
end of the solution of a worked Example by a hollow symbol ‘&’. There are
many worked examples within the text to explain the concepts being taught,
together with a good stock of problems to reinforce the teaching. The solutions
are a key part of the teaching, and tackling them on your own and then reading
our version of the solution is a key part of the learning process.
No one can learn Mathematics by simply reading – it is a ‘hands on’ activity.
The reader should not be afraid to draw pictures to illustrate what seems to be
happening to a sequence or a function, to get a feeling for their behaviour. A
wise old man once said that ‘A picture is worth a thousand words!’. A good
picture may even suggest a method of proof. However, at the same time it is
important not to regard a picture on its own as a proof of anything; it may
illustrate just one situation that can arise and miss many other possibilities!
It is important NOT to become discouraged if a topic seems difficult. It took
mathematicians hundreds of years to develop Analysis to its current polished
state, so it may take the reader a few hours at several sittings to really grasp the
more complex or subtle ideas.
Acknowledgements
The material in the Open University course on which this book is based was
contributed to in some way by many colleagues, including Phil Rippon, Robin
Wilson, Andrew Brown, Hossein Zand, Joan Aldous, Ian Harrison, Alan Best,
Alison Cadle and Roberta Cheriyan. Its eventual appearance in book form
owes much to Lynne Barber.
Without the forebearance of my family, the writing of the book would have
been impossible.
It is important to read themargin notes!
This signposting benefitsstudents greatly in theauthor’s experience.
Tackling the problems is agood use of your time, notsomething to skimp.
xii Preface
1 Numbers
In this book we study the properties of real functions defined on intervals of the
real line (possibly the whole real line) and whose image also lies on the real
line. In other words, they map R into R . Our work will be from a very precise
point of view in order to establish many of the properties of such functions
which seem intuitively obvious; in the process we will discover that some
apparently true properties are in fact not necessarily true!
The types of functions that we shall examine include:
exponential functions, such as x 7! ax;where a; x 2 R ;
trigonometric functions, such as x 7! sin x;where x 2 R ;
root functions, such as x 7!ffiffiffi
xp;where x � 0:
The types of behaviour that we shall examine include continuity, differ-
entiability and integrability – and we shall discover that functions with these
properties can be used in a number of surprising applications.
However, to put our study of such functions on a secure foundation, we need
first to clarify our ideas of the real numbers themselves and their properties. In
particular, we need to devote some time to the manipulation of inequalities,
which play a key role throughout the book.
In Section 1.1, we start by revising the properties of rational numbers and
their decimal representation. Then we introduce the real numbers as infinite
decimals, and describe the difficulties involved in doing arithmetic with such
decimals.
In Section 1.2, we revise the rules for manipulating inequalities and show
how to find the solution set of an inequality involving a real number, x, by
applying the rules. We also explain how to deal with inequalities which involve
modulus signs.
In Section 1.3, we describe various techniques for proving inequalities,
including the very important technique of Mathematical Induction.
The concept of a least upper bound, which is of great importance in
Analysis, is introduced in Section 1.4, and we discuss the Least Upper
Bound Property of R .
Finally, in Section 1.5, we describe how least upper bounds can be used to
define arithmetical operations in R .
Even though you may be familiar with much of this material we recommend
that you read through it, as we give the system of real numbers a more careful
treatment than you may have met before. The material on inequalities and least
upper bounds is particularly important for later on.
In later chapters we shall define exactly what the numbers p and e are, and
find various ways of calculating them. But, first, we examine numbers in
general.
For example, what exactly isthe number
ffiffiffi
2p
?
You may omit this sectionat a first reading.
1
1.1 Real numbers
We start our study of the real numbers with the rational numbers, and investi-
gate their decimal representations, then we proceed to the irrational numbers.
1.1.1 Rational numbers
We assume that you are familiar with the set of natural numbers
N ¼ f1; 2; 3; . . .g;and with the set of integers
Z ¼ f. . .; �2; �1; 0; 1; 2; 3; . . .g:The set of rational numbers consists of all fractions (or ratios of integers)
Q ¼ p
q: p 2 Z; q 2 N
� �
:
Remember that each rational number has many different representations as a
ratio of integers; for example
1
3¼ 2
6¼ 10
30¼ . . .:
We also assume that you are familiar with the usual arithmetical operations of
addition, subtraction, multiplication and division of rational numbers.
It is often convenient to represent rational numbers geometrically as points
on a number line. We begin by drawing a line and marking on it points
corresponding to the integers 0 and 1. If the distance between 0 and 1 is
taken as a unit of length, then the rationals can be arranged on the line with
positive rationals to the right of 0 and negative rationals to the left.
–3 –2 –1 0 1 2 332
12
52
–
negative positive
For example, the rational 32
is placed at the point which is one-half of the way
from 0 to 3.
This means that rationals have a natural order on the number-line. For
example, 1922
lies to the left of 78
because
19
22¼ 76
88and
7
8¼ 77
88:
If a lies to the left of b on the number-line, then
a is less than b or b is greater than a;
and we write
a < b or b > a:
For example
19
22<
7
8or
7
8>
19
22:
We write a� b, or b� a, if either a< b or a¼ b.
Note that 0 is not a naturalnumber.
2 1: Numbers
Problem 1 Arrange the following rational numbers in order:
0; 1;�1; 1720;�17
20; 45
53;�45
53:
Problem 2 Show that between any two distinct rational numbers there
is another rational number.
1.1.2 Decimal representation of rational numbers
The decimal system enables us to represent all the natural numbers using only
the ten integers
0; 1; 2; 3; 4; 5; 6; 7; 8 and 9;
which are called digits. We now remind you of the basic facts about the
representation of rational numbers by decimals.
Definition A decimal is an expression of the form
�a0 � a1a2a3 . . .;
where a0 is a non-negative integer and a1, a2, a3, . . . are digits.
If only a finite number of the digits a1, a2, a3, . . . are non-zero, then the
decimal is called terminating or finite, and we usually omit the tail of 0s.
Terminating decimals are used to represent rational numbers in the follow-
ing way
�a0 � a1a2a3 . . . an ¼ � a0 þa1
101þ a2
102þ � � � þ an
10n
� �
:
It can be shown that any fraction whose denominator contains only powers
of 2 and/or 5 (such as 20¼ 22� 5) can be represented by such a terminating
decimal, which can be found by long division.
However, if we apply long division to many other rationals, then the process
of long division never terminates and we obtain a non-terminating or infinite
decimal. For example, applying long division to 13
gives 0.333 . . ., and for 1922
we obtain 0.86363 . . ..
Problem 3 Apply long division to 17
and 213
to find the corresponding
decimals.
These non-terminating decimals, which are obtained by applying the long
division process, have a certain common property. All of them are recurring;
that is, they have a recurring block of digits, and so can be written in shorthand
form, as follows:
0:333 . . . ¼ 0:3;
0:142857142857 . . . ¼ 0:142857 . . .;
0:86363 . . . ¼ 0:863:
It is not hard to show, whenever we apply the long division process to a
fraction pq, that the resulting decimal is recurring. To see why we notice that
there are only q possible remainders at each stage of the division, and so one
of these remainders must eventually recur. If the remainder 0 occurs, then the
resulting decimal is, of course, terminating; that is, it ends in recurring 0s.
For example
0:8500 . . .;
13:1212 . . .;
�1:111 . . .:
For example,
0:8500 . . . ¼ 0:85:
For example
0:85 ¼ 0þ 8
101þ 5
102
¼ 85
100¼ 17
20:
Another commonly usednotation is
0:3:
or 0:1:
42857:
:
1.1 Real numbers 3
Non-terminating recurring decimals which arise from the long division of
fractions are used to represent the corresponding rational numbers. This
representation is not quite so straight-forward as for terminating decimals,
however. For example, the statement
1
3¼ 0:3 ¼ 3
101þ 3
102þ 3
103þ � � �
can be made precise only when we have introduced the idea of the sum of a
convergent infinite series. For the moment, when we write the statement 13¼ 0:3
we mean simply that the decimal 0:3 arises from 13
by the long division process.
The following example illustrates one way of finding the rational number
with a given decimal representation.
Example 1 Find the rational number (expressed as a fraction) whose decimal
representation is 0:863:
Solution First we find the fraction x such that x ¼ 0:63:If we multiply both sides of this equation by 102 (because the recurring block
has length 2), then we obtain
100x ¼ 63:63 ¼ 63þ x:
Hence
99x ¼ 63) x ¼ 63
99¼ 7
11:
Thus
0:863 ¼ 8
10þ x
10¼ 8
10þ 7
110¼ 95
110¼ 19
22: &
The key idea in the above solution is that multiplication of a decimal by 10k
moves the decimal point k places to the right.
Problem 4 Using the above method, find the fractions whose decimal
representations are:
(a) 0:231; (b) 2:281.
The decimal representation of rational numbers has the advantage that it
enables us to decide immediately which of two distinct positive rational numbers
is the greater. We need only examine their decimal representations and notice
the first place at which the digits differ. For example, to order 78
and 1922
we write
7
8¼ 0:875 . . . and
19
22¼ 0:86363 . . .;
and so
0:86#363 . . . < 0:87
#5) 19
22<
7
8:
Problem 5 Find the first two digits after the decimal point in the deci-
mal representations of 1720
and 4553
, and hence determine which of these
two rational numbers is the greater.
Warning Decimals which end in recurring 9s sometimes arise as alternative
representations for terminating decimals. For example
1 ¼ 0:9 ¼ 0:999 . . . and 1:35 ¼ 1:349 ¼ 1:34999 . . .:
We return to this topic inChapter 3.
Whenever possible, we avoidusing the form of a decimalwhich ends in recurring 9s.
4 1: Numbers
You may find this rather alarming, but it is important to realise that this is
a matter of convention. We wish to allow the decimal 0.999 . . . to represent
a number x, so x must be less than or equal to 1 and greater than each of the
numbers
0:9; 0:99; 0:999; . . .:
The only rational with these properties is 1.
1.1.3 Irrational numbers
One of the surprising mathematical discoveries made by the Ancient Greeks
was that the system of rational numbers is not adequate to describe all the
magnitudes that occur in geometry. For example, consider the diagonal of a
square of side 1. What is its length? If the length is x, then, by Pythagoras’
Theorem, x must satisfy the equation x2¼ 2. However, there is no rational
number which satisfies this equation.
Theorem 1 There is no rational number x such that x2¼ 2.
Proof Suppose that such a rational number x exists. Then we can write x¼ pq.
By cancelling, if necessary, we may assume that p and q have no common
factor. The equation x2¼ 2 now becomes
p2
q2¼ 2; so p2 ¼ 2q2:
Now, the square of an odd number is odd, and so p cannot be odd. Hence p is
even, and so we can write p¼ 2r, say. Our equation now becomes
ð2rÞ2 ¼ 2q2; so q2 ¼ 2r2:
Reasoning as before, we find that q is also even.
Since p and q are both even, they have a common factor 2, which contradicts
our earlier statement that p and q have no common factors.
Arguing from our original assumption that x exists, we have obtained two
contradictory statements. Thus, our original assumption must be false. In other
words, no such x exists. &
Problem 6 By imitating the above proof, show that there is no rational
number x such that x3¼ 2.
Since we want equations such as x2¼ 2 and x3¼ 2 to have solutions, we
must introduce new numbers which are not rational numbers. We denote these
new numbers byffiffiffi
2p
andffiffiffi
23p
, respectively; thusffiffiffi
2p� �2¼ 2 and
ffiffiffi
23p� �3¼ 2. Of
course, we must introduce many other new numbers, such asffiffiffi
3p
,ffiffiffiffiffi
115p
, and
so on. Indeed, it can be shown that, if m, n are natural numbers and xm¼ n has
no integer solution, thenffiffiffi
nmp
cannot be rational. A number which is not rational
is called irrational.
There are many other mathematical quantities which cannot be described
exactly by rational numbers. For example, the number pwhich denotes the area
of a disc of radius 1 (or half the length of the perimeter of such a disc) is
irrational, as is the number e.
x
1
1
x2 ¼ 12 þ 12 ¼ 2
This is a proof bycontradiction.
For
ð2k þ 1Þ2 ¼ 4k2 þ 4k þ 1
¼ 4ðk2 þ kÞ þ 1:
The case m¼ 2 is treated inExercise 5 for this sectionin Section 1.6.
Lambert proved that p isirrational in 1770.
1.1 Real numbers 5
It is natural to ask whether irrational numbers, such asffiffiffi
2p
and p, can
be represented as decimals. Using your calculator, you can check that
(1.41421356)2 is very close to 2, and so 1.41421356 is a very good appro-
ximate value forffiffiffi
2p
. But is there a decimal which representsffiffiffi
2p
exactly? If
such a decimal exists, then it cannot be recurring, because all the recurring
decimals correspond to rational numbers.
In fact, it is possible to represent all the irrational numbers mentioned so
far by non-recurring decimals. For example, there are non-recurring decimals
such thatffiffiffi
2p¼ 1:41421356 . . . and p ¼ 3:14159265 . . .:
It is also natural to ask whether non-recurring decimals, such as
0:101001000100001 . . . and 0:123456789101112 . . .;
represent irrational numbers. In fact, a decimal corresponds to a rational
number if and only if it is recurring; so a non-recurring decimal must corres-
pond to an irrational number.
We may summarise this as:
recurring decimal, rational number
non-recurring decimal, irrational number
1.1.4 The real number system
Taken together, the rational numbers (recurring decimals) and irrational num-
bers (non-recurring decimals) form the set of real numbers, denoted by R .
As with rational numbers, we can determine which of two real numbers is
greater by comparing their decimals and noticing the first pair of correspond-
ing digits which differ. For example
0:10#100100010000 . . . < 0:12
#3456789101112 . . .:
We now associate with each irrational number a point on the number-line.
For example, the irrational number x¼ 0.123456789101112 . . . satisfies each
of the inequalities
0:1 < x < 0:2
0:12 < x < 0:13
0:123 < x < 0:124
..
.
We assume that there is a point on the number-line corresponding to x,
which lies to the right of each of the (rational) numbers 0.1, 0.12, 0.123 . . ., and
to the left of each of the (rational) numbers 0.2, 0.13, 0.124, . . ..
As usual, negative real numbers correspond to points lying to the left of 0;
and the ‘number-line’, complete with both rational and irrational points, is
called the real line.
In fact
ð1:41421356Þ2
¼1:9999999932878736:
We prove thatffiffiffi
2p
has adecimal representation inSection 1.5.
When comparing decimals inthis way, we do not alloweither decimal to end inrecurring 9s.
6 1: Numbers
There is thus a one–one correspondence between the points on the real line
and the set R of real numbers (or decimals).
We now state several properties of R , with which you will be already
familiar, although you may not have met their names before. These properties
are used frequently in Analysis, and we do not always refer to them explicitly
by name.
Order Properties of R
1. Trichotomy Property If a, b2R , then exactly one of the following
inequalities holds
a < b or a ¼ b or a > b:
2. Transitive Property If a, b, c2R , then
a < b and b < c) a < c:
3. Archimedean Property If a2R , then there is a positive integer n
such that
n > a:
4. Density Property If a, b2R and a< b, then there is a rational number
x and an irrational number y such that
a < x < b and a < y < b:
Remark
The Archimedean Property is sometimes expressed in the following equiva-
lent way: for any positive real number a, there is a positive integer n such
that 1n< a.
The following example illustrates how we can prove the Density Property.
Example 2 Find a rational number x and an irrational number y satisfying
a < x < b and a < y < b;
where a ¼ 0:123 and b ¼ 0:12345 . . ..
Solution The two decimals
a ¼ 0:1233#
. . . and b ¼ 0:1234#5 . . .
differ first at the fourth digit. If we truncate b after this digit, we obtain the
rational number x¼ 0.1234, which satisfies the requirement that a< x< b.
To find an irrational number y between a and b, we attach to x a (sufficiently
small) non-recurring tail such as 010010001 . . . to give y¼ 0.1234j010010001 . . ..It is then clear that y is irrational (because its decimal is non-recurring) and that
a< y< b. &
Problem 7 Find a rational number x and an irrational number y such
that a< x< b and a< y< b, where a ¼ 0:3 and b ¼ 0:3401:
The first three of theseproperties are almost self-evident, but the DensityProperty is not so obvious.
1.1 Real numbers 7
Theorem 2 Density Property of R
If a, b2R and a< b, then there is a rational number x and an irrational
number y such that
a < x < b and a < y < b:
Proof For simplicity, we assume that a, b� 0. So, let a and b have decimal
representations
a ¼ a0 � a1a2a3 . . . and b ¼ b0 � b1b2b3 . . .;
where we arrange that a does not end in recurring 9s, whereas b does not
terminate (this latter can be arranged by replacing a terminating representation
by an equivalent representation that ends in recurring 9s).
Since a< b, there must be some integer n such that
a0 ¼ b0; a1 ¼ b1; . . .; an�1 ¼ bn�1; but an < bn:
Then x¼ a0 � a1a2a3 . . . an� 1bn is rational, and a< x< b as required.
Finally, since x< b, it follows that we can attach a sufficiently small non-
recurring tail to x to obtain an irrational number y for which a< y< b. &
Remark
One consequence of the Density Property is that between any two real numbers
there are infinitely many rational numbers and infinitely many irrational
numbers.
Problem 8 Prove that between any two real numbers a and b there are
at least two distinct rational numbers.
1.1.5 Arithmetic in R
We can do arithmetic with recurring decimals by first converting the decimals
to fractions. However, it is not obvious how to perform arithmetical operations
with non-recurring decimals.
Assuming that we can representffiffiffi
2p
and p by the non-recurring decimals
ffiffiffi
2p¼ 1:41421356 . . . and p ¼ 3:14159265 . . .;
can we also represent the sumffiffiffi
2pþ p and the product
ffiffiffi
2p� p as deci-
mals? Indeed, what do we mean by the operations of addition and multi-
plication when non-recurring decimals (irrationals) are involved, and do
these operations satisfy the same properties as addition and multiplication
of rationals?
It would take many pages to answer these questions fully. Therefore, we
shall assume that it is possible to define all the usual arithmetical operations
with decimals, and that they do satisfy the usual properties. For definiteness,
we now list these properties.
You may omit the followingproof on a first reading.
Here a0, b0 are non-negativeintegers, and a1, b1, a2, b2, . . .are digits.
A proof of the previousremark would involve ideassimilar to those involved intackling Problem 8.
8 1: Numbers
Arithmetic in R
Addition Multiplication
A1 If a, b 2 R , then
aþ b 2 R .
M1 If a, b 2 R , then
a� b2 R .
A2 If a 2 R , then
aþ 0¼ 0þ a¼ a.
M2 If a 2 R , then
a� 1¼ 1� a¼ a.
A3 If a 2 R , then there is a
number �a 2 R such that
aþ (�a)¼ (� a)þ a¼ 0.
M3 If a 2 R � {0}, then there is a
number a�1 2 R such that
a� a�1¼ a�1� a¼ 1.
A4 If a, b, c 2 R , then
(aþ b)þ c¼ aþ (bþ c).
M4 If a, b, c 2 R , then
(a� b)� c¼ a� (b� c).
A5 If a, b 2 R , then
aþ b¼ bþ a.
M5 If a, b 2 R , then
a� b¼ b� a.
D If a, b, c 2 R , then a� (bþ c)¼ a� bþ a� c.
To summarise the contents of this table:
� R is an Abelian group under the operation of additionþ;
� R � {0} is an Abelian group under the operation of multiplication�;
� These two group structures are linked by the Distributive Property.
It follows from the above properties that we can perform addition, subtraction
(where a� b¼ aþ (�b)), multiplication and division (where ab¼ a� b�1) in
R , and that these operations satisfy all the usual properties.
Furthermore, we shall assume that the set R contains the nth roots and
rational powers of positive real numbers, with their usual properties. In
Section 1.5 we describe one way of justifying the existence of nth roots.
1.2 Inequalities
Much of Analysis is concerned with inequalities of various kinds; the aim of
this section and the next section is to provide practice in the manipulation of
inequalities.
1.2.1 Rearranging inequalities
The fundamental rule, upon which much manipulation of inequalities is based,
is that the statement a < b means exactly the same as the statement b� a> 0.
This fact can be stated concisely in the following way:
Rule 1 For any a, b2R , a< b, b� a> 0.
Put another way, the inequalities a< b and b� a> 0 are equivalent.
There are several other standard rules for rearranging a given inequality into
an equivalent form. Each of these can be deduced from our first rule above. For
CLOSURE
IDENTITY
INVERSES
ASSOCIATIVITY
COMMUTATIVITY
DISTRIBUTATIVITY
Properties A1–A5
Properties M1–M5
Property D
Any system satisfying theproperties listed in the table iscalled a field. Both Q and R
are fields.
Recall that the symbol ‘,’means ‘if and only if’ or‘implies and is implied by’.
1.2 Inequalities 9
example, we obtain an equivalent inequality by adding the same expression to
both sides.
Rule 2 For any a, b, c2R , a< b, aþ c< bþ c.
Another way to rearrange an inequality is to multiply both sides by a non-zero
expression, making sure to reverse the inequality if the expression is
negative.
Rule 3
� For any a, b2R and any c > 0, a < b, ac < bc;
� For any a, b2R and any c < 0, a < b, ac > bc.
Sometimes the most effective way to rearrange an inequality is to take reci-
procals. However, in this case, both sides of the inequality should be positive,
and the direction of the inequality has to be reversed.
Rule 4 (Reciprocal Rule)
For any positive a, b2R , a5b, 1a> 1
b:
Some inequalities can be simplified only by taking powers. However, in order
to do this, both sides must be non-negative and must be raised to a positive
power.
Rule 5 (Power Rule)
For any non-negative a, b2R , and any p > 0, a < b, ap < bp.
For positive integers p, Rule 5 follows from the identity
bp� ap ¼ b� að Þ bp�1 þ bp�2aþ � � � þ bap�2 þ ap�1� �
;
thus, since the right-hand bracket is positive, we have
b� a > 0, bp � ap > 0;
which is equivalent to our desired result.
Remark
There are corresponding versions of Rules 1–5 in which the strict inequality
a < b is replaced by the weak inequality a � b.
Problem 1 State (without proof) the versions of Rules 1–5 for weak
inequalities.
We shal l give one more rule for rearra nging inequal ities in Sub-se ction 1.2.3.
1.2.2 Solving inequalities
Solving an inequality involving an unknown real number x means determin-
ing those values of x for which the given inequality holds; that is, finding
the solution set of the inequality. We can often do this by rewriting the
inequality in an equivalent, but simpler form, using the rules given in the
last sub-section.
For example
253, 20530 ðc ¼ 10Þ;
253, �20 > �30
ðc ¼ �10Þ:
For example
2 < 4, 1
2ð¼0:5Þ
>1
4¼0:25ð Þ:
For example
4 < 9, 412ð¼2Þ< 9
12ð¼3Þ:
We shall discuss the meaningof non-integer powers inSection 1.5.
For example,
b3� a3 ¼ b� að Þ�b2 þ baþ a2� �
:
The solution set is the set ofthose values of x for which theinequality holds.
10 1: Numbers
In this activity we frequently use the usual rules for the sign of a product, and
the fact that the square of any real number is non-negative. Also, we need to
remember the difference between the logical statements: ‘implies’, ‘is implied
by’ and ‘implies and is implied by’.
Example 1 Solve the inequality xþ2xþ4
> x�32x�1
:
Solution We rearrange this inequality to give a somewhat simpler inequality,
using Rule 1
xþ 2
xþ 4>
x� 3
2x� 1, xþ 2
xþ 4� x� 3
2x� 1> 0
, x2 þ 2xþ 10
xþ 4ð Þ 2x� 1ð Þ > 0
, xþ 1ð Þ2þ 9
xþ 4ð Þ 2x� 1ð Þ > 0:
Now, the numerator is always positive. The denominator vanishes when
x¼�4 or x ¼ 12. By examining separately the sign of the denominator when
x < �4,�4 < x < 12
and x > 12, we can deduce that the last fraction is positive
precisely when x < �4 or x > 12. Hence the solution set of the original
inequality is
x :xþ 2
xþ 4>
x� 3
2x� 1
� �
¼ �1;�4ð Þ [ 12;1
� �
: &
Example 2 Solve the inequality 12x2 þ 2
< 14:
Solution Since 2x2þ 2> 0, we have
1
2x2 þ 2<
1
4, 2x2 þ 2 > 4 ðby Rule 4Þ
, x2 þ 1 > 2 ðby Rule 3Þ, x2 � 1 > 0 ðby Rule 1Þ, x� 1ð Þ xþ 1ð Þ > 0:
This last inequality holds precisely when x<�1 or x> 1. It follows that the
solution set of the original inequality is
x :1
2x2 þ 2<
1
4
� �
¼ �1;�1ð Þ [ 1;1ð Þ: &
Problem 2 Use each of the following expressions to write down an
inequality with the given expression on its left-hand side which is equi-
valent to the inequality x> 2:
(a) xþ 3; (b) 2� x; (c) 5xþ 2; (d) �15xþ2
.
Problem 3 Solve the following inequalities:
(a) 4x�x2�7x2�1
� 3; (b) 2x2 � xþ 1ð Þ2:
� þ �
þ þ �� � þ
It is a common strategy tobring all terms to one side.
We bring everything to acommon denominator.
Here we complete the squarein the numerator, since wecannot factorise it.
This is because the finaldisplayed inequality isequivalent to the inequalitywe are solving. The logicalimplication symbols betweenthe displayed inequalitieswere all ‘implies and isimplied by’.
Here we factorise the left-hand side of the inequalityto examine the signs of itsfactors.
1.2 Inequalities 11
Note that, for the moment, weare examining only those x forwhich x� 0.
Notice the use of theTransitive Property here.
y
x
y = x
For example
j3j ¼ j�3j ¼ 3:
We sometimes write
ja� bj ¼ dða; bÞ:
For example, the distancefrom �2 to 3 is
�2ð Þ � 3j j ¼ �5j j ¼ 5:
Warning Great care is needed when solving inequalities which involve
rational powers. In particular, when applying Rule 5 both sides of the inequal-
ity must be non-negative.
Example 3 Solve the inequalityffiffiffiffiffiffiffiffiffiffiffiffiffi
2xþ 3p
> x.
Solution The expressionffiffiffiffiffiffiffiffiffiffiffiffiffi
2xþ 3p
is defined only when 2xþ 3� 0; that is,
when x � � 32. Hence we need only consider those x in �3
2,1
�
.
We can obtain an equivalent inequality by squaring, provided that bothffiffiffiffiffiffiffiffiffiffiffiffiffi
2xþ 3p
and x are non-negative. Thus, for x� 0, we obtain
ffiffiffiffiffiffiffiffiffiffiffiffiffi
2xþ 3p
> x, 2xþ 3 > x2 ðby Rule 5; with p ¼ 2Þ, x2 � 2x� 3 < 0
, x� 3ð Þ xþ 1ð Þ < 0:
So the part of the solution set in [0,1) is [0, 3).
We now examine those x for which �32� x < 0. For such x,
ffiffiffiffiffiffiffiffiffiffiffiffiffi
2xþ 3p
� 0
and x< 0, so thatffiffiffiffiffiffiffiffiffiffiffiffiffi
2xþ 3p
ð� 0Þ > x, for all such x. It follows that all these x,
namely the set ½� 32
, 0Þ, belong to the solution set too.
Combining these results, the solution set of the original inequality is
x :ffiffiffiffiffiffiffiffiffiffiffiffiffi
2xþ 3p
> x �
¼ � 3
2; 0
�
[ 0; 3½ Þ
¼ � 3
2; 3
�
: &
Problem 4 Solve the inequalityffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2x2 � 2p
> x:
1.2.3 Inequalities involving modulus signs
We now turn our attention to inequalities involving the modulus, or abso-
lute value, of a real number. Recall that, if a2R , then its modulus jaj is
defined by
aj j ¼ a; if a � 0,
�a; if a < 0.
�
It is often useful to think of jaj as the distance along the real line from 0 to a.
In the same way, ja� bj is the distance along the real line from 0 to a� b,
which is the same as the distance from a to b.
Notice also that jaþ bj ¼ ja� (�b)j is the distance from a to �b.
12 1: Numbers
For example, if a¼�2, b¼ 1,then:
1. j�2j> 0;
2. �j�2j ��2� j�2j;3. j� 2j2¼ (� 2)2;
4. j(�2) � 1j ¼ j1 � (�2)j;5. j(�2)� 1j ¼ j� 2j � j1j.
Note that, in a similar way
aj j � b, �b � a � b:
We take a¼ x� 2, b¼ 1 inRule 6.
For jx� 2j2¼ (x� 2)2.
We now list some basic properties of the modulus, which follow immediately
from the definition:
Properties of the modulus For any real numbers a and b:
1. jaj � 0, with equality if and only if a¼ 0;
2. –jaj � a� jaj;3. jaj2¼ a2;
4. ja� bj ¼ jb� aj;5. jabj ¼ jaj � jbj.
There is a basic rule for rearranging inequalities involving modulus signs:
Rule 6 For any real numbers a and b, where b> 0: aj j < b,�b < a < b.
Also, it is often possible, and sometimes easier, to use Rule 5 with p¼ 2 than to
use Rule 6. The following example illustrates the use of both rules.
Example 4 Solve the inequality jx� 2j < 1.
Solution Using Rule 6, we obtain
x� 2j j < 1, �1 < x� 2 < 1
, 1 < x < 3:
So the solution set of the original inequality is
x: x� 2j j < 1f g ¼ 1; 3ð Þ:
Alternatively, using Rule 5 (with p¼ 2), we obtain
x� 2j j < 1, x� 2ð Þ2< 1
, x2 � 4xþ 3 < 0
, x� 1ð Þ x� 3ð Þ < 0:
Again, this shows that the required solution set is (1, 3). &
Example 5 Solve the inequality jx� 2j � jxþ 1j.
Solution Using Rule 5 (with p¼ 2), we obtain
x� 2j j � xþ 1j j , x� 2ð Þ2� xþ 1ð Þ2
, x2 � 4xþ 4 � x2 þ 2xþ 1
, 3 � 6x
, 1
2� x:
So the solution set of the original inequality is
x: x� 2j j � xþ 1j jf g ¼ 1
2;1
�
: &
The inequalities in Examples 4 and 5 can easily be interpreted geometrically.
In Example 4, the inequality jx� 2j < 1 holds when the distance from x to 2
is strictly less than 1. So it holds for all points on either side of 2 at a distance
less than 1 from 2� namely, in the open interval (1, 3).
1.2 Inequalities 13
In Example 5, the inequality jx� 2j � jxþ 1j holds when the distance from x
to 2 is less than or equal to the distance from x to�1, since jxþ 1j ¼ jx� (�1)j.The mid-point of 2 and �1 (that is, the point x where the distance from x to 2
equals the distance from x to �1) is 12. So the inequality holds when x lies
in [12
, 1).
Some good ideas when tackling problems involving inequalities of these
types are:
� use your geometrical intuition, where possible, to give yourself an idea of
the sets involved;
� test one or two values of x in your final solution set to see if they are valid –
this often detects errors in manipulating inequality signs!
Problem 5 Solve the following inequalities:
(a) j2x2 � 13j < 5; (b) jx� 1j � 2jxþ 1j.
1.3 Proving inequalities
In this section we show you how to prove inequalities of various types. We
shall use the rules for rearranging inequalities given in Section 1.2, and also use
other rules which enable us to deduce new inequalities from old. We have
already met the first rule in Section 1.1, where it was called the Transitive
Property of R .
Transitive Rule a < b and b < c) a < c.
We use the Transitive Rule when we want to prove that a < c, and we know
that a < b and b < c.
The following rules are also useful:
Combination Rules
If a < b and c < d, then:
Sum Rule aþ c< bþ d;
Product Rule ac< bd (provided that a,c� 0).
There are also weak and weak/strict versions of the Transitive Rule and
Combination Rules, which we will ask you to work out and use as they arise.
Remark
It is important to appreciate that the Transitive Rule and the Combination
Rules have a different nature from Rules 1–6 in Section 1.2 . Rules 1–6 tell you
how to rearrange inequalities into equivalent forms, whereas the Transitive
Rule and the Combination Rules enable you to deduce new inequalities which
are not equivalent to the old ones.
For example, since 2< 3 and4< 5, then
2þ 4 < 3þ 5;
2� 4 < 3� 5:
For example, if a< b andc� d, then
aþ c < bþ d
and
ac < bd; provided a; c > 0:
14 1: Numbers
1.3.1 The Triangle Inequality
If a and b are both vectors in R2, then the vector aþ b is obtained from the
‘parallelogram construction’:
b
a
a + b
By elementary geometry, the length of any side of a triangle is less than or
equal to the sum of the lengths of the other two sides. In the special case of R ,
when all the vectors lie on a line, this can be interpreted as the Triangle
Inequality, which involves the absolute value of the real numbers a, b and aþ b.
Triangle Inequality If a, b 2 R , then:
1. aþ bj j � aj j þ bj j;2. a� bj j �
�
� aj j � bj j�
� (the ‘reverse form’ of the Triangle Inequality).
Proof In order to prove part 1, we use Rule 5, with p¼ 2
aþ bj j � aj j þ bj j , aþ bð Þ2� aj j þ bj jð Þ2
, a2 þ 2abþ b2 � a2 þ 2 aj j bj j þ b2
, 2ab � 2 aj j bj j:The final inequality is certainly true for all a, b 2 R , and so the first
inequality must also be true for all a, b 2 R . Hence we have proved part 1.
We prove part 2 by using the same method
a� bj j ��
� aj j � bj j�
�, a� bð Þ2� aj j � bj jð Þ2
, a2 � 2abþ b2 � a2 � 2 aj j bj j þ b2
, �2ab � �2 aj j bj j, 2ab � 2 aj j bj j;
which is again true for all a, b 2 R . &
Remarks
1. Although we have used double-headed arrows here, the actual proof
requires only the arrows going from right to left. For example, in the
proof of part 1 the important implication is
aþ bj j � aj j þ bj j ( 2ab � 2 aj j bj j:2. Part 1 of the Triangle Inequality can also be proved by using Rule 6, as
follows. By Rule 6
aþ bj j � aj j þ bj j , � aj j þ bj jð Þ � aþ b � aj j þ bj jð Þ: (1)
Here we discuss R2, rather
than R , simply because theargument is thengeometrically clearer.
In the ‘parallelogramconstruction’, you draw thevector a from the origin tosome point, then the vector bfrom that point to a final point.The vector aþ b is then thevector from the origin to thatfinal point.
For example, with a¼�1 andb¼ 3:
1. j�1þ 3j � j�1j þ j3j;2. j(�1)� 3j � j j�1 j� j3j j.
Remember that aj j2¼ a2:
Part 2 is sometimes called the‘cunning form’ or ‘backwardsform’ of the TriangleInequality.
1.3 Proving inequalities 15
Now, we know that�|a|� a� |a| and�|b|� b� |b|; so, by the Sum Rule
� aj j þ bj jð Þ � aþ b � aj j þ bj jð Þ:It follows that the left-hand inequality in (1) must also hold, as required.
3. An obvious modification of the proof in Remark 2 shows that the following
more general form of the Triangle Inequality also holds:
Triangle Inequality for n terms For any real numbers a1, a2, . . ., an
a1 þ a2 þ � � � þ anj j � a1j j þ a2j j þ � � � þ anj j:
The following example is a typical application of the Triangle Inequality.
Example 1 Use the Triangle Inequality to prove that:
(a) jaj � 1)j3þ a3j � 4; (b) jbj< 1) j3� bj> 2.
Solution
(a) Suppose that aj j � 1. The Triangle Inequality then gives
3þ a3�
�
�
� � 3j j þ a3�
�
�
�
¼ 3þ aj j3
� 3þ 1 ðsince aj j � 1Þ¼ 4:
(b) Suppose that jbj< 1. The ‘reverse form’ of the Triangle Inequality then
gives
3� bj j ��
� 3j j � bj j�
�
¼�
�3� bj j�
�
� 3� bj j:
Now jbj< 1, so that � jbj>�1. Thus
3� bj j > 3� 1
¼ 2;
and we can then deduce from the previous chain of inequalities that
j3� bj> 2, as desired. &
Remarks
1. The results of Example 1 can also be stated in the form:
(a) j3þ a3j � 4, for jaj � 1;
(b) j3� bj> 2, for jbj< 1.
2. The reverse implications
3þ a3�
�
�
� � 4) aj j � 1 and 3� bj j > 2) bj j < 1
are FALSE. For example, try putting a ¼ �32
and b¼�2!
Problem 1 Use the Triangle Inequality to prove that:
(a) aj j � 12) aþ 1j j � 3
2; (b) bj j < 1
2) b3 � 1�
�
�
� > 78.
Note the use of the TransitiveRule here.
Again, we use the TransitiveRule.
16 1: Numbers
1.3.2 Inequalities involving n
In Analysis we often need to prove inequalities involving an integer n. It is a
common convention in mathematics that the symbol n is used to denote an
integer (frequently a natural number).
It is often possible to deal with inequalities involving n by using the
rearrangement rules given in Section 1.2. Here is such an example.
Example 2 Prove that 2n2 � nþ 1ð Þ2; for n � 3:
Solution Rearranging this inequality into an equivalent form, we obtain
2n2 � nþ 1ð Þ2 , 2n2 � nþ 1ð Þ2� 0
, n2 � 2n� 1 � 0
, n� 1ð Þ2� 2 � 0 ðby ‘completing the square’Þ, n� 1ð Þ2� 2:
This final inequality is clearly true for n� 3, and so the original inequality
2n2� (nþ 1)2 is true for n� 3. &
Remarks
1. In Problem 3 of Section 1.2, we asked you to solve the inequality 2x2� (xþ 1)2;
its solution set was �1; 1�ffiffiffi
2p
� �
[ 1þffiffiffi
2p
;1 �
. In Example 2, above,
we found those natural numbers n lying in this solution set.
2. An alternative solution to Example 2 is as follows
2n2 � nþ 1ð Þ2 , 2 � nþ 1
n
� 2
ðby Rule 3Þ
,ffiffiffi
2p� 1þ 1
nðby Rule 5, with p ¼ 1
2Þ;
and this final inequality certainly holds for n� 3.
Problem 2 Prove that 3nn2þ2
< 1; for n > 2.
1.3.3 More on inequalities
We now look at a number of inequalities and methods for proving inequalities
that will be useful later on.
Example 3 Prove that ab � aþ b2
� �2, for a, b 2R .
Solution We tackle this inequality using the various rearrangement rules and
a chain of equivalent inequalities until we obtain an inequality that we know
must be true
ab � aþ b
2
� 2
, ab � a2 þ 2abþ b2
4
, 4ab � a2 þ 2abþ b2
, 0 � a2 � 2abþ b2
, 0 � a� bð Þ2:
n 1 2 3 4
2n2 2 8 18 32
(nþ 1)2 4 9 16 25
This has the followinggeometric interpretation: Thearea of a rectangle withsides of length a and b isless than or equal to the areaof a square with sides oflength aþb
2.
a
b
1.3 Proving inequalities 17
This final inequality is certainly true, since all squares are non-negative. It
follows that the original inequality ab � aþb2
� �2is also true, for a, b 2 R . &
Remark
A close examination of the above chain of equivalent statements shows that in
fact ab ¼ aþb2
� �2if and only if a¼ b.
Problem 3 Prove that aþbffiffi
2p �
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ b2p
; for a, b 2 R .
Problem 4 Suppose that a >ffiffiffi
2p
. Prove the following inequalities:
(a) 12
aþ 2a
� �
< a; (b) 12
aþ 2a
� �� �2> 2:
Hint: In part (b), use the result of Example 3 and the subsequent remark.
Example 4 Prove thatffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ b2p
� aþ b; for a; b � 0:
Solution We tackle this inequality using the various rearrangement rules and
a chain of equivalent inequalities until we obtain an inequality that we know
must be trueffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ b2p
� aþ b, a2 þ b2 � aþ bð Þ2
, a2 þ b2 � a2 þ 2abþ b2
, 0 � 2ab:
This final inequality is certainly true, since a, b� 0. It follows that the
original inequalityffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ b2p
� aþ b is also true, for a, b� 0. &
Problem 5 Use the result of Example 4 to prove thatffiffiffiffiffiffiffiffiffiffiffi
cþ dp
�ffiffiffi
cpþ
ffiffiffi
dp
; for c; d � 0:
Example 5 Prove thatffiffiffi
ap�
ffiffiffi
bp�
�
�
� �ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a� bj jp
, for a, b� 0.
Solution Notice first that interchanging the roles of a and b leaves the
inequality unaltered. It follows that it is sufficient to prove the inequality
under the assumption that a� b.
So, assume that a� b. Then we know thatffiffiffi
ap�
ffiffiffi
bp
and a� bj j ¼ a� b:Hence
ffiffiffi
ap�
ffiffiffi
bp�
�
�
�
�
��
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a� bj jp
,ffiffiffi
ap�
ffiffiffi
bp�
ffiffiffiffiffiffiffiffiffiffiffi
a� bp
,ffiffiffi
ap�
ffiffiffiffiffiffiffiffiffiffiffi
a� bp
þffiffiffi
bp
:
This final inequality is certainly true, and is obtained from the result of
Problem 5 by simply substituting a� b in place of c and b in place of d. It
follows that the original inequalityffiffiffi
ap�
ffiffiffi
bp�
�
�
� �ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a� bj jp
is also true, for
a, b� 0. &
Problem 6 Prove thatffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
aþ bþ cp
�ffiffiffi
apþ
ffiffiffi
bpþ
ffiffiffi
cp; for a; b; c � 0:
We often use the Binomial Theorem and the Principle of Mathematical
Induction (see Appendix 1) to prove inequalities.
Example 6 Prove the following inequalities, for n� 1:
(a) 2n� 1þ n; (b) 21n � 1þ 1
n:
In the formffiffiffiffiffi
abp
� aþb2
thisinequality is sometimes calledthe Arithmetic–GeometricMean Inequality for a, b.
This has the followinggeometric interpretation: Thelength of the hypotenuse of aright-angled triangle whoseother sides are of lengthsa and b is less than or equal tothe sum of the lengths of thosetwo sides.
ba2 + b2
a
This will simplify the detailsof our chain of inequalities.
We avoid one modulus as aresult of our simplifyingassumption!
Never be ashamed to utiliseevery tool at your disposal!(Why do the same worktwice?)
n 1 2 3 4
2n 2 4 8 16
1þ n 2 3 4 5
18 1: Numbers
Solution
(a) By the Binomial Theorem for n� 1
1þ xð Þn ¼ 1þ nxþ n n� 1ð Þ2!
x2 þ � � � þ xn
� 1þ nx; for x � 0:
Then, if we substitute x¼ 1 in this last inequality, we get
2n � 1þ n; for n � 1:
(b) We start by rewriting the required result in an equivalent form
21n � 1þ 1
n, 2 � 1þ 1
n
� n
ðby the Power RuleÞ:
Now, if we substitute x ¼ 1n
in the Binomial Theorem for (1þ x)n, we get
1þ 1
n
� n
¼ 1þ n1
n
�
þ n n� 1ð Þ2!
1
n
� 2
þ � � � þ 1
n
� n
� 1þ 1 ¼ 2:
Since the inequality 2 � 1þ 1n
� �nis true, it follows that the original
inequality 21n � 1þ 1
n, for n� 1, is also true, as required. &
Problem 7 Prove the inequality 1þ 1n
� �n� 52� 1
2n; for n� 1.
Hint: consider the first three terms in the binomial expansion.
Example 7 Prove that 2n� n2, for n� 4.
Solution Let P(n) be the statement
PðnÞ : 2n � n2:
First we show that P(4) is true: 24� 42.
STEP 1 Since 24¼ 16 and 42¼ 16, P(4) is certainly true.
STEP 2 We now assume that P(k) holds for some k� 4, and deduce that
P(kþ 1) is then true.
So, we are assuming that 2k� k2. Multiplying this inequality by 2
we get
2kþ1 � 2k2;
so it is therefore sufficient for our purposes to prove that 2k2� (kþ 1)2.
Now
2k2 � k þ 1ð Þ2 , 2k2 � k2 þ 2k þ 1
, k2 � 2k � 1 � 0 ðby ‘completing the square’Þ
, k � 1ð Þ2�2 � 0:
This last inequality certainly holds for k� 4, and so 2kþ1� (kþ 1)2
also holds for k� 4.
In other words: P(k) true for some k� 4)P(kþ 1) true.
It follows, by the Principle of Mathematical Induction, that 2n � n2, for
n� 4. &
Problem 8 Prove that 4n> n4, for n� 5.
n 1 2 3 4
21n 2 1.41 1.26 1.19
1þ 1n
2 1.5 1.33 1.25
We decrease the sum byomitting subsequentnon-negative terms.
We decrease the sum byomitting all but the first twoterms.
n 1 2 3 4 5
2n 2 4 8 16 32
n2 1 4 9 16 25
This assumption is just P(k).
Since P(kþ 1) is:2kþ1� (kþ 1)2.
1.3 Proving inequalities 19
The value of this result willcome from making suitablechoices of x and n forparticular purposes.
We prove the result usingMathematical Induction.
This assumption is P(k).This multiplication is validsince 1þ xð Þ � 0.
We decrease the expressionif we omit the final non-negative term.
You saw in part (b) of
Example 6 that 21n � 1þ 1
n:
We give the proof ofTheorem 2 at the end of thesub-section.
Three important inequalities in Analysis
Our first inequality, called Bernoulli’s Inequality, will be of regular use in later
chapters.
Theorem 1 Bernoulli’s Inequality
For any real number x��1 and any natural number n, (1þ x)n� 1þ nx.
Remark
In part (a) of Example 6, you saw that (1þ x)n� 1þ nx, for x> 0 and n a
natural number. Theorem 1 asserts that the same result holds under the weaker
assumption that x��1.
Proof Let P(n) be the statement
PðnÞ : 1þ xð Þn� 1þ nx; for x � �1:
STEP 1 First we show that P(1) is true: (1þ x)1� 1þ x. This is obviously
true.
STEP 2 We now assume that P(k) holds for some k� 1, and prove that P(kþ 1)
is then true.
So, we are assuming that 1þ xð Þk � 1þ kx; for x��1. Multiplying
this inequality by (1þ x), we get
1þ xð Þkþ1 � 1þ xð Þ 1þ kxð Þ¼ 1þ k þ 1ð Þxþ kx2
� 1þ k þ 1ð Þx:
Thus, we have 1þ xð Þkþ1� 1þ k þ 1ð Þx; in other words the statement
P(kþ 1) holds.
So, P(k) true for some k� 1)P(kþ 1) true.
It follows, by the Principle of Mathematical Induction, that 1þ xð Þn�1þ nx, for x � �1, n � 1: &
Problem 9 By applying Bernoulli’s Inequality with x ¼ � 1ð2nÞ, prove
that 21n � 1þ 1
2n�1, for any natural number n.
Our second inequality is of considerable use in various branches of
Analysis. In Problem 3 you proved that aþbffiffi
2p �
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ b2p
, for a, b2R . We
can rewrite this inequality in the equivalent form aþ bð Þ2� 2 a2 þ b2ð Þ or
aþ bð Þ2� a2 þ b2ð Þ 12 þ 12ð Þ. The Cauchy–Schwarz Inequality is a general-
isation of this result to 2n real numbers.
Theorem 2 Cauchy–Schwarz Inequality
For any real numbers a1; a2; . . .; an and b1; b2; . . .; bn; we have
a1b1þ a2b2 þ � � � þ anbnð Þ2
� a21 þ a2
2 þ � � � þ a2n
� �
b21 þ b2
2 þ � � � þ b2n
� �
:
20 1: Numbers
For example, with n¼ 3
1þ 2þ 3ð Þ 1
1þ 1
2þ 1
3
�
¼ 6� 11
6
¼ 11 � 32:
We give the proof ofTheorem 3 at the end of thesub-section.
For example, with n¼ 3
1þ 1
3
� 3
¼ 64
27¼ 2:37 . . .
5 1þ 1
4
� 4
¼ 625
256¼ 2:44 . . .
You may omit these proofs ata first reading.
Note that we will use the �notation to keep the argumentbrief.
A is non-zero, by assumption,so that 1/A makes sense.
Problem 10 Use Theorem 2 to prove that for any positive real numbers
a1, a2, . . ., an, then a1 þ a2 þ � � � þ anð Þ 1a1þ 1
a2þ � � � þ 1
an
� �
� n2:
Our final result also has many useful applications. In Example 3 you proved
that ab � aþ b2
� �2, for a, b2R ; it follows that, if a and b are positive, then
abð Þ12� aþb
2. The Arithmetic Mean–Geometric Mean Inequality is a general-
isation of this result for two real numbers to n real numbers.
Theorem 3 Arithmetic Mean–Geometric Mean Inequality
For any positive real numbers a1, a2, . . ., an, we have
a1a2 . . . anð Þ1n� a1 þ a2þ � � �þ an
n:
Problem 11 Use Theorem 3 with the nþ 1 positive numbers 1, 1þ 1n,
1þ 1n, . . ., 1þ 1
nto prove that, for any positive integer n
1þ 1
n
� n
� 1þ 1
nþ 1
� nþ1
:
Proofs of Theorems 2 and 3
Theorem 2 Cauchy–Schwarz Inequality
For any real numbers a1, a2, . . ., an and b1, b2, . . ., bn, we have
a1b1þ a2b2 þ � � � þanbnð Þ2� a21 þ a2
2 þ � � � þ a2n
� �
b21 þ b2
2 þ � � � þ b2n
� �
:
Proof If all the as are zero, the result is obvious; so we need only examine
the case when not all the as are zero. It follows that, if we denote the
sumPn
k¼1a2k ¼ a2
1 þ a22 þ � � � þ a2
n by A, then A> 0. Also, denotePn
k¼1b2k¼b2
1þb22þ���þb2
n by B andPn
k¼1akbk¼ a1b1þ a2b2þ���þanbn by C.
Now, for any real number l, we have that lak þ bkð Þ2� 0, so that
X
n
k¼1
l2a2k þ 2lakbk þ b2
k
� �
¼X
n
k¼1
lak þ bkð Þ2 � 0;
which we may rewrite in the form
l2Aþ 2lC þ B � 0:
But this inequality is equivalent to the inequality
lAþ Cð Þ2þAB � C2; for any real number l:
Since A is non-zero, we may now choose l ¼ � CA. It follows from the last
inequality that AB�C2, which is exactly what we had to prove. &
Remark
If not all the as are zero, equality can only occur ifP
n
k¼1
lak þ bkð Þ2¼ 0; that is,
if all the numbers ak are proportional to all the numbers bk, 1 � k � n:
1.3 Proving inequalities 21
Theorem 3 Arithmetic Mean–Geometric Mean Inequality
For any positive real numbers a1, a2, . . ., an, we have
a1a2 . . . anð Þ1n� a1 þ a2þ . . .þan
n: (2)
Proof Since the ai are positive, we can rewrite (2) in the equivalent form
a1a2 . . . anð Þ1n
a1 þ a2þ � � � þanð Þ=n� 1: (3)
Now, replacing each term ai by lai for any non-zero number l does not alter
the left-hand side of the inequality (3). It follows that it is sufficient to prove the
inequality (2) in the special case when the product of the terms ai is 1. Hence it
is sufficient to prove the following statement P(n) for each natural number n:
P(n): For any positive real numbers ai with a1a2 . . . an¼ 1, then
a1þ a2þ � � � þ an� n.
First, the statement P(1) is obviously true.
Next, we assume that P(k) holds for some k� 1, and prove that P(kþ 1) is
then true.
Now, if all the terms a1, a2, . . ., akþ1 are equal to 1, the result P(kþ 1)
certainly holds. Otherwise, at least two of the terms differ from 1, say a1 and a2,
such that a1> 1 and a2< 1. Hence
a1 � 1ð Þ � a2 � 1ð Þ � 0;
which after some manipulation we may rewrite as
a1 þ a2 � 1þ a1a2: (4)
We are now ready to tackle P(kþ 1). Then
a1 þ a2þ � � � þ akþ1 � 1þ a1a2 þ a3 þ a4 þ � � � þ akþ1
� k þ 1;
since we may apply the assumption that P(k) holds to the k quantities a1a2,
a3, a4, . . ., akþ1. This last inequality is simply the statement that P(kþ 1) is
indeed true.
It follows by the Principle of Mathematical Induction that P(n) holds for all
natural numbers n, and so the inequality (2) must also hold. &
Remark
A careful examination of the proof of Theorem 3 shows that equality can only
occur if all the terms ai are equal.
1.4 Least upper bounds and greatest lower bounds
1.4.1 Upper and lower bounds
Any finite set {x1, x2, . . ., xn} of real numbers obviously has a greatest element
and a least element, but this property does not necessarily hold for infinite sets.
We denote the typical term byai rather than ak to avoidconfusion with a different useof the letter k in theMathematical Inductionargument below.
We will prove this byMathematical Induction.
The argument is exactly thesame whichever two termsactually differ from 1.
You should check thisyourself.
By (4).
That isa1a2ð Þ a3a4 . . . akþ1 ¼ 1
) a1a2ð Þ þ a3 þ a4 þ � � �þ akþ1 � k:
22 1: Numbers
For example, the interval (0, 2] has greatest element 2, but neither of the sets
N ¼ {1, 2, 3, . . .} nor [0, 2) has a greatest element. However the set [0, 2) is
bounded above by 2, since all points of [0, 2) are less than or equal to 2.
Definitions A set ER is bounded above if there is a real number, M
say, called an upper bound of E, such that
x � M; for all x 2 E:
If the upper bound M belongs to E, then M is called the maximum element
of E, denoted by max E.
Geometrically, the set E is bounded above by M if no point of E lies to the
right of M on the real line.
For example, if E ¼ [0, 2), then the numbers 2, 3, 3.5 and 157.1 are all upper
bounds of E, whereas the numbers 1.995, 1.5, 0 and �157.1 are not upper
bounds of E. Although it seems obvious that [0, 2) has no maximum element,
you may find it difficult to write down a formal proof. The following example
shows you how to do this:
Example 1 Determine which of the following sets are bounded above, and
which have a maximum element:
(a) E1 ¼ [0, 2); (b) E2 ¼ f1n
: n ¼ 1; 2; . . .g; (c) E3 ¼ N .
Solution
(a) The set E1 is bounded above. For example, M ¼ 2 is an upper bound
of E1, since
x � 2; for all x 2 E1:
However, E1 has no maximum element. For each x in E1, we have x< 2,
and so there is some real number y such that
x < y < 2;
by the Density Property of R .
Hence y2E1, and so x cannot be a maximum element.
(b) The set E2 is bounded above. For example, M¼ 1 is an upper bound of E2,
since
1
n� 1; for all n ¼ 1; 2; . . .:
Also, since 12E2
max E2 ¼ 1:
(c) The set E3 is not bounded above. For each real number M, there is a
positive integer n such that n>M, by the Archimedean Property of R .
Hence M cannot be an upper bound of E3.
This also means that E3 cannot have a maximum element. &
Problem 1 Sketch the following sets, and determine which are bounded
above, and which have a maximum element:
(a) E1¼ (�1, 1]; (b) E2 ¼ f1� 1n
: n ¼ 1; 2; . . .g;(c) E3¼ {n2: n¼ 1, 2, . . .}.
For example, y can be of theform 1.99. . . 9 or y¼ 1
2(xþ 2).
2 is not a maximum element,since 2 =2E1.
1.4 Least upper bounds and greatest lower bounds 23
Similarly, we define lower bounds. For example, the interval (0, 2) is
bounded below by 0, since
0 � x; for all x 2 0; 2ð Þ:However, 0 does not belong to (0, 2), and so 0 is not a minimum element of
(0, 2). In fact, (0, 2) has no minimum element.
Definitions A set ER is bounded below if there is a real number, m say,
called a lower bound of E, such that
m � x; for all x 2 E:
If the lower bound m belongs to E, then m is called the minimum element of
E, denoted by min E.
Geometrically, the set E is bounded below by m if no point of E lies to the
left of m on the real line.
Problem 2 Determine which of the following sets are bounded below,
and which have a minimum element:
(a) E1¼ (�1, 1]; (b) E2 ¼ f1� 1n: n ¼ 1; 2; . . .g;
(c) E3¼ {n2: n¼ 1, 2, . . .}.
The following terminology is also useful:
Definition A set ER is bounded if it is bounded above and bounded
below.
For example, the set E2 ¼ f1� 1n: n ¼ 1; 2; . . .g is bounded, but the sets
E1¼ (�1, 1] and E3¼ {n2: n¼ 1, 2, . . .} are not bounded.
Similar terminology applies to functions.
Definitions A function ƒ defined on an interval IR is said to:
� be bounded above by M if f(x)�M, for all x2 I; M is an upper bound of ƒ;
� be bounded below by m if f(x)�m, for all x2 I; m is a lower bound of ƒ;
� have a maximum (or maximum value) M if M is an upper bound of ƒ and
f(x)¼M, for at least one x2 I;
� have a minimum (or minimum value) m if m is a lower bound of ƒ and
f(x)¼m, for at least one x2 I.
Example 2 Let ƒ be the function defined by f (x)¼ x2, x 2 12
, 3 �
. Determine
whether ƒ is bounded above or below, and any maximum or minimum value
of ƒ.
Solution First, ƒ is increasing on the interval 12
, 3Þ
, so that since 12� x < 3 it
follows that 14� f ðxÞ < 9. Hence ƒ is bounded above and bounded below.
Next, since f 12
� �
¼ 14
and 14
is a lower bound for ƒ on the interval 12
, 3Þ
, it
follows that ƒ has a minimum value of 14
on this interval.
Finally, 9 is an upper bound for ƒ on the interval 12
, 3Þ
but there is no point
x in 12
, 3Þ
for which ƒ(x)¼ 9. So 9 cannot be a maximum of ƒ on the interval.
However, if y is any number in (8, 9) there is a number x>ffiffiffi
yp
inffiffiffi
yp
, 3� �
2ffiffiffi
2p
, 3� �
12
, 3 �
such that ƒ(x)¼ x2> y, so that no number in (8, 9) will serve
Strictly speaking, M and m arethe upper bound and lowerbound of the image set{ƒ(x): x 2 I}.
2√2 y 312
24 1: Numbers
as a maximum of ƒ on the interval. It follows that ƒ has no maximum value
on 12
, 3Þ
. &
Problem 3 Let ƒ be the function defined by f xð Þ ¼ 1x2 ; x 2 �3;�2½ Þ.
Determine whether ƒ is bounded above or below, and any maximum or
minimum value of ƒ.
1.4.2 Least upper bounds and greatest lower bounds
We have seen that the interval [0, 2] has a maximum element 2, but [0, 2) has
no maximum element. However, the number 2 is ‘rather like’ a maximum
element of [0, 2), because 2 is an upper bound of [0, 2) and any number less
than 2 is not an upper bound of [0, 2). In other words, 2 is the least upper
bound of [0, 2).
Definition A real number M is the least upper bound, or supremum, of a
set ER if:
1. M is an upper bound of E;
2. if M0<M, then M0 is not an upper bound of E.
In this case, we write M¼ sup E.
If E has a maximum element, max E, then sup E¼max E. For example,
the closed interval [0, 2] has least upper bound 2. We can think of the least
upper bound of a set, when it exists, as a kind of ‘generalised maximum
element’.
If a set does not have a maximum element, but is bounded above, then we may
be able to guess the value of its least upper bound. As in the case E¼ [0, 2), there
may be an obvious ‘missing point’ at the upper end of the set. However it is
important to prove that your guess is correct. We now show you how to do this.
Example 3 Prove that the least upper bound of [0, 2) is 2.
Solution We know that M¼ 2 is an upper bound of [0, 2), because
x � 2; for all x 2 ½0; 2Þ:To show that 2 is the least upper bound, we must prove that each number
M0< 2 is not an upper bound of [0, 2). To do this, we must find an element
x in [0, 2) which is greater than M0. But, if M0< 2, then there is a real number
x such that
M0 < x < 2
and also
0 < x < 2:
Since x2 [0, 2), the number M0 cannot be an upper bound of [0, 2). Hence
M¼ 2 is the least upper bound, or supremum, of [0, 2). &
Although the conclusion of Example 3 may seem painfully obvious, we have
written out the solution in detail because it illustrates the strategy for determin-
ing the least upper bound of a set, if it has one.
Part 1 says that M is an upperbound.Part 2 says that no smallernumber can be an upperbound.
For example, x can be of theform 1.99 . . . 9 for a suitablylarge number of digits, or it
can be 12
M0 þ 2ð Þ since
M0< 12
M0 þ 2ð Þ < 2:
1.4 Least upper bounds and greatest lower bounds 25
Strategy Given a subset E of R , to show that M is the least upper bound, or
supremum, of E, check that:
1. x�M, for all x2E;
2. if M0<M, then there is some x2E such that x>M0.
Notice that, if M is an upper bound of E and M2E, then part 2 is auto-
matically satisfied, and so M¼ sup E¼max E.
Example 4 Determine the least upper bound of E ¼ f1� 1n2 : n ¼ 1; 2; . . .g.
Solution We guess that the least upper bound of E is M¼ 1. Certainly, 1 is
an upper bound of E, since
1� 1
n2� 1; for n ¼ 1; 2; . . .:
To check part 2 of the strategy, we need to show that, if M0< 1, then there
is some natural number n such that
1� 1
n2> M0: (1)
However
1� 1
n2> M0 , 1�M0 >
1
n2
, 1
1�M0< n2 ðsince 1�M0 > 0Þ
,ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
1�M0
r
< n ðsince1
1�M0> 0
and n > 0Þ:We can certainly choose n so that this final inequality holds, by the
Archimedean Property of R , and so we can choose n so that inequality (1) holds.
Hence 1 is the least upper bound of E. &
Remark
Although we used double-headed arrows in this solution, the actual proof
required only the implications going from right to left. In other words, the
proof uses only the fact that
1� 1
n2> M0 (
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
1�M0
r
< n:
Problem 4 Determine sup E, if it exists, for each of the following sets:
(a) E1¼ (�1, 1]; (b) E2 ¼ f1� 1n
: n ¼ 1, 2, . . .g;(c) E3¼ {n2: n¼ 1, 2, . . .}.
Similarly, we define the notion of a greatest lower bound.
Definition A real number m is the greatest lower bound, or infimum, of
a set ER if:
1. m is a lower bound of E;
2. if m0>m, then m0 is not a lower bound of E.
In this case, we write m¼ inf E.
GUESS the value of M, thenCHECK parts 1 and 2.
That is, 1¼ sup E.
Part 1 says that m is a lowerbound.Part 2 says that no largernumber can be a lower bound.
26 1: Numbers
If E has a minimum element, min E, then inf E¼min E. For example, the
closed interval [0, 2] has greatest lower bound 0. We can think of the greatest
lower bound of a set, when it exists, as a kind of ‘generalised minimum
element’.
The strategy for establishing that a number is the greatest lower bound of
a set is very similar to that for proving that a number is the least upper bound
of a set.
Strategy Given a subset E of R , to show that m is the greatest lower
bound, or infimum, of E, check that:
1. x�m, for all x2E;
2. if m0>m, then there is some x2E such that x<m0.
Notice that, if m is a lower bound of E and m2 E, then part 2 is automatically
satisfied, and so m¼ inf E¼min E.
Problem 5 Determine inf E, if it exists, for each of the following sets:
(a) E1¼ (1, 5]; (b) E2 ¼ f 1n2 : n ¼ 1; 2; . . .g.
Remarks
1. For any subset E of R , inf E� sup E. This follows from the fact that, for
any x2E, we have inf E� x� sup E.
2. For any bounded interval I of R , let a be its left end-point and b its right
end-point. Then inf I¼ a and sup I¼ b.
Least upper bounds and greatest lower boundsof functions
Similar terminology applies to bounds for functions.
Definitions Let ƒ be a function defined on an interval IR . Then:
� A real number M is the least upper bound, or supremum, of ƒ on I if:
1. M is an upper bound of ƒ(I);
2. if M0<M, then M0 is not an upper bound of ƒ(I).
In this case, we write M ¼ sup f or supI
f or sup f xð Þ : x 2 If g or supx2I
f xð Þ:� A real number m is the greatest lower bound, or infimum, of ƒ on I if:
1. m is a lower bound of ƒ(I);
2. if m0>m, then m0 is not a lower bound of ƒ(I).
In this case, we write m ¼ inf f or infI
f or inf f xð Þ : x 2 If g or infx2I
f xð Þ.
Notice, for instance, that:
� if M is an upper bound for f on I, then supI
f � M,
� if m is a lower bound for f on I, then infI
f � m.
GUESS the value of m, thenCHECK parts 1 and 2.
There are similar definitionsfor the least upper bound andthe greatest lower bound of fon a general set S in R .
These are really thedefinitions for the least upperbound or the greatest lowerbound of the set {ƒ(x): x 2 I}.
Forsup
I
f is the least upper bound,
and infI
f is the greatest lower
bound, for f on I.
1.4 Least upper bounds and greatest lower bounds 27
The strategies for proving that M is the least upper bound or m the greatest
lower bound of ƒ on I are similar to the corresponding strategies for the least
upper bound or the greatest lower bound of a set E.
Strategies Let ƒ be a function defined on an interval IR . Then:
� To show that m is the greatest lower bound, or infimum, of f on I, check
that:
1. f(x)�m, for all x2 I;
2. if m0>m, then there is some x2 I such that f(x)<m0.
� To show that M is the least upper bound, or supremum, of f on I, check that:
1. f(x)�M, for all x2 I;
2. if M0<M, then there is some x2 I such that f(x)>M0.
Example 5 Let ƒ be the function defined by f xð Þ ¼ x2; x 2 12; 3
�
. Determine
the least upper bound and the greatest lower bound of ƒ on 12; 3
�
.
Solution We have already seen that 9 is an upper bound for ƒ on 12; 3
�
, and
that no smaller number will serve as an upper bound. It follows that 9 must be
the least upper bound of ƒ on 12; 3
�
.
Similarly, we have already seen that 14
is the minimum value of ƒ on 12; 3
�
;
it follows that 14
is the greatest lower bound of ƒ on 12; 3
�
, and this is actually
attained ðat the point 12Þ. &
Problem 6 Let ƒ be the function defined by f xð Þ ¼ 1x2 ; x 2 ½1; 4Þ:
Determine the least upper bound and the greatest lower bound of ƒ on ½1, 4).
Remark
For any interval I of R , infI
f � supI
f . This follows from the fact that, for
any x2 I, we have infI
f � f xð Þ � supI
f .
The least upper bound and the greatest lower bound of a function on an
interval will be particularly significant in our later work on continuity and
integrability of functions.
1.4.3 The Least Upper Bound Property
In the examples in the previous sub-section, it was easy to guess the values of
sup E and inf E. At times, however, we shall meet sets for which these values
are not so easy to determine. For example, if
E ¼ 1þ 1
n
� n
: n ¼ 1; 2; . . .
� �
;
then it can be shown that E is bounded above by 3, but it is not easy to guess the
least upper bound of E.
In such circumstances, it is reassuring to know that sup E does exist, even
though it may be difficult to find. This existence is guaranteed by the following
fundamental result.
You saw this in Example 2.
Example 2.
Chapters 4 and 7.
We will study this set closelyin Section 2.5.
28 1: Numbers
The Least Upper Bound Property of R Let E be a non-empty subset
of R . If E is bounded above, then E has a least upper bound.
We leave the proof of the Least Upper Bound Property of R to the next sub-
section. However, the Property itself is intuitively obvious. If the set E lies
entirely to the left of some number M, then you can imagine moving M steadily
to the left until you meet E. At this point, sup E has been reached.
The Least Upper Bound Property of R can be used to show that R does
include decimals which represent irrational numbers such asffiffiffi
2p
, as we
claimed in Section 1.1.
In Sections 1.2 and 1.3 we have taken for granted the existence of rational
powers and their properties, without giving formal definitions. How can we
supply these definitions? For example, how can we defineffiffiffi
2p
as a decimal?
Consider the set
E ¼ x 2 Q : x > 0; x2< 2 �
:
This is the set of positive rational numbers whose squares are less than 2.
Intuitively,ffiffiffi
2p
lies on the number line to the right of the numbers in E, but
‘only just’! In fact, we should expectffiffiffi
2p
to be the least upper bound of E.
Certainly E has a least upper bound, by the Least Upper Bound Property,
because E is bounded above, by 1.5 for example. Thus it seems likely that sup E
is the decimal representation offfiffiffi
2p
. But how can we prove that (sup E)2¼ 2?
We shall prove this in Section 1.5, once we have described how to do
arithmetic with real numbers (decimals).
Finally, note that there is a corresponding result about lower bounds.
The Greatest Lower Bound Property of R Let E be a non-empty subset
of R . If E is bounded below, then E has a greatest lower bound.
1.4.4 Proof of the Least Upper Bound Property
We know that E is a non-empty set, and we shall assume for simplicity that E
contains at least one positive number. We also know that E is bounded above.
The following procedure gives us the successive digits in a particular decimal,
which we then prove to be the least upper bound of E.
Procedure to find a¼ a0 � a1a2 . . .¼ sup E
Choose in succession:
� the greatest integer a0 such that a0 is not an upper bound of E;
� the greatest digit a1 such that a0 � a1 is not an upper bound of E;
� the greatest digit a2 such that a0 � a1a2 is not an upper bound of E;
..
.
� the greatest digit an such that a0 � a1a2 . . . an is not an upper bound of E;
..
.
The Least Upper BoundProperty of R is an exampleof an existence theorem, onewhich asserts that a realnumber exists having a certainproperty. Analysis containsmany such results whichdepend on the Least UpperBound Property of R . Whilethese results are often verygeneral, and their proofselegant, they do not alwaysprovide the most efficientmethods of calculating goodapproximate values for thenumbers in question.
You may omit this proof at afirst reading.
For example, if
E ¼fx 2 Q : x > 0;
x2 < 2g;then
a0 ¼ 1; since 12 < 2 < 22;
a0 � a1 ¼ 1:4; since
1:42 < 2 < 1:52;
a0 � a1a2 ¼ 1:41; since
1:412 < 2 < 1:422;
..
.
1.4 Least upper bounds and greatest lower bounds 29
Thus, at the nth stage we choose the digit an so that:
� a0 � a1a2 . . . an is not an upper bound of E;
� a0 � a1a2 . . . an þ 110n is an upper bound of E.
We now prove that the least upper bound of E is a¼ a0 � a1a2 . . ..First, we have to prove that a is an upper bound of E. To do this, we prove
that, if x> a, then x =2E (this is equivalent to proving, that, if x2E, then x� a).
We begin by representing x as a non-terminating decimal x¼ x0 � x1x2 . . ..Since x> a, there is an integer n such that
a < x0 � x1x2 . . . xn:
Hence
x0 � x1x2 . . . xn � a0 � a1a2 . . . an þ1
10n;
and so, by our choice of an, x¼ x0 � x1 x2 . . . xn is an upper bound of E. Since
x> x0 � x1x2 . . . xn, we have that x =2E, as required.
Next, we have to show that, if x< a, then x is not an upper bound of E. Since
x< a, there is an integer n such that
x < a0 � a1a2 . . . an;
and so x is not an upper bound of E, by our choice of an.
Thus we have proved that a is the least upper bound of E. &
Remark
Notice that this proof does not use any arithmetical properties of the real
numbers but only their order properties, together with the arithmetical proper-
ties of rational numbers. In the next section, we use the Least Upper Bound
Property to define some of the arithmetical operations on R .
1.5 Manipulating real numbers
1.5.1 Arithmetic in R
At the end of Section 1.1 we discussed the decimalsffiffiffi
2p¼ 1:41421356 . . . and p ¼ 3:14159265 . . .;
and asked whether it is possible to add and multiply these numbers to obtain
another real number. We now explain how this can be done, using the Least
Upper Bound Property of R .
A natural way to obtain a sequence of approximations to the sumffiffiffi
2pþ p
is to truncate each of the above decimals, and form the sums of the
truncations.
If each of the decimals is truncated at the same decimal place, this gives a
sequence of approximations which is increasing:
Here we are assuming thatffiffiffi
2p
and p can be represented asdecimals.
30 1: Numbers
ffiffiffi
2p
pffiffiffi
2pþ p
1 3 4
1.4 3.1 4.5
1.41 3.14 4.55
1.414 3.141 4.555
1.4142 3.1415 4.5557... ..
. ...
Intuitively, we should expect that the sumffiffiffi
2pþ p is greater than each of the
numbers in the right-hand column, but ‘only just’! To accord with our intuition,
therefore, we define the sumffiffiffi
2pþ p to be the least upper bound of the set of
numbers in the right-hand column; that isffiffiffi
2pþ p ¼ sup 4; 4:5; 4:55; 4:555; 4:5557; . . .f g:
To be sure that this definition makes sense, we need to show that this set
is bounded above. But all the truncations offfiffiffi
2p
are less than 1.5, and all the
truncations of p are less than, say, 4. Hence, all the sums in the right-hand
column are less than 1.5þ 4¼ 5.5. So, by the Least Upper Bound Property, the
set of numbers in the right-hand column does have a least upper bound, and we
can defineffiffiffi
2pþ p in this way.
This method can be used to define the sum of any pair of positive real
numbers.
Let us check that this method of adding decimals gives the correct answer
when we use it in a familiar case. Consider the simple calculation
1
3þ 2
3¼ 0:333 . . .þ 0:666 . . .:
Truncating each of these decimals and forming the sums, we obtain the set
0; 0:9; 0:99; 0:999; . . .f g:
The supremum of this set is, of course, the number 0.999. . .¼ 1, which is the
correct answer.
Similarly, we can define the product of any two positive real numbers. For
example, to defineffiffiffi
2p� p, we can form the sequence of products of their
truncations:
ffiffiffi
2p
pffiffiffi
2p� p
1 3 3
1.4 3.1 4.34
1.41 3.14 4.4274
1.414 3.141 4.441374
1.4142 3.1415 4.4427093... ..
. ...
We do not expect you to usethis method to add decimals!
1.5 Manipulating real numbers 31
As before, we defineffiffiffi
2p� p to be the least upper bound of the set of numbers
in the right-hand column.
Similar ideas can be used to define the operations of subtraction and
division.
Thus we can define arithmetic with real numbers in terms of the familiar
arithmetic with rationals, using the Least Upper Bound Property of R .
Moreover, it can be proved that these operations in R satisfy all the usual
properties of a field.
1.5.2 The existence of roots
Just as we usually take for granted the basic arithmetical operations with real
numbers, so we usually assume that, given any positive real number a, there is
a unique positive real number b ¼ffiffiffi
ap
such that b2¼ a. We now discuss the
justification for this assumption.
First, here is a geometrical justification. Given line segments of lengths 1
and a, we can construct a semi-circle with diameter aþ 1 as shown.
a
b
1
Using similar triangles, we see that
a
b¼ b
1;
and so
b2 ¼ a:
This shows that there should be a positive real number b such that b2¼ a,
so that the length of the vertical line segment in the figure can be described
exactly by the expressionffiffiffi
ap
. But does b ¼ffiffiffi
ap
exist exactly as a real number?
In fact it does, and a more general result is true.
Theorem 1 For each positive real number a and each integer n> 1, there
is a unique positive real number b such that
bn ¼ a:
We call this number b the nth root of a, and we write b ¼ffiffiffi
anp
. We also defineffiffiffi
0np¼ 0, since 0n¼ 0, and if n is odd we define
ffiffiffiffiffiffiffiffiffiffi
�að Þnp
¼ �ffiffiffi
anp
, since
�ffiffiffi
anp
ð Þn¼ �a if n is odd.
Let us illustrate Theorem 1 with the special case a¼ 2 and n¼ 2. In this case,
Theorem 1 asserts the existence of a real number b such that b2¼ 2. In other
words, it asserts the existence of a decimal b which can be used to defineffiffiffi
2p
precisely.
Here is a direct proof of Theorem 1 in this special case. We choose the
numbers 1, 1.4, 1.41, 1.414, . . . to satisfy the inequalities
We omit the details.
These properties were listedin Sub-section 1.1.5.
For each positive integer n,we can also construct
ffiffiffi
np
asfollows:
1
1
1 1
1
1
1
2
345
6
7
We shall prove Theorem 1 inSub-section 4.3.3.
For example,ffiffiffiffiffiffiffiffiffiffi
�8ð Þ3p
¼ �2:
32 1: Numbers
12< 2 < 22
ð1:4Þ2< 2 < ð1:5Þ2
ð1:41Þ2< 2 < ð1:42Þ2
ð1:414Þ2< 2 < ð1:415Þ2
..
.
(1)
This process gives an infinite decimal
b ¼ 1:414 . . .;
and we claim that
b2 ¼ ð1:414 . . .Þ2 ¼ 2:
This can be proved using our method of multiplying decimals:
b b b2
1 1 1
1.4 1.4 1.96
1.41 1.41 1.9881
1.414 1.414 1.999396... ..
. ...
We have to prove that the least upper bound of the set E of numbers in the right-
hand column is 2, in other words that
sup E ¼ sup f1; ð1:4Þ2; ð1:41Þ2; ð1:414Þ2; . . .g ¼ 2:
To do this, we employ the strategy given in Sub-section 1.4.2.
First, we check that M¼ 2 is an upper bound of E. This follows from the left-
hand inequalities in (1).
Next, we check that, if M0< 2, then there is a number in E which is greater
than M0. To prove this, put
x0 ¼ 1; x1 ¼ 1:4; x2 ¼ 1:41; x3 ¼ 1:414; . . .:
Then, by the right-hand inequalities in (1), we have that
xn þ1
10n
� 2
> 2:
Also
xn þ1
10n
� 2
� x2n ¼
1
10n2xn þ
1
10n
�
51
10n2� 2þ 1ð Þ ¼ 5
10n;
and so
x2n > xn þ
1
10n
� 2
� 5
10n
> 2� 5
10n
¼ 1:99 . . . 95:n digits
Notice that
b ¼ 1:414 . . .
is the decimal that weobtained as the least upperbound of the set
fx 2 Q : x > 0; x2 < 2gin Sub-section 1.5.1.
For example, if n¼ 1, then
1:4þ 1
10
� 2
¼ 1:52 > 2:
For example, if n¼ 2, then
ð1:41Þ2 > 1:95:
1.5 Manipulating real numbers 33
So, if M0< 2, then we can choose n so large that xn2>M0 (while still having
xn2E). This proves that the least upper bound of E is 2, and so (1.414. . .)2¼ 2.
Thus we can defineffiffiffi
2p¼ 1:414 . . .;
which justifies our earlier claim thatffiffiffi
2p
can be represented exactly by a
decimal.
1.5.3 Rational powers
Having discussed nth roots, we are now in a position to define the expression
ax, where a is positive and x is rational.
Definition If a> 0, m2Z and n2N , then
amn ¼
ffiffiffi
anp� �m
:
For example, for a> 0, with m¼ 1 we have a1n ¼
ffiffiffi
anp
, and with m¼ 2 and
n¼ 3 we have a23 ¼
ffiffiffi
a3pð Þ2:
This notation is particularly useful, because rational powers (or rational expo-
nents) satisfy the following exponent laws (whose proofs depend on Theorem 1):
Exponent Laws
� If a, b> 0 and x2Q , then axbx¼ (ab)x.
� If a> 0 and x, y2Q , then ax ay¼ axþ y.
� If a> 0 and x, y2Q , then axð Þy¼ axy.
If x and y are integers, these laws actually hold for all non-zero real numbers
a and b. However, if x and y are not integers, then we must have a, b> 0. For
example, �1ð Þ12 is not defined as a real number.
However, if a is a negative real number, then amn can be defined whenever
m2Z, n2N and mn
is reduced to its lowest terms with n odd, as follows
amn ¼
ffiffiffi
anp� �m
:
Finally, you may have wondered why we did not mention that each positive
number has two nth roots when n is even. For example, 22¼ (�2)2¼ 4. We
shall adopt the convention that, for a> 0,ffiffiffi
anp
and a1n always refer to the positive
nth root of a. If we wish to refer to both roots (for example, when solving
equations), we write �ffiffiffi
anp
.
1.5.4 Real powers
We conclude this section by briefly discussing the meaning of ax when a> 0
and x is an arbitrary real number. We have defined this expression when x is
rational, but the same definition does not work if x is irrational. However, it
is common practice to write down expressions such asffiffiffi
2p ffiffi
2p
, and even to apply
the Exponent Laws to give equalities such as
For example
212 � 3
12 ¼ 6
12;
212 � 2
13 ¼ 2
56;
212
� �13¼ 2
16:
This extends our abovedefinition of a
mn ; for instance,
it defines a1n whenever n2N
and n is odd. For example,�8ð Þ
23¼ 4.
34 1: Numbers
ffiffiffi
2p ffiffi
2p�
ffiffi
2p
¼ffiffiffi
2p� �
ffiffi
2p�ffiffi
2p
¼ffiffiffi
2p� �2
¼ 2:
Can such manipulations be justified?
In fact, it is possible to define ax, for a> 0 and x2R , using the Least Upper
Bound Property of R , but it is then rather tricky to check that the Exponent
Laws work. In Chapters 2 and 3 we shall explain how to define the expression
ex, and in Chapter 4 we use ex to define the real powers in general and show that
the Exponent Laws hold. For the time being, whenever the expression ax
appears, you should assume that x is rational.
1.6 Exercises
Section 1.1
1. Arrange the following numbers in increasing order:
(a) 736; 3
20; 1
6; 7
45; 11
60;
(b) 0:465, 0:465, 0:465, 0.4655, 0.4656.
2. Find the fractions whose decimal expansions are:
(a) 0:481; (b) 0:481.
3. Let x ¼ 0:21 and y ¼ 0:2. Find xþ y and xy (in decimal form).
4. Find a rational number x and an irrational number y in the interval
(0.119, 0.12).
5. Prove that, if n is a positive integer which is not a perfect square, thenffiffiffi
np
is irrational.
Hint: If p, q are positive integers such that�
pq
�2 ¼ n, and k is the positive
integer such that k < pq< k þ 1 (why does such a positive integer k exist?),
show that 0< p� kq< q and nq�kpp�kq
¼ pq, and hence obtain a contradiction.
Section 1.2
1. Solve the following inequalities:
(a) x�1x2þ4
< xþ1x2�4
; (b)ffiffiffiffiffiffiffiffiffiffiffiffiffi
4x� 3p
> x;
(c) 17� 2x4�
�
�
� � 15; (d) xþ 1j j þ x� 1j j < 4.
Section 1.3
1. Use the Triangle Inequality to prove that
aj j � 1) a� 3j j � 2:
2. Prove that (a2þ b2)(c2þ d2)� (acþ ad)2, for any a, b, c, d2R .
3. Prove the inequality 3n� 2n2þ 1, for n¼ 1, 2, . . .:
1.6 Exercises 35
(a) by using the Binomial Theorem, applied to (1þ x)n with x¼ 2;
(b) by using the Principle of Mathematical Induction.
4. Use the Principle of Mathematical Induction to prove that, for n¼ 1, 2, . . . :
(a) 12 þ 22 þ 32 þ � � � þ n2 ¼ n nþ1ð Þ 2nþ1ð Þ6
;
(b)
ffiffiffiffiffiffiffiffi
5=4
4nþ1
q
� 1�3�5� ... � 2n�1ð Þ2�4�6� ... � 2nð Þ �
ffiffiffiffiffiffiffiffi
3=4
2nþ1
q
.
5. Apply Bernoulli’s Inequality, first with x ¼ 2n
and then with x ¼ �2ð3nÞ to
prove that1þ 2
3n� 2� 3
1n � 1þ 2
n; for n ¼ 1; 2; . . . :
6. By applying the Arithmetic Mean–Geometric Mean Inequality to the nþ 1
positive numbers 1, 1� 1n
, 1� 1n
, 1� 1n
, . . . , 1� 1n, prove that
1� 1n
� �n� 1� 1nþ1
� �nþ1
; for n ¼ 1; 2; . . . :
7. Use the Cauchy–Schwarz Inequality to prove that, if a1, a2, . . . , an are
positive numbers with a1þ a2þ � � � þ an¼ 1, thenffiffiffi
ap
1 þffiffiffi
ap
2 þ � � � þffiffiffi
ap
n �ffiffiffi
np
:
8. Use the Cauchy–Schwarz Inequality to prove that
3ffiffiffiffiffiffiffiffiffiffi
cos x4p
þ 4
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1�ffiffiffiffiffiffiffiffiffiffi
cos xpq
� 5; for x 2 0; p2
�
:
Section 1.4
In Exerc ises 1–4 , take E1¼ {x: x2Q , 0� x< 1} and E2 ¼
1þ 1n
� �2:
n ¼ 1; 2; . . .g:1. Prove that each of the sets E1 and E2 is bounded above. Which of them has
a maximum element?
2. Prove that each of the sets E1 and E2 is bounded below. Which of them has
a minimum element?
3. Determine the least upper bound of each of the sets E1 and E2.
4. Determine the greatest lower bound of each of the sets E1 and E2.
5. For each of the following functions, determine whether it has a maximum
or a minimum, and determine its supremum and infimum:
(a) f xð Þ ¼ 11þx2 ; x 2 0; 1½ Þ; (b) f xð Þ ¼ 1� xþ x2; x 2 0; 2½ Þ.
6. Prove that, for any two numbers a, b2R
minfa; bg ¼ 12
aþ b� ja� bjð Þ and
maxfa; bg ¼ 12
aþ bþ jaþ bjð Þ:
36 1: Numbers
2 Sequences
This chapter deals with sequences of real numbers, such as
1;1
2;1
3;1
4;1
5;1
6; . . .;
0; 1; 0; 1; 0; 1; . . .;
1; 2; 4; 8; 16; 32; . . .:
It describes in detail various properties that a sequence may possess, the most
important of which is convergence. Roughly speaking, a sequence is conver-
gent, or tends to a limit, if the numbers, or terms, in the sequence approach
arbitrarily close to a unique real number, which is called the limit of the
sequence. For example, we shall see that the sequence
1;1
2;1
3;1
4;1
5;1
6; . . .
is convergent with limit 0. On the other hand, the terms of the sequence
0; 1; 0; 1; 0; 1; . . .
do not approach arbitrarily close to any unique real number, and so this
sequence is not convergent. Likewise, the sequence
1; 2; 4; 8; 16; 32; . . .
is not convergent.
A sequence which is not convergent is called divergent. The sequence
0; 1; 0; 1; 0; 1; . . .;
is a bounded divergent sequence. The sequence
1; 2; 4; 8; 16; 32; . . .
is unbounded; its terms become arbitrarily large and positive, and we say that it
tends to infinity.
Intuitively, it seems plausible that some sequences are convergent, whereas
others are not. However, the above description of convergence, involving the
phrase ‘approach arbitrarily close to’, lacks the precision required in Pure
Mathematics. If we wish to work in a serious way with convergent sequences,
prove results about them and decide whether a given sequence is convergent,
then we need a rigorous definition of the concept of convergence.
Historically, such a definition emerged only in the late nineteenth century,
when mathematicians such as Cantor, Cauchy, Dedekind and Weierstrass sought
to place Analysis on a rigorous non-intuitive footing. It is not surprising, there-
fore, that the definition of convergence seems at first sight rather obscure, and it
may take you a little time to master the logic that it involves.
Three dots are used to indicatethat the sequence continuesindefinitely.
37
In Section 2.1 we show how to picture the behaviour of a sequence by
drawing a sequence diagram. We also introduce monotonic sequences; that is,
sequences which are either increasing or decreasing.
In Section 2.2 we explain the definition of a null sequence; that is, a
sequence which is convergent with limit 0. We then establish various proper-
ties of null sequences, and list some basic null sequences.
In Section 2.3 we discuss general convergent sequences (that is, sequences
which converge but whose limit is not necessarily 0), together with techniques
for calculating their limits.
In Section 2.4 we study divergent sequences, giving particular emphasis to
sequences which tend to infinity or tend to minus infinity. We also show that
convergent sequences are bounded; it follows that unbounded sequences are
necessarily divergent.
In Section 2.5 we prove the Monotone Convergence Theorem, which states
that any increasing sequence which is bounded above must be convergent, and,
similarly, that any decreasing sequence which is bounded below must be con-
vergent. We use this theorem to study simple examples of sequences defined by
recurrence formulas, and particular sequences which converge to p and e.
2.1 Introducing sequences
2.1.1 What is a sequence?
Ever since learning to count you have been familiar with the sequence of
natural numbers
1; 2; 3; 4; 5; 6; . . .:
You have also encountered many other sequences of numbers, such as
2; 4; 6; 8; 10; 12; . . .;
1
2;1
4;1
8;
1
16;
1
32;
1
64; . . .:
We begin our study of sequences with a definition and some notation.
Definition A sequence is an unending list of real numbers
a1; a2; a3; . . .:
The real number an is called the nth term of the sequence, and the sequence
is denoted by
anf g:
In each of the sequences above, we wrote down the first few terms and left
you to assume that subsequent terms were obtained by continuing the pattern in
an obvious way. It is sometimes better, however, to give a precise description
of a typical term of a sequence, and we do this by stating an explicit formula for
the nth term. Thus the expression {2n� 1} denotes the sequence
1; 3; 5; 7; 9; 11; . . .;
Alternative notations areanf g11 and anf g1n¼1:
38 2: Sequences
and the sequence {an} defined by the statement
an ¼ �1ð Þn; n ¼ 1; 2; . . .;
has terms
a1 ¼ �1; a2 ¼ 1; a3 ¼ �1; a4 ¼ 1; a5 ¼ �1; . . .:
Problem 1
(a) Calculate the first five terms of each of the following sequences:
(i) {3nþ 1}; (ii) {3�n}; (iii) {(� 1)nn}.
(b) Calculate the first five terms of each of the following sequences {an}:
(i) an ¼ n!; n ¼ 1; 2; . . .;
(ii) an ¼ 1þ 1n
� �n; n ¼ 1; 2; . . . (to 2 decimal places).
Sequences often begin with a term corresponding to n¼ 1. Sometimes,
however, it is necessary to begin a sequence with some other value of n. We
indicate this by writing, for example, anf g13 to represent the sequence
a3; a4; a5; . . .:
Sequence diagrams
It is often helpful to picture how a given sequence {an} behaves by drawing a
sequence diagram; that is, a graph of the sequence in R2. To do this, we mark
the values n¼ 1, 2, 3, . . . on the x-axis and, for each value of n, we plot the
point (n,an). For example, the sequence diagrams for the sequences {2n� 1},1n
� �
and {(� 1n)} are as follows:
Problem 2 Draw a sequence diagram, showing the first five points, for
each of the following sequences:
(a) {n2}; (b) {3}; (c) 1þ 1n
�ng��
; (d)ð�1Þn
n
n o
.
(In part (c), use the result of Problem 1, part (b).)
2.1.2 Monotonic sequences
Many of the sequences considered so far have the property that, as n increases,
their terms are either increasing or decreasing. For example, the sequence
{2n� 1} has terms 1, 3, 5, 7, . . ., which are increasing, whereas the sequence
For example, the sequence1
n!�n
� �
cannot begin withn¼ 1 or n¼ 2.
In Figure (a), the pointsplotted all lie on the straightline y¼ 2x� 1.
In Figure (b), they all lie onthe hyperbola y ¼ 1
x:
2.1 Introducing sequences 39
1n
� �
has terms 1; 12; 1
3; 1
4; . . .; which are decreasing. The sequence {(� 1)n} is
neither increasing nor decreasing. All this can be seen clearly on the above
sequence diagrams.
We now give a precise meaning to these words increasing and decreasing,
and introduce the term monotonic.
Definition A sequence {an} is:
� constant, if anþ1 ¼ an; for n ¼ 1; 2; . . .;
� increasing, if anþ1 � an; for n ¼ 1; 2; . . .;
� decreasing, if anþ1 � an; for n ¼ 1; 2; . . .;
� monotonic, if {an} is either increasing or decreasing.
Remarks
1. Note that, for a sequence {an} to be increasing, it is essential that anþ1� an,
for all n� 1. However, we do not require strict inequalities, because we
wish to describe sequences such as
1; 1; 2; 2; 3; 3; 4; 4; . . . and 1; 2; 2; 3; 4; 4; 5; 6; 6; . . .
as increasing. One slightly bizarre consequence of the definition is that
constant sequences are both increasing and decreasing!
2. A sequence {an} is said to be:
strictly increasing, if anþ1 > an; for n ¼ 1; 2; . . .;
strictly decreasing, if anþ1 < an; for n ¼ 1; 2; . . .;
strictly monotonic, if {an} is either strictly increasing or strictly
decreasing.
3. A diagram does NOT constitute a proof! In our first example we formally
establish the monotonicity properties of our three sequences.
Example 1 Determine which of the following sequences {an} are
monotonic:
(a) an¼ 2n�1, n¼ 1, 2, . . .;
(b) an ¼ 1n
, n ¼ 1, 2, . . .;
(c) an¼ (� 1)n, n¼ 1, 2, . . .
Solution
(a) The sequence {2n� 1} is monotonic because
an ¼ 2n� 1 and anþ1 ¼ 2 nþ 1ð Þ � 1 ¼ 2nþ 1;
so that
anþ1 � an ¼ 2nþ 1ð Þ � 2n� 1ð Þ ¼ 2 > 0; for n ¼ 1; 2; . . .:
Thus {2n� 1} is increasing.
(b) The sequence 1n
� �
is monotonic because
an ¼1
nand anþ1 ¼
1
nþ 1;
so that
In fact, strictly increasing.
40 2: Sequences
anþ1�an¼1
nþ1�1
n¼ n� nþ1ð Þ
nþ1ð Þn ¼�1
nþ1ð Þn< 0; for n¼ 1; 2; . . .:
Thus 1n
� �
is decreasing.
Alternatively, since an> 0, for all n, and
anþ1
an
¼ n
nþ 1< 1; for n ¼ 1; 2; . . .;
it follows that
anþ1 < an; for n ¼ 1; 2; . . .:
Thus {an} is decreasing.
(c) The sequence {(�1)n} is not monotonic. In fact, a1¼�1, a2¼ 1 and
a3¼�1.
Hence a3< a2, which means that {an} is not increasing.
Also, a2> a1, which means that {an} is not decreasing.
Thus {(�1)n} is neither increasing nor decreasing, and so is not mono-
tonic. &
Example 1 illustrates the use of the following strategies:
Strategy To show that a given sequence {an} is monotonic, consider the
expression anþ1� an.
� If anþ1 � an � 0; for n ¼ 1; 2; . . .; then anf g is increasing.
� If anþ1 � an � 0; for n ¼ 1; 2; . . .; then anf g is decreasing.
If an> 0 for all n, it may be more convenient to use the following version
of the strategy:
Strategy To show that a given sequence of positive terms, {an}, is mono-
tonic, consider the expression anþ1
an.
� If anþ1
an� 1; for n ¼ 1; 2; . . .; then anf g is increasing.
� If anþ1
an� 1; for n ¼ 1; 2; . . .; then anf g is decreasing.
Problem 3 Show that the following sequences {an} are monotonic:
(a) an ¼ n!; n ¼ 1; 2; . . .;
(b) an ¼ 2�n; n ¼ 1; 2; . . .;
(c) an ¼ nþ 1n; n ¼ 1; 2; . . .:
It is often possible to guess whether a sequence given by a specific formula is
monotonic by calculating the first few terms. For example, consider the
sequence {an} given by
an ¼ 1þ 1
n
� �n
; n ¼ 1; 2; . . .:
In Problem 1 you found that the first five terms of this sequence are
approximately
2; 2:25; 2:37; 2:44 and 2:49:
In fact, strictly decreasing.
A single counter-example issufficient to show that{(�1)n} is not increasing;similarly a (different!) singlecounter-example is sufficientto show that {(�1)n} is notdecreasing.
We will study this importantsequence in detail inSection 2.5.
2.1 Introducing sequences 41
These terms suggest that the sequence {an} is increasing, and in fact it is.
However, the first few terms of a sequence are not always a reliable guide to
the sequence’s behaviour. Consider, for example, the sequence
an ¼10n
n!; n ¼ 1; 2; . . .:
The first five terms of this sequence are approximately
10; 50; 167; 417 and 833;
this suggests that the sequence {an} is increasing. However, calculation of
more terms shows that this is not so, and the sequence diagram for {an} looks
like this:
Simplifying anþ1
an, we find that
anþ1
an
¼ 10nþ1
nþ 1ð Þ!
�
10n
n!¼ 10
nþ 1:
But 10nþ1� 1, for n¼ 9, 10, . . ., so that anþ1
an� 1, for n¼ 9, 10, . . .; it follows
that the sequence {an} is decreasing, if we ignore the first eight terms.
In a situation like this, when a given sequence has a certain property
provided that we ignore a finite number of terms, we say that the sequence
eventually has the property. Thus we have just seen that the sequence 10n
n!
� �
is
eventually decreasing.
Another example of this usage is the following statement:
the terms of the sequence {n2} are eventually greater than 100.
This statement is true because n2> 100, for all n> 10.
Problem 4 Classify each of the following statements as TRUE or
FALSE, and justify your answers (that is, if a statement is TRUE, prove it;
if a statement is FALSE, give a specific counter-example):
(a) The terms of the sequence {2n} are eventually greater than 1000.
(b) The terms of the sequence {(�1)n} are eventually positive.
(c) The terms of the sequence 1n
� �
are eventually less than 0.025.
(d) The sequence n4
4n
n o
is eventually decreasing.
The first two terms certainlyprove that the sequence isneither decreasing norconstant.
In fact
a6 ’ 1389; a7 ’ 1984;
a8 ’ 2480; a9 ’ 2756;
a10 ’ 2756; a11 ’ 2505;
a12 ’ 2088:
In particular, the fact thata12< a11 shows that thesequence {an} cannot beincreasing.
In fact, a10
a9¼1 and anþ1
an< 1;
for n ¼ 10; 11; . . .:
Note this general approach to‘TRUE’ and ‘FALSE’.
42 2: Sequences
2.2 Null sequences
2.2.1 What is a null sequence?
In this sub-section we give a precise definition of a null sequence (that is, a
sequence which converges to 0) and introduce some properties of null
sequences.
We shall frequently use the rules for rearranging inequalities which you met
in Chapter 1, so you may find it helpful to reread that section quickly before
starting here. We shall also use the following inequalities which were proved in
Sub-section 1.3.3
2n � 1þ n; for n ¼ 1; 2; . . .;
and
2n � n2; for n � 4:
Problem 1 For each of the following statements, find a number X such
that the statement is true:
(a) 1n< 1
100; for all n >X; (b) 1
n< 3
1000; for all n > X:
Problem 2 For each of the following statements, find a number X such
that the statement is true:
(a)�1ð Þnn2
< 1
100; for all n> X; (b)
�1ð Þnn2
< 3
1000; for all n> X:
The solutions of Problems 1 and 2 both suggest that the larger and larger we
choose n, the closer and closer to 0 the terms of the sequences 1n
� �
and�1ð Þnn2
n o
become.
We can express this in terms of the Greek letter " (pronounced ‘epsilon’),
which we introduce to denote a positive number that may be as small as we
please in any given particular instance. In terms of the sequence diagrams for1n
� �
and�1ð Þnn2
n o
, this means that the terms of these sequences eventually lie
inside a horizontal strip in the sequence diagram from�" up to ". However, the
smaller we choose ", the further to the right we have to go before we can be sure
that all the terms of the sequence from that point onwards lie inside the strip.
That is, the smaller we choose " the larger we have to choose X if we wish to have
1
n< "; for all n > X; or
�1ð Þn
n2
< "; for all n > X:
Definition A sequence {an} is a null sequence if:
for each positive number ", there is a number X such that
anj j < "; for all n > X: (1)
Using this terminology, it follows from our previous discussion that the
sequences 1n
� �
and�1ð Þnn2
n o
are both null.
We can interpret our finding of a suitable number X for (1) to hold as an
‘"�X game’ in which player A chooses a positive number " and challenges
player B to find some number X such that the property (1) holds.
Section 1.2
In fact, we use
2n � n; for n ¼ 1; 2; . . .:
Thus in Problems 1 and 2 wechose the particular examplesof 1
100and 3
1000in place of the
‘general’ positive number ".
ε
–εX n
Note that here X need not bean integer; any appropriatereal number will serve. X doesNOT depend on n, but does ingeneral depend on ".
2.2 Null sequences 43
This is the case because, if
n > 1ffiffi
"p ½¼ X�,
then we have n2 > 1" or " > 1
n2;
hence, for n>X, we have
anj j ¼ 1n2 <
1X2 ¼ ":
All that happens is that thedefinition (1) remains validwith a possibly different valueof X having to be chosen.
This is often expressed in thefollowing memorable way:‘a finite number of terms donot matter’.
For example, if 7þ p willserve as a suitable value for X,then so will 12 or 37; but 10might not.
The term inside the modulusis positive.
Here we use the ReciprocalRule for inequalities that youmet in Sub-section 1.2.1.
Here we use the Power Rulefor inequalities that you metin Sub-section 1.2.1.
For example, consider the sequence {an} where an ¼ �1ð Þnn2 ; n ¼ 1; 2; . . ..
This sequence is null. For
anj j ¼�1ð Þn
n2
¼ 1
n2;
so that, if we make the choice X ¼ 1ffiffi
"p , it is certainly true that
anj j < "; for all n > X:
In terms of the "�X game, this simply means that whatever choice of " is
made by player A, player B can always win by making the choice X ¼ 1ffiffi
np !
Remarks
1. In the sequence�1ð Þnn2
n o
, the signs of the terms made no difference to
whether the sequence was null, for they disappeared immediately we took
the modulus of the terms in order to examine whether the definition (1) of a
null sequence was satisfied. Indeed, in general, the signs of terms in a
sequence make no difference to whether a sequence is null. We can express
this formally as follows:
A sequence {an} is null if and only if the corresponding sequence {janj} is null.
2. A null sequence {an} remains null if we add, delete or alter a finite number
of terms in the sequence.
Similarly, a non-null sequence remains non-null if we add, delete or alter
a finite number of terms.
3. If one number serves as a suitable value of X for the inequality in (1) to hold,
then any larger number will also serve as a suitable X. Hence, for simplicity
in some proofs, we may assume if we wish that our initial choice of X in (1)
is a positive integer.
Example 1 Prove that the sequence 1n3
� �
is a null sequence.
Solution We have to prove that
for each positive number ", there is a number X such that
1
n3
< "; for all n > X: (2)
In order to find a suitable value of X for (2) to hold, we rewrite the inequality1n3
< " in various equivalent ways until we spy a value for X that will suit
our purpose. Now
1
n3
< " , 1
n3< "
, n3 >1
"
, n >1ffiffiffi
"3p :
So, let us choose X to be 1ffiffi
"3p . With this choice of X, the above chain of
equivalent inequalities shows us that, if n>X (the last line in the chain), then1n3
< " (in the first line). Thus, with this choice of X, (2) holds; so 1n3
� �
is
indeed null. &
44 2: Sequences
However, any value of "greater than 1 providesno information!
Example 2 Prove that the following sequence is not null
an ¼1; if n is odd;0; if n is even:
�
Solution To prove that the sequence is not null, we have to show that it does
not satisfy the definition. In other words, we must show that the following
statement is not true:
for each positive number ", there is a number X such thatjanj5 ", for all n4X:
So, what we have to show is that the following is true:
for some positive number ", whatever X one choosesjanj 65", for all n4X:
which we may rephrase as:
there is some positive number ", such that whatever X one choosesjanj 65 ", for all n4X:
2
n
–2
1
Sequence diagram
1
n–1/2
1/2
Strip of half-width 12
So, we need to find some positive number " with such a property. The
sequence diagram provides the clue! If we choose " ¼ 12, then the point (n, an)
lies outside the strip � 12; 1
2
� �
for every odd n. In other words, whatever X one
then chooses, the statement
anj j < "; for all n > X;
is false. It follows that the sequence is not a null sequence. &
Note
Notice that any positive value of " less than 1 will serve for our purpose here;
there is nothing special about the number 12.
These two examples illustrate the following strategy:
Strategy for using the definition of null sequence
1. To show that {an} is null, solve the inequality janj<" to find a number X
(generally depending on ") such that janj<", for all n>X.
2. To show that {an} is not null, find ONE value of " for which there is NO
number X such that janj<", for all n>X.
2.2 Null sequences 45
Problem 3 Use the above strategy to determine which of the following
sequences are null:
(a) 12n�1
� �
; (b)�1ð Þn10
n o
; (c)�1ð Þn
n4þ1
n o
:
We now look at a number of Rules for ‘getting new null sequences from old’.
Power Rule If {an} is a null sequence, where an� 0, for n¼ 1, 2, . . ., and
p> 0, then apn
� �
is a null sequence.
Thus, for example, the sequence�
1ffiffi
np�
is null – we simply apply the Power
Rule to the sequence 1n
� �
that we saw earlier to be null, using the positive
power p ¼ 12.
Remark
Notice that the Power Rule also holds without the requirement that an � 0, so
long as apn is defined – for example, if p ¼ 1
mwhere m is a positive integer.
Combination Rules If {an} and {bn} are null sequences, then the follow-
ing are also null sequences:
Sum Rule {anþ bn};
Multiple Rule {lan}, for any real number l;
Product Rule {anbn}.
Thus, for example, we may use known examples of null sequences to verify
that the following sequences are also null:
1nþ 1
n3
� �
– by applying the Sum Rule to the null sequences 1n
� �
and 1n3
� �
;47pn3
� �
– by applying the Multiple Rule to the null sequence 1n3
� �
with
l¼ 47p;
f 1n 2n�1ð Þg – by applying the Product Rule to the null sequences 1
n
� �
and
12n�1
� �
:
Problem 4 Use the above rules to show that the following sequences
are null:
(a) 1
2n�1ð Þ3n o
; (b) 6ffiffi
n5p þ 5
2n�1ð Þ7n o
; (c) f 1
3n4 2n�1ð Þ13g:
Our next rule, the Squeeze Rule, also enables us to ‘get new null sequences
from old’ – but in a slightly different way. To illustrate this rule, we look first at
the sequence diagrams of the two sequences�
1ffiffi
np�
and�
11þffiffi
np�
.
11 + n}}
1n }}
n
1
. . .
. . .
√
√
We shall prove these Ruleslater, in Sub-section 2.2.2.
Recall that apn ¼ anð Þp:
For 1ffiffi
np ¼ 1
n
� �12:
Note that the number l has tobe a fixed number that doesnot depend on n.
In your solution, you may useany of the sequences that youhave proved to be null so farin this sub-section.
46 2: Sequences
The points corresponding to the sequence�
11þffiffi
np�
are squeezed in between
the horizontal axis and the points corresponding to the null sequence�
1ffiffi
np�
,
since 11þffiffi
np < 1
ffiffi
np , for n¼ 1, 2, . . .. Hence, if from some point onwards all the
points corresponding to�
1ffiffi
np�
lie in a narrow strip in the sequence diagram of
half-width " about the axis, then (from the same point onwards) all the points
corresponding to�
11þffiffi
np�
will also lie in the same strip. So, since " may be any
positive number, it certainly looks from this sequence diagram argument that
the sequence�
11þffiffi
np�
must be a null sequence too.
Squeeze Rule If {bn} is a null sequence and
anj j � bn; for n ¼ 1; 2; . . .;
then {an} is a null sequence.
The trick in using the Squeeze Rule to prove that a given sequence {an} is
null is to think of a suitable sequence {bn} that dominates {an}and is itself null.
Thus, for example, since the sequence�
1ffiffi
np�
is null and 11þffiffi
np < 1
ffiffi
np , it follows
from the Squeeze Rule that�
11þffiffi
np�
is also a null sequence.
Proof of the Squeeze Rule We want to prove that {an} is null; that is: for
each positive number ", there is a number X such that
janj < "; for all n > X: (3)
We know that {bn} is null, so there is some number X such that
jbnj < "; for all n > X: (4)
We also know that janj � bn, for n¼ 1, 2, . . ., and hence it follows from (4) that
janj ð< jbnjÞ < "; for all n > X:
Thus inequality (3) holds, as required. &
Remark
In applying the Squeeze Rule, it is often useful to remember the following fact:
The behaviour of a finite number of terms does not matter: it is sufficient to
check that janj � bn holds eventually.
Example 3 Use the Squeeze Rule to prove that the sequence 12
� �n� �
is null.
Proof We want to prove that 12
� �n� �
is null.
What sequence can we find that dominates 12
� �n� �
? Well, we saw at the start
of this sub-section that 2n� n, for n¼ 1, 2, . . ., and it follows from this, by the
Reciprocal Rule for inequalities, that 12n � 1
n; for n ¼ 1; 2; . . .:
Thus the sequence 1n
� �
dominates the sequence 12
� �n� �
, and is itself null. It
follows from the Squeeze Rule that 12
� �n� �
is null. &
Problem 5 Use the inequality 2n� n2, for n� 4, and the Squeeze Rule
to prove that the sequence n 12
� �n� �
is null.
Example 4 Use the Squeeze Rule to prove that the sequence 10n
n!
� �
is null.
Solution We want to prove that 10n
n!
� �
is null.
With a bit of inspiration, we guess that the sequence 10n
n!
� �
is eventually
dominated by ln
� �
, for some constant l.
That is, {an} is dominated by{bn}.
And this is the X that we shalluse to verify (3).
That is, that {an} is eventuallydominated by {bn}.
2.2 Null sequences 47
Instances of these that wehave seen are
1ffiffiffi
np� �
;
1
2
� �n� �
;
n1
2
� �n� �
;
10n
n!
� �
;
n10
n!
� �
:
We suggest that you omitthese proofs on a first reading,and return to them when youare confident that youunderstand the basic ideas.
Recall that X need not be aninteger; any appropriate realnumber will serve.
Recall that anp¼ (an)p.
Writing out the expression 10n
n! in full, we see that
10n
n!¼ 10
1
� �
10
2
� �
. . .10
10
� �
� 10
11
� �
. . .10
n�1
� �
10
n
� �
< 3000� 10
n; for all n > 10;
¼ 30000
n:
In other words, 10n
n!
� �
is eventually dominated by ln
� �
, for l¼ 30000. This latter
sequence is null, by the Multiple Rule applied to the null sequence 1n
� �
:It follows, from the Squeeze Rule and the fact that 30000
n
� �
is a null sequence,
that the sequence 10n
n!
� �
is also null. &
Problem 6 Prove that the following sequences are null:
(a)�
1n2þn
�
; (b)�1ð Þnn!
n o
; (c) sin n2
n2þ2n
n o
:
We can now list a good number of generic types of null sequences, of which we
have seen specific instances already in this sub-section. We call these basic
null sequences, since we shall use them commonly together with the
Combination Rules and other rules for null sequences in order to prove that
particular sequences that we meet are themselves null.
Basic null sequences
(a) 1np
� �
; for p > 0;
(b) fcng; for jcj < 1;
(c) fnpcng; for p > 0 and jcj < 1;
(d) cn
n!
� �
; for any real c;
(e) np
n!
� �
; for p > 0:
2.2.2 Proofs
We now give a number of proofs which were omitted from the previous sub-
section so as not to slow down your gaining an understanding of the key ideas
there. First, recall the definition of a null sequence.
Definition A sequence {an} is null if
for each positive number ", there is a number X such that
janj < "; for all n > X:
Proofs of the Power Rule and the Combination Rules
In the previous sub-section we proved the Squeeze Rule, but did not prove the
Power Rule or the Combination Rules. We now supply these proofs.
Power Rule If {an} is a null sequence, where an� 0, for n¼ 1, 2, . . ., and,
if p>0, then apn
� �
is a null sequence.
48 2: Sequences
Proof We want to prove that apn
� �
is null; that is:
for each positive number ", there is a number X such that
apn < "; for all n > X: (5)
We know that {an} is null, so there is some number X such that
an < "1p; for all n > X: (6)
Taking the pth power of both sides of (6), we obtain the desired result (5), with
the same value of X. &
Remark
Notice how we used the (positive) number "1p in place of " in (6) in order to
obtain " in the final result (5). In the proofs which follow, we again apply the
definition of null sequence, using positive numbers which depend in some way
on ", in order to obtain " in the inequality that we are aiming to prove.
Sum Rule If {an} and {bn} are null sequences, then {anþ bn} is a null
sequence.
Proof We want to prove that the sum {anþ bn} is null; that is:
for each positive number ", there is a number X such that
an þ bnj j < "; for all n > X: (7)
We know that {an} and {bn} are null, so there are numbers X1 and X2 such that
anj j <1
2"; for all n > X1;
and
bnj j <1
2"; for all n > X2:
Hence, if X¼max {X1, X2}, then both the two previous inequalities hold; so
if we add them we obtain, by the Triangle Inequality, that
an þ bnj j � anj j þ bnj j
<1
2"þ 1
2" ¼ "; for all n > X:
Thus inequality (7) holds, with this value of X. &
Before going on, first a comment about the number 12
that appears several
times in the above proof. It is used twice in expressions 12" at the start of the
proof in order to end up with a final inequality that says that some expression is
‘<"’. While this means that we end up with an inequality that shows at once
that the desired result holds, in fact it is not strictly necessary to end up with
precisely ‘<"’, as the following result shows:
Lemma The ‘Ke Lemma’ Let {an} be a sequence. Suppose that, for
each positive number ", there is a number X such that
anj j < K"; for all n > X;
where K is a positive real number that does not depend on " or X. Then {an}
is a null sequence.
For janpj ¼ an
p, since an� 0.
Here we use the number "1p in
place of " in the definition ofnull sequence.
We use 12" here rather than ",
in order to end up with thesymbol " on its owneventually in the desiredinequality (7).
You met the TriangleInequality in Sub-section 1.3.1.
In this case, the result that aparticular sequence is null.
Loosely speaking, we mayexpress this result as ‘K" isjust as good as "’ in thedefinition of null sequence.
For example, K might be 2 or p7
or 259, but it could not be 2nor X
259.
2.2 Null sequences 49
Proof We want to prove that the sequence {an} is null; that is:
for each positive number ", there is a number X such that
anj j < "; for all n > X:
Now, whatever this positive number " may be, the number "K
is also a
positive number. It follows from the hypothesis stated in the Lemma that
therefore there is some number X such that
anj j < K � "
K
¼ "; for all n > X:
This is precisely the condition for {an} to be null. &
From time to time we shall use this Lemma in order to avoid arithmetic
complexity in our proofs.
Multiple Rule If {an} is a null sequence, then {lan} is a null sequence for
any real number l.
Proof We want to prove that the multiple {lan} is null; that is:
for each positive number ", there is a number X such that
lanj j < "; for all n > X: (8)
If l¼ 0, this is obvious, and so we may assume that l 6¼ 0.
We know that {an} is null, so there is some number X such that
anj j <1
lj j "; for all n > X:
Multiplying both sides of this inequality by the positive number jlj, this gives
us that
lanj j < "; for all n > X:
Thus the desired result (8) holds. &
Product Rule If {an} and {bn} are null sequences, then {anbn} is a null
sequence.
Proof We want to prove that the product {anbn} is null; that is:
for each positive number ", there is a number X such that
anbnj j < "; for all n > X: (9)
We know that {an} and {bn} are null, so there are numbers X1 and X2 such
that
anj j <ffiffiffi
"p; for all n > X1;
and
bnj j <ffiffiffi
"p; for all n > X2:
You may omit this proof on afirst reading.
We use 1lj j " here rather than "
in order to end up eventuallywith the symbol " on its ownin the desired inequality (8).
We useffiffiffi
"p
here rather than "in order to end up eventuallywith the symbol " on its ownin the desired inequality (9).
50 2: Sequences
Hence, if X¼max {X1, X2}, then both the two previous inequalities hold; so
if we multiply them we obtain that
anbnj j ¼ anj j � bnj j<
ffiffiffi
"p�
ffiffiffi
"p¼ "; for all n > X:
Thus inequality (9) holds with this value of X. &
Basic null sequences
At the end of the previous sub-section we gave a list of basic null sequences.
We end this sub-section by proving that these sequences are indeed null
sequences.
Basic null sequences The following sequences are null sequences:
(a) 1np
� �
; for p > 0;
(b) cnf g; for cj j < 1;
(c) npcnf g; for p > 0; cj j < 1;
(d) cn
n!
� �
; for any real c;
(e) np
n!
� �
; for p > 0.
Proof
(a) To prove that 1np
� �
is null, for p> 0, we simply apply the Power Rule to the
sequence 1n
� �
, which we know is null.
(b) To prove that {cn} is null, for jcj< 1, it is sufficient to consider only the
case 0� c< 1. If c¼ 0, the sequence is obviously null; so we may assume
that 0< c< 1.
With this assumption, we can write c in the form
c ¼ 1
1þ a; for some number a > 0:
Now, by the Binomial Theorem
1þ að Þn � 1þ na
� na; for n ¼ 1; 2; . . .;
and hence
cn ¼ 1
1þ að Þn
� 1
na; for n ¼ 1; 2; . . .:
Since 1n
� �
is null, we deduce that f 1nag is also null by the Multiple Rule;
hence {cn} is null, by the Squeeze Rule.
(c) To prove that {npcn} is null, for p > 0 and jcj < 1, we may again assume
that 0 < c < 1. Hence
c ¼ 1
1þ a; for some number a > 0:
For example
1
n10
� �
;
0:9ð Þnf g;
n3 0:9ð Þn� �
;
10n
n!
� �
;
n10
n!
� �
You met the BinomialTheorem in Sub-section 1.3.3.
Here we take l ¼ 1a
in theMultiple Rule.
2.2 Null sequences 51
First, we deal with the case p¼ 1. By the Binomial Theorem
1þ að Þn � 1þ naþ 1
2n n� 1ð Þa2
� 1
2n n� 1ð Þa2; for n ¼ 2; 3; . . .;
and hence
ncn ¼ n
1þ að Þn
� n12
n n� 1ð Þa2¼ 2
n� 1ð Þa2; for n ¼ 2; 3; . . .:
Since 1n�1
� �12
is null, we deduce that 2ðn�1Þa2
n o1
2is also null, by the
Multiple Rule. Hence {ncn} is null, by the Squeeze Rule. This proves
part (c) in the case p¼ 1.
To deduce that {npcn} is null for any p> 0 and 0< c< 1, we note that
npcn ¼ ndnð Þp; for n ¼ 1; 2; . . .;
where d ¼ c1p. Since 0< d< 1, we know that {ndn} is null, and so {npcn} is
null, by the Power Rule.
(d) To prove that cn
n!
� �
is null for any real c, we may assume that c> 0. If we
choose any integer m such that mþ 1> c, then we have, for n>mþ 1, that
cn
n!¼ c
1
� c
2
�
. . .c
m
�
� c
mþ 1
� �
. . .c
n� 1
� c
n
�
� c
1
� c
2
�
. . .c
m
�
� c
n
¼ K � c
n;
where K ¼ cm
m! is a constant.
Since 1n
� �
is null, we deduce that Kcn
� �
is also null, by the Multiple Rule; it
follows that cn
n!
� �
is also null, by the Squeeze Rule.
(e) To prove that np
n!
� �
is null, for p > 0, we write
np
n!¼ np
2n
� �
� 2n
n!
� �
; for n ¼ 1; 2; . . .:
Since np
2n
� �
and 2n
n!
� �
are both null sequences, by parts (c) and (d) respec-
tively, we deduce that np
n!
� �
is null, by the Product Rule. &
2.3 Convergent sequences
2.3.1 What is a convergent sequence?
In the previous section we looked at null sequences; that is, sequences which
converge to 0. We now turn our attention to sequences which converge to
limits other than 0.
Here we use the special caseof part (c) that we havealready proved,
For c is a constant and m issome constant, and hence cm
m! isalso some constant. Note thatwhat is varying is n, nothingelse.
52 2: Sequences
Problem 1 Consider the sequence an ¼ nþ1n; n ¼ 1; 2; . . ..
(a) Draw the sequence diagram for {an}, and describe (informally) how
this sequence behaves.
(b) What can you say (formally) about the behaviour of the sequence
bn ¼ an � 1; n ¼ 1; 2; . . .?
The terms of the sequence {an} in Problem 1 appear to get arbitrarily close to 1;
that is, the sequence {an} appears to converge to 1. If we subtract 1 from each
term an to form the sequence {bn}, then we obtain a null sequence. This
example suggests the following definition of a convergent sequence:
Definition The sequence {an} is convergent with limit ł, or converges to
the limit ł, if {an� ‘} is a null sequence. In this case, we say that {an}
converges to ł, and we write:
EITHER limn!1
an ¼ ‘,
OR an! ‘ as n!1.
The following are examples of convergent sequences:
every null sequence converges to 0;
every constant sequence {c} converges to c;
the sequencen
nþ1n
o
is convergent with limit 1.
Problem 2 For each of the following sequences {an}, draw its seque-
nce diagram and show that {an} converges to ‘ by considering an� ‘:(a) an ¼ n2�1
n2þ1; ‘ ¼ 1; (b) an ¼ n3þ �1ð Þn
2n3 ; ‘ ¼ 12:
The definition of convergence of a sequence is often given in the following
equivalent form:
Equivalent definition The sequence {an} is convergent with limit ł if:
for each positive number ", there is a number X such that
an � ‘j j < "; for all n > X: (1)
Remarks
1. In terms of the sequence diagram for {an}, this definition states that, for
each positive number ", the terms an eventually lie inside the horizontal
strip from ‘� " up to ‘þ ".
� + ε{an}
|an – �| < ε
� – ε < an < � + ε
n
� – ε
X
�
These statements are read as:‘the limit of an, as n tends toinfinity, is ‘’ and ‘an tends to‘, as n tends to infinity’.Often, we omit ‘as n!1’.Do not let this use of thesymbol1 tempt you to thinkthat1 is a real number.
See Problem 1 above.
Note that here X need not bean integer; any appropriatereal number will serve.
2.3 Convergent sequences 53
2. Just as for null sequences, we can interpret ‘an! ‘ as n!1’ as an "�X
game in which player A chooses a positive number " and challenges player
B to find some number X such that (1) holds.
For example, consider the sequence {an} where an ¼ 8nþ62n
, n ¼ 1, 2, . . ..This sequence is convergent, with limit 4. For
an � 4j j ¼ 8nþ 6
2n� 4
¼ 6
2n¼ 3
n;
so that, if we make the choice X ¼ 3", it is certainly true that
an � 4j j < "; for all n > X:
In terms of the "�X game, this simply means that whatever choice
of " is made by player A, player B can always win by making the choice
X ¼ 3"!
B winsBA
I'll play ANY positive
ε
I'll play
X = 3ε
3. If a sequence is convergent, then it has a unique limit. We can see this by
using a sequence diagram. Suppose that the sequence {an} has two limits, ‘and m, where ‘ 6¼m. Then it is possible to choose a positive number " such
that horizontal strips from ‘� " up to ‘þ " and from m� " up to mþ " do
not overlap. For example, we can take " ¼ 13‘� mj j.
m + ε
� + ε
� – ε
� – m
n
m – ε
�
m
However, since ‘ and m are both limits of {an}, the terms an must eventually
lie inside both strips, and this is impossible.
A formal proof of this fact is given in the Corollary to Theorem 3, later in
this section.
4. If a given sequence converges to ‘, then this remains true if we add, delete
or alter a finite number of terms. This follows from the corresponding result
for null sequences.
5. Not all sequences are convergent. For example, the sequence {(�1)n} is not
convergent.
In this section we restrict our attention to sequences which do converge.
This is the case because, if
n >3
"½¼ X�;
then we have " > 3n
or 3n< ";
hence, for n>X, we have
an � 4j j ¼ 3
n<
3
X¼ ":
The vertical distance between‘ and m is j‘�mj, so that anychoice of " that is less thanhalf of this quantity will serve.
In other words ‘altering afinite number of terms doesnot matter’. See Sub-section 2.2.1.
We discuss non-convergentsequences in Section 2.4.
54 2: Sequences
2.3.2 Combination Rules for convergent sequences
So far you have tested the convergence of a given sequence {an} by calculating
an� ‘ and showing that {an� ‘} is null. This presupposes that you know in
advance the value of ‘. Usually, however, you are not given the value of ‘. You
are given only a sequence {an} and asked to decide whether or not it converges
and, if it does, to find its limit. Fortunately many sequences can be dealt with
by using the following Combination Rules, which extend the Combination
Rules for null sequences:
Theorem 1 Combination Rules
If limn!1
an ¼ ‘ and limn!1
bn ¼ m, then:
Sum Rule limn!1
an þ bnð Þ ¼ ‘þ m;
Multiple Rule limn!1
lanð Þ ¼ l‘; for any real number l;
Product Rule limn!1
anbnð Þ ¼ ‘m;
Quotient Rulelim
n!1
an
bn
� �
¼ ‘
m; provided that m 6¼ 0:
Remarks
1. In applications of the Quotient Rule, it may happen that some of the terms bn
take the value 0, in which case an
bnis not defined. However, we shall see (in
Lemma 1) that this can happen only for finitely many bn (because m 6¼ 0), and
so {bn} is eventually non-zero. Thus the statement of the rule does make sense.
2. The following rule is a special case of the Quotient Rule:
Corollary 1 If limn!1
an ¼ ‘ and ‘ 6¼ 0, then:
Reciprocal Rule limn!1
1
an
¼ 1
‘.
We shall prove the Combination Rules at the end of this sub-section, but first
we illustrate how to apply them.
Applying the Combination Rules
Example 1 Show that each of the following sequences {an} is convergent,
and find its limit:
(a) an ¼ 2nþ 1ð Þ nþ 2ð Þ3n2 þ 3n
; (b) an ¼ 2n2 þ 10n
n!þ 3n3 .
Solution Although the expressions for an are quotients, we cannot apply the
Quotient Rule immediately, because the sequences defined by the numerators
and denominators are not convergent. In each case, however, we can rearrange
the expressions for an and then apply the Combination Rules.
(a) In this case we divide both the numerator and denominator by n2 to give
an ¼2nþ 1ð Þ nþ 2ð Þ
3n2 þ 3n¼
2þ 1n
� �
1þ 2n
� �
3þ 3n
:
Sub-section 2.2.1.
A finite number of termsdo not matter.
2.3 Convergent sequences 55
Since 1n
� �
is a basic null sequence, we find by the Combination Rules that
limn!1
an ¼2þ 0ð Þ 1þ 0ð Þ
3þ 0¼ 2
3:
(b) This time we divide both the numerator and denominator by n! to give
an ¼2n2 þ 10n
n!þ 3n3¼
2n2
n! þ 10n
n!
1þ 3n3
n!
Since�
n2
n!
�
, 10n
n!
� �
and�
n3
n!
�
are all basic null sequences, we find by the
Combination Rules that
limn!1
an ¼0þ 0
1þ 0¼ 0: &
Remark
We simplified each of the above sequences by dividing both numerator and
denominator by the dominant term. In part (a), we divided by n2, which is the
highest power of n in the expression. In part (b), the choice of dominant term was
a little harder, but the choice of n! ensured that the resulting quotients in the
numerator and denominator were all typical terms of convergent sequences.
In choosing the dominant term the following ordering is often useful:
Domination HierarchyA factorial term n! dominates a power term cn for any c 2 R .
A power term cn dominates a term np for p> 0, jcj< 1.
The above examples illustrate the following general strategy:
Strategy To evaluate the limit of a complicated quotient:
1. Identify the dominant term, bearing in mind the basic null sequences.
2. Divide both the numerator and the denominator by the dominant term.
3. Apply the Combination Rules.
Problem 3 Show that each of the following sequences {an} is con-
vergent, and find its limit:
(a) an ¼ n3 þ 2n2 þ 32n3 þ 1
; (b) an ¼ n2 þ 2n
3n þ n3 ; (c) an ¼ n!þ �1ð Þn2n þ 3n! :
Proofs of the Combination Rules We prove the Sum Rule, the Multiple
Rule and the Product Rule by using the corresponding Combination Rules for
null sequences. Remember that
limn!1
an ¼ ‘
means that
an ! ‘f g is a null sequence:
See the list of basic nullsequences (introduced inSub-section 2.2.1) which isrepeated in the margin below.
Basic null sequences:1
np
� �
; for p> 0;
cnf g; for cj j < 1;
npcnf g; for p > 0; c > 1;
cn
n!
� �
; for c 2 R ;
np
n!
� �
; for p > 0:
Warning Notice that ‘3n!’means ‘3� (n!)’ and not‘(3n)!’.
You may omit these proofs ata first reading.
56 2: Sequences
Sum Rule If limn!1
an ¼ ‘ and limn!1
bn ¼ m, then
limn!1
an þ bnð Þ ¼ ‘þ m:
Proof By assumption, {an� ‘} and {bn�m} are null sequences; since
an þ bnð Þ � ‘þ mð Þ ¼ an � ‘ð Þ þ bn � mð Þwe deduce that {(anþ bn)� (‘þm)} is null, by the Sum Rule for null
sequences. &
Product Rule If limn!1
an ¼ ‘ and limn!1
bn ¼ m, then
limn!1
anbnð Þ ¼ ‘m:
Proof The idea here is to express anbn� ‘m in terms of an� ‘ and bn�m
anbn � ‘m ¼ an � ‘ð Þ bn � mð Þ þ m an � ‘ð Þ þ ‘ bn � mð Þ:Since {an� ‘} and {bn�m} are null, we deduce that {anbn� ‘m} is null, by
the Combination Rules for null sequences. &
Remark
Note that the Multiple Rule is just a special case of the Product Rule in which
the sequence {bn} is a constant sequence.
To prove the Quotient Rule, we need to use the following lemma, which will
also be needed in the next sub-section:
Lemma 1 If limn!1
an ¼ ‘ and ‘> 0, then there is a number X such that
an >1
2‘, for all n > X:
Proof Since 12‘ > 0, the terms an must eventually lie within a distance 1
2‘ of
the limit ‘.
{an}
X n
|an – �| < 21�
21
23
� < an < �
�
23�
21�
In other words, there is a number X such that
an � ‘j j < 1
2‘; for all n > X:
Hence
� 1
2‘ < an � ‘ <
1
2‘; for all n > X;
and so the left-hand inequality gives1
2‘ < an; for all n > X;
as required. &
Here we are taking " ¼ 12‘ in
the definition of convergence.
2.3 Convergent sequences 57
Quotient Rule If limn!1
an ¼ ‘ and limn!1
bn ¼ m, then
limn!1
an
bn
� �
¼ ‘
m, provided that m 6¼ 0:
Proof We assume that m> 0; the proof for the case m< 0 is similar. Once
again the idea is to write the required expression in terms of an� ‘ and bn�m
an
bn
� ‘
m¼ m an � ‘ð Þ � ‘ bn � mð Þ
bnm:
Now, however, there is a slight problem: {m(an� ‘)� ‘(bn�m)} is certainly a
null sequence, but the denominator is rather awkward. Some of the terms bn
may take the value 0, in which case the expression is undefined.
However, by Lemma 1, we know that that for some X we have
bn >1
2m; for all n > X:
Thus, for all n>X
an
bn
� ‘
m
¼ m an � ‘ð Þ � ‘ bn � mð Þj jbnm
� m an � ‘ð Þ � ‘ bn � mð Þj j12
m2
� mj j � an � ‘j j þ ‘j j � bn � mj j12
m2:
Since this last expression defines a null sequence, it follows by the Squeeze
Rule that�
an
bn� ‘
m
�
is null. &
2.3.3 Further properties of convergent sequences
There are several other theorems about convergent sequences which will be
needed in later chapters. The first is a general version of the Squeeze Rule.
Theorem 2 Squeeze Rule
If:
1. bn� an� cn, for n¼ 1, 2, . . .,
2. limn!1
bn ¼ limn!1
cn ¼ ‘,then lim
n!1an ¼ ‘:
Proof By the Combination Rules
limn!1
cn � bnð Þ ¼ ‘� ‘ ¼ 0 ;
so that cn � bnf g is a null sequence. Also, by condition 1
0 � an � bn � cn � bn; for n ¼ 1; 2; . . .;
and so an � bnf g is null, by the Squeeze Rule for null sequences.
In particular, this implies thatthe terms of {bn} areeventually positive.
Here we use Lemma 1.
Here we apply the TriangleInequality to the numerator.
You met the Squeeze Rulefor null sequences inSub-section 2.2.1.
That is, an � bnf g is squeezedby cn � bnf g.
58 2: Sequences
Now we write an in the form
an ¼ an � bnð Þ þ bn :
Hence by the Combination Rules
limn!1
an ¼ limn!1
an � bnð Þ þ limn!1
bn
¼ 0þ ‘ ¼ ‘: &
Remark
In applications of the Squeeze Rule, it is sufficient to check that condition 1
applies eventually. This is because the values of a finite number of terms do not
affect convergence.
The following example and problem illustrate the use of the Squeeze Rule and
the Binomial Theorem in the derivation of two important limits.
Example 2
(a) Prove that, if c> 0, then
1þ cð Þ1n� 1þ c
n; for n ¼ 1; 2; . . .:
(b) Deduce that
limn!1
a1n¼ 1; for any positive number a:
Solution
(a) Using the rules for inequalities, we obtain
1þ cð Þ1n� 1þ c
n, 1þ c � 1þ c
n
�n
; since c > 0 :
The right-hand inequality holds, because
1þ c
n
�n
� 1þ nc
n
�
¼ 1þ c ;
by the Binomial Theorem. It follows that the required inequality also holds.
(b) We consider the cases a> 1, a¼ 1, a< 1 separately.
If a> 1, then we can write a¼ 1þ c, where c> 0. Then, by part (a)
1 � a1n ¼ 1þ cð Þ
1n � 1þ c
n; for n ¼ 1; 2; . . .:
Since limn!1
1þ cn
� �
¼ 1, we deduce that
limn!1
a1n ¼ 1;
by the Squeeze Rule.
If a¼ 1, then a1n ¼ 1, for n¼ 1, 2, . . ., so
limn!1
a1n ¼ 1:
If 0< a< 1, then 1a> 1, so that lim
n!11a
� �1n¼ 1 by the first case in part (b).
Hence, by the Reciprocal Rule
limn!1
a1n ¼ 1
limn!1
1a
� �1n
¼ 1
1¼ 1: &
A finite number of terms donot matter.
We proved this inequality forthe case c ¼ 1 in Example 6(b)of Sub-section 1.3.3.
All the terms in the BinomialTheorem expansion ofð1þ c
nÞn are positive, so we
get a smaller number if weignore all the terms after thefirst two.
In this application of theSqueeze Rule, the ‘lower’sequence is {1}.
2.3 Convergent sequences 59
Problem 4
(a) Use the Binomial Theorem to prove that
n1n �1þ
ffiffiffiffiffiffiffiffiffiffiffi
2
n� 1
r
; for n ¼ 2; 3; . . .:
Hint: 1þ xð Þn� n n�1ð Þ2! x2; for n � 2; x � 0:
(b) Use the Squeeze Rule to deduce that
limn!1
n1n ¼ 1:
Next we show that taking limits preserves weak inequalities.
Theorem 3 Limit Inequality Rule
If limn!1
an ¼ ‘ and limn!1
bn ¼ m, and also
an � bn; for n ¼ 1; 2; . . .;
then ‘�m.
Proof Suppose that an! ‘ and bn!m, where an� bn for n¼ 1, 2, . . ., but
that it is not true that ‘�m. Then ‘>m and so, by the Combination Rules
limn!1
an � bnð Þ ¼ ‘� m > 0:
Hence, by Lemma 1, there is an X such that
an � bn >1
2‘� mð Þ; for all n > X: (2)
However, we assumed that an� bn� 0, for n¼ 1, 2, . . ., so statement (2) is a
contradiction.
Hence it is true that ‘�m. &
Warning Taking limits does NOT preserve strict inequalities. That is, if
limn!1
an ¼ ‘ and limn!1
bn ¼ m, and also an< bn, for n¼ 1, 2, . . ., then it follows
from Theorem 3 that ‘�m – but it does NOT necessarily follow that ‘<m.
For example, 12n< 1
n, for n¼ 1, 2, . . ., but lim
n!11
2n¼ lim
n!11n¼ 0.
We can now give the formal proof, promised earlier, that a convergent
sequence has only one limit.
Corollary 2 If limn!1
an ¼ ‘ and limn!1
an ¼ m, then ‘¼m.
Proof Applying Theorem 3 with bn¼ an, we deduce that ‘�m and m� ‘.Hence ‘¼m. &
In Sub-section 2.2.1, we saw that a sequence {an} is null if and only if the
sequence jan jf g is null. Our next result is a partial generalisation of this fact.
Theorem 4 If limn!1
an ¼ ‘, then limn!1
anj j ¼ ‘j j.
This is a ‘proof bycontradiction’.
60 2: Sequences
Proof Using the ‘reverse form’ of the Triangle Inequality, we obtain
anj j � ‘j j
� an � ‘j j; for n ¼ 1; 2; . . .:
Since {an� ‘} is null, we deduce from the Squeeze Rule for null sequences that
anj j � ‘j jf g is null, as required. &
Remark
Theorem 4 is only a partial generalisation of the earlier result about null
sequences, because the converse statement does not hold. That is, if janj! j‘j,it does NOT necessarily follow that an! ‘. For example, consider the sequence
an¼ (�1)n, for n¼ 1, 2, . . .; in this case, janj! 1, but {an} does not even
converge.
2.4 Divergent sequences
2.4.1 What is a divergent sequence?
We have commented several times that not all sequences are convergent. We
now investigate the behaviour of sequences which do not converge.
Definition A sequence is divergent if it is not convergent.
Here are the sequence diagrams for �1ð Þnf g, {2n} and �1ð Þnnf g. Each of
these sequences is divergent but, as you can see, they behave differently.
It is rather tricky to prove, directly from the definition, that these sequences
are divergent.
The aim of this section is to obtain criteria for divergence, which avoid
having to argue directly from the definition. At the end of the section, we give
a strategy involving two criteria which deal with all types of divergence.
We obtain these criteria by establishing certain properties, which are neces-
sarily possessed by a convergent sequence; if a sequence does not have these
properties, then it must be divergent.
You met this in Sub-section1.3.1, expressed in the forma� bj j �
aj j � bj j
. Here weare writing an in place of a and‘ in place of b.
For example, to show that thesequence ð�1Þnf g isdivergent, we have to showthat ð�1Þnf g is notconvergent; that is, for everyreal number ‘, the sequenceð�1Þn � ‘f g is not null.
2.4 Divergent sequences 61
2.4.2 Bounded and unbounded sequences
One property possessed by a convergent sequence is that it must be bounded.
Definition A sequence {an} is bounded if there is a number K such that
anj j � K; for n ¼ 1; 2; . . .:
A sequence is unbounded if it is not bounded.
Thus a sequence {an} is bounded if all the terms an lie on the sequence diagram
in a horizontal strip from �K up to K, for some positive number K.
K
–K
{an}⏐an⏐ ≤ K
–K ≤ an ≤ Kn
For example, the sequence ð�1Þnf g is bounded, because
ð�1Þnj j � 1; for n ¼ 1; 2; . . .:
However the sequences {2n} and {n2} are unbounded, since, for each number K,
we can find terms of these sequences whose absolute values are greater than K.
Problem 1 Classify the following sequences as bounded or unbounded:
(a) 1þð�1Þnf g; (b) ð�1Þn nf g; (c) 2nþ1n
� �
; (d) 1� 1n
� �n� �
.
The sequence ð�1Þnf g shows that:
a bounded sequence is not necessarily convergent.
However we can prove that:
a convergent sequence is necessarily bounded.
Theorem 1 Boundedness Theorem
If {an} is convergent, then {an} is bounded.
Proof We know that an! ‘, for some real number ‘. Thus {an� ‘} is a null
sequence, and so there is a number X such that
an � ‘j j < 1; for all n > X:
For simplicity in the rest of the proof, we shall now assume that our initial
choice of X is a positive integer.
Nowanj j ¼ ðan � ‘Þ þ ‘j j� an � ‘j j þ ‘j j; by the Triangle Inquality:
It follows that
anj j � 1þ ‘j j; for all n > X:
This is the type of inequality needed to prove that {an} is bounded, but it does
not include the terms a1, a2, . . ., aX. To complete the proof, we let K be the
maximum of the numbers ja1j, ja2j, . . ., jaXj, 1þ j‘j.
Take "¼ 1 in the definition ofa null sequence.
K
X n
� + 1
� – 1
–K
�
That is, K¼max{ja1j,ja2j, . . ., jaXj, 1þ j‘j}.
62 2: Sequences
It follows that
anj j � K; for n ¼ 1; 2; . . .;
as required. &
From Theorem 1, we obtain the following test for the divergence of a
sequence:
Corollary 1 If {an} is unbounded, then {an} is divergent.
For example, the sequences {2n} and �1ð Þnnf g are both unbounded, so they
are both divergent.
Problem 2 Classify the following sequences as convergent or diver-
gent, and as bounded or unbounded:
(a)ffiffiffi
npf g; (b) n2 þ n
n2 þ 1
n o
; (c) �1ð Þnn2� �
; (d) n �1ð Þn� �
:
2.4.3 Sequences which tend to infinity
Although the sequences {2n} and �1ð Þnnf g are both unbounded (and hence
divergent), there is a marked difference in their behaviour. The terms of both
sequences become arbitrarily large, but those of the sequence {2n} become
arbitrarily large and positive. To make this precise, we must explain what we
mean by ‘arbitrarily large and positive’.
Definition The sequence {an} tends to infinity if:
for each positive number K, there is a number X such that
an > K; for all n > X:
In this case, we write
an !1 as n!1 :
Remarks
1. In terms of the sequence diagram for {an}, this definition states that,
for each positive number K, the terms an eventually lie above the line at
height K.
K
X n
{an}
2. If a sequence tends to infinity, then it is unbounded – and hence divergent,
by Corollary 1.
Often, we omit ‘as n!1’.
2.4 Divergent sequences 63
3. If a sequence tends to infinity, then this remains true if we add, delete or
alter a finite number of terms.
4. In the definition we can replace the phrase ‘for each positive number K’ by
‘for each number K’; for, if the inequality ‘an>K, for all n>X’ holds for each
number K, then in particular it certainly holds for each positive number K.
There is a version of the Reciprocal Rule for sequences which tend to
infinity. This enables us to use our knowledge of null sequences to identify
sequences which tend to infinity.
Theorem 2 Reciprocal Rule
(a) If the sequence {an} satisfies both of the following conditions:
1. {an} is eventually positive,
2. f 1ang is a null sequence,
then an!1.
(b) If an!1, then 1an! 0.
Proof of part (a) To prove that an!1, we have to show that:
for each positive number K, there is a number X such that
an > K; for all n > X: (1)
Since {an} is eventually positive, we can choose a number X1 such that
an > 0; for all n > X1:
Since f 1ang is null, we can choose a number X2 such that
1
an
<1
K; for all n > X2:
Now, let X¼max {X1X2}; then
0 <1
an
<1
K; for all n > X:
This statement is equivalent to the statement (1), so an!1, as required. &
Example 1 Use the Reciprocal Rule to prove that the following sequences
tend to infinity:
(a) n3
2
n o
; (b) n!þ 10nf g; (c) n!� 10nf g
Solution
(a) Each term of the sequence�
n3
2
�
is positive and 1n3=2¼ 2
n3. Now, 1n3
� �
is a
basic null sequence and so 2n3
� �
is null, by the Multiple Rule. Hence�
n3
2
�
tends to infinity, by the Reciprocal Rule.
(b) Each term of the sequence n!þ 10nf g is positive and
limn!1
1
n!þ 10n¼ lim
n!1
1n!
1þ 10n
n!
¼ 0
1þ 0¼ 0;
by the Combination Rules.
Hence n!þ 10nf g tends to infinity, by the Reciprocal Rule.
A finite number of terms donot matter.
We prove only part (a): theproof of part (b) is similar.
Here we are taking " ¼ 1K
in thedefinition of a null sequence.
We make this choice of X sothat BOTH of the precedinginequalities hold for all n>X.
Notice that, in parts (b) and(c), n! is the dominant term.
Alternatively, since1
n!þ 10n � 1n!, the sequence
�
1n!þ 10n
�
is null, by the
Squeeze Rule for nullsequences.
64 2: Sequences
(c) First note that
n!� 10n ¼ n! 1� 10n
n!
� �
; for n ¼ 1; 2; . . .:
Since 10n
n!
� �
is a basic null sequence, we know that 10n
n! is eventually less
than 1, and so n!� 10n is eventually positive. Also
limn!1
1
n!� 10n¼ lim
n!1
1n!
1� 10n
n!
¼ 0
1� 0¼ 0;
by the Combination Rules.
Hence {n!� 10n} tends to infinity, by the Reciprocal Rule. &
There are also versions of the Combination Rules and Squeeze Rule for
sequences which tend to infinity. We state these without proof.
Theorem 3 Combination Rules
If {an} tends to infinity and {bn} tends to infinity, then:
Sum Rule {anþ bn} tends to infinity;
Multiple Rule {lan} tends to infinity, for l> 0;
Product Rule {anbn} tends to infinity.
Theorem 4 Squeeze Rule
If {bn} tends to infinity, and
an � bn; for n ¼ 1; 2; . . .;
then {an} tends to infinity.
Problem 3 For each of the following sequences {an}, prove that an!1:
(a) 2n
n
� �
; (b) 2n � n100� �
; (c) 2n
nþ 5n100
� �
; (d) 2n þ n2
n10 þ n
n o
.
We can also define {an} tends to minus infinity.
Definition The sequence {an} tends to minus infinity if
�an !1 as n!1:In this case, we write
an ! �1 as n!1:
For example, the sequences {�n2} and 10n � n!f g both tend to minus infinity,
because {n2} and n!� 10nf g tend to infinity. Sequences which tend to minus
infinity are unbounded, and hence divergent. However, the sequence �1ð Þnnf gshows that an unbounded sequence need not tend to infinity or to minus infinity.
2.4.4 Subsequences
We now give two useful criteria for establishing that a sequence diverges; both
of them involve the idea of a subsequence. For example, consider the bounded
divergent sequence �1ð Þnf g. This sequence splits naturally into two:
This follows by taking "¼ 1in the definition of a nullsequence.
This follows from the fact thatsequences which tend toinfinity are unbounded.
2.4 Divergent sequences 65
the even terms a2, a4, . . ., a2k, . . ., each of which equals 1;
the odd terms a1, a3, . . ., a2k�1, . . ., each of which equals �1.
Both of these are sequences in their own right, and we call them the evensubsequence {a2k} and the odd subsequence {a2k� 1}.
In general, given a sequence {an} we may consider many different subse-
quences, such as:
{a3k}, comprising the terms a3, a6, a9, . . .;
{a4kþ1}, comprising the terms a5, a9, a13, . . .;
{a2k!}, comprising the terms a2, a4, a12, . . ..
Definition The sequence ankf g is a subsequence of the sequence {an} if
{nk} is a strictly increasing sequence of positive integers; that is, if
n1 < n2 < n3 < . . .:
For example, the subsequence {a5kþ2} corresponds to the sequence of
positive integers
nk ¼ 5k þ 2; k ¼ 1; 2; . . .:
The first term of {a5kþ 2} is a7, the second is a12, the third is a17, and so on.
Notice that any strictly increasing sequence {nk} of positive integers must
tend to infinity, since it can be proved by Mathematical Induction that
nk � k; for k ¼ 1; 2; . . .:
Problem 4
(a) Let an¼ n2, for n¼ 1, 2, . . .. Write down the first five terms of each
of the subsequences ankf g, where:
(i) nk¼ 2k; (ii) nk¼ 4k� 1; (iii) nk¼ k2.
(b) Write down the first three terms of the odd and even subsequences of
the following sequence: an ¼ n �1ð Þn ; n ¼ 1; 2; . . .:
Our next theorem shows that certain properties of sequences are inherited by
their subsequences.
Theorem 5 Inheritance Property of Subsequences
For any subsequence ankf g of {an}:
(a) if an! ‘ as n!1, then ank! ‘ as k!1;
(b) if an!1 as n!1, then ank!1 as k!1.
Proof of part (a) We want to show that for each positive number ", there is a
number K such that
ank� ‘j j < "; for all k > K: (2)
However, since {an� ‘} is null, we know that there is a number X such that
an � ‘j j < "; for all n > X:
{(–1)n}
2 3 4 5 6 7 n1
1
–1 {a2k – 1}
{a2k}
The sequence ankf g is the
sequence
an1; an2
; an3; . . .:
Note that the sequence {an} isa subsequence of itself.
A result similar to part (b)holds for sequences wherean!�1.
You may omit this proof at afirst reading.We prove onlypart (a): the proof of part (b) issimilar.
66 2: Sequences
For simplicity in the rest of the proof, we shall now assume that our initial
choice of X is a positive integer.
So, if we take K so large that
nK � X ;then
nk > nK � X; for all k > K;
and soank� ‘j j < "; for all k > K;
which proves (2). &
The following criteria for establishing that a sequence is divergent are
immediate consequences of Theorem 5, part (a):
Corollary 2
1. First Subsequence Rule The sequence {an} is divergent if it has two
convergent subsequences with different limits.
2. Second Subsequence Rule The sequence {an} is divergent if it has a
subsequence which tends to infinity or a subsequence which tends to
minus infinity.
We can now formulate the strategy promised at the beginning of this section.
Strategy To prove that a sequence {an} is divergent:
EITHER
1. show that {an} has two convergent subsequences with different limits
OR
2. show that {an} has a subsequence which tends to infinity or a subse-
quence which tends to minus infinity.
For example, the sequence �1ð Þnf g has two convergent subsequences
which have different limits: namely, the even subsequence with limit 1 and
the odd subsequence with limit �1. So the sequence �1ð Þnf g is divergent, by
the First Subsequence Rule.
On the other hand, the sequence n �1ð Þn� �
has a subsequence (the even
subsequence) which tends to infinity. So n �1ð Þn� �
is divergent by the Second
Subsequence Rule.
Remark
In order to apply the above strategy successfully to prove that a sequence is
divergent, you need to be able to spot convergent subsequences with different
limits or subsequences which tend to infinity or to minus infinity. It is not
always easy to do this, and some experimentation may be required! If the
formula for an involves the expression (�1)n, it is a good idea to consider the
odd and even subsequences, although this may not always work. It may be
helpful to calculate the values of the first few terms in order to try to spot some
subsequences whose behaviour can be identified.
Problem 5 Use the above strategy to prove that each of the following
sequences {an} is divergent:
(a) �1ð Þnþ 1n
� �
; (b) 13
n� 13
n� �� �
; (c) n sin 12
np� �� �
.
We will occasionally makethis assumption if we want torefer to terms such as aX ratherthan mess about with nastyexpressions such as a[X],where [X] is the integral partof X.
Though we do not prove thefact, ANY divergent sequenceis of one (or both) of these twotypes.
The fact that this strategy willalways apply is simply areformulation of the resultmentioned in the abovemargin note.
In part (b) the square bracketsdenote ‘the integer part’function.
2.4 Divergent sequences 67
We end this section by giving a result about subsequences which will be
needed in later chapters.
Theorem 6 If the odd and even subsequences of {an} both tend to the
same limit ‘, then
limn!1
an ¼ ‘:
Proof We want to show that:
for each positive number ", there is a number X such that
an � ‘j j < "; for all n > X: (3)
We know that there are integers K1 and K2 such that
a2k�1 � ‘j j < "; for all k > K1;
and
a2k � ‘j j < "; for all k > K2:
So we now let
X ¼ maxf2K1 � 1; 2K2g:With this choice of X, each n>X is either of the form 2k� 1, with k>K1, or of
the form 2k, with k>K2; it follows then that (3) holds with this value of X. &
2.5 The Monotone Convergence Theorem
2.5.1 Monotonic sequences
In Section 2.3, we gave various techniques for finding the limit of a convergent
sequence. As a result, you may be under the impression that, if we know that a
sequence converges, then there is some way of finding its limit explicitly.
However, it is sometimes possible to prove that a sequence is convergent, even
though we do not know its limit. For example, this situation occurs with a given
sequence {an}, which has the following two properties:
1. {an} is an increasing sequence;
2. {an} is bounded above; that is, there is a real number M such that
an � M; for n ¼ 1; 2; . . .:
Likewise, if {an} is a sequence which is decreasing and bounded below, then
{an} must be convergent.
We combine these two results into one statement.
Theorem 1 Monotone Convergence Theorem
If the sequence {an} is:
EITHER
1. increasing and 2. bounded above
OR
1. decreasing and 2. bounded below,
then {an} is convergent.
{a2k�1} and {a2k} are the oddand even subsequences of {an}.
68 2: Sequences
Proof of the first case Let {an} be a sequence that is increasing and bounded
above. Since {an} is bounded above, the set {an: n¼ 1, 2, . . .} has a least upper
bound, ‘ say; this is true by the Least Upper Bound Property of R . We now
prove that limn!1
an ¼ ‘.We want to show that:
for each positive number ", there is a number X such that:
an � ‘j j < "; for all n > X: (1)
We know that, if "> 0, then, since ‘ is the least upper bound of the set
{an: n¼ 1, 2, . . .}, there is an integer X such that
aX > ‘� ":Since {an} is increasing, an� aX, for n>X, and so
an > ‘� "; for all n > X:
Thus
an � ‘j j ¼ ‘� an < "; for all n > X;
which proves (1).
Hence {an} converges to ‘. &
The Monotone Convergence Theorem tells us that a sequence such as
1� 1n
� �
, which is increasing and bounded above (by 1, for example), must
be convergent. In this case, of course, we know already that 1� 1n
� �
is
convergent (with limit 1) without using the Monotone Convergence Theorem.
The Monotone Convergence Theorem is often used when we suspect that a
sequence is convergent, but cannot find the actual limit. It can also be used to
give precise definitions of numbers, such as p, about which we have only an
intuitive idea, and we do this later in this section.
For completeness, we point out that:
If {an} is increasing but is not bounded above, then an!1 as n!1.
Indeed, for any real number K, we can find an integer X such that
aX > K;
because {an} is not bounded above. Since {an} is increasing, an� aX for n>X,
and so
an > K; for all n > X:
Hence an!1 as n!1.
Similarly, we have the following result:
If {an} is decreasing but is not bounded below, then an!�1 as n!1.
We summarise these results about monotonic sequences in the following
useful theorem:
Theorem 2 Monotonic Sequence Theorem
If the sequence {an} is monotonic, then:
EITHER {an} is convergent
OR an!�1.
We prove only the increasingversion; the proof of thedecreasing version is similar.You met this Property inSub-section 1.4.3.
� + ε
� – ε
�
X n
This follows from thedefinition of least upperbound.
jan� ‘j ¼ ‘� an, becausean� ‘.
We will use the MonotoneConvergence Theorem inprecisely this way in Chapter 3.
Sub-section 2.5.4.
The expression ‘an!�1’ isshorthand for ‘either an!1or an!�1’.
2.5 The Monotone Convergence Theorem 69
Next we note a consequence of the Monotone Convergence Theorem that is
sometimes useful.
Corollary 1
(a) If a sequence {an} is increasing and bounded above, then it tends to its
least upper bound, sup {an: n ¼ 1, 2, . . .}.
(b) If a sequence {an} is decreasing and bounded below, then it tends to its
greatest lower bound, inf {an: n ¼ 1, 2, . . .}.
Problem 1 Prove part (b) of the Corollary.
We meet applications of these results in the rest of this section, to functions
defined recursively and to the study of the numbers e and p. But our first
application is to a result of considerable value in Analysis, Topology and other
parts of Mathematics.
The Bolzano–Weierstrass Theorem
We have seen that if a sequence is convergent, then certainly it must be
bounded. However if a sequence is bounded, it does not follow that it is
necessarily convergent. For example, the sequence {an}, where an ¼ 1n, is
bounded (by 1, since janj ¼ j 1n j � 1, 2, . . .) and is also convergent (to the
limit 0). Yet the sequence {an}, where an¼ (�1)n, is also bounded (by 1, since
anj j ¼
�1ð Þn
¼ 1, for n¼ 1, 2, . . .) but is divergent (since its odd and even
subsequences tend to different limits).
Our next result shows that, if a sequence is bounded, then, even though it
may diverge, it cannot behave ‘too badly’!
Theorem 3 Bolzano–Weierstrass Theorem
A bounded sequence must contain a convergent subsequence.
That is, if a sequence {an} is such that janj �M, for all n, then there exist
some number ‘ in [�M, M] and some subsequence ankf g such that ank
f gconverges to ‘ as k!1.
For example, the sequence {sin n} is bounded. With the tools at our disposal,
it is not at all clear what the behaviour of this sequence might be as n!1!
However, the Bolzano–Weierstrass Theorem asserts that there is at least one
number ‘ in [�1, 1] such that some subsequence {sin nk} converges to ‘. This is
itself a surprising result! In fact, however, using quite sophisticated mathe-
matics we can prove that every number ‘ in [�1, 1] has this property!
The proof of the Bolzano–Weierstrass Theorem that we give is of interest in
its own right; it depends on the notion of repeated bisection, which is a
standard technique in many areas of Analysis.
Proof We shall assume that the bounded sequence {an} has the property that
janj � M, for all n. Then all the terms an lie in the closed interval [�M, M],
which we will denote as [A1, B1].
Next, denote by p the midpoint of the interval [A1, B1]. Then at least one of
the two intervals [A1, p] and [p, B1] must contain infinitely many terms in the
The proof of part (a) iscontained in the proof ofTheorem 1 above.
This proof is similar to theearlier discussion.
We shall use it in Section 4.2.
This was the BoundednessTheorem – Theorem 1,Sub-section 2.4.2.
For jsin nj � 1.
Sadly, this is beyond the rangeof this book.
A1 ¼ �M and B1 ¼ M.
p ¼ 12
A1 þ B1ð Þ.
70 2: Sequences
sequence {an} – for otherwise the whole sequence would only contain finitely
many terms. If only one of the two intervals has this property, denote that
interval by the notation [A2, B2]; if both intervals have this property, choose the
left one (that is, [A1, p]) to be [A2, B2].
In either case, we obtain:
1. A2, B2½ � A1, B1½ �;2. B2 � A2 ¼ 1
2B1 � A1ð Þ;
3. [A1, B1] and [A2, B2] both contain infinitely many terms in the sequence {an}.
We now repeat this process indefinitely often, bisecting [A2, B2] to obtain
[A3, B3], and so on. This gives a sequence of closed intervals {[An, Bn]} for
which:
1. Anþ1;Bnþ1½ � An;Bn½ �, for each n2N;
2. Bn � An ¼ 12
� �n�1B1 � A1ð Þ, for each n2N;
3. each interval [An, Bn] contains infinitely many terms in the sequence {an}.
Property 1 implies that the sequence {An} is increasing and bounded above by
B1 ¼ M. Hence, by the Monotone Convergence Theorem, {An} is convergent;
denote by A its limit. By the Limit Inequality Rule, we must have that A � M.
Similarly, Property 1 implies that the sequence {Bn} is decreasing and
bounded below by A1¼�M. Hence, by the Monotone Convergence
Theorem, {Bn} is convergent; denote by B its limit. By the Limit Inequality
Rule, we must have that B� �M.
We may then deduce, by letting n!1 in Property 2 and using the
Combination Rules for sequences, that
B� A ¼ limn!1
Bn � limn!1
An ¼ limn!1
Bn � Anð Þ
¼ limn!1
1
2
� �n�1
B1 � A1ð Þ
¼ B1 � A1ð Þ limn!1
1
2
� �n�1
¼ 0:
In other words, the sequences {An} and {Bn} both converge to a common limit.
Denote this limit by ‘. Since ‘¼A¼B, we must have �M� ‘�M.
We find a suitable subsequence of {an} as follows. Choose any an1that lies in
[A1, B1]; this is possible since [A1, B1] contains infinitely terms in {an}. Next,
choose a term an2in the sequence, with n2> n1, such that an2
2 A2;B2½ �; this is
possible since [A2, B2] contains infinitely terms in {an}, by Property 3. And so on.
In this way, we construct a subsequence ankf g of {an}, with nkþ 1> nk, for
each k. Since Ak � ank� Bk, it follows from the Squeeze Rule for sequences
(that is, by letting k!1) that ankf g converges to ‘, as required. &
2.5.2 Sequences defined by recursion formulas
As we have seen, often sequences are defined by formulas; that is, for a given n,
we substitute that value of n into a formula and obtain the term an in the sequence
{an} Another way of specifying a sequence is to define its terms ‘inductively’, or
‘recursively’; here we specify the first term (or several terms) in the sequence,
and then have a formula that enables us to calculate all successive terms.
If both intervals containinfinitely many terms in thesequence, it does not matterwhich choice we take; but wespecify the left interval justfor definiteness.
Theorem 3, Sub-section 2.3.3.
That is, ‘2 [�M, M].
Since nkþ 1> nk, we musthave nk� k, for each k.
For Ak!A¼ ‘ andBk!B¼ ‘.
2.5 The Monotone Convergence Theorem 71
For example, the following sequences are defined recursively:
{an}, where a1¼ 2 and anþ1 ¼ 14
a2n þ 3
� �
, for n� 1;
{an}, where a1¼ 2, a2¼ 4 and an ¼ 12
an�1 þ an�2ð Þ, for n� 3.
Do these sequences converge? If they do converge, what are their limits? If we
choose a different value for a1 (or for a1 and a2, in the second sequence), do
their behaviours change? We can use the Monotone Convergence Theorem to
answer many such questions.
First, consider the sequence with recursion formula
anþ1 ¼1
4a2
n þ 3� �
; for n � 1: (2)
If indeed {an} is convergent, what value could its limit take? If we let ‘ denote
limn!1
an and let n!1 in equation (2), we obtain ‘ ¼ 14‘2 þ 3ð Þ. This can be
rearranged as 4‘¼ ‘2þ 3, and so as ‘2� 4‘þ 3¼ 0. Hence (‘� 1)(‘� 3)¼ 0,
so that ‘ = 1 or ‘ = 3.
This suggests that we should look at the differences anþ1� 1 and anþ1� 3.
Using equation (2), we deduce that
anþ1 � 1 ¼ 1
4a2
n � 1� �
¼ 1
4an þ 1ð Þ an � 1ð Þ; for n � 1; (3)
anþ1 � 3 ¼ 1
4a2
n � 9� �
¼ 1
4an þ 3ð Þ an � 3ð Þ; for n � 1: (4)
Next, it is useful to examine whether the sequence is monotonic by looking
at the difference anþ1� an
anþ1 � an ¼1
4a2
n � 4an þ 3� �
¼ 1
4an � 1ð Þ an � 3ð Þ: (5)
So the sign of anþ1� an depends on where an lies on the number-line in relation
to the numbers 1 and 3.
For example, it follows from (5) that
if 1 < an < 3, for some n � 1, then anþ1 � an < 0,
so that anþ1 < an:(6)
If an¼ 1, for some number n� 1, it follows from equation (3) that anþ1¼ 1.
Consequently, if our initial term is a1¼ 1, then it follows that an¼ 1, for all
n� 1; in other words, the sequence is simply the constant sequence {1}.
Similarly, if an¼ 3, for some number n� 1, it follows from equation (3) that
anþ1¼ 3. Consequently, if our initial term is a1¼ 3, then it follows that an¼ 3,
for all n� 1; in other words, the sequence is simply the constant sequence {3}.
Next, suppose that 1< an< 3, for some number n� 1. It follows from
equation (3) that anþ1� 1 is positive, so that 1< anþ1. On the other hand, it
follows from equation (4) that anþ1� 3 is negative, so that anþ1< 3.
Thus, if 1< an< 3, for some number n� 1, then 1< anþ 1< 3.
We can then prove by induction that, if 1< a1< 3, then 1< an< 3, for all
n> 1. (We omit the details.)
Next, it follows, from this last fact and (6) above that, if 1< a1< 3, then the
sequence {an} is strictly decreasing.
Hence, if 1< a1< 3 the sequence {an} is strictly decreasing and is bounded
below. Hence, by the Monotone Convergence Theorem, {an} is convergent.
Of course we have not yetproved that {an} isconvergent. This has been a‘what if?’ discussion so far,but a useful one.
Because 1 and 3 are thepossible limits.
We shall use this fact below.
In particular this discussioncovers the case a1¼ 2 that wementioned at the start of thissub-section.
Now comes the coup degrace!
72 2: Sequences
Since {an} is decreasing, whatever its limit might be (and we know that the only
two possibilities for the limit are 1 and 3), its limit must be less than or equal to its
first term a1. It follows that the limit of the sequence must be 1.
Problem 2 Let the sequence {an} be defined by the recursion formula
anþ1 ¼ 14
a2n þ 3
� �
, for n� 1.
(a) Prove that, if a1> 3, then an!1 as n!1.
(b) In the case that 0� a1< 1, determine whether {an} is convergent
and, if so, to what limit.
(c) Describe the behaviour of the sequence {an} in the case that a1< 0.
2.5.3 The number e
We will define e to be the limit of the sequence 1þ 1n
� �n� �
. If we plot the first
few terms of this sequence on a sequence diagram, then it seems that the
sequence is increasing and converges to a limit, which is less than 3.
To show that 1þ 1n
� �n� �
is convergent, using the Monotone Convergence
Theorem, we prove that the sequence is increasing and is bounded above.
1. 1þ 1n
� �n� �
is increasing
By the Binomial Theorem
1þ 1
n
� �n
¼ 1þ n1
n
� �
þ n n� 1ð Þ2!
1
n
� �2
þ . . . þ 1
n
� �n
:
The general term in this expansion is
n n� 1ð Þ . . . n� k þ 1ð Þk!
1
n
� �k
¼ 1
k!1� 1
n
� �
1� 2
n
� �
. . . 1� k � 1
n
� �
: (7)
If k is fixed, then each of the factors
1� 1
n
� �
; 1� 2
n
� �
; . . . ; 1� k � 1
n
� �
is increasing, and so the general term (7) increases as n increases. Since this
is true for each fixed k, the sequence ð1þ 1nÞn
� �
is increasing.
2. 1þ 1n
� �n� �
is bounded above
Here, note that the general term (7) satisfies the inequality
1
k!1� 1
n
� �
1� 2
n
� �
. . . 1� k � 1
n
� �
� 1
k!;
We saw that 1 and 3 were theonly possible limits at the startof the discussion.
For example, to three decimalplaces the first two terms are 2and 2.25, the sixth term is2.521, and the hundredth termis 2.704.
2.5 The Monotone Convergence Theorem 73
since each of the brackets is at most 1. Hence
1þ 1
n
� �n
� 1þ 1þ 1
2!þ 1
3!þ . . . þ 1
n!
� 1þ 1þ 1
21þ 1
22þ . . . þ 1
2n�1;
since k!¼ k (k� 1) . . . 2.1� 2k�1.
Now
1þ 1
21þ 1
22þ þ 1
2n�1¼ 1�
1� 12
� �n
1� 12
¼ 2� 1� 1
2
� �n� �
¼ 2� 1
2n�1;
and so
1þ 1
n
� �n
� 3� 1
2n�1; for n ¼ 1; 2; . . .:
Thus the sequence 1þ 1n
� �n� �
is bounded above, by 3.
It follows, by the Monotone Convergence Theorem, that the sequence
1þ 1n
� �n� �
is convergent with limit at most 3. This allows us to make the
following definition:
Definition e ¼ limn!1
1þ 1n
� �n:
For larger and larger values of n, the terms 1þ 1n
� �ngive better and better
approximate values for e. Unfortunately, the sequence 1þ 1n
� �n� �
converges
to e rather slowly, and we need to take very large integers n to get a reasonable
approximation to e¼ 2.71828. . ..Now we would like in general to make the definition of ex as
ex ¼ limn!1
1þ xn
� �n, but first we need to know that this limit actually exists.
Problem 3 Let x> 0.
(a) Prove that ð1þ xnÞn
� �
is an increasing sequence, by adapting the
method above where x¼ 1.
(b) Verify that 1þ kn� ð1þ 1
nÞk, for k¼ 1, 2, . . .. Using this fact, prove
that ð1þ xnÞn
� �
is bounded above.
(c) Deduce that ð1þ xnÞn
� �
is convergent.
Problem 4 By considering the product of the first n terms of the
sequence ð1þ 1nÞn
� �
, prove that n! > nþ1e
� �n, for n¼ 1, 2, . . ..
We now examine the convergence of the sequence ð1þ xnÞn
� �
, for x< 0.
Recall that, when x< 0, then �x> 0; in particular, from the above discus-
sion, the sequence ð1� xnÞn
� �
converges. We shall use Bernoulli’s inequality
(1þ c)n� 1þ nc, for c��1.
In investigating the convergence of the sequence ð1þ xnÞn
� �
, we are
only interested in what happens when n is large. So, we need consider
only the situation when n>�x. Then n2> x2, so that 1n2 <
1x2; thus x2
n2 < 1,
or � x2
n2 > �1.
For the sum of the geometricprogression aþ arþ ar2þ . . .
þ arn�1 is a 1�rn
1�r, if r 6¼ 1;
here we have a¼ 1, r ¼ 12:
For example
1þ 11000
� �1000¼ 2:716 . . .:
Then e¼ e1
So limn!1
1þ xn
� �nexists
for x> 0. Note that the limit isat least 1, and so cannot bezero.
You met Bernoulli’sInequality in Sub-section 1.3.3.
Recall that � x> 0.
74 2: Sequences
It follows that we may substitute � x2
n2 for c in Bernoulli’s Inequality. This
gives that
1� x2
n2
� �n
� 1þ n � x2
n2
� �
¼ 1� x2
n; for n > �x:
It follows that
1 � 1� x2
n2
� �n
� 1� x2
n; for n > �x: (8)
Now the sequence 1� x2
n
n o
converges to the limit 1, by the Combination
Rules. Hence, by applying the Squeeze Rule to inequality (8), the sequence
1� x2
n2
�nn o
converges to the limit 1 as n!1But
1þ x
n
�n
¼1þ x
n
� �n1� x
n
� �n
1� xn
� �n
¼1� x2
n2
�n
1� xn
� �n : (9)
We have just seen that the numerator is convergent, and we saw above that the
denominator is convergent (to a non-zero limit) for�x> 0; it follows from (9),
by the Quotient Rule, that the sequence ð1þ xnÞn
� �
is convergent.
So whatever value we choose for the real number x, the sequence ð1þ xnÞn
� �
is convergent. This means that we can now make the following legitimate
definition:
Definition For any real value of x,
ex ¼ limn!1
1þ x
n
�n
:
We can deduce more than this from equation (9), if we rewrite it in the
convenient form
1þ x
n
�n
1� x
n
�n
¼ 1� x2
n2
� �n
:
Letting n!1, we deduce from this last equation that
exe�x ¼ 1:
We have shown that this holds for x< 0; however, by simply interchanging x
and �x, it is clear that this holds for x> 0 also.
Theorem 4 Inverse Property of ex
For any real value of x, ex e�x¼ 1
The next property of the exponential function that we need to verify is that
ex ey¼ exþ y for any real values of x and y. We shall prove this fact later.
Here we use the fact that f1ng is
a basic null sequence.
Recall that here we areconsidering the case x< 0.
Note that when x¼ 0, thesequence is simply theconstant sequence {1}.
The functionx 7! ex, for x 2 R , is calledthe exponential function.
Here we also use the fact that
limn!1
1� x2
n2
� �n
¼ 1;
that we proved above.
In Sub-section 3.4.3.
2.5 The Monotone Convergence Theorem 75
2.5.4 The number p
One of the oldest mathematical problems is to determine the area of a disc of
radius r, and the length of its perimeter. It is well known that these magnitudes
are pr2 and 2pr, respectively. But what exactly is p? Is there a real number pwhich makes these formulas correct? – and, if so, how is it formally defined?
We now give a precise definition of p as the area of a disc of radius 1, using a
method originally devised by Archimedes to calculate approximate values for
the area of this disc. The idea is to calculate the areas of regular polygons
inscribed in the disc. Archimedes found an easy way to calculate the areas of
such regular polygons with 6 sides, 12 sides, 24 sides and so on. The results of
parts (a) and (b) of the following problem help to give the first two of these areas.
Problem 5 Verify that the following triangles have the stated areas:
Hint: In part (d), use the half-angle formula tan# ¼ 2 tan 12#
1�tan2 12#:
Let sn denote the number of sides of the nth such inner polygon (so s1¼ 6,
s2¼ 12, s3¼ 24, and, in general, sn¼ 3� 2n) and let an denote the area of the
nth inner polygon. Then
an ¼1
2sn sin
2psn
� �
; for n ¼ 1; 2; . . .: (10)
Geometrically, it is obvious that each time we double the number of sides of
the inner polygon the area increases, and so
a1 < a2 < a3 < . . . < an < anþ1 < . . .:
Hence {an} is (strictly) increasing. But is {an} convergent?
Here ‘inscribed’ means thatall the vertices of the polygonlie on the circle, so that theinside of the polygon iscontained in the inside of thecircle.
Here p is used only as asymbol to represent angles; itsvalue is not required.
For example, to three decimalplaces
a1 ¼ 2:598,
a2 ¼ 3,
. . .
a6 ¼ 3:141:
76 2: Sequences
Notice that each of the polygons lies inside a square of side 2, which has
area 4. This means that
an � 4; for n ¼ 1; 2; . . .;
and so {an} is increasing and bounded above (by 4). Hence, by the Monotone
Convergence Theorem, {an} is convergent.
Our intuitive idea of the area of the disc suggests that its area is greater than
each of the areas an, but ‘only just’! Put another way, the area of the disc should
be the limit of the increasing sequence {an}. This leads us to make the
following definition:
Definition p ¼ limn!1
an:
We shall explain in a moment how to calculate the terms an without
assuming a value for p.
First, however, we describe how to estimate the area of the disc using outer
polygons. Once again we start with a regular hexagon and repeatedly double
the number of sides. The results of part (c) and (d) of the last problem help to
give the first two such areas.
As before, let sn¼ 3� 2n, for n¼ 1, 2, . . ., and let bn denote the area of the
nth outer polygon. This nth outer polygon consists of sn isosceles triangles,
each of height 1 and base 2 tanðpsnÞ. Thus
bn ¼ sn tanpsn
� �
; for n ¼ 1; 2; . . .: (11)
Geometrically, it is obvious that each time we double the number of sides of
the outer polygon the area decreases, and so
b1 > b2 > b3 > > bn > bnþ1 > :
So the sequence {bn} is (strictly) decreasing and bounded below (by 0,
for example). Thus, by the Monotone Convergence Theorem, {bn} is also
convergent.
Intuitively, we expect that {bn} has the same limit as {an}, which we have
defined to be p. But how can we prove this?
This will enable us to squeezep between the area of theinner polygons and the area ofthe outer polygons.
For example, to three decimalplaces
b1 ¼ 3:464;
b2 ¼ 3:215;
. . .
b6 ¼ 3:142:
2.5 The Monotone Convergence Theorem 77
The terms an and bn can be calculated by using the following equations,
which are known jointly as the Euclidean algorithm
anþ1 ¼ffiffiffiffiffiffiffiffiffi
anbn
p
; for n ¼ 1; 2; . . .; (12)
and
bnþ1 ¼2anþ1bn
anþ1þ bn
; for n ¼ 1; 2; . . .: (13)
Starting with a1 ¼ 32
ffiffiffi
3p¼ 2:598 . . . and b1 ¼ 2
ffiffiffi
3p¼ 3:464 . . ., we use
these equations iteratively to calculate first a2 ¼ffiffiffiffiffiffiffiffiffi
a1b1
p, then b2 ¼ 2a2b1
a2þb1, and
so on. Here are (approximations to) the first few values of each sequence
obtained in this way:
sn 6 12 24 48 96 192
an 2.598 3 3.106 3.133 3.139 3.141
bn 3.464 3.215 3.160 3.146 3.143 3.142
It does appear that both sequences do converge to a common limit, namely p.
Remark
We have defined p using the areas of approximating polygons. An alternative
approach uses the perimeters of these polygons.
2.5.5 Proofs
We now prove several of the results referred to in the discussion of p in the
previous sub-section.
First, we prove equations (12) and (13)
anþ1 ¼ffiffiffiffiffiffiffiffiffi
anbn
p
; for n ¼ 1; 2; . . .; (12)
and
bnþ1 ¼2anþ1bn
anþ1 þ bn
; for n ¼ 1; 2; . . .: (13)
Using the half-angle formulas for sin # and cos #, it is easy to check that
sin1
2# ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
2sin# tan
1
2#
r
and
tan1
2# ¼ sin# tan#
sin#þ tan#:
Hence, since snþ1¼ 2sn
anþ1 ¼1
2snþ1 sin
2psnþ1
� �
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
2sn sin
2psn
� �
� sn tanpsn
� �
s
¼ffiffiffiffiffiffiffiffiffi
anbn
p
:
Equations (12) and (13)can be verified fromequations (10) and (11); wegive the details in Sub-section 2.5.5 below.
We prove this in Sub-section 2.5.5 below.
You may omit this sub-section at a first reading.
These comprise the Euclideanalgorithm.
We omit the details in bothcases.
This proves equation (12).
78 2: Sequences
Similarly
bnþ1 ¼ snþ1 tanp
snþ1
� �
¼2sn sin p
sn
�
� sn tan psn
�
sn sin psn
�
þ sn tan psn
�
¼ 2anþ1bn
anþ1 þ bn
:
Finally we prove that limn!1
bn ¼ limn!1
an.
To do this, we rearrange equation (12) as follows
bn ¼a2
nþ1
an
:
Then, by the Combination Rules, we can deduce that
limn!1
bn ¼limn!1
anþ1
�2
limn!1
an
¼ p2
p¼ p:
Then, since we defined p to be precisely limn!1
an, it follows that limn!1
bn ¼ limn!1
an,
as required.
Definition Let an denote the area of the regular polygon with 3� 2n sides
inscribed in a disc of radius 1, and bn the area of the regular polygon with
3� 2n sides circumscribing the disc. Then
p is the common value of the limits limn!1
an and limn!1
bn.
Remark
It can be shown that every two applications of equations (12) and (13) give an
extra decimal place in the decimal expansion for p. This decimal expansion is
now known to many millions of decimal places, using sequences which
converge to p much more rapidly than {an} or {bn}. Since p is irrational,
there is no possibility of the decimal expansion of p recurring.
2.6 Exercises
Section 2.1
1. Calculate the first five terms of each of the following sequences, and draw a
sequence diagram in each case:
(a) {n2� 4nþ 4}; (b)�1ð Þnþ1
n!
n o
; (c) sin 14
np� �� �
.
This proves equation (13).
We shall return in Section 8.5to the question ofapproximating p, by the use ofpower series.
2.6 Exercises 79
2. Determine which of the following sequences are monotonic:
(a)�
nþ 1nþ 2
�
; (b)�1ð Þn
n
n o
; (c)�
21n
�
.
3. Prove that the following sequences are eventually monotonic:
(a) 5n
n!
� �
; (b) n þ 8n
� �
.
Section 2.2
1. For each of the following sequences {an} and numbers ", find a number X
such that janj<", for all n>X:
(a) an ¼ �1ð Þnn5 , "¼ 0.001; (b) an ¼ 1
2nþ 1ð Þ2, "¼ 0.002.
2. Use the definition of null sequence to prove that the two sequences in the
previous exercise are null.
3. Prove that the following sequences are not null:
(a)ffiffiffi
npf g; (b) 1 þ �1ð Þn
n
n o
.
4. Assuming that 1n
� �
is null, deduce that the following sequences are null.
State which rules you use.
(a) 2ffiffi
np þ 3
n7
n o
; (b) sin nn2 þ 1
n o
; (c)tan�1 2nð Þ
nþ 1þ sin n
n o
.
5. Prove thatffiffiffiffiffiffiffiffiffiffiffi
nþ 1p
�ffiffiffi
np� �
is null.
Hint: Use the fact that a� b ¼ a2 � b2
aþ b, for aþ b 6¼ 0.
6. Use the list of basic null sequences to prove that the following sequences are
null. State which rules you use.
(a) 34n þ 2n
3n
� �
; (b)�
6n10
n!
�
; (c)�
n1010n
n!
�
.
7. Prove that, if the sequence {an} of positive numbers is null, thenffiffiffiffiffi
anp� �
is
null.
Section 2.3
1. Show that the following sequences converge to 1, by calculating an� 1 in
each case:
(a)�
n� 1nþ 3
�
; (b)�
n2
n2þnþ1
�
.
2. Use the Combination Rules to find the limits of the following sequences:
(a) n3
2n3 þ 3nþ 4n
n o
; (b)�
n10 � 2n
2n þ n100
�
; (c)�
5n!þ 5n
n100 þ n!
�
.
3. (a) Prove that, if the sequence {an} of positive numbers is convergent with
limit ‘, where ‘> 0, thenffiffiffiffiffi
anp� �
is convergent with limitffiffi
‘p� �
.
You met this list in Sub-section 2.2.1.
80 2: Sequences
Hint: Use the fact that a� b ¼ a2 � b2
aþ b, for aþ b 6¼ 0.
(b) Prove that, if the sequence {ak} is convergent with limit ‘, ‘ 6¼ 0, and a1n
k
is defined, where n2N , then a1n
k
n o
is convergent with limit ‘1n.
Hint: Use the identity aq� bq¼ (a� b)(aq�1þ aq�2bþ aq�3b2þ þbq�1) with suitable choices for a, b and q.
Section 2.4
1. Classify the following sequences as convergent or divergent, and as
bounded or unbounded:
(a)�
n14
�
; (b)ð�1Þn100n
n!
n o
.
2. Use the Reciprocal Rule to prove that the following sequences tend to
infinity:
(a) n!n3
� �
; (b) n2 þ 2n� �
; (c) n2 � 2n� �
; (d) n!� n3 � 3n� �
.
3. Use the Subsequence Rules to prove that the following sequences are
divergent:
(a) �1ð Þn2nf g; (b)�1ð Þnn2
2n2 þ 1
n o
.
4. Each of the following general statements concerning sequences {an} and {bn}
is true or false. For each statement that is true, prove it. For each statement that
is false, write down examples of sequences to illustrate your assertion:
(a) If {an} and {bn} are divergent, then {anþ bn} is divergent.
(b) If {an} and {bn} are divergent, then {anbn} is divergent.
(c) If {an} is convergent and {bn} is divergent, then {anbn} is divergent.
(d) If a2n
� �
is convergent, then {an} is convergent.
(e) If a2n
� �
is convergent, and an> 0, then {an} is convergent.
(f) If {bn} is convergent, where bn¼ anþ1an, then {an} is convergent.
(g) If an! 0 as n!1, then anþ1þanþ2þþa2nffiffi
np
n o
is convergent.
5. Prove that every sequence has a monotonic subsequence.
Section 2.5
1. Prove that, if {an} is increasing and has a subsequence ankf g which is
convergent, then {an} is convergent.
2. Let anf g1n¼1 be a sequence for which anþ1 ¼ 3an þ 1anþ3
, for n� 1.
(a) Prove that, if a1>�1, the sequence {an} converges.
Hint: consider the cases �1< a1< 1, a1¼ 1 and a1> 1 separately.
(b) If a1��1, does the sequence converge or diverge?
The Power Rule in Sub-section 2.2.1 and thesubsequent remark there showthat this result holds in thecase that ‘¼ 0.
2.6 Exercises 81
3. Let anf g1n¼1 be a sequence for which anþ1 ¼ 12
an þ ‘2
an
�
, for n� 1, where
a1> 0 and ‘> 0.
(a) Prove that {an} is decreasing for n� 2, and hence that an! ‘ as n!1.
(b) With ‘2¼ 2 and a1¼ 1.5, calculate the terms a2, a3, a4 and a5.
4. Obtain formulas for the perimeters of the inner and outer polygons intro-
duced in Sub-section 2.5.4. Prove that these sequences tend to 2p as n!1.
5. Bernoulli’s Inequality states that (1þ x)n� 1þ nx, for x��1, n¼ 1, 2, . . ..
(a) Apply Bernoulli’s Inequality, with x ¼ �1n2 , to give an alternative proof
that 1þ 1n
� �n� �
is increasing.
(b) Apply Bernoulli’s Inequality, with x ¼ 1n2�1
, to prove that 1þ 1n
� �nþ1n o
is decreasing.
(c) Deduce from parts (a) and (b) that
1þ 1
n
� �n
� e � 1þ 1
n
� �nþ1
; for n ¼ 1; 2; . . .:
6. Let an; bn½ �f g1n¼1 be a sequence of closed intervals with the property that
anþ1; bnþ1½ � � an; bn½ �, for n¼ 1, 2, . . ..
(a) Prove that there exists some point c that lies in all the intervals [an, bn],
for n¼ 1, 2, . . ..
(b) Prove that, if bn� an! 0 as n!1, then there exists only one such
point c.
You met this in Sub-section 1.3.3.
This result is sometimescalled The Nested IntervalsTheorem.
82 2: Sequences
3 Series
The Ancient Greek philosopher Zeno of Elea proposed a number of paradoxes
of the infinite, which are said to have had a profound influence on Greek
mathematics. For example, Zeno claimed that it is impossible for an object to
travel a given distance, since it must first travel half the distance, then half
of the remaining distance, then half of what remains, and so on. There must
always remain some distance left to travel, and so the journey cannot be
completed.
This paradox relies partly on the intuitive feeling that it is impossible to add
up an infinite number of positive quantities and obtain a finite answer.
However, the following illustration of the paradox suggests that such an
infinite sum is plausible.
0 1
12
14
18
78
34
12
116
1516
The distance from 0 to 1 can be split up into the infinite sequence of
distances 12
, 14
, 18, . . ., and so it seems reasonable to write
1
2þ 1
4þ 1
8þ 1
16þ � � � ¼ 1:
This chapter is devoted to the study of such expressions, which are called
infinite series.
The following example shows that infinite series need to be treated with
care. Suppose that it is possible to add up 2, 4, 8, . . ., and that the answer is s
2þ 4þ 8þ 16þ � � � ¼ s:
If we multiply through by 12, we obtain
1þ 2þ 4þ 8þ � � � ¼ 1
2s;
which we can rewrite in the form
1þ s ¼ 1
2s:
It follows from this equation that s¼�2, which is obviously non-sensical.
We can avoid reaching such absurd conclusions by performing arithmetical
operations only with convergent infinite series.
In Section 3.1 we define the concept of convergent infinite series in terms
of convergent sequences, and give some examples. We also describe various
properties which are common to all convergent series.
Because the left-hand side ofthe previous equation can beexpressed as
1þ ð2þ 4þ 8þ � � �Þ¼ 1þ s:
83
Section 3.2 is devoted to series with non-negative terms, and we give several
tests for the convergence of such series.
In Section 3.3 we deal with the much harder problem of determining
whether a series is convergent when it contains both positive and negative
terms. We also look at what can happen if we rearrange the terms of a
convergent series – does the new series still converge? This section is quite
long, so you might wish to tackle it in several sessions.
In Section 3.4 we explain how ex can be represented as an infinite series of
powers of x, and we use this representation to prove that the number e is
irrational, and that exey ¼ exþy.
3.1 Introducing series
3.1.1 What is a convergent series?
We begin by defining precisely what is meant by the statement that
1
2þ 1
4þ 1
8þ 1
16þ � � � ¼ 1:
Let sn be the sum of the first n terms on the left-hand side. Then
s1 ¼1
2;
s2 ¼1
2þ 1
4¼ 3
4;
s3 ¼1
2þ 1
4þ 1
8¼ 7
8; . . .; and so on:
In general, using the formula for the sum of a geometric progression, with
a ¼ r ¼ 12, we obtain
sn ¼1
2þ 1
4þ 1
8þ � � � þ 1
2n¼ 1
2þ 1
22þ 1
23þ � � � þ 1
2n
¼ 1
2�
1� 12
� �n
1� 12
¼ 1� 1
2
� �n
:
The sequence 12
� �n� �
is a null sequence, and so
limn!1
sn ¼ limn!1
1� 1
2
� �n� �
¼ 1� limn!1
1
2
� �n
¼ 1:
It is the precise mathematical statement that sn! 1 as n!1which justifies
our earlier (intuitive) statement that
1
2þ 1
4þ 1
8þ 1
16þ � � � ¼ 1:
We use this approach to define a convergent infinite series.
You first met ex in Section 2.5.
The sum of a geometricprogression aþ arþar 2 þ � � � þ arn�1 is a 1�rn
1�r,
if r 6¼ 1.
Here we use the CombinationRules for sequences.
84 3: Series
Definition Given a sequence {an} of real numbers, the expression
a1 þ a2 þ a3 þ � � �is called an infinite series, or simply a series. The nth partial sum of this
series is
sn ¼ a1 þ a2 þ a3 þ � � � þ an:
The behaviour of the infinite series a1þ a2þ a3þ � � � is determined by the
behaviour of {sn}, its sequence of partial sums.
Definition The series a1þ a2þ a3þ � � � is convergent with sum s if the
sequence {sn} of partial sums converges to s. In this case, we say that the
series converges to s, and we write
a1 þ a2 þ a3 þ � � � ¼ s:
The series diverges if the sequence {sn} diverges.
Thus we prove results about series by applying results for sequences proved
in Chapter 2 to the sequence of partial sums {sn}.
Example 1 For each of the following infinite series, calculate its nth partial
sum, and determine whether the series is convergent or divergent:
(a) 1þ 1þ 1þ � � �; (b) 13þ 1
32 þ 133 þ � � �; (c) 2þ 4þ 8þ � � �.
Solution
(a) In this case
sn ¼ 1þ 1þ � � � þ 1 ¼ n:
The sequence {sn} tends to infinity, and so this series is divergent.
(b) Using the formula for summing a finite geometric progression, with
a ¼ r ¼ 13, we obtain
sn ¼1
3þ 1
32þ 1
33þ � � � þ 1
3n
¼ 1
3�
1� 13
� �n
1� 13
¼ 1
21� 1
3
� �n� �
:
Since 13
� �n� �
is a basic null sequence
limn!1
sn¼1
2;
and so this series is convergent, with sum 12.
(c) In this case
sn ¼ 2þ 4þ 8þ � � � þ 2n
¼ 22n � 1
2� 1
� �
¼ 2nþ1 � 2:
The sequence {2nþ1� 2} tends to infinity, and so this series is divergent.&
3.1 Introducing series 85
Sigma notation
Next, we explain how to use the sigma notation to represent infinite series.
Recall that a finite sum such as
1
2þ 1
4þ 1
8þ � � � þ 1
1024¼ 1
21þ 1
22þ 1
23þ � � � þ 1
210
can be represented using sigma notation as
X
10
n¼1
1
2n:
This notation can be adapted to represent infinite series, as follows
X
1
n¼1
an¼ a1 þ a2 þ a3 þ � � �:
Convention When using the sigma notation to represent the nth partial
sum sn of a series, we write
sn ¼ a1 þ a2 þ a3 þ � � � þ an ¼X
n
k¼1
ak:
We have written k as the ‘dummy variable’ here, to avoid using n for two
different purposes in the same expression. We could have used any letter (other
than n) for the dummy variable; for example, the expressionsP
n
i¼1
ai andP
n
r¼1
ar
also stand for a1þ a2þ a3þ � � � þ an.
If we need to begin a series with a term other than a1, then we write, for example
X
1
n¼0
an¼ a0 þ a1 þ a2 þ � � � orX
1
n¼3
an¼ a3 þ a4 þ a5 þ � � �:
For any series, the nth partial sum sn is obtained by adding the first n terms. For
example, the nth partial sums of the above two series are
sn ¼ a0 þ a1 þ a2 þ � � � þ an�1 ¼X
n�1
k¼0
ak and
sn ¼ a3 þ a4 þ a5 þ � � � þ anþ2 ¼X
nþ2
k¼3
ak:
Problem 1 For each of the following infinite series, calculate the nth
partial sum sn, and determine whether the series is convergent or divergent:
(a)P
1
n¼1
�13
� �n; (b)
P
1
n¼0
�1ð Þn; (c)P
1
n¼�1
12
� �n:
Remark
Notice that inserting into a series, or omitting or altering, a finite number of
terms does not affect the convergence of the series, but may affect the sum. For
example, since the series 12þ 1
4þ 1
8þ � � � converges with limit 1, it follows that
� is the Greek capital letter‘sigma’.
For exampleX
1
n¼1
1
2n¼ 1
2þ 1
4þ 1
8þ � � � ;
X
1
n¼1
1
n¼ 1þ 1
2þ 1
3þ 1
4þ � � � :
For the above examples
sn¼1
2þ1
4þ1
8þ�� �þ 1
2n
¼X
n
k¼1
1
2k;
sn¼ 1þ1
2þ1
3þ1
4þ�� �þ1
n
¼X
n
k¼1
1
k:
For example, we writeX
1
n¼3
1
n!� n
for the series
1
3!� 3þ 1
4!� 4þ � � �:
Recall that the convergence ordivergence of sequences wasnot affected by omitting oraltering a finite number ofterms in the sequence.
86 3: Series
the series 8þ 4þ 2þ 1þ 12þ 1
4þ 1
8þ � � � ¼ 8þ 4þ 2þ 1ð Þ þ 1
2þ 1
4þ 1
8þ � � �
converges with limit (8þ 4þ 2þ 1)þ 1¼ 16.
3.1.2 Geometric series
All the series considered so far are examples of geometric series. The standard
geometric series with first term a and common ratio r isX
1
n¼0
arn¼ aþ ar þ ar2 þ � � �:
The following theorem enables us to decide whether any given geometric
series is convergent or divergent.
Theorem 1 Geometric Series
(a) If jrj< 1, thenP
1
n¼0
arn is convergent, with sum a1�r
.
(b) If jrj � 1 and a 6¼ 0, thenP
1
n¼0
arn is divergent.
Proof
(a) If r 6¼ 1, then the nth partial sum sn is given by
sn ¼ aþ ar þ ar2 þ � � � þ arn�1 ¼ a1� rn
1� r:
Now, if rj j < 1, then rnf g is a basic null sequence, and so
limn!1
sn ¼ limn!1
a1� rn
1� r¼ a
1� rlim
n!11� rnð Þ
¼ a
1� r;
by the Combination Rules for sequences. Thus, if jrj< 1, thenP
1
n¼0
arn is
convergent, with sum a1�r
.
(b) Part (b) can be deduced immediately from the Non-null Test, which
appears later in this section. We defer the proof until then. &
Decimal representation of rational numbers
Geometric series provide another interpretation of the decimal representation
of rational numbers. Recall that rational numbers have terminating, or recur-
ring, decimal representations. For example
3
10¼ 0:3 and
1
3¼ 0:333 . . . ¼ 0:3:
Another way to interpret the symbol 0.333 . . . is as an infinite series
3
101þ 3
102þ 3
103þ � � �:
This is a geometric series, with first term a ¼ 310
and common ratio r ¼ 110
.
Since 0 < 110< 1, this series is convergent, with sum
a
1� r¼
310
1� 110
¼ 3
9¼ 1
3:
For 12þ 1
4þ 1
8þ � � � ¼ 1:
This value for sn is easilyverified by MathematicalInduction.
Sub-section 3.1.5, Example 3.
Sub-section 1.1.3.
3.1 Introducing series 87
Thus we have another method of finding the fraction which is equal to a given
recurring decimal, by calculating the sum of the corresponding geometric series.
Problem 2 Interpret the following decimals as infinite series, and
hence represent them as fractions:
(a) 0.111 . . .; (b) 0.86363 . . .; (c) 0.999 . . ..
3.1.3 Telescoping series
Geometric series are easy to deal with because it is possible to find a formula
for the nth partial sum sn. The next problem deals with another series for which
we can calculate sn explicitly.
Problem 3 Calculate the first four partial sums of the following series,
giving your answers as fractions
X
1
n¼1
1
n nþ 1ð Þ ¼1
1� 2þ 1
2� 3þ 1
3� 4þ � � �:
The results obtained in Problem 3 suggest the general formula
sn ¼1
1� 2þ 1
2� 3þ 1
3� 4þ � � � þ 1
n nþ 1ð Þ ¼n
nþ 1:
This formula can be proved by Mathematical Induction, or we can use the identity
1
n nþ 1ð Þ ¼1
n� 1
nþ 1; for n ¼ 1; 2; . . .;
which implies that
sn ¼1
1� 2þ 1
2� 3þ 1
3� 4þ � � � þ 1
n� 1ð Þnþ1
n nþ 1ð Þ
¼ 1
1� 1
2
� �
þ 1
2� 1
3
� �
þ 1
3� 1
4
� �
þ � � � þ 1
n� 1� 1
n
� �
þ 1
n� 1
nþ 1
� �
¼ 1� 1
nþ 1
¼ n
nþ 1:
Since
limn!1
sn ¼ limn!1
n
nþ 1¼ lim
n!1
1
1þ 1n
¼ 1;
we deduce that the given series is convergent, with sum
X
1
n¼1
1
n nþ 1ð Þ ¼ 1:
Problem 4 Find the nth partial sum of the seriesP
1
n¼1
1n nþ2ð Þ, using the
fact that
2
n nþ 2ð Þ ¼1
n� 1
nþ 2; for n ¼ 1; 2; . . .:
Deduce that this series is convergent, and find its sum.
You should at least read thesolution to part (c), even ifyou do not tackle it first.
This cancellation of adjacentterms explains why this seriesis said to be telescoping.
88 3: Series
3.1.4 Combination Rules for convergent series
At the start of this chapter we saw that performing arithmetical operations on
the divergent series 2þ 4þ 8þ � � � can lead to absurd conclusions. However,
the following result shows that there are Combination Rules for convergent
series, which follow directly from the Combination Rules for convergent
sequences:
Theorem 2 Combination Rules
IfP
1
n¼1
an ¼ s andP
1
n¼1
bn ¼ t, then:
Sum RuleP
1
n¼1
an þ bnð Þ ¼ sþ t;
Multiple RuleP
1
n¼1
lan ¼ ls, for l2R .
Proof Consider the sequences of partial sums {sn} and {tn}, where
sn ¼X
n
k¼1
ak and tn ¼X
n
k¼1
bk:
By assumption, sn! s and tn! t as n!1.
Sum Rule
The nth partial sum of the seriesP
1
n¼1
an þ bnð Þ is
X
n
k¼1
akþbkð Þ¼ a1þb1ð Þþ a2þb2ð Þþ �� �þ anþbnð Þ
¼ a1þa2þ�� �þanð Þþ b1þb2þ�� �þbnð Þ¼ snþ tn:
By the Sum Rule for sequences
limn!1
sn þ tnð Þ ¼ limn!1
sn þ limn!1
tn ¼ sþ t;
and so the sequence {snþ tn} of partial sums ofP
1
n¼1
an þ bnð Þ has limit sþ t.
Hence this series is convergent and
X
1
n¼1
an þ bnð Þ ¼ sþ t:
Multiple Rule
The nth partial sum of the seriesP
1
n¼1
lan is
X
n
k¼1
lak ¼ la1 þ la2 þ � � � þ lan
¼ l a1 þ a2 þ � � � þ anð Þ ¼ lsn:
By the Multiple Rule for sequences
limn!1
lsnð Þ ¼ l limn!1
sn¼ ls;
We may rearrange the termshere, because this is the sumof a finite number of terms.
3.1 Introducing series 89
and so the sequence {lsn} of partial sums ofP
1
n¼1
lan has limit ls. Hence this
series is convergent and
X
1
n¼1
lan ¼ ls: &
Example 2 Prove that the following series is convergent and calculate its sum
X
1
n¼1
1
2nþ 3
n nþ 1ð Þ
� �
:
Solution We know thatP
1
n¼1
12n is convergent, with sum 1, and that
P
1
n¼1
1n nþ1ð Þ is
convergent, with sum 1.
Hence, by the Sum Rule and the Multiple Rule
X
1
n¼1
1
2nþ 3
n nþ 1ð Þ
� �
is convergent; with sum 1þð3� 1Þ ¼ 4: &
Problem 5 Prove that the following series is convergent and calculate
its sum
X
1
n¼1
3
4
� �n
� 2
n nþ 1ð Þ
� �
:
3.1.5 The Non-null Test
For all the infinite series we have so far considered, it is possible to derive a
simple formula for the nth partial sum. For most series, however, this is very
difficult or even impossible.
Nevertheless, it may still be possible to decide whether such series are
convergent or divergent by applying various tests. The first test that we give
arises from the following result:
Theorem 3 IfP
1
n¼1
an is a convergent series, then {an} is a null sequence.
Proof Let sn ¼P
n
k¼1
ak denote the nth partial sum ofP
1
n¼1
an. BecauseP
1
n¼1
an is
convergent, we know that {sn} is convergent. Suppose that limn!1
sn ¼ s.
We want to deduce that {an} is null. To do this, we write
sn ¼ sn�1 þ an;
and so
an ¼ sn � sn�1:
Thus, by the Combination Rules for sequences
limn!1
an ¼ limn!1
sn � sn�1ð Þ ¼ limn!1
sn � limn!1
sn�1
¼ s� s ¼ 0;
and so {an} is a null sequence, as required. &
You met the second series inSub-section 3.1.3.
For example, consider theseries
X
1
n¼1
1
n¼ 1þ 1
2þ 1
3þ � � � ;
and
X
1
n¼1
n2
2n2 þ 1¼ 1
3þ 4
9þ 9
19þ � � � :
90 3: Series
The following useful test for divergence is an immediate corollary of
Theorem 3:
Corollary Non-null Test If {an} is not a null sequence, thenP
1
n¼1
an is
divergent.
The virtue of the Non-null Test is its ease of application. For example, it
enables us to decide immediately thatX
1
n¼1
1;X
1
n¼1
�1ð Þn andX
1
n¼1
n are divergent;
because the corresponding sequences {1}, {(�1n)} and {n} are not null.
There are various ways to show that a given sequence {an} does not tend to
zero. For example, we can show that
an ! ‘ as n!1;where ‘ 6¼ 0, or that {an} tends to infinity or to minus infinity. More generally,
we can use the result about subsequences which states that
if an ! 0 as n!1; then ank! 0 as k!1;
for any subsequence ankf g of {an}. This leads to the following strategy:
Strategy To show thatP
1
n¼1
an is divergent, using the Non-null Test:
EITHER
1. show that {an} has a convergent subsequence with non-zero limit;
OR
2. show that {an} has a subsequence which tends to infinity, or a subse-
quence which tends to minus infinity.
This strategy can be used to prove part (b) of Theorem 1, as follows:
Example 3 Prove that, if jrj � 1 and a 6¼ 0, thenP
1
n¼0
arn is divergent.
Solution We want to show that, if jrj � 1 and a 6¼ 0, then {arn} is not a null
sequence. Now, the even subsequence
ar2k� �
¼ a; ar2; ar4; ar6; . . .� �
converges to a 6¼ 0 if jrj ¼ 1, and tends to infinity or to minus infinity if jrj � 1.
Thus, {ar2k} is not null, and so {arn} is not null.
Hence, by the Non-null Test,P
1
n¼0
arn is divergent if jrj> 1 and a 6¼ 0. &
Warning The converse of the Non-null Test is FALSE. In other words:
If {an} is a null sequence, it is not necessarily true thatP
1
n¼1
an is convergent.
For example, the sequence f1ng is a null sequence, but
P
1
n¼1
1n¼1þ 1
2þ 1
3þ � � �
is divergent.
You met this InheritanceProperty of subsequences inSub-section 2.4.4, Theorem 5,for any limit ‘.
It tends to infinity if a> 0 andto minus infinity if a< 0.
We shall prove this rathersurprising fact in Sub-section 3.2.1.
3.1 Introducing series 91
The important thing to remember is that you can never use the Non-null Test
to prove that a series is convergent, although you may be able to use it to prove
that a series is divergent.
Problem 6 Prove that the seriesP
1
n¼1
n2
2n2 þ 1is divergent.
3.2 Series with non-negative terms
In this section we consider only seriesP
1
n¼1
an with non-negative terms. In other
words we assume that an� 0, for n¼ 1, 2, . . ..
It follows that the partial sums ofP
1
n¼1
an, which are given by
s1 ¼ a1; s2 ¼ a1 þ a2; s3 ¼ a1 þ a2 þ a3; . . .;
sn ¼ a1 þ a2 þ a3 þ � � � þ an; . . .;
form an increasing sequence {sn}.
The fact that {sn} is increasing makes it easier to deal with series having
non-negative terms. If we can prove that the sequence {sn} of partial sums is
bounded above, then {sn} is convergent, by the Monotone Convergence
Theorem, and soP
1
n¼1
an is convergent. On the other hand, if we can prove
that the sequence {sn} of partial sums is not bounded above, then {sn} cannot
be convergent, and so we must have that sn!1 as n!1. We can rephrase
these facts in the following convenient way:
Boundedness Theorem for series A seriesP
1
n¼1
an of non-negative terms is
convergent if and only if its sequence {sn} of partial sums is bounded above.
We shall use this, for example, to prove the Cauchy Condensation Test in
Sub-section 3.2.1.
3.2.1 Tests for convergence
We now explore several tests for the convergence of series with non-negative
terms.
Problem 1 Use your calculator to find the first eight partial sums of
each of the following series
X
1
n¼1
1
n2¼ 1þ 1
22þ 1
32þ � � � and
X
1
n¼1
1
n¼ 1þ 1
2þ 1
3þ � � �;
giving your answers to 2 decimal places. Plot your answers on a
sequence diagram.
Example 1 Prove that the seriesP
1
n¼1
1n2 ¼ 1þ 1
22 þ 132 þ � � � is convergent.
Sub-section 2.5.1, Theorem 1.
This is essentially the sameresult as that of theBoundedness Theorem forsequences, Sub-section 2.4.2,Theorem 1.
92 3: Series
Solution All the terms of the series are positive, so we shall use the
Monotone Convergence Theorem.
Let sn be the nth partial sum of the series. Then
sn ¼ 1þ 1
22þ 1
32þ 1
42þ � � � þ 1
n2
< 1þ 1
1� 2þ 1
2� 3þ 1
3� 4þ � � � þ 1
n� 1ð Þ � n
¼ 1þ 1
1� 1
2
� �
þ 1
2� 1
3
� �
þ 1
3� 1
4
� �
þ � � � þ 1
n� 1� 1
n
� �
¼ 2� 1
n< 2:
It follows that {sn} is both increasing and bounded above (by 2), so that by
the Monotone Convergence Theorem {sn} is convergent. Hence the series
itself is also convergent. &
Remark
In fact the sum of the series isP
1
n¼1
1n2 ¼ p2
6, surprisingly. However a proof of this
goes beyond the scope of this book.
Not all series are convergent, though!
Example 2 Prove that the seriesP
1
n¼1
1n¼ 1þ 1
2þ 1
3þ � � � is divergent.
Solution All the terms of the series are positive.
Let sn be the nth partial sum of the series. Then
sn ¼ 1þ 1
2þ 1
3þ 1
4þ 1
5þ 1
6þ 1
7þ 1
8þ 1
9þ 1
10þ � � �
¼ 1þ 1
2þ 1
3þ 1
4
� �
þ 1
5þ 1
6þ 1
7þ 1
8
� �
þ � � �
� � � þ 1
2k þ 1þ 1
2k þ 2þ � � � þ 1
2k þ 2k
� �
þ � � �:
It follows that a subsequence of partial sums is
s1 ¼ 1;
s2 ¼ 1þ 1
2;
s4 ¼ 1þ 1
2þ 1
3þ 1
4
� �
> 1þ 1
2þ 1
2;
s8 ¼ 1þ 1
2þ 1
3þ 1
4
� �
þ 1
5þ 1
6þ 1
7þ 1
8
� �
> 1þ 1
2þ 1
2þ 1
2;
..
.
s2k > 1þ 1
2þ 1
2þ � � � þ 1
2|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}
¼ 1þ 1
2k:
k terms
For any k> 1, 1k2< 1
ðk�1Þk.
Here we have telescopingcancellation.
The proof depends on use oftrigonometric series or ofComplex Analysis.
This series is often called theharmonic series, since itsterms are proportional to thelengths of strings that produceharmonic tones in music.
We think of n as being fairlylarge, in order to see thepattern.
Here each bracket adds up tomore than 1
2.
3.2 Series with non-negative terms 93
Hence the sequence {sn} is increasing and not bounded above. Therefore it
cannot be convergent, so that by the Monotone Convergence Theorem the
sequence {sn} is divergent. Hence the series itself is also divergent. &
We can extend this type of domination approach to test further series for
convergence, in the following way:
Theorem 1 Comparison Test
(a) If 0 � an � bn, for n ¼ 1, 2, . . ., andP
1
n¼1
bn is convergent, thenP
1
n¼1
an is
convergent.
(b) If 0 � bn � an, for n¼ 1, 2, . . ., andP
1
n¼1
bn is divergent, thenP
1
n¼1
an is
divergent.
Remark
As with the Squeeze Rule for non-negative null sequences, it is sufficient to
prove that the necessary inequalities in parts (a) and (b) hold eventually.
In applications, we use this result in the following way:
Strategy To test a seriesP
1
n¼1
an of non-negative terms for convergence
using the Comparison Test:
EITHER find a convergent seriesP
1
n¼1
bn of non-negative terms where the bn
dominate the an,
OR find a divergent seriesP
1
n¼1
bn of non-negative terms where the an
dominate the bn.
Example 3 Prove that the seriesP
1
n¼1
1n3 is convergent.
Solution We shall use the Comparison Test.
Let bn ¼ 1n2. Then 0 � 1
n3 � bn, for n¼ 1, 2, . . ., and we know that the seriesP
1
n¼1
1n2 is convergent (this was Example 1 above). It follows from part (a) of the
Comparison Test that the seriesP
1
n¼1
1n3 is convergent. &
Example 4 Prove that the seriesP
1
n¼1
1ffiffi
np is divergent.
Solution We shall use the Comparison Test.
Let bn ¼ 1n. Then 0 � bn � 1
ffiffi
np , for n¼ 1, 2, . . ., and we know that the series
P
1
n¼1
1n
is divergent (this was Example 2 above). It follows from part (b) of the
Comparison Test that the seriesP
1
n¼1
1ffiffi
np is divergent. &
Next, consider the seriesP
1
n¼1
1ffiffi
npþ3
ffiffi
n4pþ1
. Its terms are somewhat similar to
those of the seriesP
1
n¼1
1ffiffi
np , since we can express 1
ffiffi
npþ3
ffiffi
n4pþ1
in the form
In the proof of part (a), in Sub-section 3.2.2, we shall in factsee that
X
1
n¼1
an �X
1
n¼1
bn:
That is, it is sufficient that0� an� bn or 0� bn� an forall n>X, for some number X.
That is, where an� bn.
That is, where bn� an.
For 1n3 � 1
n2 :
For 1n� 1
ffiffi
np :
In other words, the termffiffiffi
np
dominates the term 3ffiffiffi
n4pþ 1.
94 3: Series
1ffiffi
np � 1
1þ3n�1
4þn�1
2
, where the second fraction tends to 1 as n!1. So, since the
seriesP
1
n¼1
1ffiffi
np is divergent, it seems likely that the series
P
1
n¼1
1ffiffi
npþ3ffiffi
n4pþ1
is also
divergent.
We can pin down the underlying idea here in the following useful result:
Theorem 2 Limit Comparison Test
Suppose thatP
1
n¼1
an andP
1
n¼1
bn have positive terms, and that
an
bn
! L as n!1; where L 6¼ 0:
(a) IfP
1
n¼1
bn is convergent, thenP
1
n¼1
an is convergent.
(b) IfP
1
n¼1
bn is divergent, thenP
1
n¼1
an is divergent.
So, take an ¼ 1ffiffi
npþ3ffiffi
n4pþ1
and bn ¼ 1ffiffi
np , for n¼ 1, 2, . . .. Then both an and bn
are positive, and
an
bn
¼ 1=ffiffiffi
npþ 3
ffiffiffi
n4pþ 1ð Þð Þ
1=ffiffiffi
np
ð Þ
¼ffiffiffi
np
ffiffiffi
npþ 3
ffiffiffi
n4pþ 1
¼ 1
1þ 3n�14 þ n�
12
! 1 as n!1:
It then follows from part (b) of the Limit Comparison Test that, since the
seriesP
1
n¼1
1ffiffi
np is divergent, the series
P
1
n¼1
1ffiffi
npþ3ffiffi
n4pþ1
is also divergent.
Example 5 Use the Limit Comparison Test to prove that the seriesP
1
n¼1
n2�3nþ45n4�n
is convergent.
Solution Set an ¼ n2�3nþ45n4�n
and bn ¼ 1n2, for n¼ 1, 2, . . .. Then both an and bn
are positive, and
an
bn
¼ n2 � 3nþ 4
5n4 � n� n2
1
¼ n4 � 3n3 þ 4n2
5n4 � n
¼ 1� 3n�1 þ 4n�2
5� n�3! 1
5as n!1:
:
Since 156¼ 0 and the series
P
1
n¼1
1n2 is known to be convergent, then the series
P
1
n¼1
n2�3nþ45n4�n
is also convergent. &
By Example 4.
In other words, bn behaves‘rather like an’ for large n.Note that it is IMPORTANT thatL is non-zero.
Since 1 6¼ 0.
By Example 4.
We make this choice for bn
since, for large n, the
expression n2�3nþ45n4�n
is ‘more or
less the same as’ n2
5n4 ¼ 15n2 –
where the factor ‘5’ will notaffect our argument.
Dividing numerator anddenominator by the dominantterm n4.
By Example 1.
3.2 Series with non-negative terms 95
Problem 2 Use the Comparison Test or the Limit Comparison Test to
determine the convergence of the following series:
(a)P
1
n¼1
1n3þn
; (b)P
1
n¼1
1nþffiffi
np ; (c)
P
1
n¼1
nþ42n3�nþ1
; (d)P
1
n¼1
cos2 2nð Þn3 :
Our next test is motivated in part by the geometric series – recall that the
geometric series aþ ar þ ar2 þ � � � ¼P
1
n¼0
arn, where a 6¼ 0, is convergent if
jrj< 1 but divergent if jrj �1. The reason that this converges, if jrj< 1, is that
the terms are then forced to tend to 0 so quickly that the partial sums sn increase
so slowly as n increases that they remain bounded above. On the other hand, if
jrj � 1, the terms do not tend to 0, so that the series must diverge.
Theorem 3 Ratio Test
Suppose thatP
1
n¼1
an has positive terms, and that anþ1
bn! ‘ as n!1.
(a) If 0� ‘< 1, thenP
1
n¼1
an is convergent.
(b) If ‘> 1, thenP
1
n¼1
an is divergent.
Remark
If ‘¼ 1, then the test gives us no information on whether the series converges.
For example, if an ¼ 1n, then anþ1
an¼ n
nþ1¼ 1
1þ1n
! 1 as n!1; we have seen
that the seriesP
1
n¼1
an ¼P
1
n¼1
1n
diverges. On the other hand, if an ¼ 1n2 then
anþ1
an¼ n2
nþ1ð Þ2 ¼1
1þ2nþ 1
n2
! 1 as n!1; but the seriesP
1
n¼1
an ¼P
1
n¼1
1n2 converges.
Example 6 Use the Ratio Test to determine the convergence of the following
series:
(a)P
1
n¼1
n2n; (b)
P
1
n¼1
10n
n! .
Solution
(a) Let an ¼ n2n, n¼ 1, 2, . . .. Then an is positive, and
anþ1
an
¼ nþ 1
2nþ1� 2n
n
¼1þ 1
n
2! 1
2as n!1:
Since 0 < 12< 1, it follows from the Ratio Test that the series
P
1
n¼1
an ¼P
1
n¼1
n2n is convergent.
(b) Let an ¼ 10n
n! , n¼ 1, 2, . . .. Then an is positive, and
anþ1
an
¼ 10nþ1
nþ 1ð Þ!�n!
10n
¼ 10
nþ 1! 0 as n!1:
By the Non-null Test.
Note that with the Ratio Test,we concentrate on the seriesP
1
n¼1
an itself and do not need to
‘think of’ some other seriesP
1
n¼1
bn.
Part (b) includes the caseanþ1
an!1.
That is, the Ratio Test isinconclusive if ‘¼ 1.
By Example 2.
By Example 1.
96 3: Series
It follows from the Ratio Test that the seriesP
1
n¼1
an ¼P
1
n¼1
10n
n! is
convergent. &
Problem 3 Use the Ratio Test to determine whether the following
series are convergent
(a)P
1
n¼1
n3
n!; (b)P
1
n¼1
n22n
n! ; (c)P
1
n¼1
2nð Þ!nn .
We have now seen examples of many different convergent and divergent
series, including examples of various generic types. We call these basic series,
since we shall use them commonly together with various other tests in order to
prove that particular series are themselves convergent or divergent.
Basic series The following series are convergent:
(a)P
1
n¼1
1np ; for p � 2;
(b)P
1
n¼1
cn; for 0 � c < 1;
(c)P
1
n¼1
npcn; for p > 0; 0 � c < 1;
(d)P
1
n¼1
cn
n!; for c � 0.
The following series is divergent:
(e)P
1
n¼1
1np ; for p � 1.
Another test for convergence
When we examined the convergence or divergence of the seriesP
1
n¼1
1n, we found
it convenient to group the terms into blocks that contained in turn 1, 2, 4, 8, . . .successive individual terms of the series. In fact, a similar technique applies to
a wide range of series, and the following result is thus often invaluable in order
to save repeating that same type of argument:
Theorem 4 Cauchy Condensation Test
Let {an} be a decreasing sequence of positive terms. Then, if bn ¼ 2na2n , for
n¼ 1, 2, . . .
X
1
n¼1
an is convergent if and only ifX
1
n¼1
bn is convergent:
For example, consider the seriesP
1
n¼1
1n. Here, let an ¼ 1
n, so that
bn ¼ 2na2n ¼ 2n � 1
2n¼ 1:
Then the Condensation Test tells us that the seriesP
1
n¼1
1n
is convergent if and
only if the seriesP
1
n¼1
bn ¼P
1
n¼1
1 is convergent. This is divergent, by the Non-
null Test, so that the original seriesP
1
n¼1
1n
must itself have been divergent.
Instances of these that wehave seen areP
1
n¼1
1n2;
Geometric Series,P
1
n¼1
12
� �n;
P
1
n¼1
n2n ¼
P
1
n¼1
n 12
� �n;
P
1
n¼1
10n
n!;
P
1
n¼1
1ffiffi
np ¼
P
1
n¼1
1n1=2
.
You will see this techniqueused in the proof ofTheorem 4, in Sub-section 3.2.2.
For, if an ¼ 1n, then a2n ¼ 1
2n.
3.2 Series with non-negative terms 97
This use of the Condensation Test is much simpler than setting out to
perform a grouping and estimation exercise on such occasions! The
Condensation Test is also often useful when the individual terms of a series
include expressions such as loge n.
Example 7 Use the Condensation Test to determine the convergence of the
seriesP
1
n¼2
1
nffiffiffiffiffiffiffiffi
loge np .
Solution Let an ¼ 1
nffiffiffiffiffiffiffiffi
loge np , n¼ 2, . . .. Then an is positive; and, since
nffiffiffiffiffiffiffiffiffiffiffiffi
loge np� �
is an increasing sequence, anf g ¼n
1
nffiffiffiffiffiffiffiffi
loge np
o
is a decreasing
sequence.
Next, let bn ¼ 2na2n ; thus
bn ¼ 2n � 1
2nffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
logeð2nÞp
¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
n loge 2p ¼ 1
ffiffiffiffiffiffiffiffiffiffiffiffi
loge 2p � 1
n12
:
SinceP
1
n¼2
1
n12
is a basic divergent series, it follows by the Multiple Rule that
P
1
n¼2
bn must be divergent. Hence by the Condensation Test, the original series
P
1
n¼2
1
nffiffiffiffiffiffiffiffi
loge np must also be divergent. &
Problem 4 Use the Condensation Test to determine the convergence
of the following series:
(a)P
1
n¼2
1n loge n
; (b)P
1
n¼2
1
n loge nð Þ2.
3.2.2 Proofs
In the previous sub-section we gave a number of tests for the convergence of
series with non-negative terms. We now supply the proofs of these tests.
Theorem 1 Comparison Test
(a) If 0� an� bn, for n¼ 1, 2, . . ., andP
1
n¼1
bn is convergent, thenP
1
n¼1
an is
convergent.
(b) If 0� bn� an, for n¼ 1, 2, . . ., andP
1
n¼1
bn is divergent, thenP
1
n¼1
an is
divergent.
Proof
(a) Consider the nth partial sums
sn ¼ a1 þ a2 þ � � � þ an; for n ¼ 1; 2; . . .;
Note that here we only sumfrom n¼ 2, since it makes no
sense to talk about 1
1ffiffiffiffiffiffiffiffi
loge 1p , as
loge 1 ¼ 0. Again we areusing the fact that theconvergence of a series is notaffected if we add or alter afinite of terms.
You may omit these proofs ata first reading.
98 3: Series
and
tn ¼ b 1 þ b2 þ � � � þ bn ; for n ¼ 1; 2; . . .:
We know that
a1 � b1 ; a2 � b2 ; . . .; an � bn ;
and so
sn � tn ; for n ¼ 1; 2; . . .:
We also know thatP
1
n ¼ 1
bn is convergent , and so the incr easing seque nce
{tn} is conver gent, with limi t t, say. Hence
sn � tn � t ; for n ¼ 1; 2; . . .;
and so the increasing sequence {sn} i s b ou nd e d a b ov e b y t. By t he M ono tone
C o nv er g e nc e T h eo r e m , { sn} i s t he refore also c onv erg ent, and soP
1
n ¼ 1
an is a
con verg ent se ri es.
Note : Moreover , by the Limit Inequ ality Rule
X
1
n¼ 1
an �X
1
n ¼ 1
bn :
(b) We can deduce part (b) from part (a), as follows. Suppose thatP
1
n ¼ 1
an is
conver gent. Th en, by part (a),P
1
n ¼ 1
bn mus t also be conver gent. Howeve r,P
1
n ¼ 1
bn is assumed to be dive rgent, and soP
1
n¼ 1
an mus t also be dive rgent. &
The orem 2 Limit Compari son Test
Sup pose thatP
1
n ¼ 1
an andP
1
n¼ 1
bn have posi tive terms , and that
an
bn
! L as n !1; where L 6¼ 0:
(a) IfP
1
n¼ 1
bn is convergent , thenP
1
n ¼ 1
an is conver gent.
(b) IfP
1
n ¼ 1
bn is divergen t, thenP
1
n¼ 1
an is divergen t.
Proof
(a) Becaus e�
an
bn
�
is convergent , it mus t be bounded . Thus there is a const ant K
such that
an
bn
� K ; for n ¼ 1 ; 2 ; . . .;
and so
an � Kbn ; for n ¼ 1 ; 2; . . .:
By the Multiple Rule,P
1
n¼1
Kbn is convergent. Hence, by the Comparison
Test,P
1
n¼1
an is also convergent.
By adding up the previousinequalities.
You met this Rule inSub-section 2.3.3, Theorem 3.This additional inequality willsometimes be useful.
Such ‘proofs bycontradiction’ can sometimessave a great deal of detailedarguments.
See Sub-section 2.4.2,Theorem 1 (the BoundednessTheorem).
3.2 Series with non-negative terms 99
(b) We can deduce part (b) from part (a), as follows.
Suppose thatP
1
n¼1
an is convergent. Then, by part (a),P
1
n¼1
bn must also be
convergent, because
bn
an
! 1
Las n!1;
by the Quotient Rule for sequences.
However,P
1
n¼1
bn is assumed to be divergent, and soP
1
n¼1
an must also be
divergent. &
Theorem 3 Ratio Test
Suppose thatP
1
n¼1
an has positive terms, and that anþ1
an! ‘ as n!1.
(a) If 0� ‘< 1, thenP
1
n¼1
an is convergent.
(b) If ‘> 1, thenP
1
n¼1
an is divergent.
Proof
(a) Because ‘< 1, we can choose "> 0 so that
r ¼ ‘þ " < 1:
Hence there is a positive number X, which we may assume to be a positive
integer, such that
anþ1
an
� r; for all n � X:
Thus, for n�X, we have
an
aX
¼ an
an�1
� �
an�1
an�2
� �
. . .aXþ1
aX
� �
� rn�X;
since each of the n�X brackets is less than or equal to r. Hence
an � aXrn�X; for all n � X: (1)
Now
X
1
n¼X
aXrn�X ¼ aX þ aXr þ aXr2 þ � � �;
which is a geometric series with first term aX and common ratio r. Since
0< r< 1, this series is convergent, and so, by inequality (1) and the
Comparison Test,P
1
n¼X
an is also convergent. HenceP
1
n¼1
an is convergent.
(b) Since
anþ1
an
! ‘ as n!1;
and ‘> 1, there is a positive number X, which we may assume to be a
positive integer, such that
anþ1
an
� 1; for all n � X:
We are following a ‘proof bycontradiction’ approach here.
Remember that L 6¼ 0.
Remember that part (b)includes the case anþ1
an!1.
For example, take" ¼ 1
21� ‘ð Þ.
Recall that we occasionallymake the convenientassumption that X is an integerto avoid notationalcomplications.
1
X n
� + ε
� – ε�
�1
X n
100 3: Series
(This also holds if anþ1
an!1 as n!1.)
Thus, for n�X, we have
an
aX
¼ an
an�1
� �
an�1
an�2
� �
. . .aXþ1
aX
� �
� 1;
since each of the brackets is greater than or equal to 1. Hence
an � aX > 0; for all n � X;
and so {an} cannot be a null sequence. Thus, by the Non-null Test,P
1
n¼X
an is
divergent. HenceP
1
n¼1
an is also divergent. &
Theorem 4 Cauchy Condensation Test
Let {an} be a decreasing sequence of positive terms. Then, if bn ¼ 2na2n for
n¼ 1, 2, . . .
X
1
n¼1
an is convergent if and only ifX
1
n¼1
bn is convergent:
Proof Denote by sn and tn the nth partial sums of the seriesP
1
n¼1
an andP
1
n¼1
bn,
respectively
sn ¼ a1 þ a2 þ � � � þ an and tn ¼ 21a2 þ 22a4 þ � � � þ 2na2n :
Since the terms an are decreasing and a1� 0, it follows that
s2n ¼ a1 þ a2ð Þ þ a3 þ a4ð Þ þ � � � þ a2n�1þ1 þ � � � þ a2nð Þ� 0þ a2ð Þ þ a4 þ a4ð Þ þ � � � þ a2n þ � � � þ a2nð Þ¼ 0þ a2ð Þ þ 2a4ð Þ þ � � � þ 2n�1a2n
� �
¼ 1
2tn: ð2Þ
We may also write s2n in another way
s2n ¼ a1 þ a2 þ a3ð Þ þ a4 þ a5 þ a6 þ a7ð Þ þ � � �þ a2n�1 þ � � � þ a2n�1ð Þ þ a2n
� a1 þ a2 þ a2ð Þþ a4 þ a4 þ a4 þ a4ð Þ þ � � � þ a2n�1 þ � � � þ a2n�1ð Þ þ a2n
¼ a1 þ 2a2ð Þ þ 4a4ð Þ þ � � � þ 2n�1a2n�1
� �
þ a2n
� a1 þ tn: ð3Þ
Now suppose thatP
1
n¼1
an is convergent. Then, by the Boundedness Theorem for
series, its sequence of partial sums {sn} must be bounded above and so the sequence
s2nf g must also be bounded above – since it is a subsequence of {sn}. It therefore
follows from inequality (2) that the sequence 12
tn
� �
must be bounded above, so the
sequence {tn} must also be bounded above. Another application of the Boundedness
Theorem shows that therefore the seriesP
1
n¼1
bn must be convergent.
The kth bracket contains 2k�1
terms.
The kth bracket contains 2k�1
occurrences of a2k.
The kth bracket contains 2k
terms.
The kth bracket contains 2k
occurrences of a2k.
In obtaining (3), we havereplaced the term a2n in the pre-vious line by a larger term 2na2n.
The Boundedness Theoremfor series appeared inSection 3.2, above.
3.2 Series with non-negative terms 101
Next, suppose thatP
1
n¼1
an is divergent. Then, by the Boundedness Theorem
for series, its sequence of partial sums {sn} must be unbounded above, and so
the sequence s2nf gmust also be unbounded above – since it is a subsequence of
{sn}. It therefore follows from inequality (3) that the sequence {a1þ tn} must
be unbounded above, so the sequence {tn} must also be unbounded above.
Another application of the Boundedness Theorem shows that therefore the
seriesP
1
n¼1
bn must be divergent. &
We end this section by proving the convergence/divergence of the basic
series given earlier.
Basic series The following series are convergent:
(a)P
1
n¼1
1np; for p � 2;
(b)P
1
n¼1
cn; for 0 � c < 1;
(c)P
1
n¼1
npcn; for p > 0; 0 � c < 1;
(d)P
1
n¼1
cn
n!; for c � 0.
The following series is divergent:
(e)P
1
n¼1
1np; for p � 1.
Proof
(a) This series is convergent, by the Comparison Test, since, if p� 2, then
1
np� 1
n2; for n ¼ 1; 2; . . .;
and the seriesP
1
n¼1
1n2 is convergent.
(b) The seriesP
1
n¼1
cn is a geometric series with common ratio c, and so it
converges if 0� c< 1.
(c) Let an¼ npcn, for n¼ 1, 2, . . .. Then, for c 6¼ 0
anþ1
an
¼ nþ 1ð Þpcnþ1
npcn¼ 1þ 1
n
� �p
c:
Thus, if k is any integer greater than or equal to p, then
c � anþ1
an
� 1þ 1
n
� �k
c:
Now, by the Product Rule for sequences
1þ 1
n
� �k
! 1 as n!1;
In Sub-section 3.2.1.
In Chapter 7 we shall prove
thatP
1
n¼1
1np is convergent, for
all p> 1.
Sub-section 3.2.1, Example 1.
Sub-section 3.1.2.
If c¼ 0, the series is clearlyconvergent.
We introduce the integer khere, so that we can use theProduct Rule for sequences.
102 3: Series
and so, by the Squeeze Rule for sequencesanþ1
an
! c as n!1:
Since 0� c< 1, we deduce, from the Ratio Test, thatP
1
n¼1
npcn is
convergent.
(d) Let an ¼ cn
n!, for n¼ 1, 2, . . .. Then, for c 6¼ 0,
anþ1
an
¼ cnþ1
nþ 1ð Þ!
cn
n!¼ c
nþ 1:
Thus anþ1
an! 0 as n!1, and we deduce from the Ratio Test that
P
1
n¼1
cn
n! is
convergent.
(e) We saw earlier that the seriesP
1
n¼1
1n
is divergent. It follows that the seriesP
1
n¼1
1np is also divergent, by the Comparison Test, since if p� 1 then 1
np � 1n,
for n¼ 1, 2, . . .. &
3.3 Series with positive and negative terms
The study of seriesP
1
n¼1
an, with an� 0 for all values of n, is relatively straight-
forward, because the sequence of partial sums {sn} is increasing. Similarly, if
an� 0 for all values of n, then {sn} is decreasing.
However, it is harder to determine the behaviour of a series with both
positive and negative terms, because {sn} is neither increasing nor decreasing.
However, if the sequence {an} contains only finitely many negative terms, then
the sequence {sn} is eventually increasing, and we can apply the methods of
Section 3.2. Similarly, if {an} contains only finitely many positive terms, then
the sequence {sn} is eventually decreasing, and we can again apply the
methods of Section 3.2, after making a sign change.
In this section we look at series such as
1� 1
2þ 1
3� 1
4þ 1
5� 1
6þ � � � and 1þ 1
22� 1
32þ 1
42þ 1
52� 1
62þ � � �;
which contain infinitely many terms of either sign. We give several methods
which can often be used to prove that such series are convergent.
3.3.1 Absolute convergence
Suppose that we want to determine the behaviour of the infinite series
X
1
n¼1
�1ð Þnþ1
n2¼ 1� 1
22þ 1
32� 1
42þ 1
52� 1
62þ � � �: (1)
We know that the series
X
1
n¼1
1
n2¼ 1þ 1
22þ 1
32þ 1
42þ 1
52þ 1
62þ � � � (2)
is convergent. Does this mean that the series (1) is also convergent? In fact it
does, as we now prove.
If c¼ 0, the series is clearlyconvergent.
In Sub-section 3.2.1,Example 2.
For example, the convergenceof
1þ 2þ 3� 1
42� 1
52� 1
62� � � �
follows from that ofP
1
n¼1
1n2 :
The series (2) is a basicconvergent series.
3.3 Series with positive and negative terms 103
Consider the two related series
1þ 0þ 1
32þ 0þ 1
52þ 0þ � � �
and
0þ 1
22þ 0þ 1
42þ 0þ 1
62þ � � � :
Each of these series is dominated by the series (2), and so they are both
convergent, by the Comparison Test. Applying the Multiple Rule with l¼�1,
and the Sum Rule, we deduce that the series
1� 1
22þ 1
32� 1
42þ 1
52� 1
62þ � � � is convergent:
The argument just given is the basis for a concept called absolute conver-
gence, which we now define.
Definitions
A seriesP
1
n¼1
an is absolutely convergent ifP
1
n¼1
anj j is convergent.
A seriesP
1
n¼1
an that is convergent but not absolutely convergent is condi-
tionally convergent.
For example, the series (1) is absolutely convergent, because the seriesP
1
n¼1
1n2
is convergent.
However, the series
X
1
n¼1
�1ð Þnþ1
n¼ 1� 1
2þ 1
3� 1
4þ 1
5� 1
6þ � � �
is not absolutely convergent, because the seriesP
1
n¼1
1n
is divergent; the series is,
in fact, conditionally convergent.
Theorem 1 Absolute Convergence Test
IfP
1
n¼1
an is absolutely convergent, thenP
1
n¼1
an is convergent.
It follows that the series (1) is convergent (as we have already seen), and also
that the series
1þ 1
22� 1
32þ 1
42þ 1
52� 1
62þ � � �
is convergent, because the seriesP
1
n¼1
1n2 is convergent. Indeed, no matter how
we distribute the plus and minus signs among the terms of 1n2
� �
, the resulting
series is convergent.
However, the Absolute Convergence Test tells us nothing about the
behaviour of the series
1� 1
2þ 1
3� 1
4þ 1
5� 1
6þ � � � (3)
and
1þ 1
2� 1
3þ 1
4þ 1
5� 1
6þ 1
7þ 1
8� 1
9þ � � � : (4)
If the terms an are all non-negative, then absoluteconvergence and convergencehave the same meaning.
We shall examine thebehaviour of conditionallyconvergent series later, inSub-sections 3.3.2 and 3.3.3.
You saw this in Example 2,Sub-section 3.2.1.
The proofs of the results inthis sub-section appear inSub-section 3.3.6.
In fact, series (3) is convergent(by the Alternating Test,which we introduce in Sub-section 3.3.2), and series (4) isdivergent (see Exercise 2(a)for Section 3.3).
104 3: Series
The seriesP
1
n¼1
1n
is divergent, and so the two series (3) and (4) are not
absolutely convergent.
Example 1 Prove that the following series are convergent:
(a)P
1
n¼1
�1ð Þnþ1
n3 ; (b) 1þ 12� 1
4þ 1
8þ 1
16� 1
32� � �.
Solution
(a) Let an¼ �1ð Þnþ1
n3 , for n¼ 1, 2, . . .; then anj j ¼ 1n3. We know that
P
1
n¼1
1n3 is
convergent, so it follows thatP
1
n¼1
�1ð Þnþ1
n3 is absolutely convergent. Hence,
by the Absolute Convergence Test,P
1
n¼1
�1ð Þnþ1
n3 is convergent.
(b) The seriesP
1
n¼0
12n is a convergent geometric series, so that the series
1þ 1
2� 1
4þ 1
8þ 1
16� 1
32� � �
is absolutely convergent. Then, by the Absolute Convergence Test, this
series is also convergent. &
Problem 1 Prove that the following series are convergent:
(a)P
1
n¼1
ð�1Þnþ1n
n3þ1; (b)
P
1
n¼1
cos n2n .
The Absolute Convergence Test states that, ifP
anj j is convergent, thenP
an is also convergent, but it does not indicate any explicit connection
between the sums of these two series. Clearly,P
an is less thanP
anj j if any
of the terms an are negative.
For example,P
1
n¼1
12n ¼ 1
2þ 1
4þ 1
8þ � � � ¼ 1, whereas
P
1
n¼1
�1ð Þnþ1
2n ¼12� 1
4þ 1
8� � � � ¼ 1
3:
The following result relates the values ofP
an andP
anj j:
Triangle Inequality (infinite form)
IfP
1
n¼1
an is absolutely convergent, thenP
1
n¼1
an
�
�
�
�
�
�
�
�
�P
1
n¼1
anj j.
Problem 2 Show that the series 12þ 1
4� 1
8þ 1
16þ 1
32� 1
64þ � � � is con-
vergent, and that its sum lies in [�1,1]. (You do NOT need to find the sum
of the series.)
3.3.2 The Alternating Test
Suppose that we want to determine the behaviour of the following infinite
series, in which the terms have alternating signsX
1
n¼1
�1ð Þnþ1
n¼ 1� 1
2þ 1
3� 1
4þ 1
5� 1
6þ � � �:
ForP
1
n¼1
1n3 is an example of a
basic convergent series.
Recall that you met theTriangle Inequality fora1; a2; . . .; an
X
n
k¼1
ak
�
�
�
�
�
�
�
�
�
�
�X
n
k¼1
akj j;
in Sub-section 1.3.1.
3.3 Series with positive and negative terms 105
The Absolute Convergence Test does not help us with this series, becauseP
1
n¼1
1n
is divergent.
In fact, the seriesP
1
n¼1
�1ð Þnþ1
nis convergent. To see why, we first calculate
some of its partial sums and plot them on a sequence diagram
s1 ¼ 1;
s2 ¼ 1� 1
2¼ 0:5;
s3 ¼ 1� 1
2þ 1
3¼ 0:83;
s4 ¼ 1� 1
2þ 1
3� 1
4¼ 0:583;
s5 ¼ 1� 1
2þ 1
3� 1
4þ 1
5¼ 0:783;
s6 ¼ 1� 1
2þ 1
3� 1
4þ 1
5� 1
6¼ 0:616:
The sequence diagram for {sn} suggests that
s1 � s3 � s5 � . . . � s2k�1 � . . .
and
s2 � s4 � s6 � . . . � s2k � . . . ;
for all k. In other words:
the odd subsequence {s2k�1} is decreasing
and:
the even subsequence {s2k} is increasing.
Also, the terms of {s2k�1} all exceed the terms of {s2k}, and both subsequences
appear to converge to a common limit s, which lies between the odd and even
partial sums.
To prove this, we write the even partial sums {s2k} as follows
s2k ¼ 1� 1
2
� �
þ 1
3� 1
4
� �
þ � � � þ 1
2k � 1� 1
2k
� �
:
All the brackets are positive, and so the subsequence {s2k} is increasing.
We can also write
s2k ¼ 1� 1
2� 1
3
� �
� 1
4� 1
5
� �
� � � � � 1
2k � 2� 1
2k � 1
� �
� 1
2k:
Again, all the brackets are positive, and so {s2k} is bounded above, by 1.
Hence {s2k} is convergent, by the Monotone Convergence Theorem.
Let
limk!1
s2k ¼ s:
Since
s2k ¼ s2k�1 �1
2k
and 12k
� �
is null, we have
You met subsequencesearlier, in Sub-section 2.4.4.
106 3: Series
limk!1
s2k�1 ¼ limk!1
s2k þ1
2k
� �
¼ limk!1
s2k þ limk!1
1
2k
� �
¼ sþ 0 ¼ s;
by the Combination Rules for sequences. Thus the odd and even subsequences
of {sn} both tend to the same limit s, and so {sn} itself tends to s. Hence
X
1
n¼1
�1ð Þnþ1
n¼ 1� 1
2þ 1
3� 1
4þ 1
5� 1
6þ � � � is convergent, with sum s:
The same method can be used to prove the following general result:
Theorem 2 Alternating Test
If
an ¼ �1ð Þnþ1bn; n ¼ 1; 2; . . .;
where {bn} is a decreasing null sequence with positive terms, then
X
1
n¼1
an ¼ b1 � b2 þ b3 � b4 þ � � � is convergent:
When we apply the Alternating Test, there are a number of conditions which
must be checked. We now describe these in the form of a strategy.
Strategy To prove thatP
1
n¼1
an is convergent, using the Alternating Test,
check that
an ¼ �1ð Þnþ1bn; n ¼ 1; 2; . . .;
where:
1. bn� 0, for n¼ 1, 2, . . .;
2. {bn} is a null sequence;
3. {bn} is decreasing.
Remark
It is often easiest to check that {bn} is decreasing by verifying that�
1bn
�
is
increasing.
Here are some examples.
Example 2 Determine which of the following series are convergent:
(a)P
1
n¼1
�1ð Þnþ1
ffiffi
np ; (b)
P
1
n¼1
�1ð Þnþ1
n4 ; (c)P
1
n¼1
�1ð Þnþ1.
Solution
(a) The sequence�1ð Þnþ1
ffiffi
np
n o
has terms of the form an ¼ �1ð Þnþ1bn, where
bn ¼1ffiffiffi
np ; n ¼ 1; 2; . . .:
By Theorem 6 ofSub-section 2.4.4.
In fact, s¼ loge 2 ’ 0.69.
This test is sometimes calledthe Leibniz Test.
To show that {bn} is null, usethe techniques introduced inSub-section 2.2.1.
3.3 Series with positive and negative terms 107
Now:
1. bn ¼ 1ffiffi
np � 0, for n¼ 1, 2, . . .;
2. bnf g ¼�
1ffiffi
np�
is a basic null sequence;
3. bnf g ¼�
1ffiffi
np�
is decreasing, because�
1bn
�
¼ffiffiffi
npf g is increasing.
Hence, by the Alternating Test,P
1
n¼1
�1ð Þnþ1
ffiffi
np is convergent.
(b) The sequence�1ð Þnþ1
n4
n o
has terms of the form an ¼ �1ð Þnþ1bn, where
bn ¼1
n4; n ¼ 1; 2; . . .:
Now:
1. bn ¼ 1n4 � 0, for n¼ 1, 2, . . .;
2. bnf g ¼ 1n4
� �
is a basic null sequence;
3. bnf g ¼ 1n4
� �
is decreasing, because�
1bn
�
¼�
n4�
is increasing.
Hence, by the Alternating Test,P
1
n¼1
�1ð Þnþ1
n4 is convergent.
(c) The sequence�
�1ð Þnþ1�
is not a null sequence. Hence, by the Non-null
TestX
1
n¼1
�1ð Þnþ1is divergent. &
Problem 3 Determine which of the following series are convergent:
(a)P
1
n¼1
�1ð Þnþ1
n3 ; (b)P
1
n¼1
�1ð Þnþ1 nnþ2
; (c)P
1
n¼1
�1ð Þnþ1
n1=3 þ n1=2.
3.3.3 Rearrangement of a series
In the last sub-section we saw that the series 1� 12þ 1
3� 1
4þ 1
5� 1
6þ � � � is
convergent. Let us denote by ‘ the sum of this series.
If we temporarily ignore the need for careful mathematical proof and simply
‘move terms about’, look what we get
‘ ¼ 1� 1
2þ 1
3� 1
4þ 1
5� 1
6þ � � � (5)
¼ 1� 1
2� 1
4þ 1
3� 1
6� 1
8þ 1
5� 1
10� 1
12þ 1
7� 1
14� 1
16þ � � � ð6Þ
¼ 1� 1
2
� �
� 1
4þ 1
3� 1
6
� �
� 1
8þ 1
5� 1
10
� �
� 1
12
þ 1
7� 1
14
� �
� 1
16þ � � �
Alternatively, we can showthat this series is convergentby the Absolute ConvergenceTest.
Notice that the fact that thenth term includes a factor(�1)nþ1 does not imply at allthat a series is convergent!
In fact this series isconditionally convergent,since the harmonic series1þ 1
2þ 1
3þ 1
4þ 1
5þ 1
6þ � � � is
divergent.
This is the definition of ‘.
The pattern in (6) is to followthe next available positiveterm by the two next availablenegative terms.
Here we insert some brackets.
108 3: Series
¼ 1
2� 1
4þ 1
6� 1
8þ 1
10� 1
12þ 1
14� 1
16þ � � �
¼ 1
21� 1
2þ 1
3� 1
4þ 1
5� 1
6þ 1
8� � � �
� �
¼ 1
2‘:
It follows from the fact that ‘ ¼ 12‘ that ‘ ¼ 0. However this is impossible,
since the partial sum s2 of the original series is 12, and the even partial sums s2k
are increasing.
What has gone wrong? We assumed that operations that are valid for sums of
a finite number of terms also hold for sums of an infinite number of terms – we
rearranged (infinitely many of) the terms in the series (5) to obtain the series
(6) – without any justification. This rearrangement operation is not valid in
general!
We now give a precise definition of what we mean by a rearrangement.
Loosely speaking, the seriesP
1
n¼1
bn ¼ b1 þ b2 þ � � � is a rearrangement of the
seriesP
1
n¼1
an ¼ a1 þ a2 þ � � � if precisely the same terms appear in the
sequences {bn}¼ {b1, b2, . . .} and {an}¼ {a1, a2, . . .}, though they may occur
in a different order.
Definition A seriesP
1
n¼1
bn ¼ b1 þ b2 þ � � � is a rearrangement of the seriesP
1
n¼1
an ¼ a1 þ a2 þ � � � if there is a bijection ƒ such that
f : N ! N
bn 7! af ðnÞ:
Example 2 Assuming that the sum of the seriesP
1
n¼1
�1ð Þnþ1
n¼ 1� 1
2þ 1
3� 1
4þ
15� 1
6þ � � � is loge 2, prove that the series
P
1
n¼1
an¼ 1þ 13� 1
2þ 1
5þ 1
7� 1
4þ 1
9þ
111� 1
6þ � � � converges to 3
2loge 2.
Solution Let sn and tn denote the nth partial sums of the series
1þ 13� 1
2þ 1
5þ 1
7� 1
4þ 1
9þ 1
11� 1
6þ � � � and 1� 1
2þ 1
3� 1
4þ 1
5� 1
6þ � � �,
respectively.
We now introduce an extra piece of notation, Hn; we denote by Hn the nth
partial sumP
n
k¼1
1k¼ 1þ 1
2þ � � � þ 1
nof the harmonic series.
Now, the terms of the seriesP
1
n¼1
an come ‘in a natural way’ in threes, so it
seems sensible to look at the partial sums s3n
s3n ¼ 1þ 1
3� 1
2þ 1
5þ 1
7� 1
4þ 1
9þ 1
11� 1
6þ � � �
þ 1
4n� 3þ 1
4n� 1� 1
2n
Here we insert the value ofeach bracket.
Here we pull out a commonfactor of 1
2.
Recall that, since the seriesconverges to ‘, then s2k
converges to ‘ also.
In fact the series (6) doesconverge; we ask you tosupply a careful proof of thisin Exercise 2 on Section 3.3.
Recall that a bijection is aone–one onto mapping.
This will enable us to tacklethings much more easily!
From the definition of s3n.
3.3 Series with positive and negative terms 109
¼ 1þ 1
3� 1
2
� �
þ 1
5þ 1
7� 1
4
� �
þ 1
9þ 1
11� 1
6
� �
þ � � �
þ 1
4n� 3þ 1
4n� 1� 1
2n
� �
¼ 1þ 1
3þ 1
5þ � � � þ 1
4n� 1
� �
� 1
2þ 1
4þ 1
6þ � � � þ 1
2n
� �
¼ 1þ 1
2þ 1
3þ � � � þ 1
4n
� �
� 1
2þ 1
4þ 1
6þ � � � þ 1
4n
� ��
� 1
2Hn
¼ 1þ 1
2þ 1
3þ � � � þ 1
4n
� �
� 1
21þ 1
2þ 1
3þ � � � þ 1
2n
� ��
� 1
2Hn
¼ H4n �1
2H2n �
1
2Hn: (7)
Furthermore
t2n ¼ 1� 1
2þ 1
3� 1
4þ 1
5� 1
6þ � � � þ 1
2n� 1� 1
2n
¼ 1þ 1
2þ 1
3þ � � � þ 1
2n
� �
� 21
2þ 1
4þ 1
6þ � � � þ 1
2n
� �
¼ H2n � 1þ 1
2þ 1
3þ � � � þ 1
n
� �
¼ H2n � Hn: (8)
We now eliminate the H’s from equations (7) and (8) as follows
s3n¼ H4n �1
2H2n �
1
2Hn
¼ H4n� H2nð Þ þ 1
2H2n� Hnð Þ
¼ t4n þ1
2t2n: (9)
But it follows from the hypotheses of the example that tn! loge 2, and so
t2n! loge 2 and t4n! loge 2, as n!1. Hence, letting n!1 in equation (9),
we see that
s3n! loge 2þ 1
2loge 2
¼ 3
2loge 2:
Next
s3n�1¼ s3n þ1
2n
! 3
2loge 2
and
s3n�2¼ s3n þ1
2n� 1
4n� 1
! 3
2loge 2 as n!1:
It follows from the above three results that sn! 32
loge 2 as n!1: &
We insert some brackets in afinite sum, for convenience.
We separate out the positiveand negative terms.
We add some terms to thecontents of the first bracket,then subtract them again; thelast bracket is simply 1
2Hn.
By the definition of t2n.
We insert some positive terms,then remove them again.
By the definition of H2n.
By the definition of Hn.
110 3: Series
Problem 4 Prove that the seriesP
1
n¼1
an ¼ 1� 12� 1
4þ 1
3� 1
6� 1
8þ 1
5�
110� 1
12þ 1
7� 1
14� 1
16þ � � � converges to 1
2loge 2. [You may assume that
the sum of the seriesP
1
n¼1
�1ð Þnþ1
n¼1� 1
2þ 1
3� 1
4þ 1
5� 1
6þ � � � is loge 2.]
In order to better understand the behaviour of series with positive and
negative terms, we introduce some notation. For a conditionally convergent
series
X
1
n¼1
an ¼ a1 þ a2 þ � � �;
we define the quantities aþn and a�n as follows
aþn ¼an; if an � 0,
0; if an < 0,
�
and a�n ¼0; if an � 0,
�an; if an < 0.
�
In other words, the sequence aþn� �
picks out the non-negative terms and the
sequence a�n� �
picks out the non-positive terms (and discards their sign).
In particular, an¼ aþn � a�n , and aþn � 0 and a�n � 0.
Next, since the seriesP
1
n¼1
an is convergent, an! 0 as n!1 . It follows that
aþn ¼1
2anj j þ anð Þ ! 0 as n!1
and
a�n ¼1
2anj j � anð Þ ! 0 as n!1:
Now, sinceP
1
n¼1
an is conditionally convergent, it must contain infinitely
many negative terms, since otherwiseP
1
n¼1
an andP
1
n¼1
anj j would be the same
apart from a finite number of terms. Since altering a finite number of terms
does not affect the convergence of a series, it would then follow that the seriesP
1
n¼1
anj j would be convergent – which we know is not the case.
Similarly, there must be infinitely many positive terms inP
1
n¼1
an.
Finally, we show that, if the seriesP
1
n¼1
an is conditionally convergent, then
the corresponding seriesP
1
n¼1
aþn (the ‘positive part’ ofP
1
n¼1
an) andP
1
n¼1
a�n (the
‘negative part’ ofP
1
n¼1
an) are both divergent.
For, suppose thatP
1
n¼1
aþn were convergent. Then, since an¼ aþn � a�n , we
have a�n ¼ aþn � an; it follows that the seriesP
1
n¼1
a�n must also be convergent.
And then, since anj j ¼ aþn þ a�n , it would also follow thatP
1
n¼1
anj j would be
convergent – which we know is not the case.
A similar argument shows thatP
1
n¼1
a�n cannot be convergent.
In particular, it follows from the divergence of the seriesP
1
n¼1
aþn andP
1
n¼1
a�n ,
whose terms are non-negative, that their partial sums must tend to 1 as
n!1.
For example, for the series
X
1
n¼1
an¼1� 1
2þ 1
3� 1
4
þ 1
5� 1
6þ � � �;
we have
X
1
n¼1
aþn ¼ 1� 0þ 1
3� 0
þ 1
5� 0þ � � �
and
X
1
n¼1
a�n ¼ 0þ 1
2þ 0þ 1
4
þ 0þ 1
6� � �:
For the seriesP
1
n¼1
an is
assumed to be onlyconditionally convergent.
For the seriesP
1
n¼1
an is
assumed to be onlyconditionally convergent.
By the Boundedness Theoremfor series.
3.3 Series with positive and negative terms 111
In general it is not true that any rearrangement of a conditionally convergent
series converges. The situation is much more interesting!
Theorem 3 Riemann’s Rearrangement Theorem
Let the seriesP
1
n¼1
an ¼ a1 þ a2 þ � � � be conditionally convergent, with nth
partial sum sn. Then there are rearrangements of the series that have the
following properties:
1. For any given number s, the rearranged series converges to s;
2. For any given numbers x and y, one subsequence of the sn converges to x
and another subsequence of the sn converges to y;
3. The sequence sn converges to1.
4. One subsequence of the sn converges to1 and another subsequence of
the sn converges to �1.
We will not prove Theorem 3. Instead, we illustrate case (1) using the facts that
we have already discovered about conditionally convergent series. Using
similar ideas, we can prove the various parts of the theorem.
Example 3 Find a rearrangement of the seriesP
1
n¼1
�1ð Þnþ1
n¼1� 1
2þ 1
3� 1
4þ
15� 1
6þ � � � that converges to the sum 3.
Solution Since the ‘positive’ part of the seriesP
1
n¼1
�1ð Þnþ1
nhas partial sums that
tend to1, we start to construct the desired rearranged series as follows. Take
enough of the ‘positive’ terms 1, 13
, 15
, . . ., 12N1�1
so that the sum
1þ 13þ 1
5þ � � � þ 1
2N1�1is greater than 3, choosing N1 so that it is the first
integer such that this sum is greater than 3. Then these N1 terms will form
the first N1 terms in our desired rearranged series.
Next, take enough of the ‘negative’ terms 12
, 14
, 16
, . . ., 12N2
so that the
expression
1þ 1
3þ 1
5þ � � � þ 1
2N1 � 1
� �
� 1
2þ 1
4þ 1
6þ � � � þ 1
2N2
� �
is less than 3, choosing N2 so that it is the first integer such that this sum is less
than 3. These N1þN2 terms form the first N1þN2 terms in our desired rear-
ranged series.
Now add in some more ‘positive’ terms 12N1þ1
, 12N1þ3
, . . ., 12N3�1
so that
the sum
1þ 1
3þ 1
5þ � � � þ 1
2N1 � 1
� �
� 1
2þ 1
4þ 1
6þ � � � þ 1
2N2
� �
þ 1
2N1 þ 1þ 1
2N1 þ 3þ � � � þ 1
2N3 � 1
� �
is greater than 3, choosing N3 so that it is the first available integer such that
this sum is greater than 3. Then these N2þN3 terms will form the first N2þN3
terms in our desired rearranged series.
Some rearrangementsconverge, some diverge.
This is not a complete list ofall the possibilities that canoccur!
Recall that the sum of thisseries is loge 2 orapproximately 0.69.
This is possible since
1þ 1
3þ 1
5þ � � � ! 1;
as n!1.
112 3: Series
We then add in just enough ‘negative’ terms to make the next sum of
N3þN4 terms less than 3, and so on indefinitely.
In each set of two steps in this process we must use at least one of the
‘positive’ terms and one of the ‘negative’ terms, so that eventually all the
‘positive’ terms and all the ‘negative’ terms of the original series will be taken
exactly once in the new series, which we denote byP
1
n¼1
bn. So certainly the
seriesP
1
n¼1
bn is a rearrangement of the original seriesP
1
n¼1
�1ð Þnþ1
n.
But how do we know that the sum of the rearranged series is 3? As we keep
making the partial sums ofP
1
n¼1
bn swing back and forth from one side of 3 to the
other, we need the ‘radius’ of the swings to tend to 0. We achieve this by
always changing the sign of the terms fromP
1
n¼1
�1ð Þnþ1
nthat we choose next, once
the partial sum ofP
1
n¼1
bn has crossed beyond the value 3. For this ensures that
(from 12N2
onwards) the difference between a partial sum and 3 will always be
less than the absolute value of the last term 12Nk�1
or 12Nk
used, and we know that
this last term tends to 0 as we go further out along the original series.
It follows that the rearranged seriesP
1
n¼1
bn converges to 3, as required. &
Problem 5 Find a rearrangement of the seriesP
1
n¼1
�1ð Þnþ1
n¼1� 1
2þ 1
3�
14þ 1
5� 1
6þ � � � whose partial sums tend to1.
The situation is very much simpler, though, in the case of absolutely con-
vergent series.
Theorem 4 Let the seriesP
1
n¼1
an ¼ a1 þ a2 þ � � � be absolutely convergent.
Then any rearrangementP
1
n¼1
bn of the series also converges absolutely, and
P
1
n¼1
bn ¼P
1
n¼1
an.
For example, the series
1� 1
22þ 1
32� 1
42þ 1
52� 1
62þ 1
72� 1
82þ 1
92� � � �
is absolutely convergent. It follows from Theorem 4 that the following series is
also absolutely convergent – and to the same sum
1|{z}
� 1
22� 1
42|fflfflfflfflfflffl{zfflfflfflfflfflffl}
þ 1
32þ 1
52þ 1
72|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}
� 1
62� 1
82� 1
102� 1
122|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
þ � � �
1 term 2 terms 3 terms 4 terms
:
3.3.4 Multiplication of series
We now look at the question of how we might multiply together two infinite
seriesP
1
n¼0
an andP
1
n¼0
bn to obtain a single seriesP
1
n¼0
cn with the property that
At each step there are always‘positive’ terms or ‘negative’terms left to choose, sincethere are infinitely many ofeach.
For clearly 1Nk� 1
k, since we
use at least 1 term in theoriginal series at each step ofthe rearrangement process.
For the seriesP
1
n¼1
�1ð Þnþ1
n2
�
�
�
�
�
�¼P
1
n¼1
1n2 , which is
a basic convergent series.
In this sub-section we sum ourseries from n¼ 0 rather thanfrom n¼ 1, simply because inmany applications ofTheorem 5 (below) this is theversion that we shall require.
3.3 Series with positive and negative terms 113
X
1
n¼0
cn ¼X
1
n¼0
an
!
�X
1
n¼0
bn
!
:
Now, a first thought might be that cn¼ anbn, so thatP
1
n¼0
anbn ¼P
1
n¼0
an
� �
�P
1
n¼0
bn
� �
: However this is not the case!
For example, let an ¼ 12
� �nand bn ¼ 1
3
� �n. Then
X
1
n¼0
an ¼X
1
n¼0
1
2
� �n
¼ 2 andX
1
n¼0
bn ¼X
1
n¼0
1
3
� �n
¼ 3
2;
butX
1
n¼0
anbn ¼X
1
n¼0
1
6
� �n
¼ 6
5:
We can get a clue to what the correct formula for cn might be by, tempora-
rily, abandoning our rigorous approach and simply ‘pushing symbols about’!
If we multiply out both brackets and collect terms in a convenient way, we get
X
1
n¼0
an
!
�X
1
n¼0
bn
!
¼ a0þa1þa2þa3þ�� �ð Þ� b0þb1þb2þb3þ���ð Þ
¼ a0 b0 þ b1 þ b2 þ b3 þ � � �ð Þ þ a1 b0 þ b1 þ b2 þ b3 þ � � �ð Þþ a2 b0 þ b1 þ b2 þ b3 þ � � �ð Þ þ a3 b0 þ b1 þ b2 þ b3 þ � � �ð Þ þ � � �
¼ a0b0 þ a0b1 þ a0b2 þ a0b3 þ � � � þ a1b0 þ a1b1 þ a1b2 þ a1b3 þ � � �þ a2b0 þ a2b1 þ a2b2 þ a2b3 þ � � � þ a3b0 þ a3b1 þ a3b2 þ a3b3 þ � � �
¼ a0b0 þ a0b1 þ a1b0ð Þ þ a0b2 þ a1b1 þ a2b0ð Þþ a0b3 þ a1b2 þ a2b1 þ a3b0ð Þ þ � � �:
Notice that every term that we would expect to be in such a product appears
exactly once in this final expression. We have grouped the terms in the final
expression in a particularly convenient way, so that we can see a pattern
emerging from this non-rigorous argument.
We can formalise this discussion as follows:
Theorem 5 Product Rule
Let the seriesP
1
n¼0
an andP
1
n¼0
bn be absolutely convergent, and let
cn ¼ a0bn þ a1bn�1 þ a2bn�2 þ � � � þ an b0 ¼X
n
k¼0
akbn�k:
Then the seriesP
1
n¼0
cn is absolutely convergent, and
X
1
n¼0
cn ¼X
1
n¼0
an
!
�X
1
n¼0
bn
!
:
Let us now return to the two seriesP
1
n¼0
an ¼P
1
n¼0
12
� �nand
P
1
n¼0
bn ¼P
1
n¼0
13
� �nthat
we considered earlier, whose sums were 2 and 32, respectively. With the for-
mula in Theorem 5 for cn, we have
These are both geometricseries.
Note that 656¼ 2� 3
2!
The pattern here is that in thekth bracket, the sum of thesubscripts of each term isprecisely k.
114 3: Series
cn ¼X
n
k¼0
akbn�k ¼X
n
k¼0
1
2k� 1
3n�k
¼ 1
3n
X
n
k¼0
3
2
� �k
¼ 1
3n�
1� 32
� �nþ1
1� 32
¼ 1
3n� �2ð Þ � 1� 3
2
� �nþ1 !
¼ 3
2n� 2
3n:
It follows that
X
1
n¼0
cn ¼ 3�X
1
n¼0
1
2n� 2�
X
1
n¼0
1
3n
¼ 3� 2� 2� 3
2
� �
¼ 6� 3 ¼ 3:
Since 2� 32¼ 3; we see that the sum of the series
P
1
n¼0
cn is what Theorem 5
predicts it should be.
Remark
Notice that the theorem requires that the seriesP
1
n¼0
an andP
1
n¼0
bn are absolutely
convergent. Without this assumption, the result may be false.
For example, let an ¼ bn ¼ �1ð Þnþ1
ffiffiffiffiffiffi
nþ1p . Then both
P
1
n¼0
an andP
1
n¼0
bn are conver-
gent, by the Alternating Test – but they are not absolutely convergent. Now, by
the formula for cn, we have
cn ¼X
n
k¼0
akbn�k ¼X
n
k¼0
�1ð Þkþ1
ffiffiffiffiffiffiffiffiffiffiffi
k þ 1p � �1ð Þn�kþ1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
n� k þ 1p
¼ �1ð ÞnX
n
k¼0
1ffiffiffiffiffiffiffiffiffiffiffi
k þ 1p
�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
n� k þ 1p ;
so that
cnj j ¼X
n
k¼0
1ffiffiffiffiffiffiffiffiffiffiffi
k þ 1p
�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
n� k þ 1p
�X
n
k¼0
1ffiffiffiffiffiffiffiffiffiffiffi
nþ 1p
�ffiffiffiffiffiffiffiffiffiffiffi
nþ 1p ¼ 1:
Since we therefore do not have cn! 0 as n!1, it follows that the seriesP
1
n¼0
cn
is divergent.
We shall use Theorem 5 in Section 3.4 to give a definition of the exponential
function x 7! ex in terms of series and to examine its properties, and in
Section 8.4 on power series.
Yet again, this demonstratesthat the hypotheses of ourtheorems really do matter!
3.3 Series with positive and negative terms 115
3.3.5 Overall strategy for testing for convergence
We now give an overall strategy for testing a seriesP
an for convergence, as a
flow chart.
We suggest that, given a series which you wish to test for convergence or
absolute convergence, you use the tests in the order indicated in the chart.
Naturally, you may be able to short-circuit this strategy in certain cases.
Is {an} null?
NO
NO
YES
YES
YES Does Alternating Test
apply?
Try usingFirst
Principles
NO or
CAN’T DECIDE
Try using:Basic seriesRatio TestLimit Comparison TestComparison TestCondensation Test
Is Σ⏐an⏐ convergent?
Σ an is (absolutely) convergent
Σ an is convergent
(Alternating Test)
Σan is divergent
(Non-null Test)
Remark
We have labelled the final box ‘First Principles’, to indicate that, if the various
tests do not produce a result, then it may be possible to work directly with the
sequence of partial sums {sn}.
Problem 6 Test the following series for convergence and absolute
convergence:
(a)P
1
n¼1
12
n; (b)P
1
n¼1
5nþ2n
3n ; (c)P
1
n¼1
32n3�1
;
(d)P
1
n¼1
�1ð Þnþ1
n13
; (e)P
1
n¼1
�1ð Þnþ1n2
n2þ1; (f)
P
1
n¼1
�1ð Þnþ1n
n3þ5;
(g)P
1
n¼1
2n
n6; (h)P
1
n¼1
�1ð Þnþ1n
n2þ2; (i)
P
1
n¼2
1
n loge nð Þ34
:
There is a variety of further tests for convergence or divergence which can
be applied to series with non-negative terms. In this book we give only one
further test, called the Integral Test. In particular it will enable us to prove that
X
1
n¼1
1
npis convergent; for all p > 1:
For example, if you wish toexamine a series whose termsalternate in sign, it is best togo straight to the AlternatingTest.
You will meet the IntegralTest in Section 7.4.
116 3: Series
3.3.6 Proofs
We now supply the proofs omitted earlier in this section.
Theorem 1 Absolute Convergence Test
IfP
1
n¼1
an is absolutely convergent, thenP
1
n¼1
an is convergent.
Proof We know thatP
1
n¼1
anj j is convergent, and we want to prove thatP
1
n¼1
an
is convergent.
To do this, we define two new sequences
aþn ¼an; if an � 0,
0; if an < 0,
�
and a�n ¼0; if an � 0,
�an; if an < 0.
�
Both the sequences aþn� �
and a�n� �
are non-negative, and
an ¼ aþn � a�n ; for n ¼ 1; 2; . . .:
Also
aþn � anj j; for n ¼ 1; 2; . . .; (10)
and
a�n � anj j; for n ¼ 1; 2; . . .: (11)
SinceP
1
n¼1
anj j is convergent, we deduce from (10) and (11) thatP
1
n¼1
aþn and
P
1
n¼1
a�n are convergent, by the Comparison Test. Thus
X
1
n¼1
an ¼X
1
n¼1
aþn � a�n� �
¼X
1
n¼1
aþn �X
1
n¼1
a�n (12)
is convergent, by the Combination Rules for series. &
Triangle Inequality (infinite form)
IfP
1
n¼1
an is absolutely convergent, thenP
1
n¼1
an
�
�
�
�
�
�
�
�
�P
1
n¼1
anj j.
Proof Let
sn ¼ a1 þ a2 þ � � � þ an; n ¼ 1; 2; . . .;
and
tn ¼ a1j j þ a2j j þ � � � þ anj j; n ¼ 1; 2; . . .:
Then, by the Absolute Convergence Test
limn!1
sn ¼X
1
n¼1
an exists;
also
limn!1
tn ¼X
1
n¼1
anj j:
You may omit these proofs ona first reading.
For example, ifX
1
n¼1
an ¼ 1� 1
22þ 1
32
� 1
42þ � � �;
then
X
1
n¼1
aþn ¼ 1þ 0þ 1
32þ 0þ �� �
and
X
1
n¼1
a�n ¼ 0þ 1
22þ 0
þ 1
42þ�� �:
Sub-section 3.1.4.
Alternatively, the infiniteform of the TriangleInequality can be deducedfrom statements (10), (11) and(12) in the proof of theAbsolute Convergence Test.
3.3 Series with positive and negative terms 117
Now, by the Triangle Inequality
snj j ¼ a1 þ a2 þ � � � þ anj j� a1j j þ a2j j þ � � � þ anj j ¼ tn;
and so
�tn � sn � tn:
Thus, by the Limit Inequality Rule for sequences
� limn!1
tn � limn!1
sn � limn!1
tn;
that is
�X
1
n¼1
anj j �X
1
n¼1
an �X
1
n¼1
anj j;
and so
X
1
n¼1
an
�
�
�
�
�
�
�
�
�
�
�X
1
n¼1
anj j: &
Theorem 2 Alternating Test
If
an ¼ �1ð Þnþ1bn; n ¼ 1; 2; . . .;
where {bn} is a decreasing null sequence with positive terms, then
X
1
n¼1
an ¼ b1 � b2 þ b3 � b4 þ � � � is convergent:
Proof We can write the even partial sums s2k ofP
1
n¼1
an as follows
s2k ¼ b1 � b2ð Þ þ b3 � b4ð Þ þ � � � þ b2k�1 � b2kð Þ:
Since {bn} is decreasing, all the brackets are non-negative, and so the even
subsequence of partial sums, {s2k}, is increasing.
We can also write the even partial sums s2k as
s2k ¼ b1 � b2 � b3ð Þ � b4 � b5ð Þ � � � � � b2k�2 � b2k�1ð Þ � b2k:
Again, all the brackets are non-negative, and so s2k is bounded above, by b1.
Hence {s2k} is convergent, by the Monotone Convergence Theorem.
Now let
limk!1
s2k ¼ s:
Since s2k ¼ s2k�1 � b2k so that s2k�1 ¼ s2k þ b2k, and {bn} is null, we have
limk!1
s2k�1 ¼ limk!1
s2k þ b2kð Þ
¼ limk!1
s2k þ limk!1
b2k ¼ s;
In Remark 3 following theproof of the TriangleInequality in Sub-section 1.3.1,we commented that theTriangle Inequality for twonumbers, a and b, can beextended in the obvious way toany finite sum of numbers.
You met this in Sub-section 2.3.3, Theorem 3.
118 3: Series
by the Sum Rule for sequences. Thus the odd and even subsequences of {sn}
both tend to the same limit s, and so {sn} tends to s. Hence
X
1
n¼1
an ¼ b1� b2þ b3� b4þ � � � is convergent; with sum s: &
Theorem 4 Let the seriesP
1
n¼1
an ¼ a1 þ a2 þ � � � be absolutely conver-
gent. Then any rearrangementP
1
n¼1
bn of the series also converges absolutely,
andP
1
n¼1
bn ¼P
1
n¼1
an:
Proof First of all, we prove the result in the special case that ak� 0 for
all k.
Choose any positive integer n. Then choose a positive integer N such that
b1, b2, . . ., bn all occur among the terms a1, a2, . . ., aN of the original series.
It follows that
X
n
k¼1
bk �X
N
k¼1
ak
�X
1
k¼1
ak: (13)
Hence the partial sumsP
n
k¼1
bk of the rearranged seriesP
1
k¼1
bk are bounded
above, so that the rearranged series must be convergent. It also follows from
(13) thatP
1
k¼1
bk �P
1
k¼1
ak.
By reversing the roles of the terms a1, a2, . . . and b1, b2, . . . in the above
argument, the same argument shows thatP
1
k¼1
ak �P
1
k¼1
bk. It thus follows that
the two seriesP
1
k¼1
ak andP
1
k¼1
bk must in fact have the same sum.
To complete our proof, we now drop the condition that ak� 0 for all k.
We then define the quantities aþk , a�k , bþk and b�k as follows
aþk ¼ak; if ak � 0,
0; if ak < 0,
�
a�k ¼0; if ak � 0,
�ak; if ak < 0,
�
bþk ¼bk; if bk � 0,
0; if bk < 0,
�
b�k ¼0; if bk � 0,
�ak; if bk < 0.
�
Now,P
1
k¼1
aþk is convergent, since aþk ¼ 12
akj j þ akð Þ and bothP
1
k¼1
ak
�
�
�
� andP
1
k¼1
ak
converge.P
1
k¼1
bþk is a rearrangement ofP
1
k¼1
aþk , and both series have non-
negative terms. It follows from the first part of the proof that both series
converge, and have the same sum.
Similarly,P
1
k¼1
a�k andP
1
k¼1
b�k both converge, and have the same sum.
By Theorem 6 , Sub-section 2.4.4.
In particular, N� n.
For the right-hand side issimply the left-hand side withpossibly some more non-negative terms inserted.
For the partial sums of a seriesof non-negative terms arealways less than or equal thesum of the whole series.
By the Combination Rules forseries.
3.3 Series with positive and negative terms 119
But bk
�
�
�
� ¼ bþk þ b�k , and so the seriesP
1
k¼1
bk
�
�
�
� converges, by the Sum Rule
for series. In other words, the rearranged seriesP
1
k¼1
bk is absolutely convergent.
Finally
X
1
k¼1
bk ¼X
1
k¼1
bþk �X
1
k¼1
b�k
¼X
1
k¼1
aþk �X
1
k¼1
a�k
¼X
1
k¼1
ak: &
Theorem 5 Product Rule
Let the seriesP
1
n¼0
an andP
1
n¼0
bn be absolutely convergent, and let
cn ¼ a0bn þ a1bn�1 þ a2bn�2 þ � � � þ anb0 ¼X
n
k¼0
akbn�k:
Then the seriesP
1
n¼0
cn is absolutely convergent, and
X
1
n¼0
cn ¼X
1
n¼0
an
!
�X
1
n¼0
bn
!
:
Proof First, we introduce some notation, for n� 0
sn ¼P
n
k¼0
ak; tn ¼P
n
k¼0
bk; un ¼P
n
k¼0
ck;
s0n ¼P
n
k¼0
akj j; t0n ¼P
n
k¼0
bkj j; u0n ¼P
n
k¼0
ckj j;
s ¼P
1
k¼0
ak; t ¼P
1
k¼0
bk;
s0 ¼P
1
k¼0
akj j; t0 ¼P
1
k¼0
bkj j:
In particular, we know that sn! s, tn! t, sn0 ! s0 and tn
0 ! t0 as n!1.
Also, by the Product Rule for sequences, sn tn! st and sn0 tn0 ! s0t0, as n!1.
Hence, from the definition of limit, it follows that, for any positive number ",there is some integer N for which
sntn � stj j < 1
3" and s0nt0n � s0t0
�
�
�
� <1
3"; for all n � N:
It follows that, for n�N
s0nt0n � s0Nt0N�
�
�
� ¼ s0nt0n � s0t0� �
� s0Nt0N � s0t0� �
�
�
�
�
� s0nt0n � s0t0�
�
�
�þ s0Nt0N � s0t0�
�
�
�
<1
3"þ 1
3"
¼ 2
3":
And, in factP
1
k¼1
bk
�
�
�
� ¼P
1
k¼1
bþk þP
1
k¼1
b�k .
Here we use an integer Nrather than a general numberX and a weak inequality n�Nrather than a strict inequality;this simplifies the notation inthe rest of the argument alittle.
120 3: Series
Now let n be such that n� 2N, and consider the terms ai bj that occur in the
expression un� sNtN. Every such term has iþ j� n, since un consists of all such
terms; but none of these terms has both i�N and j�N.
Hence, for every term ai bj in un� sNtN, a corresponding term jaibjjwill occur in the expression sn
0 tn0 � sN
0 tN0 – for this last expression consists
of all terms jaibjj with i� n and j� n, but not with both i�N and j�N.
(Note that we made the requirement that n� 2N in order that every term in
sNtN appears in un.)
a0b0 a0b1 a0b2 ...
a1b0 a1b1 a1b2 ...
a2b0 a2b1 a2b2 ...
...
0
N
N 2N n
2N
n
sNtN
un – sNtN
In the above diagram, we set out the terms ai bj in rows and columns, where
the term ai bj occurs in the (iþ 1)th row and the ( jþ 1)th column. Then the
small square contains all the terms ai bj with 0� i�N and 0� j�N; these add
up to sNtN.
Hence, for all n� 2N, we have
un � stj j ¼ un � sNtNð Þ þ sNtN � stð Þj j� un � sNtNj j þ sNtN � stj j� s0nt0n � s0Nt0N�
�
�
�þ sNtN � stj j
<2
3"þ 1
3"
¼ ";it follows, from this inequality, that un! st as n!1.
Finally, we have
u0n � s0nt0n� s0t0:
For terms with both i�N andj�N are all in sNtN.
By the Triangle Inequality.
By the discussions aboveconcerning un� sNtN and
sn0 tn0 � sN
0 tN0.
For the product sn0 tn0 contains
more non-negative terms thandoes un
0 ; and the sequences{sn0} and {tn
0} are increasing.
3.3 Series with positive and negative terms 121
Hence, the increasing sequence {un0} is bounded above, and so tends to a
limit as n!1. Thus the seriesP
1
n¼0
cn is absolutely convergent. &
3.4 The exponential function x j!ex
Earlier, we defined e¼ 2.71828 . . . to be the limit
e ¼ limn!1
1þ 1
n
� �n
;
and we defined ex to be the limit
ex ¼ limn!1
1þ x
n
� �n
; for any real x:
Here we show that the formula
ex ¼X
1
n¼0
xn
n!; for any real x;
is an equivalent definition of the quantity ex.
Remark
The seriesP
1
n¼0
xn
n! converges for all values of x. For
xnþ1
nþ 1ð Þ!
�
�
�
�
�
�
�
�
xn
n!
�
�
�
�
�
�
�
�
¼ xj jnþ 1
! 0 as n!1;
so that, by the Ratio Test, the series is absolutely convergent for all x, and so
is convergent for all x.
3.4.1 The definition of ex as a power series, for x> 0
If we plot the partial sum functions of the infinite series of powers of x
X
1
n¼0
xn
n!¼ 1þ xþ x2
2!þ x3
3!þ � � �; (1)
the resulting graph appears to be that of ex. (A series of multiples of increasing
powers of x is called a power series.) In particular, when x ¼ 1, the sum of
the series
X
1
n¼0
1
n!¼ 1þ 1þ 1
2!þ 1
3!þ � � �
is approximately 2.71828 . . ..
In Sub-section 2.5.3.
This series is a basic series oftype (d), in the case that x� 0.
122 3: Series
y
e
y = ex
y = 1 + x
y = 1
x
y = 1 + x + x2
2!
y = 1 + x + x2
2!x3
3!+
10–1
However the fact that the sequence of partial sums of the series (1) appears
to converge to ex does not constitute a proof of this fact. Our first aim is to
supply this proof in the particular case that x> 0.
Theorem 1 If x> 0, thenP
1
n¼0
xn
n! ¼ ex.
Proof We defined ex to be ex ¼ limn!1
1þ xn
� �n, and so we have to show that
X
1
n¼0
xn
n!¼ lim
n!11þ x
n
� �n
; for x > 0:
The (n + 1)th partial sum of the above series is
snþ1 ¼ 1þ xþ x2
2!þ x3
3!þ � � � þ xn
n!:
By the Binomial Theorem
1þ x
n
� �n
¼ 1þ nx
n
� �
þ n n� 1ð Þ2!
x
n
� �2
þ � � � þ x
n
� �n
:
A typical term in this expansion is
n n�1ð Þ . . . n� kþ1ð Þk!
x
n
� �k
¼ xk
k!1�1
n
� �
1�2
n
� �
� � � 1� k�1
n
� �
� xk
k!;
since each bracket is less than 1.
Thus
1þ x
n
� �n
� 1þ xþ x2
2!þ x3
3!þ � � � þ xn
n!
¼ snþ1;
and so, by the Limit Inequality Rule for sequences
limn!1
1þ x
n
� �n
� limn!1
snþ1;
We shall deal with the casex< 0 in Sub-section 3.4.3.
Hence, for x> 0, the seriesP
1
n¼0
xn
n! gives an equivalent
definition of ex.
You may omit this proof at afirst reading.
3.4 The exponential function x j! ex 123
so that
ex �X
1
n¼0
xn
n!: (2)
On the other hand, for any integers m and n with m� n, we have
1þ x
n
� �n
� 1þ nx
n
� �
þ n n� 1ð Þ2!
x
n
� �2
þ� � � þ n n� 1ð Þ . . . n�mþ 1ð Þm!
x
n
� �m
¼ 1þ xþ x2
2!1� 1
n
� �
þ � � � þ xm
m!1� 1
n
� �
1� 2
n
� �
. . . 1�m� 1
n
� �
:
Now keep m fixed and let n!1. By the Limit Inequality Rule for
sequences, we obtain
limn!1
1þ x
n
� �n
� 1þ xþ x2
2!þ x3
3!þ � � � þ xm
m!;
and so
ex � smþ1:
Now let m!1; by the Limit Inequality Rule for sequences, we obtain
ex � limm!1
smþ1;
so that
ex �X
1
n¼0
xn
n!(3)
Combining inequalities (2) and (3), we obtain
ex ¼X
1
n¼0
xn
n!; for x > 0: &
Problem 1 Estimate e2 (to 3 decimal places) by calculating the seventh
partial sum of the series (1) when x = 2.
3.4.2 Calculating e
The representation of e by the infinite series
e ¼X
1
n¼0
1
n!¼ 1þ 1þ 1
2!þ 1
3!þ � � � (4)
provides a much more efficient way of calculating approximate values for e
than the equation e ¼ limn!1
1þ 1n
� �n. This is illustrated by the following table of
approximate values:
n 1 2 3 4 5
1þ 1n
� �n2 2.25 2.37 2.44 2.49
P
n
k¼0
1k!
2 2.50 2.67 2.71 2.717
This follows since {sn} and{snþ1} both tend to the samelimit, the sum of the infiniteseries.
The calculation of e via thelimit is a new calculation eachtime, whereas via the seriesinvolves only adding oneextra term to the previousapproximation.
124 3: Series
We can estimate how quickly the sequence of partial sums
sn ¼X
n�1
k¼0
1
k!¼ 1þ 1þ 1
2!þ 1
3!þ � � � þ 1
n� 1ð Þ! ; n ¼ 1; 2; . . .;
converges to e as follows. The difference between e and sn is given by
e� sn ¼X
1
k¼n
1
k!¼ 1
n!þ 1
nþ 1ð Þ!þ1
nþ 2ð Þ!þ � � �
¼ 1
n!1þ 1
nþ 1ð Þ þ1
nþ 1ð Þ nþ 2ð Þ þ � � ��
<1
n!1þ 1
nþ 1
n2þ � � �
�
:
The last expression in square brackets is a geometric series with first term
1 and common ratio 1n, and so its sum is 1
1�1n
¼ nn�1
. Hence
0 < e� sn <1
n!� n
n� 1
¼ 1
n� 1ð Þ!�1
n� 1; for n ¼ 1; 2; . . .: (5)
For example, this estimate shows that
0 < e� s6 <1
5!� 1
5¼ 1
600¼ 0:0016:
Since the partial sum
s6 ¼ 1þ 1þ 1
2!þ 1
3!þ 1
4!þ 1
5!
¼ 1þ 1þ 1
2þ 1
6þ 1
24þ 1
120¼ 163
60¼ 2:716;
we deduce very easily that
2:716 < e < 2:716þ 0:0016 ¼ 2:7183 :
Problem 2 Estimate how many terms in the series (4) are needed to
determine e to ten decimal places.
The inequality (5) can also be used to show that e is irrational.
Theorem 2 The number e is irrational.
Proof Suppose that e ¼ mn, where m and n are positive integers.
Now, by the above estimate (5)
0 < e� snþ1<1
n!� 1
n;
and so
0 < n! e� snþ1ð Þ< 1
n:
Here we replace various termsby larger terms (because theirdenominators are smaller); sothe new sum is greater.
In fact,
e ¼ 2:71828182845 . . .:
This is a proof bycontradiction.
3.4 The exponential function x j! ex 125
Since e ¼ mn, we have
0 < n!m
n� 1þ 1þ 1
2!þ 1
3!þ � � � þ 1
n!
� ��
<1
n:
But n! mn� 1þ 1þ 1
2!þ 13!þ � � � þ 1
n!
� �� �
is an integer, since n! times each
expression in the square bracket is itself an integer. So we have found an
integer which lies strictly between 0 and 1. This is clearly impossible, and so
e cannot be rational after all. &
3.4.3 The definition of ex as a power series, for all real x
Earlier we saw thatP
1
n¼0
xn
n!¼ ex, for all x> 0. We now show that this formula is
valid for all real values of x, and we will also verify the fundamental property
of the exponential function, namely that exey¼ exþy for all real values of x
and y. The Product Rule for series will be our crucial tool in this work.
First we check the following.
Lemma 1 For any x 2 R ,P
1
n¼0
xn
n! �P
1
n¼0
�xð Þnn! ¼ 1:
Proof By the Product Rule for series,P
1
n¼0
xn
n! �P
1
n¼0
�xð Þnn! ¼
P
1
n¼0
cn, where, for
n� 1, we have
cn ¼X
n
k¼0
xk
k!� �xð Þn�k
n� kð Þ!
¼X
n
k¼0
xk �xð Þn�k
k! n� kð Þ!
¼ 1
n!
X
n
k¼0
n!
k! n� kð Þ!xk �xð Þn�k
¼ 1
n!xþ �xð Þð Þn¼ 0:
Since c0¼ 1� 1¼ 1, the result then follows. &
We can then use the result in Lemma 1 to prove thatP
1
n¼0
xn
n!¼ ex, for x< 0 as
well as x> 0. Since, obviously,P
1
n¼0
xn
n! ¼ ex when x¼ 0, this completes the
proof thatP
1
n¼0
xn
n!¼ ex, for all x 2 R .
Theorem 3 If x< 0, thenP
1
n¼0
xn
n! ¼ ex.
Proof For x< 0, the following chain of equalities holds
X
1
n¼0
xn
n!¼ 1
,
X
1
n¼0
�xð Þn
n!ðby Lemma 1Þ
Here we have simplysubstituted m
nfor e in the
previous expression.
This was Theorem 1 inSub-section 3.4.1.
This was Theorem 5 inSub-section 3.3.4.
We may apply the Product
Rule since the seriesP
1
n¼0
xn
n! and
P
1
n¼0
�xð Þnn! are both absolutely
convergent.
Here we apply the BinomialTheorem to xþ �xð Þð Þn.
In other words, the definitionof ex as a power series isequivalent to its definitionas a limit.
Here the definition of ex onthe right is that
ex ¼ limn!1
1þ x
n
� �n
.
126 3: Series
¼ 1.
limn!1
1þ�x
n
� �n
ðby Theorem 1; since �x > 0Þ
¼ 1=e�x ðthis is the definition of e�xÞ¼ ex ðby the Inverse Property of exÞ:
This completes the proof. &
In order to be crystal clear where we have reached, we now combine the results
of Theorem 1 and Theorem 3 into the following:
Theorem 4 For all x 2 R ,P
1
n¼0
xn
n!¼ lim
n!11þ x
n
� �n¼ ex.
Finally, we verify the Fundamental Property of the exponential function,
which was left as unfinished business at the end of Sub-section 2.5.3.
Theorem 5 Fundamental Property of the exponential function
For all x, y2R , exey ¼ exþy.
Proof By the Product Rule for series, ex � ey ¼P
1
n¼0
xn
n! �P
1
n¼0
yn
n! ¼P
1
n¼0
cn,
where, for n � 1, we have
cn ¼X
n
k¼0
xk
k!� yn�k
n� kð Þ!
¼X
n
k¼0
xkyn�k
k! n� kð Þ!
¼ 1
n!
X
n
k¼0
n!
k! n� kð Þ! xkyn�k
¼ xþ yð Þn
n!:
Since c0¼ 1� 1¼ 1, the result then follows. &
3.5 Exercises
Section 3.1
1. Prove thatP
1
n¼1
�34
� �nis convergent, and find its sum.
2. Interpret 0:12 as an infinite series, and hence find the value of 0:12 as a
fraction.
The Inverse Property of ex
was Theorem 4 of Sub-section 2.5.3.
3.5 Exercises 127
3. Prove that
1
4n2 � 1¼ 1
2
1
2n� 1� 1
2nþ 1
� �
; for n ¼ 1; 2; . . .;
and deduce that
X
1
n¼1
1
4n2 � 1¼ 1
2:
4. Determine whether the following series converge:
(a)P
1
n¼1
45
� �nþ 4n nþ2ð Þ
� �
; (b)P
1
n¼1
1þ 12
� �n� �
; (c)P
1
n¼1
1ffiffi
np � 1
ffiffiffiffiffiffi
nþ1p
� �
.
Section 3.2
1. Determine whether the following series converge:
(a)P
1
n¼1
cos 1=nð Þ2n2þ3
; (b)P
1
n¼1
n2
2n3�n; (c)
P
1
n¼1
ffiffiffiffi
2np
4n3þnþ2;
(d)P
1
n¼1
nþ1ð Þ52n ; (e)
P
1
n¼1
n23n
n! ; (f)P
1
n¼1
n!ð Þ22nð Þ!.
2. (a) Use the Ratio Test to prove thatP
1
n¼1
2nn!nn converges, but that
P
1
n¼1
3nn!nn
diverges.
(b) For which positive values of c can you use the Ratio Test to prove
thatP
1
n¼1
cnn!nn is convergent?
3. Prove that
1ffiffiffi
np � 1
ffiffiffiffiffiffiffiffiffiffiffi
nþ 1p ¼ 1
ffiffiffi
np ffiffiffiffiffiffiffiffiffiffiffi
nþ 1p ffiffiffiffiffiffiffiffiffiffiffi
nþ 1p
þffiffiffi
np� � ; for n ¼ 1;2; . . .;
and use Exercise 4(c) on Section 3.1 to deduce thatP
1
n¼1
1
n32
is convergent.
4. Determine whether the following series converge:
(a)P
1
n¼3
1n logenð Þ loge logenð Þð Þ; (b)
P
1
n¼2
ffiffiffiffiffiffi
nþ1p
2n2�3nþ1ð Þ logenþ logenð Þ2ð Þ.
Section 3.3
1. Test the following series for convergence and absolute convergence:
(a)P
1
n¼1
�1ð Þnþ1
1þffiffi
np ; (b)
P
1
n¼1
sin nn2 ; (c)
P
1
n¼1
�1ð Þnþ1n!
n4þ3; (d)
P
1
n¼1
nþ2n
3nþ5.
2. (a) Prove that
1
3n� 2þ 1
3n� 1� 1
3n>
1
3n; for n ¼ 1; 2; . . .;
and deduce that
1þ 1
2� 1
3þ 1
4þ 1
5� 1
6þ � � � is divergent:
128 3: Series
(b) Prove that
1
n� 1
2n� 1
2nþ 2
�
�
�
�
�
�
�
�
<1
n2; for n ¼ 1; 2; . . .;
and deduce that
1� 1
2� 1
4þ 1
3� 1
6� 1
8þ 1
5� 1
10� 1
12þ 1
7� 1
14� 1
16þ � � �
is convergent:
(c) Prove that
1ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2n� 1p � 1
2n� 1
2n; for n ¼ 1; 2; . . .;
and deduce that
1� 1
2þ 1
ffiffiffi
3p � 1
4þ 1
ffiffiffi
5p � 1
6þ � � �
is divergent:
3. Prove that there exists a rearrangement of the seriesP
1
n¼1
�1ð Þnþ1
n¼ 1� 1
2þ 1
3�
14þ 1
5� 1
6þ � � � for which one subsequence of its partial sums tends to 1 and
another subsequence of its partial sums tends to�1. Alternatively, prove that
no such rearrangement exists.
3.5 Exercises 129
4 Continuity
The graphs of many functions that we come across look to be smooth curves,
without any jumps.
1y = ex
– 12 π
–1
1 y = sin x
12 π
On the other hand, the graphs of some functions that arise naturally, or that
we construct for various purposes, do not appear to be smooth, but to contain
jumps.
y = [x]
1
–2 –1 1
–1
221– π
21 π
y = tan x
However there are some functions where the graphs appear to have an
unreasonable number of jumps!
How can we describe this complicated situation? The key underlying idea is
that of continuity. Loosely speaking, a function is said to be continuous if we
can draw its graph without taking our pencil off the page. This is the same as
the graph ‘having no jumps’. The concept is an important one in Analysis
since, in many situations, the crucial step in proving that a function has some
property is to prove that the function is continuous. This is the first of two
chapters that study continuous functions.
In Section 4.1, we define the phrase:
the function f is continuous at the point c,
and we give a number of rules which state, for example, that various combina-
tions and compositions of continuous functions are themselves continuous.
130
Using these rules, together with a list of basic continuous functions, we can
deduce that many functions are continuous at each point of their domains. For
example, the functions x 7! xþ 1x
and x 7! x sin 1x
are continuous at each point
of R � {0}, and the trigonometric and exponential functions are continuous
at each point of their domains.
Section 4.2 is devoted to the properties of continuous functions. The two
fundamental properties of continuous functions are the Intermediate Value
Theorem and the Extreme Value Theorem. We shall see a useful applica-
tion of the Intermediate Value Theorem to finding the zeros of various
functions.
In Section 4.3, we discuss the Inverse Function Rule; this rule gives condi-
tions under which a continuous function f has a continuous inverse function
f�1. We then use the Inverse Function Rule to obtain the inverses of standard
functions. Some of these inverse functions will be familiar to you already, but
the Inverse Function Rule enables us to establish their properties by a rigorous
argument.
Finally, in Section 4.4, we shall provide a rigorous definition of the expo-
nential function x 7! ax, where x2R and a> 0.
4.1 Continuous functions
4.1.1 What is continuity?
To accord with our intuitive idea of what we should mean by ‘the function f is
continuous at the point c’, we wish to define this concept in such a way that the
following two functions are continuous at the point c:
y1.
c x
f (c)
y2.
c x
f (c)
On the other hand, we wish to formulate our definition so that the following
two functions are not continuous at the point c:
y3.
c x
f (c)
y4.
f (c)
c x
In particular, to findingsolutions of polynomialequations.
4.1 Continuous functions 131
So, our definition must say, in a precise way, that:
if x tends to c, then f (x) tends to f (c).
There are several ways of making this idea precise. In this chapter we adopt a
definition which involves the convergence of sequences, as this will enable us
to use the results about sequences that we met in Chapter 2.
Definitions A function f defined on an interval I that contains c as an
interior point is continuous at c if:
for each sequence {xn} in I such that xn! c, then f (xn)! f (c).
If f is not continuous at the point c in I, then it is discontinuous at c.
Thus, for example, in graph 3 above, if {xn} is a strictly decreasing
sequence that tends to c as n!1, then {f (xn)} is a strictly decreasing
sequence that tends to {f (c)}. On the other hand, if {xn} is a strictly
increasing sequence that tends to c as n!1, then {f (xn)} is a strictly
increasing sequence that tends to a limit as n!1 – but that limit is some
number less than f (c).
However, in graph 4, if {xn} is a strictly increasing sequence that tends to c
as n!1, then {f (xn)} is a strictly increasing sequence that tends to f (c); but, if
{xn} is a strictly decreasing sequence that tends to c as n!1, then the
sequence {f (xn)} may not tend to any limit as n!1.
On the other hand, in graphs 1 and 2, no matter how the sequence {xn} tends
to c, then for sure we do have that {f (xn)} tends to f (c).
For these examples, then, the definition agrees with what we believe the
essence of continuity should be.
Example 1 Prove that the function f (x)¼ x3, x2R , is continuous at the point 12.
Solution Let {xn} be any sequence in R that converges to 12; that is, xn ! 1
2.
Then, by the Combination Rules for sequences, it follows that
f xnð Þ ¼ x3n !
1
2
� �3
¼ 1
8as n!1;
while f 12
� �
¼ 18. In other words, {f(xn)} converges to f 1
2
� �
as n!1.
It follows that f is continuous at 12, as required. &
Example 2 Prove that the function f xð Þ ¼ 1; x < 0;2; x � 0;
�
is discontinuous at 0.
Solution Let {xn} be any sequence in R that converges to 0 from the right. By
looking at the graph y¼ f(x), it is clear that for such a sequence f(xn)! 2¼ f(0).
This will not help us to show that f is discontinuous at 0!
However, if we let {xn} be any non-constant sequence in R that converges
to 0 from the left, the situation will be very different. By looking at the
graph y¼ f(x), it is clear that for such a sequence f(xn)! 1 6¼ f(0). We now
make this precise.
In Chapter 5 we shall meet adifferent definition, and seethat the two definitions are infact equivalent.
18
12
y
y = x3
x
y
2
y = {1, x < 0,2, x ≥ 0.
x
1
To prove that a function isdiscontinuous, all we need todo is to find just one sequencefor which our definition ofcontinuity at the relevantpoint does not hold.
132 4: Continuity
Choose the sequence xnf g ¼ � 1n
� �
; n ¼ 1; 2; . . .. For this sequence
f xnð Þ ¼ 1! 1 6¼ f 0ð Þ as n!1:It follows that the function f cannot be continuous at 0. &
Problem 1
(a) Determine whether the function f xð Þ ¼ x3 � 2x2; x 2 R , is contin-
uous at 2.
(b) Determine whether the function f xð Þ ¼ x½ �; x 2 R , is continuous at 1.
Problem 2
(a) Prove that the function f xð Þ ¼ 1; x 2 R , is continuous on R .
(b) Prove that the function f xð Þ ¼ x; x 2 R , is continuous on R .
However, as we know, not every function is defined on the whole of R – but we
still want to be able to discuss the continuity of such functions. For example:
� f xð Þ ¼ffiffiffi
xp
, where the domain of f is the interval [0,1);
� f xð Þ ¼ x�12 þ ð1� xÞ
12, where the domain of f is the half-closed interval (0, 1];
� f xð Þ ¼ 1x, where the domain of f is R � {0}.
How can we deal with these various different situations?
The first two situations are dealt with by introducing the notion of one-sided
continuity.
Suppose that f is defined on some set S that contains an interval [c, cþ r) for
some r> 0, but where f is not necessarily defined on any interval (c� r, cþ r)
that contains c as an interior point. Then it seems reasonable to say that f is
continuous on one side of c. In particular, that f is continuous on the right at the
point c of S.
Similarly, if f is defined on some set S that contains an interval (c� r, c] for
some r> 0, but f is not necessarily defined on any interval (c� r, cþ r) that
contains c as an interior point, then it seems reasonable to say that f is
continuous on one side of c. In particular, that f is continuous on the left at
the point c of S.
The case of the function f xð Þ ¼ 1x, where the domain of f is R – {0}, is
different. The domain of f is the whole of R , with just one point omitted. That
is, it is the union of two open intervals (�1, 0) and (0,1) – to both of which
the definition of continuity applies in its originally stated form. So it makes
sense to simply drop the original restriction that the domain of f must be an
open interval in R .
These considerations lead us to making the following less restrictive defini-
tion of continuity than our original definition:
Definitions A function f defined on a set S in R that contains a point c is
continuous at c if:
for each sequence {xn} in S such that xn! c, then f(xn)! f(c).
If f is continuous on the whole of its domain, we often simply say that f is
continuous (without explicit mention of the points of continuity).
If f is not continuous at the point c in S, then it is discontinuous at c.
Here we make a specificchoice for the sequence {xn}to prove that f cannot becontinuous at 0.
Here [�] is the integer partfunction.
By ‘continuous on R’, wemean ‘continuous at eachpoint of R’.
y = x1
y
x
Here f is continuous on S.
4.1 Continuous functions 133
In addition, we have the following related definitions:
Definitions A function f whose domain contains an interval [c, cþ r) for
some r> 0 is continuous on the right at c if:
for each sequence {xn} in [c, cþ r) such that xn! c, then f(xn)! f(c).
A function f whose domain contains an interval (c� r, c] for some r> 0 is
continuous on the left at c if:
for each sequence {xn} in (c� r, c] such that xn! c, then f(xn)! f(c).
A function f whose domain contains an interval I is continuous on I if it is
continuous at each interior point of I, continuous on the right at the left end-
point of I (if this belongs to I), and continuous on the left at the right end-
point of I (if this belongs to I).
The connection between the definitions of continuity and of one-sided
continuity is rather obvious.
Theorem 1 A function f whose domain contains an interval I that contains
c as an interior point is continuous at c if and only if f is both continuous on
the left at c and continuous on the right at c.
Example 3 Determine whether the function f given by f xð Þ ¼ffiffiffi
xp; x � 0, is
continuous on x : x 2 R ; x � 0f g.
Solution The domain of f is the interval I ¼ x : x � 0f g.Thus, we have to show that for each c in I:
for each sequence {xn} in I such that xn! c, thenffiffiffiffiffi
xnp !
ffiffiffi
cp
.
First, if c¼ 0, then we know already that, for any null sequence {xn} in I,ffiffiffiffiffi
xnp� �
is also a null sequence.
Next, let c> 0. We have to prove that if xn � cf g is a null sequence, then so
isffiffiffiffiffi
xnp �
ffiffiffi
cp� �
. Now, from the identity
ffiffiffiffiffi
xn
p �ffiffiffi
cp¼ xn � c
ffiffiffiffiffi
xnp þ
ffiffiffi
cp ;
we see that, since c 6¼ 0
ffiffiffiffiffi
xn
p �ffiffiffi
cp! 0
2ffiffiffi
cp ¼ 0 as n!1:
In other words,ffiffiffiffiffi
xnp �
ffiffiffi
cp� �
is indeed a null sequence. This completes the
proof. &
Problem 3 Prove that the nth root function
f xð Þ ¼ x1n; where n 2 N and
x � 0; if n is even;x 2 R ; if n is odd;
�
is continuous.
We omit a proof of thisstraight-forward result.
In fact, f is continuous on theright at 0, and continuous atall other points of its domain.
y
x0
y = √x
Recall that c 6¼ 0, so that 1ffiffi
cp is
indeed defined.
You may omit this Problem ifyou are short of time.
We defined a1n in
Sub-section 1.5.3.
134 4: Continuity
Hints: Use the result of Exercise 3(b) on Section 2.3, in Section 2.6, and
the identity aq � bq ¼ a� bð Þ aq�1 þ aq�2bþ aq�3b2 þ � � � þ bq�1ð Þwith suitable choices for a, b and q. In your solution, be careful not to
use the letter n for two different purposes.
Example 4 Determine whether the function f xð Þ ¼ 1x; x 2 R � 0f g, is
continuous.
Solution The domain of f is the set R � {0}, the union of the two open
intervals (�1, 0) and (0,1).
Let c be any point of R � {0}, and let {xn} be any sequence in R � {0} that
converges to c. Then, by the Quotient Rule for sequences, it follows that
f xnð Þf g ¼�
1xn
�
! 1c
as n!1; in other words, that f xnð Þ ! f cð Þ ¼ 1c
as
n!1. So f is continuous at c.
Since c is an arbitrary point of R � {0}, it follows that f is continuous
on R � {0}. &
Our work so far on continuity illustrates the following general strategy:
Strategy for continuity
� To prove that a function f : S!R is continuous at a point c of S, prove
that:
for each sequence {xn} in S such that xn! c, then f(xn)! f(c).
� To prove that a function f : S!R is discontinuous at a point c of S:
find one sequence {xn} in S such that xn! c but f(xn) 6! f(c).
Problem 4 Prove that the following functions are continuous on R :
(a) f(x)¼ c, x2R ;
(b) f(x)¼ xn, x2R , n2N;
(c) f(x)¼ jxj, x2R .
Problem 5 Determine the points of continuity and discontinuity of the
signum function
f xð Þ ¼�1; x < 0,
0; x ¼ 0,
1; x > 0.
8
<
:
Remarks
The definition of function that we are using involves a mapping that we call
f (say) from a set in R , the domain A (say), to another set in R , the codomain
B (say).
1. Let f be continuous at a point c in A, and assume that, for some set A0 � A,
c2A0. Suppose that another function g has domain A0 on which g(x)¼ f(x).
Technically g is a different function from f, for sure. However if f is
continuous at c, then it is a simple matter of some definition checking to
Example 3 above was thespecial case of Problem 3,where n¼ 2.
Sub-section 2.3.2.
Recall that just one suchsequence suffices.
y
x
y = f(x)
–1
1
You can think of the mappingf as being a formula x 7! y thatmaps a point x of A onto to apoint y of B, and we writey¼ f (x) to indicate this.
4.1 Continuous functions 135
verify that g too is continuous at c. So, restriction of a function to a smaller
domain does not affect its continuity at a point.
2. Let f and g be functions defined on sets in R that contain an open interval I,
and c2 I. Then, if f(x)¼ g(x) for all x2 I, f is continuous at c if g is
continuous at c, and f is discontinuous at c if g is discontinuous at c.
Again, we may simply ignore any difference between the domains of the
functions when we are studying continuity at a point.
3. The underlying point here is that continuity at a point is a local property. It
is only the behaviour of the function near that point that determines whether
it is continuous at the point.
4.1.2 Rules for continuous functions
We have seen how to recognise whether a given function is continuous at a
point. However, it would be tedious to have to go back to first principles to
determine on each occasion whether a complicated function is continuous.
As usual, we avoid such problems by having a set of rules that enable us to
construct continuous functions.
Combination Rules for continuous functions
If f and g are functions that are continuous at a point c, then so are:
Sum Rule fþ g;
Multiple Rule lf, for l2R ;
Product Rule fg;
Quotient Rule fg, provided that f (c) 6¼ 0.
For example, any polynomial p(x)¼ a0þ a1xþ � � � þ anxn, x2R , is contin-
uous at all points of R since we can build up the expression for p by successive
applications of the Combination Rules for continuous functions.
Similarly, any rational function rðxÞ ¼ pðxÞqðxÞ, where p and q are polynomials
and the domain of r is R minus the points where q vanishes, is continuous at all
points of its domain, since we can build up the expression for r by successive
applications of the Combination Rules for continuous functions.
The Combination Rules above are all natural analogues of the corresponding
results for sequences. However, we can combine functions in more ways than
we can combine sequences – for example, we can compose functions f and g to
obtain the function g � f . This approach too will often enable us to obtain new
continuous functions.
Composition Rule Let f be continuous on a set S1 that contains a point c,
and g be continuous on a set S2 that contains the point f(c). Then g � f is
continuous at c.
For example, we know that the function f(x)¼ x2þ 1, x2R , is continuous on R
and that the function g xð Þ ¼ffiffiffi
xp; x � 0, is continuous on I¼ {x : x� 0}. It
follows from the Composition Rule that the function g � f : x 7!ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 1p
is
For instance, the function
p xð Þ ¼ x17�34x5þ5x� 8
is continuous on R .
For instance, the function
r xð Þ ¼ 1� 2x
x2 � 4;
x 2 R � 2f g; is continuouson its domain.
Functions obtained in thisway are called compositefunctions.
136 4: Continuity
continuous at all points c of R (the domain of f) whose image f(c) lies in I. But
all such points f(c) lie in I; so it follows that in fact the composite g � f is
continuous on R .
Problem 6 Prove that the function f given by f xð Þ ¼ x32; x � 0; is
continuous.
Finally, just as we had a Squeeze Rule for convergent sequences, we have a
corresponding Squeeze Rule for continuous functions.
Theorem 1 Squeeze Rule
Let the functions f, g and h be defined on an open interval I, and c2 I. If:
1. g(x) f(x) h(x), for all x2 I,
2. g(c)¼ f(c)¼ h(c),
3. g and h are continuous at c,
then f is also continuous at c.
Example 5 Prove that the function f given by f xð Þ ¼ x2 sin 1x
� �
; x 6¼ 0,
0; x ¼ 0,
�
is continuous at 0.
Solution The diagram in the margin suggests that we should find functions
g and h that squeeze f near 0.
So, we define g(x)¼�x2, x2R , and h(x)¼ x2, x2R . With these two
chosen, we check the conditions of the Squeeze Rule.
The first condition is of course vital! Now we know that
�1 sin1
x
� �
1; for any x 6¼ 0:
It follows that
�x2 x2 sin1
x
� �
x2; for any x 6¼ 0;
so that
g xð Þ ¼ �x2 �
f xð Þ x2 ¼ �
h xð Þ; for any x 2 R :
So condition 1 of the Squeeze Rule is satisfied.
Next, the functions f, g and h all take the value 0 at the point 0. Thus
condition 2 of the Squeeze Rule is satisfied.
Finally, the functions g and h are polynomials, and so in particular they are
continuous at 0. So condition 3 of the Squeeze Rule is satisfied.
It follows then from the Squeeze Rule that f is continuous at 0, as required.&
To test your understanding of these techniques, try the following problems.
Problem 7 Prove that the following function is continuous on R ,
stating each rule or fact about continuity that you are using
f xð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ xþ 1p
� 5x
1þ x2; x 2 R :
We state this rule only in thecase of an interior point of anopen interval. However moregeneral versions also exist!
y
Ic x
y = h(x)
y = f (x)y = g (x)
We shall study thetrigonometric functions indetail in Sub-section 4.1.3.The only property that youneed here is that, for any realnumber x, jsin xj 1.
y = x2 sin 1x
x
y
0
These inequalities areobviously true for x¼ 0, aswell as for x 6¼ 0.
Recall that all polynomials arecontinuous on R .
4.1 Continuous functions 137
Problem 8 Using the elementary properties of the trigonometric func-
tions, determine whether the following functions are continuous at 0:
(a) f ðxÞ ¼ x sinð1xÞ; x 6¼ 0,
0; x = 0;
�
(b) f ðxÞ ¼ sinð1xÞ; x 6¼ 0,
0; x = 0.
�
Proofs
We now give the proofs of the Combination, Composition and Squeeze Rules.
First, recall the Combination Rules.
Combination Rules for continuous functions
If f and g are functions that are continuous at a point c, then so are:
Sum Rule fþ g;
Multiple Rule lf, for l2R ;
Product Rule fg;
Quotient Rule fg, provided that f(c) 6¼ 0.
The proofs of all these are very similar, and depend on the corresponding
results for sequences. We prove only the Sum Rule.
Proof of the Sum Rule We want to prove that fþ g is continuous at c.
Suppose that f and g have domains S1 and S2, respectively. Then the domain
of fþ g is S1 \ S2, and this set contains the point c.
Thus, we have to show that:
for each sequence {xn} in S1 \ S2 such that xn! c, then
f xnð Þ þ g xnð Þ ! f cð Þ þ g cð Þ:
We know that the sequence {xn} lies in S1 and in S2, and that both functions
f and g are continuous at c. Hence
f xnð Þ ! f cð Þ and g xnð Þ ! g cð Þ;
and it follows from the Sum Rule for Sequences that f(xn)þ g(xn)!f(c)þ g(c), as required. &
Next, recall the Composition Rule.
Composition Rule Let f be continuous on a set S1 that contains a point c,
and g be continuous on a set S2 that contains the point f(c). Then g � f is
continuous at c.
Proof We want to prove that g � f is continuous at c.
Now, we know that g � f is certainly defined on the set S¼ {x : x2 S1 and
f(x)2 S2}, and this set contains the point c.
Thus, we have to show that:
for each sequence {xn} in S such that xn! c, then g f xnð Þð Þ ! g f cð Þð Þ.We know that the sequence {xn} lies in S1, and that f is continuous at c.
Hence, we have that f(xn)! f(c).
You may omit the rest of thisSub-section at a first reading.
138 4: Continuity
We also know that {f(xn)} lies in S2, and that g is continuous at f(c). It
follows that g f xnð Þð Þ ! g f cð Þð Þ, as required. &
Finally, recall the Squeeze Rule.
Theorem 1 Squeeze Rule
Let the functions f, g and h be defined on an open interval I, and c2 I. If:
1. g(x) f(x) h(x), for all x2 I,
2. g(c)¼ f(c)¼ h(c),
3. g and h are continuous at c,
then f is also continuous at c.
Proof We have to show that f is continuous at c.
Thus, we have to prove that:
for each sequence {xn} in the domain of f such that xn! c, thenf (xn)! f (c).
Now, since xn! c there is some number X such that xn2 I, for all n>X.
Hence, by condition 1 we have that
g xnð Þ f xnð Þ h xnð Þ; for all n > X: (1)
So, if we now let n!1 and use conditions 2 and 3, we get that
limn!1
g xnð Þ ¼ g cð Þ ¼ f cð Þ and limn!1
h xnð Þ ¼ h cð Þ ¼ f cð Þ: (2)
It follows, from (1), (2) and the Squeeze Rule for sequences, that
limn!1
f xnð Þ ¼ f cð Þ;
as required. &
4.1.3 Trigonometric functions and the exponentialfunction
Trigonometric functions
For the moment we are assuming that you have a knowledge of the trigono-
metric functions arising from your study of trigonometry.
We will prove that the trigonometric functions are continuous on the
whole of their domains. But, first, we need a basic inequality for the sine
function.
Lemma 1 0 sin x x, for 0 x p2:
Proof If x¼ 0, then sin 0¼ 0; so there is an equality.
Suppose next that 0 < x p2, and consider the following diagram, which
represents a quarter circle, centred at the origin, with radius 1.
This follows from thedefinition of a sequenceconverging to c.
Sub-section 2.3.3.
4.1 Continuous functions 139
Since the circle has radius 1, the arc AB has length x and the perpendicular AC
has length sin x. Hence
0 < sin x x; for 0 < x p2: &
We can now extend the inequality in Lemma 1 to obtain a more general result.
Theorem 2 The Sine Inequality
sin xj j xj j; for x 2 R :
Proof We saw in Lemma 1 that the inequality holds for 0 x p2:
For x > p2, we have
sin xj j 1 <1
2p < x ¼ xj j;
and so the desired inequality is also true in this case.
Finally, the inequality also holds for x< 0, since
sin �xð Þj j ¼ sin xj j and �xj j ¼ xj j: &
This is the key tool that we need to prove the continuity of the trigonometric
functions.
Theorem 3 The trigonometric functions (sine, cosine and tangent) are
continuous.
Proof To prove that the sine function is continuous at each point c2R , we
need to show that:
for each sequence fxng in R such that xn ! c; then sin xn ! sin c: (3)
We use the formula
sin xn � sin c ¼ 2 cos1
2xn þ cð Þ
� �
sin1
2xn � cð Þ
� �
;
it follows that
For the shortest distance fromthe point A to the line BC isthe perpendicular from Ato BC.
y
1
0–π
y = ⎜x ⎜
y = ⎜sin x ⎜
⎜sin x ⎜≤ ⎜x ⎜, for x ∈π x
Recall that this means they arecontinuous at each point oftheir domains.
This is a standardtrigonometric formula.
140 4: Continuity
sin xn � sin cj j ¼ 2 cos1
2xn þ cð Þ
� ��
�
�
�
�
�
�
�
� sin1
2xn � cð Þ
� ��
�
�
�
�
�
�
�
2 sin1
2xn � cð Þ
� ��
�
�
�
�
�
�
�
21
2xn � cð Þ
�
�
�
�
�
�
�
�
¼ xn � cj j:
Hence, if {xn� c} is null, then {sin xn� sin c} is also null, by the Squeeze
Rule for null sequences. In other words, the result (3) holds.
The continuity of the cosine and tangent functions now follows from the
formulas
cos x ¼ sin xþ 1
2p
� �
and tan x ¼ sin x
cos x;
using the Combination Rules and the Composition Rule. &
Problem 9 Prove that the following function is continuous on R ,
stating each rule or fact about continuity that you are using
f xð Þ ¼ x2 þ 1þ 3 sin x2 þ 1� �
; for x 2 R :
Problem 10 Prove that the function f xð Þ ¼ sin p2
cos x� �
is continuous
on R .
The exponential function x 7! ex
We now prove that the exponential function is continuous on R . But first we
start with some inequalities that we will need to do this.
Theorem 4 The Exponential Inequalities
(a) ex> 1þ x, for x> 0; and ex� 1þ x, for x� 0;
(b) ex 11�x
; for 0 x < 1;
(c) 1þ x ex 11�x
; for xj j < 1:
Proof We prove inequalities (a) and (b) using the exponential series
ex ¼ 1þ xþ x2
2!þ x3
3!þ � � �; for x � 0:
(a) For x> 0, we have x2
2! > 0, x3
3! > 0, and so on. Hence
ex > 1þ x; for x > 0:
It follows from this that ex� 1þ x, for x� 0, since e0¼ 1
(b) For x� 0, we have x2
2! x2, x3
3! x3, and so on. Hence
ex 1þ xþ x2 þ x3 þ � � �:The series on the right is a geometric series; it converges to the sum 1
1�x,
for 0 x 1. Hence
ex 1
1� x; for 0 x < 1:
By taking moduli.
For cos 12
xn þ cð Þ�
�
�
� 1:
Here we use the SineInequality, Theorem 2.
You met this series inSub-section 3.4.1.
4.1 Continuous functions 141
Section 4.4.
For example
f xð Þ ¼ 3x7 � 4x2 þ 5;
f xð Þ ¼ 3x
x4 � 6;
f xð Þ ¼ xj j;f xð Þ ¼ x
12; x � 0;
f xð Þ ¼ x13; x 2 R ;
f xð Þ ¼ sin x; cos x; tan x;
f xð Þ ¼ ex:
(c) We have just seen that these inequalities hold for 0 x< 1.
For �1< x< 0, we have that 0<�x< 1, so that, by parts (a) and (b)
1þ �xð Þ e�x 1
1� �xð Þ ; for �1 < x < 0:
By taking reciprocals and reversing the inequalities, we may reformulate
this in the form
1þ x ex 1
1� x; for �1 < x < 0:
This completes the proof of the desired result. &
Remark
Notice that, when x 6¼ 0, the results of Theorem 4 hold with strict inequalities.
We are now able to prove the continuity of the exponential function.
Theorem 5 The exponential function x 7! ex, x2R , is continuous.
Proof To prove that the exponential function is continuous at each point
c2R , we need to show that:
for each sequence fxng in R such that xn ! c; then exn ! ec: (4)
We use the formula exn � ec ¼ ec exn�c � 1ð Þ. If we apply Theorem 4, part (c),
with xn� c in place of x, we obtain
1þ xn � cð Þ exn�c 1
1� xn � cð Þ ; for xn � cj j < 1:
Thus, if {xn� c} is null, then jxn� cj< 1 eventually, and so exn�c ! 1 as
n!1, by the Squeeze Rule for sequences. Hence exn ! ec, so that the desired
result (4) holds. &
Remark
Later we shall give a rigorous definition of the general exponential function
x 7! ax, for x2R and a> 0, and we shall show that all exponential functions are
continuous on R .
Problem 11 Prove that the following function is continuous on R ,
stating each rule or fact about continuity that you are using
f xð Þ ¼ cos x5 � 5x2� �
þ 7e�x2
:
We end this section by listing the various types of functions that we have
found to be continuous on their domains.
Basic continuous functions The following functions are continuous:
� polynomials and rational functions;
� modulus function;
� nth root function;
� trigonometric functions (sine, cosine and tangent);
� the exponential function.
142 4: Continuity
4.2 Properties of continuous functions
4.2.1 The Intermediate Value Theorem
In Section 4.3 we shall prove that the function f(x)¼ x5þ x� 1, x2R , is a
one-function on R . From the graph of f it certainly looks as though f maps R
onto R ; how can we prove this?
For example, is there a value of x for which f(x)¼ 0? In other words,
is there a root of the equation x5þ x� 1¼ 0? The shape of the graph
y¼ x5þ x� 1 certainly suggests that such a number x exists; since f(0)¼�1
and f(1)¼ 1, we would expect there to be some number x in the interval (0, 1)
such that f(x)¼ 0. However we do not have a formula for solving the equation
to find x.
Now, we introduced the concept of continuity on the grounds that it would
enable us to pin down precisely the idea that the graph of a ‘well-behaved’
function does not have ‘gaps’ or ‘jumps’, and this is the key to proving that
such a number x exists.
Theorem 1 Intermediate Value Theorem
Let f be a continuous function on [a, b], and let k be any number lying
(strictly) between f(a) and f(b). Then there exists a number c in (a, b) such
that f(c)¼ k.
This result is illustrated below in the two possible cases:
As the graph above on the right shows, there may be more than one possible value
of c such that f(c)¼ k. All that we claim is that there is at least one such point c.
The requirement in Theorem 1 that f be continuous at each point of [a, b] is
essential. For example, the function
f ðxÞ ¼1x; � 1 x < 0,
0; x ¼ 0,1x; 0 < x 1,
8
<
:
is continuous on [�1, 1] except at 0 – where it is discontinuous. For this
function, f(�1)¼�1 and f(1)¼ 1, but there is no number c in (�1, 1) such that
f ðcÞ ¼ 12
(for example).
The following example shows a typical application of the Intermediate
Value Theorem.
This is one of the mainexistence theorems inAnalysis.
Note that f(a) 6¼ f(b) anda< c< b.
4.2 Properties of continuous functions 143
Example 1 Prove that there is a number c in (0, 1) such that c5þ c� 1¼ 0.
Solution Consider the function f(x)¼ x5þ x� 1 on the interval [0, 1]. Then
f is continuous on [0, 1] since it is a basic continuous function, and f(0)¼�1
and f(1)¼ 1.
Since f(0)< 0< f(1), it then follows from the Intermediate Value Theorem
that there is a number c in (0, 1) such that f(c)¼ 0; that is, such that
c5þ c� 1¼ 0. &
If f is a function and c is a real number such that f(c)¼ 0, then c is called a
zero of the function f. We often show that an equation has a solution by proving
that a related continuous function has a zero (by using the Intermediate Value
Theorem with k¼ 0).
Problem 1 Prove that there is a real number c in (0, 1) such that
cos c¼ c.
Problem 2 Let the function f: [0, 1]! [0, 1] be continuous. Prove that
there is a real number c in [0, 1] such that f(c)¼ c.
Proof of Theorem 1 We use the method of repeated bisection.
We shall assume that f(a)< f(b). If, in fact, f(a)> f(b), the proof is very
similar.
First, denote the closed interval [a, b] as [A1, B1]. Then, denote by p the
midpoint of the interval [A1, B1]. Notice that, if f(p)¼ k, then the proof is
complete, since we can take c¼ p.
Otherwise, we define one of the two intervals [A1, p] and [p, B1] to be
[A2, B2] in the following way
A2;B2½ � ¼ A1; p½ �; if f ( p) > 0,
p;B1½ �; if f ( p) < 0.
�
In either case, we obtain:
1. [A2, B2]� [A1, B1];
2. B2 � A2 ¼ 12
B1 � A1ð Þ;3. f(A2)< k< f(B2).
We now repeat this process indefinitely often, bisecting [A2, B2] to obtain
[A3, B3], and so on. If, at any stage, we encounter a bisection point p such
that f(p)¼ k, then the proof is complete.
Otherwise, we obtain a sequence of closed intervals {[An, Bn]} with the
following properties:
1. [Anþ1, Bnþ1]� [An, Bn], for each n2N;
2. Bn � An ¼ 12
� �n�1B1 � A1ð Þ, for each n2N;
3. f(An)< k< f(Bn), for each n2N .
Property 1 implies that the sequence {An} is increasing and bounded above by
B1¼ b. Hence by the Monotone Convergence Theorem, {An} is convergent;
denote by A its limit. By the Limit Inequality Rule for sequences, we must have
that A b.
Similarly, Property 1 implies that the sequence {Bn} is decreasing and
bounded below by A1¼ a. Hence by the Monotone Convergence Theorem,
For f is a polynomial.
The function f is strictlyincreasing on [0, 1], and so thenumber c must be unique inthis case.
We do not consider thepossibility of complex zerosin this book.
For examplef xð Þ ¼ 1
3þ 1
2x sin p
2x
� �
:
You should at least skim thisproof on a first reading, as themethod is an important one.
p ¼ 12
A1 þ B1ð Þ:
Theorem 3, Sub-section 2.3.3.
144 4: Continuity
{Bn} is convergent; denote by B its limit. By the Limit Inequality Rule for
sequences, we must have that B� a.
We may then deduce, by letting n!1 in Property 2 and using the
Combination Rules for sequences, that
B� A ¼ limn!1
Bn � limn!1
An ¼ limn!1
Bn � Anð Þ
¼ limn!1
12
� �n�1B1 � A1ð Þ
¼ B1 � A1ð Þ limn!1
12
� �n�1¼ 0:
In other words, the sequences {An} and {Bn} both converge to a common limit.
Denote this limit by c; then we must have a c b.
Now we use the fact that f is continuous at c. It follows that
limn!1
f Anð Þ ¼ f cð Þ and limn!1
f Bnð Þ ¼ f cð Þ:
Then, by Property 3, f(An)< k, for n¼ 1, 2, . . .; so that, by letting n!1, we
obtain f(c) k by the Limit Inequality Rule for sequences. Similarly, we can
deduce from the fact that k< f(Bn) that f(c)� k. It follows that f(c)¼ k.
Finally, notice that, since f(a)< k< f(b), we cannot have either c¼ a or
c¼ b; so that, in fact, a< c< b as required. &
The method of repeated bisection is very powerful, and of wide application
in Mathematics.
Now, we saw above that the continuous function f(x)¼ x5þ x� 1, x2 [0, 1],
has a zero in (0, 1). Now, since f 12
� �
¼ 132þ 1
2� 1 ¼ � 15
32< 0 and f(1)¼ 1 > 0,
we can make the stronger statement that f must have a zero in 12; 1
� �
, by the
Intermediate Value Theorem.
Problem 3 Use the method of repeated bisection to find an interval of
length 18
that contains a zero of the function f(x)¼ x5þ x� 1, x2 [0, 1].
Antipodal points
Two points on the surface of the Earth are called antipodal points if the line
between them passes through the centre of the Earth. (In this sense, antipodal
points are ‘opposite each other’.)
The following result is a rather interesting application of the Intermediate
Value Theorem! It makes the (physically reasonable) assumption that tem-
perature is a continuous function of position on the Earth’s surface.
Theorem 2 Antipodal Points Theorem
There is always a pair of antipodal points on the Equator of the Earth at
which the temperature is the same.
To prove this, we must set up the situation as a mathematical problem.
Let f(�) denote the temperature at a point on the Equator at an angle � radians
East of Greenwich, for 0 �< 2p, and extend f to be defined on [0, 2p] by
That is, c2 [a, b].
Recall that taking limits‘flattens inequalities’.
That is, c2 (a, b).
In fact the same result holdsfor any ‘Great Circle’ on theEarth’s surface – that is, theintersection of a plane throughthe Earth’s centre with theEarth’s surface.
This process is often calledMathematical modelling.
4.2 Properties of continuous functions 145
requiring that f(2p)¼ f(0). Then Theorem 2 can be rephrased in the following
equivalent way:
Theorem 20 Antipodal Points Theorem
Let f : [0, 2p]!R be continuous, with f(0)¼ f(2p). Then there exists a
number c in [0, p] such that f(c)¼ f(cþ p).
Proof of Theorem 20 Notice, first, that if f(0)¼ f(p) then we can take c¼ 0.
We now prove the result under the assumption that f(0)< f(p). (The proof in
the case that f(0)> f(p) is very similar.)
Next, define the function g as follows
g �ð Þ ¼ f �ð Þ � f �þ pð Þ; for � 2 0; p½ �:Then, since f is continuous on [0, 2p], it follows that g is continuous on [0, p],
by the Combination Rules.
But
g 0ð Þ ¼ f 0ð Þ � f pð Þ < 0
and
g pð Þ ¼ f pð Þ � f 2pð Þ ¼ f pð Þ � f 0ð Þ> 0:
It then follows from the Intermediate Value Theorem, with k¼ 0, that there
exists some number c in (0, p) such that g(c)¼ 0; in other words, such that
f(c)� f(cþ p)¼ 0. This completes the proof. &
4.2.2 Zeros of polynomials
You will have already met a standard method for solving a polynomial
equation of degree 2, and possibly equations of degrees 3 and 4 too. How-
ever there exists no method of solving a general polynomial equation of
degree 5 or higher by means of formulas. How many zeros can a polynomial
equation have?
In fact, a polynomial equation of degree n can have at most n roots in R .
Theorem 3 Fundamental Theorem of Algebra
Let p(x)¼ anxnþ an�1xn�1þ � � � þ a1xþ a0, x2R , where an 6¼ 0. Then
the equation p(x)¼ 0 has at most n roots in R .
Remark
In fact, in its most general form the Fundamental Theorem of Algebra states
that, if p(x)¼ anxnþ an�1xn�1þ � � � þ a1xþ a0, where the coefficients ak may
For, if f(c)¼ f(cþ p), then cand cþ p are antipodal pointswith the same temperature.
Quite often in mathematicswe obtain our results byapplying a standard result tosome cunningly chosenauxiliary function!
These are called quadratic,cubic and quartic equations,respectively.
This is a quite difficult resultto prove!
We do not prove this result,which requires methodsbeyond the scope of this book.
146 4: Continuity
be real or complex numbers and an 6¼ 0, then the equation p(x)¼ 0 has exactly n
roots in C, the set of all complex numbers.
Now, it is sometimes straight-forward to locate zeros of a polynomial if we
have some idea of where to look for them in the first place.
Problem 4 Let p xð Þ ¼ x6 � 4x4 þ xþ 1, x 2 R . Prove that p has a
zero in each of the intervals (�1, 0), (0, 1) and (1, 2).
However to follow this approach for finding zeros (if any) of a given
polynomial, we need first to have an inkling where to look for them. We
usually start by applying the following result, which gives an interval in
which the zeros must lie:
Theorem 4 Zeros Localisation Theorem
Let p xð Þ ¼ xn þ an�1xn�1 þ � � � þ a1xþ a0; x 2 R , be a polynomial. Then
all the zeros of p (if there are any) lie in the open interval (�M, M), where
M ¼ 1þmax an�1j j; . . .; a1j j; a0j jf g:
Example 2 Prove that the polynomial p xð Þ ¼ x4 � 2x2 � xþ 1, x 2 R , has
at least two zeros in R .
Solution We will apply the Zeros Localisation Theorem to p, since its
leading coefficient is 1. Since
M ¼ 1þmax �2j j; �1j j; 1j jf g¼ 3;
it follows that all the zeros of p lie in (�3, 3).
We now compile a table of values of p(x), for x¼�3, �2, �1, 0, 1, 2, 3:
x �3 �2 �1 0 1 2 3
p(x) 67 11 1 1 �1 7 61
We find that p(0) and p(1) have opposite signs, as do p(1) and p(2), so p
must have a zero in each of the intervals (0, 1) and (1, 2), by the Intermediate
Value Theorem.
Thus we have proved that p has at least two zeros in R . &
Problem 5 Prove that the polynomial p xð Þ ¼ x5 þ 3x4 � x� 1, x 2 R ,
has at least three zeros in R .
Although we cannot at this stage prove the Fundamental Theorem of
Algebra, nevertheless we can use the Zeros Localisation Theorem and the
Intermediate Value Theorem to prove the following:
Theorem 5 Every real polynomial of odd degree has at least one zero in R .
Thus, for example, for sufficiently large x the polynomial p xð Þ ¼xn þ an�1xn�1 þ � � � þ a1xþ a0, x 2 R is essentially dominated by its leading
term xn, and so is positive for large positive values of x and is negative for large
negative values of x.
Notice that the coefficient ofxn, the leading coefficient of p,has been set as 1.
The proof appears at the endof the sub-section.
Often, it is not necessary tocompute the values of p(x) forall integers x in [�M, M].
In fact this polynomial hasexactly two zeros in R .
For example,p xð Þ ¼ x5 þ 3x4 � x� 1:
Since n is odd.
4.2 Properties of continuous functions 147
No corresponding result holds for polynomials of even degree. Thus, for
instance, the polynomial p xð Þ ¼ x2 þ 1 has no zeros in R .
Proofs of Theorems 4 and 5
We supply these proofs which were omitted earlier, so as not to disturb the flow
of the text.
Theorem 4 Zeros Localisation Theorem
Let p xð Þ ¼ xn þ an�1xn�1 þ � � � þ a1xþ a0; x 2 R , be a polynomial. Then
all the zeros of p (if there are any) lie in the open interval (�M, M), where
M ¼ 1þmax an�1j j; . . .; a1j j; a0j jf g:
Proof In order to concentrate on the dominating term xn, we define the
function r as follows
r xð Þ ¼ p xð Þxn� 1 ¼ an�1
xþ � � � þ a1
xn�1þ a0
xn; x 2 R � 0f g:
Then, by using the Triangle Inequality, we obtain that, for jxj> 1
r xð Þj j ¼ an�1
xþ � � � þ a1
xn�1þ a0
xn
�
�
�
�
�
�
an�1
x
�
�
�
�
�
�þ � � � þ a1
xn�1
�
�
�
�
�
�þ a0
xn
�
�
�
�
�
�
max an�1j j; . . .; a1j j; a0j jf g � 1
xj j þ � � � þ1
xj jn�1þ 1
xj jn
!
< M � 1
xj j þ � � � þ1
xj jn�1þ 1
xj jn þ � � � !
¼ M �1xj j
1� 1xj j¼ M
xj j � 1:
It follows that, if xj j � M ¼ 1þmax an�1j j; . . .; a1j j; a0j jf g, then r xð Þj j < 1:Now, from the definition of r(x) we see that
p xð Þ ¼ xn 1þ r xð Þð Þ; for xj j � M:
But since r xð Þj j < 1, we certainly have that 1þ r xð Þ > 0. It follows from the
above expression for p(x) in terms of r(x) that p(x) must have the same sign as
xn, for xj j � M:It follows that any zero of p must lie in (�M, M). &
Theorem 5 Every polynomial of odd degree has at least one zero in R .
Proof By dividing the polynomial by its leading coefficient, we may assume
that the polynomial is of the form
p xð Þ ¼ xn þ an�1xn�1 þ � � � þ a1xþ a0; x 2 R , where n is odd.
We can then define M and r(x) as in Theorem 4, and all the statements in the
proof of Theorem 4 then hold here too.
Since n is odd, xn is positive for x�M and negative for x�M. It follows
from the arguments in the proof of Theorem 4 that p(x) must be positive for
x�M and negative for x�M. In particular, p(M)> 0 and p(�M)< 0.
Then, by the Intermediate Value Theorem, p must have a zero in (�M, M).&
You may omit these at a firstreading.
Recall that the coefficient ofxn, the leading coefficientof p, is 1.
Here the modulus of eachcoefficient is at most themaximum value over allcoefficients.
jxj> 1, so that 1xj j < 1:
By summing the geometricseries.
r xð Þj j51, �15r xð Þ51:
So we have certainly got goodvalue from the arguments inthe previous proof!
148 4: Continuity
4.2.3 The Extreme Values Theorem
We now look at whether continuous functions on closed intervals are bounded
or unbounded. It turns out that they are bounded, and we use this fact a great
deal throughout Analysis!
Theorem 6 The Extreme Values Theorem
A continuous function on a closed interval possesses a maximum value and
a minimum value on the interval. In other words, if f is a function that is
continuous on a closed interval [a, b], then there exist points c and d in [a, b]
such that
f cð Þ f xð Þ f dð Þ; for x 2 a; b½ �:
If f is continuous on an interval that is not closed, then it may not be
bounded. For example, the function f xð Þ ¼ 1x, x 2 0; 1ð �, is continuous on (0,1]
but is not bounded above (and so it is not bounded), but it is bounded below
(by 1).
Similarly, if a function is not continuous on a closed interval, then it may not
be bounded. For example, the function
f ðxÞ ¼ 1; x ¼ 0;1x; 0 < x 1;
�
takes the values 1 when x¼ 0 and 1x
when 0< x 1. f is bounded below, by 0,
but is not bounded above (and so it is not bounded).
Problem 6
(a) Determine the maximum and the minimum of the function
f xð Þ ¼ x2, x 2 �1; 2½ �, on [�1, 2]. Specify all points in [�1, 2]
where these are attained.
(b) Determine the maximum and the minimum of the function
g xð Þ ¼ sin x, x 2 0, 2p½ �, on [0, 2p]. Specify all points in [0, 2p]
where these are attained.
Since a function is bounded if and only it is both bounded above and
bounded below, we sometimes use the following version of Theorem 6 in
applications:
Corollary 1 The Boundedness Theorem
A continuous function on a closed interval is bounded. In other words, if f is
a function that is continuous on the closed interval [a, b], then there exist a
number M such that
f xð Þj j M; for x 2 a; b½ �:
On other occasions, we shall find the following consequence of Theorem 6
and the Intermediate Value Theorem useful:
Corollary 2 The Interval Image Theorem
The image of a closed interval under a continuous function is a closed
interval.
We discussed bounds,suprema, infima, maxima andminima of functions earlier, inSub-section 1.4.2.
We prove this at the end of thesub-section.
For it will often be sufficientfor our purposes.
4.2 Properties of continuous functions 149
Proof of Theorem 6 Without some systematic approach to studying con-
tinuous functions, we would never be able to prove this theorem, and would be
reduced to much hand-waving arguments and loose assertions. In this case we
shall use our earlier work on sequences, and the following related result:
Lemma 1 Let f be a function defined on an interval I.
(a) If f is bounded above on I and sup f xð Þ: x 2 If g ¼ M, then there exists
some sequence {xn} in I such that f xnð Þ ! M as n!1.
(b) If f is not bounded above on I, then there exists some sequence {xn} in I
such that f xnð Þ!1 as n!1.
Proof
(a) If there is some point, c say, in I for which f(c)¼M, then the constant
sequence {c} has the desired property.
Now suppose that no such point c exists. Then, since sup f xð Þ:fx 2 Ig ¼ M, it follows from the definition of supremum that there is
some point, x1 say, in I for which
f x1ð Þ > M � 1:
Next, since f(x1)<M (from our assumption that there is no point where
f takes the value M), choose a point, x2 say, in I for which
f x2ð Þ > max f x1ð Þ; M � 1
2
�
;
notice, in particular, that this last inequality ensures that x2 6¼ x1.
Continuing this process indefinitely, we obtain a sequence {xn} of
distinct points in I for which
f xnð Þ > max f x1ð Þ; f x2ð Þ; . . .; f xn�1ð Þ; M � 1
n
�
:
Since the sequence M � 1n
� �
converges to M, it follows, by the Squeeze
Rule for sequences, that f xnð Þ ! M as n!1.
(b) It follows from the definition of ‘unbounded above’ that there is some
point, x1 say, in I for which f(x1)> 1. Then, for each n> 1, we can
construct a sequence {xn} of distinct points in I for which
f xnð Þ > max f x1ð Þ; f x2ð Þ; . . .; f xn�1ð Þ; nf g:Since the sequence {n} tends to 1, it follows, by the Squeeze Rule for
sequences, that f(xn)!1 as n!1. &
We are now in a position to prove Theorem 6.
Theorem 6 The Extreme Values Theorem
A continuous function on a closed interval possesses a maximum value and
a minimum value on the interval. In other words, if f is a function that is
continuous on the closed interval [a, b], then there exist points c and d in
[a, b] such that
f cð Þ f xð Þ f dð Þ; for x 2 a; b½ �:
Notice that this is a resultabout sup for any function f onan interval; no assumption ofcontinuity is involved.
A similar result holds forfunctions that are boundedbelow on I or are unboundedbelow on I.
This is a proof bycontradiction.
So f x2ð Þ > f x1ð Þ andf x2ð Þ > M � 1
2. Also,
f x2ð Þ < M:
For M � 1n< f xnð Þ < M:
150 4: Continuity
Proof First, we shall assume that f is not bounded above, and verify that this
assumption leads to a contradiction with known facts about f.
It follows from part (b) of Lemma 1 that there exists some sequence {xn} in
[a, b] such that f(xn)!1 as n!1. Then, by the Bolzano–Weierstrass
Theorem, {xn} must contain a convergent subsequence xnkf g; denote by d
the limit of xnkf g. Since all the xnk
lie in [a, b], it follows, by the Limit
Inequality Rule for sequences, that d2 [a, b].
Since f is continuous at d and the sequence xnkf g converges to d, it follows
that f xnkð Þ ! f dð Þ. However, f xnk
ð Þf g is a subsequence of the sequence {f(xn)}
which tends to1, so that f xnkð Þ ! 1:
This is a contradiction. So f must be bounded above on [a, b] after all.
Next, denote by M the number sup f xð Þ: x 2 a; b½ �f g. It follows, from part (a)
of Lemma 1, that there exists some sequence {xn} in [a, b] such that f(xn)!M as
n!1. Then, by the Bolzano–Weierstrass Theorem, {xn} must contain a con-
vergent subsequence xnkf g; denote by d the limit of xnk
f g. Since all the xnklie in
[a, b], it follows, by the Limit Inequality Rule for sequences, that d2 [a, b].
Since f is continuous at d and the sequence xnkf g converges to d, it follows
that f xnkð Þ ! f dð Þ. Therefore, since f xnk
ð Þf g is a subsequence of the sequence
{f(xn)} which tends to M, we have M¼ f(d).
This complete the proof that there exists a point d in [a, b] such that
f xð Þ f dð Þ; for all x 2 a; b½ �:The proof of the existence of a point c in [a, b] such that f(c) f(x), for all
x2 [a, b], is similar; we omit it. &
4.3 Inverse functions
4.3.1 Existence of an inverse function
Let f be the function f(x)¼ 2x, x2R . Then, given any number y in R , we can
find a unique number x ¼ 12
y in the domain of f such that y¼ f(x)¼ 2x.
The inverse function f �1, defined by f�1 yð Þ ¼ 12
y; y 2 R , undoes the
‘effect’ of f; that is, f �1( f(x))¼ x, x2R .
Also, f undoes the effect of f �1; that is, f ( f �1(y))¼ y, y2R .
Not every function has an inverse function! For example, consider the
function
g xð Þ ¼ x2; x 2 R :
Since g(2)¼ 4¼ g(�2), we cannot define g�1(4) uniquely. Thus g fails to
have an inverse function, because it is not one–one. However the function
h xð Þ ¼ x2; x 2 ½0;1Þ;is one–one and has an inverse function
h�1 yð Þ ¼ ffiffiffi
yp; y 2 ½0;1Þ:
In general, if f : A! R is one–one, then, for each point y in the image f(A),
there is a unique point x in A such that f(x)¼ y. Thus f is a one–one correspon-
dence between A and f(A); and so we can define the inverse function f �1 by
f �1( y)¼ x, where y¼ f(x).
It is also possible to proveTheorem 6 by the method ofrepeated bisection.
Theorem 3, Sub-section 2.5.1.
xnk2 a; b½ � , a xnk
b:
Inheritance Property ofSubsequences, Theorem 5,Sub-section 2.4.4.
M must exist, since we haveshown that the setf xð Þ: x 2 a; b½ �f g is bounded.
4.3 Inverse functions 151
Definition Let f : A! R be a one–one function. Then the inverse functionf �1 has domain f(A) and is specified by
f�1 yð Þ ¼ x; where y ¼ f xð Þ; x 2 A:
For some functions f, we can find the inverse function f �1 directly, by
solving the equation y¼ f(x) algebraically to obtain x in terms of y.
Example 1 Prove that the function f defined by f xð Þ ¼ 11�x
; x 2 ð�1; 1Þ,has an inverse function defined on (0,1).
Solution First, we solve the equation y ¼ 11�x
to give x in terms of y. By
simple manipulation, we obtain
y ¼ 1
1� x, x ¼ 1� 1
y:
Now, for each x2 (�1, 1), we have x< 1, and so f xð Þ ¼ 11�x
> 0; thus
f �1,1ð Þð Þ � 0,1ð Þ. Also, for each y2 (0,1), we have
x ¼ 1� 1
y2 �1; 1ð Þ;
and so f is a one–one correspondence between (�1, 1) and (0,1). Hence
f�1 yð Þ ¼ 1� 1
y; y 2 ð0;1Þ: &
Remark
Usually, when defining a function we write x for the domain variable. To
conform with this practice, we may rewrite the inverse function f �1 in
Example 1 as follows
f�1 xð Þ ¼ 1� 1
x; x 2 ð0;1Þ:
The graph y¼ f �1 (x) is obtained by reflecting the graph y¼ f(x) in the line
y¼ x. This reflection interchanges the x- and y-axes.
Proving that a function f is one–one
We have seen that, if f : A! R is one–one, then f has an inverse function f �1
with domain f(A). For the function f considered in Example 1, it is possible to
determine f �1 explicitly by solving the equation y¼ f (x) to obtain x in terms
of y. Unfortunately, it is generally not possible to solve the equation y¼ f (x) in
this way.
Nevertheless, it may still be possible to prove that f has an inverse function
f �1 by showing that f is one–one in some other way. For example, f is one–one
if it is either strictly increasing or strictly decreasing; that is, if f is strictly
monotonic.
Definitions A function f defined on an interval I is:
� increasing on I if x15x2 ) f x1ð Þ f x2ð Þ; for x1; x2 2 I;
� strictly increasing on I if x15x2 ) f x1ð Þ5 f x2ð Þ; for x1; x2 2 I;
� decreasing on I if x15x2 ) f x1ð Þ � f x2ð Þ; for x1; x2 2 I;
y
y = 1 1
0 1 x(– ∞, 1)
1 – x
152 4: Continuity
� strictly decreasing on I if x15x2 ) f x1ð Þ > f x2ð Þ; for x1; x2 2 I;
� monotonic on I if f is either increasing on I or decreasing on I;
� strictly monotonic on I if f is either strictly increasing on I or strictly
decreasing on I.
The most powerful technique for proving that a function f is strictly mono-
tonic is to compute the derivative f 0 of f and examine the sign of f 0(x). We shall
not study differentiation in detail until Chapter 6, so here we consider only
functions which can be proved to be strictly monotonic by manipulating
inequalities rather than by Calculus.
For example, if n2N , then the function f(x)¼ xn, x2 [0, 1), is strictly
increasing; and, if n is odd, the function f(x)¼ xn, x2R , is strictly increasing.
Similarly, if n2N , then the function f(x)¼ x�n, x2 (0, 1), is strictly
decreasing.
Example 2 Prove that the function f xð Þ ¼ x5 þ x� 1; x 2 R ; is one–one.
Solution If x1< x2, then x51< x5
2. Hence
x51 þ x1 � 1 < x5
2 þ x2 � 1;
so f is strictly increasing, and thus one–one. &
Problem 1 Prove that the following functions are one–one:
(a) f xð Þ ¼ x4 þ 2xþ 3; x 2 0;1½ Þ;(b) f xð Þ ¼ x2 � 1
x; x 2 0;1ð Þ:
4.3.2 The Inverse Function Rule
If the function f: A!R is strictly monotonic, then f is one–one, and so f has
an inverse function f �1 with domain f(A). However, it is not always easy to
determine f(A). However, if f is known to be continuous on A, then the
following result simplifies the problem immediately.
Theorem 1 Inverse Function Rule
Let f : I! J, where I is an interval and J is the image f(I), be a function such that:
1. f is strictly increasing on I;
2. f is continuous on I.
Then J is an interval, and f has an inverse function f� 1: J! I such that:
10. f �1 is strictly increasing on J;
20. f �1 is continuous on J.
Remarks
1. The interval I may be any type of interval: open or closed, half-open,
bounded or unbounded.
2. There is a similar version of the Inverse Function Rule with ‘strictly
increasing’ replaced by ‘strictly decreasing’.
We prove the InverseFunction Rule inSub-section 4.3.4.
4.3 Inverse functions 153
Example 3 Prove that the function f xð Þ ¼ x5 þ x� 1; x 2 R ; has a contin-
uous inverse function, with domain R .
Solution The domain of f is R , which is an interval.
We have already seen, in Example 2, that f is strictly increasing on R and so
has an inverse function f�1. Since f is a polynomial, it is a basic continuous
function on R . Thus f satisfies the hypotheses of the Inverse Function Rule.
Then it follows from the Rule that the image J¼ f(R) is an interval, and that
the inverse function f�1: J!R is strictly increasing and continuous on J.
It remains to check that the image J is the whole of R . Notice that J¼ f(R)
contains each of the numbers
f nð Þ ¼ n5 þ n� 1; n 2 Z:
Now, J contains each of the intervals [f (�n), f (n)]; and f �nð Þ ¼ �n5�n�1! �1 as n!1, while f nð Þ ¼ n5 þ n� 1!1 as n!1. It follows
that, in fact, J¼ f (R) must be (�1,1)¼R .
Thus f has a continuous inverse function f �1: R!R . &
As the above example shows, when we apply the Inverse Function Rule the
hardest step is to determine the image J¼ f (I). Since J is an interval, it is
sufficient to determine the end-points of J, which may be real numbers or one
of the symbols1 and�1. We must also determine whether or not these end-
points belong to J.
The following diagrams illustrate two examples:
y
d = f (b)
c = f (a)
a x
y = f (x)y = f (x) J = (c, d ]
I = (a, b]I = [a, ∞)a b x
c
y
d
J = [c, d )
Notice that, if a is an end-point of I and a2 I, then c¼ f(a) is the correspond-
ing end-point of J and c2 J.
On the other hand, if a is an end-point of I and a =2 I (this includes the
possibility that a may be 1 or �1), then it is a little harder to find the
corresponding end-point of J. However, it can be shown that, if {xn} is a
monotonic sequence in I and xn! a, then f(xn)! c, and c =2 J: (We prove this
in Sub-section 4.3.4.)
Example 4 Prove that the function f xð Þ ¼ x4 þ 2xþ 3; x 2 0; 1½ Þ, has a
continuous inverse function with domain [3,1).
Solution The domain of f is [0,1), which is an interval.
Also, we know that f is strictly increasing and continuous on [0,1), and so
conditions 1 and 2 of the Inverse Function Rule hold. It follows that the
image J¼ f([0,1)) is an interval, and f has a continuous inverse function f �1:
J! [0,1) which is strictly increasing on J.
It remains to check that the image J is [3,1).
For the end-point 0 of I, we have 02 I, so the corresponding end-point of J is
f(0)¼ 3, and 32 J.
For example:
0; 1ð � has end-points 0
and 1;
1;1½ Þ has end-points 1
and1:Do not let this use of thesymbol1 tempt you to thinkthat1 is a real number.
y
d
J = [c, d )
c = f (a)
ac ∈ J
a ∈ II = [a, ∞)
x
y = f (x)
d = f (b)
{ f (an)}
{an}a
c
b x
y = f (x)
I = (a, b]
y
J = (c, d ]
c ∉ J
a ∉ I
You saw this in Problem 1(a ),above.
154 4: Continuity
The other end-point of I is1, so to find the corresponding end-point of J we
choose the monotonic sequence {n}, which lies in I and tends to infinity. Then
f nð Þ ¼ n4 þ 2nþ 3!1 as n!1. It follows that the corresponding end-
point of J is1. Thus, J¼ [3,1), as required. &
We now summarise the strategy for establishing that a given continuous
function f has a continuous inverse function.
Strategy To prove that f : I! J, where I is an interval with end-points a
and b, has a continuous inverse f �1: J! I:
1. show that f is strictly increasing on I;
2. show that f is continuous on I;
3. determine the end-point c of J corresponding to the end-point a of I as
follows:
� if a2 I, then f(a)¼ c (and c2 J);
� if a =2 I, then f(xn)! c (and c =2 J),
where {xn} is a monotonic sequence in I such that xn! a;
4. determine the end-point d of J corresponding to the end-point b of I,
similarly.
Problem 2 Use the above strategy to prove that the function
f xð Þ ¼ x2 � 1x; x 2 0;1ð Þ, has a continuous inverse function with
domain R .
Hint: Use the result of Problem 1(b) in Sub-section 4.3.1.
4.3.3 Inverses of standard functions
We now use the Inverse Function Rule and the above strategy to define
continuous inverse functions for certain standard functions. Although you
will be familiar with these inverse functions already, we can now prove that
they exist and are continuous. We also remind you of some properties of these
inverse functions.
For each function, we give brief remarks on the four steps of the strategy. In
each case, the continuity of the function f follows directly from the results of
Section 4.1.
The nth root function
We asserted the existence of the nth root function in Sub-section 1.5.2,
Theorem 1. We can at last provide the proof of that assertion!
The nth root function For any positive integer n� 2, the function
f xð Þ ¼ xn; x 2 0;1½ Þ;has a strictly increasing continuous inverse function f�1 xð Þ ¼
ffiffiffi
xnp
with
domain 0;1½ Þ, called the nth root function.
There is a correspondingversion of this strategy for thecase that f is decreasing on I.
4.3 Inverse functions 155
In this case, the strategy is easy to apply:
1. f is strictly increasing on [0,1);
2. f is continuous on [0,1);
3. f(0)¼ 0;
4. f(k)¼ kn!1 as k!1.
It follows, from the Inverse Function Rule, that f has a strictly increasing
continuous inverse function f �1: [0,1)! [0,1).
Remark
If n is odd, then the nth root function can be extended to a continuous function
whose domain is the whole of R .
Inverse trigonometric functions
The function sin�1 The function
f xð Þ ¼ sin x; x 2 � 1
2p;
1
2p
� �
;
has a strictly increasing continuous inverse function with domain [�1, 1],
called sin�1.
In this case:
1. the geometric definition of f(x)¼ sin x shows that f is strictly increasing on
�12p; 1
2p
�
;
2. f is continuous on �12p; 1
2p
�
;
3. sin �12p
� �
¼ �1;
4. sin 12p
� �
¼ 1.
It follows that f �12p; 1
2p
�� �
¼ �1; 1½ �. Hence, by the Inverse Function
Rule, f has a strictly increasing continuous inverse function f�1:�1; 1½ � ! � 1
2p; 1
2p
�
:
The decreasing version of the strategy can be applied similarly to prove that
the cosine function has an inverse, if we restrict its domain suitably.
The function cos�1 The function
f xð Þ ¼ cos x; x 2 0; p½ �;has a strictly decreasing continuous inverse function with domain [�1, 1],
called cos�1.
The domain [0, p] of f is chosen here by convention, so that f is a strictly
monotonic restriction of the cosine function.
Similarly, to form an inverse of the tangent function we must restrict its
domain to � 12p; 1
2p
� �
, since the tangent function is strictly increasing and
continuous on this interval.
f is a basic continuousfunction.
We use {k} rather than {n}here, to avoid using n for twodifferent purposes in the sameexpression.
156 4: Continuity
The function tan�1 The function
f xð Þ ¼ tan x; x 2 � 1
2p;
1
2p
� �
;
has a strictly increasing continuous inverse function with domain R , called
tan�1.
In this case, the image set f � 12p; 1
2p
� �� �
is R , because (for example) if {xn}
is a monotonic sequence in � 12p; 1
2p
� �
and xn ! 12p as n!1, then
f xnð Þ ¼ tan xn ¼sin xn
cos xn
!1 as n!1:
Remark
Some texts use arc sin, arc cos and arc tan instead of sin�1, cos�1 and tan�1,
respectively.
Problem 3
(a) Determine the values of sin�1ð 1ffiffi
2p Þ, cos�1 �1
2
� �
and tan�1ffiffiffi
3p� �
.
(b) Prove that cos 2 sin�1 x� �
¼ 1� 2x2, for x 2 ½�1; 1�.Hint: Let y ¼ sin�1 x:
The function loge
We now discuss one of the most important inverse functions.
The function loge The function
f xð Þ ¼ ex; x 2 R ;
has a strictly increasing continuous inverse function f �1 with domain
(0,1), called loge.
In this case:
1. f is strictly increasing on R , since
x1 < x2 ) x2 � x1 > 0
) ex2�x1 > 1 ðsince ex � 1þ x > 1; for x > 0Þ) ex2 > ex1 ;
2. f is continuous on R ;
3. f nð Þ ¼ en !1 as n!1;
4. f �nð Þ ¼ e�n ! 0 as n!1:
It follows that the image of R under f is f Rð Þ ¼ 0;1ð Þ. Hence, by the
Inverse Function Rule, f has a strictly increasing continuous inverse function
f�1: 0;1ð Þ ! R .
Problem 4 Prove that loge xþ loge y ¼ loge xyð Þ, for x; y 2 0;1ð Þ.Hint: Let a ¼ loge x and b ¼ loge y.
4.3 Inverse functions 157
Inverse hyperbolic functions
The function sinh�1 The function
f xð Þ ¼ sinh x ¼ 1
2ex � e�xð Þ; x 2 R ;
has a strictly increasing continuous inverse function f �1 with domain R ,
called sinh�1.
In this case:
1. f is strictly increasing on R , since
x1 < x2 ) ex1<ex2
) �e�x1<�e�x2
) ex1 � e�x1< ex2 � e�x2
) sinh x1 < sinh x2;
2. f is continuous on R , by the Combination Rules;
3. f nð Þ ¼ 12
en � e�nð Þ ! 1 as n!1;
4. f �nð Þ ¼ 12
e�n � enð Þ ! �1 as n!1.
It follows that the image of R under f is f (R)¼R . Hence, by the Inverse
Function Rule, f has a strictly increasing continuous inverse function
f �1: R!R .
The function cosh�1 The function
f xð Þ ¼ cosh x ¼ 1
2ex þ e�xð Þ; x 2 0;1½ Þ;
has a strictly increasing continuous inverse function f �1 with domain
[1,1), called cosh�1.
In this case:
1. f is strictly increasing on [0,1), since
x1 < x2 ) sinh x1 < sinh x2
) 1þ sinh2 x1
� �
12< 1þ sinh2 x2
� �
12
) cosh x1 < cosh x2; since cosh2 x ¼ 1þ sinh2 x;
2. f is continuous on [0,1), by the Combination Rules;
3. f 0ð Þ ¼ 1;
4. f nð Þ ¼ 12
en þ e�nð Þ ! 1 as n!1.
It follows that the image of [0,1) under f is f ([0,1))¼ [1,1). Hence, by
the Inverse Function Rule, f has a strictly increasing continuous inverse func-
tion f �1: [1,1)! [0,1).
The strategy can be applied in a similar way to show that f(x)¼ tanh x is
strictly increasing and continuous on R , with f (R)¼ (�1, 1). We omit the
details.
158 4: Continuity
The function tanh�1 The function
f xð Þ ¼ tanh x ¼ sinh x
cosh x; x 2 R ;
has a strictly increasing continuous inverse function f�1 with domain
(�1, 1), called tanh�1.
The inverse hyperbolic functions can be expressed in terms of loge, as the
following example shows.
Example 5 Prove that sinh�1 x ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 1p
� �
, for x 2 R .
Solution Let y ¼ sinh�1 x, for x 2 R . Then
x ¼ sinh y ¼ 1
2ey � e�yð Þ;
and so
e2y � 2xey � 1 ¼ 0:
This is a quadratic equation in ey, so that
ey ¼ xffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 1p
:
Since ey> 0, we must choose theþ sign here, so that
y ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 1p
� �
: &
Problem 5 Prove that cosh�1 x ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � 1p� �
, for x 2 1;1½ Þ.
4.3.4 Proof of the Inverse Function Rule
We now prove the Inverse Function Rule, and justify our strategy for determin-
ing the domains of inverse functions.
Theorem 1 Inverse Function Rule
Let f: I! J, where I is an interval and J is the image f (I), be a function
such that:
1. f is strictly increasing on I;
2. f is continuous on I.
Then J is an interval, and f has an inverse function f �1: J! I such that:
10. f �1 is strictly increasing on J;
20. f �1 is continuous on J.
Proof The proof is in four parts:
J¼ f(I) is an interval.
Let y1, y22 f(I), with y1< y2, and let y2 (y1, y2). Now y1¼ f (x1) and
y2¼ f(x2), for some x1, x22 I, with x1< x2 since f is strictly increasing on I.
It follows, by the Intermediate Value Theorem, that there is some number
x2 (x1, x2) for which f(x)¼ y. Hence y2 f(I).
It follows that f(I) is an interval.
You may omit the proofs inthis sub-section at a firstreading, but you should atleast read the statement ofTheorem 2 below.
4.3 Inverse functions 159
The inverse function f �1: J! I exists.
The function f is strictly increasing and is therefore one–one; also, f maps
I onto J, so the function f �1: J! I exists, by definition.
f �1 is strictly increasing on J.
We have to show that
y1 < y2 ) f�1 y1ð Þ < f�1 y2ð Þ; for y1; y2 2 J:
Notice that
f�1 y1ð Þ � f�1 y2ð Þ ) f f�1 y1ð Þ� �
� f f�1 y2ð Þ� �
) y1 � y2:
Hence y1 < y2 ) f�1 y1ð Þ < f�1 y2ð Þ, as required.
f �1 is continuous on J.
Let y2 J, and (for simplicity) assume that y is not an end-point of J. Then
y¼ f(x), for some x2 I, and we want to prove that
yn ! y) f�1 ynð Þ ! f�1 yð Þ ¼ x:
Thus, we want to deduce that:
for each "> 0, there is some number X such that
x� " < f�1 ynð Þ < xþ "; for all n > X: (1)
Since f is strictly increasing, we know that
f x� "ð Þ < f xð Þ < f xþ "ð Þ;also, since yn! y¼ f(x), there is some number X such that
f x� "ð Þ < yn < f xþ "ð Þ; for all n > X:
If we then apply the strictly increasing function f �1 to these inequalities, we
obtain (1), as required.
This completes the proof of the Inverse Function Rule. &
In fact, a careful check of the first part of the above proof shows that a slight
change enables us to prove the following result that is of interest in its own right.
Theorem 2 The image of an interval under a continuous function is also
an interval.
Proof Let f be a continuous function on an interval I. We shall assume that f
is non-constant on I, since otherwise the result is trivial.
Let y1, y22 f(I), with y1< y2, and let y2 (y1, y2). Now y1¼ f(x1) and
y2¼ f(x2), for some x1, x22 I, and x1 6¼ x2. Let I0 denote the interval with end-
points x1 and x2. It follows, by applying the Intermediate Value Theorem to the
function f on I0, that there is some number x2 I0 for which f(x)¼ y. Hence
y2 f(I).
It follows that f(I) is an interval. &
Finally, we justify our strategy for determining the end-points of J¼ f(I).
In Corollary 2,Sub-section 4.2.3, we showedthat the image of a closedinterval under a continuousfunction is a closed interval.
We must introduce thesymbol I0 since we do notknow in general whetherx1< x2 or x2< x1.
160 4: Continuity
Strategy for finding the end-points of J If a is an end-point of the
interval I, then we can find the corresponding end-point c of J as follows:
� if a2 I, then f (a)¼ c (and c2 J);
� if a =2 I, then f(xn)! c (and c =2 J),
where {xn} is a monotonic sequence in I such that xn! a.
Proof For simplicity, we shall suppose that a is the left end-point of I.
y
c = f (a)
d = f (b)
{ f(xn)}
{xn}
a b x
y = f (x)y = f (x)
J = [c, d )
I = [a, ∞)I = (a, b]
a x
d
y
J = (c, d ]
c ∉ J
a ∉ I
c ∈ J
a ∈ I
If a2 I, then c¼ f (a)2 J and
f xð Þ � f að Þ ¼ c; for x 2 I;
and so c is the corresponding left end-point of J.
On the other hand, if a =2 I, then we select any decreasing sequence {xn} in I
such that xn! a as n!1. Then {f(xn)} is also decreasing; it therefore
follows, by the Monotonic Sequence Theorem for sequences, that
f xnð Þ ! c as n!1; (2)
where c is a real number or �1.
Now, f(xn)2 J for n¼ 1, 2, . . .; therefore, since J is an interval, it follows that
f xnð Þ; f x1ð Þ½ � � J; for n ¼ 1; 2; . . .:
Hence, by (2), we have that
c; f x1ð Þð � ¼[
1
n¼1
f xnð Þ; f x1ð Þ½ � � J:
To deduce, finally, that c is the left end-point of J, we need to show that c =2 J.
Suppose that in fact c2 J. Then c¼ f(x), for some x2 I; it follows that
f xð Þ ¼ c < f xnð Þ; for n ¼ 1; 2; . . .¼) x < xn; for n ¼ 1; 2; . . .¼) x a:
Thus x =2 I. This contradiction completes the proof. &
4.4 Defining exponential functions
4.4.1 The definition of ax
Earlier, we looked at the definition of the irrational numberffiffiffi
2p¼ 1:4142 . . ..
We have also defined ax for a> 0 when x is rational, but we have not yet
defined ax when x is irrational.
Here we consider only theend-point a of I; a similarresult holds for the otherend-point b of I.
This was Theorem 2 in Sub-section 2.5.1: if the sequence{an} is monotonic, then either{an} is convergent oran!1.
Taking limits ‘flattens’inequalities.
Sub-section 1.1.1.
Sub-section 1.5.3.
4.4 Defining exponential functions 161
One possible method for defining the irrational power 2ffiffi
2p
involves the
decimal representation offfiffiffi
2p
. Each of the truncations offfiffiffi
2p¼ 1:4142 . . . is
a rational number, and the corresponding rational numbers
21; 21:4; 21:41; 21:414; 21:4142; . . . (1)
are defined, and form an increasing sequence which is bounded above by
22¼ 4 (for example). Hence, by the Monotone Convergence Theorem for
sequences, the sequence (1) is convergent, and the limit of this sequence can
be taken as the definition of 2ffiffi
2p:
We can define ax for a> 0 and x2R similarly. However, with this definition
it is difficult to establish the properties of ax, such as the Exponential Laws. It is
more convenient to define ax by using the exponential function x 7! ex, whose
properties we have already discussed.
Recall that
ex ¼ limn!1
1þ x
n
� �n
¼X
1
n¼0
xn
n!; for x 2 R ;
ex ¼ e�xð Þ�1; for x 2 R ;
and
exþy ¼ ex � ey; for x; y 2 R : (2)
Recall too that the function x 7! ex is strictly increasing and continuous; and
hence, by the Inverse Function Rule, it has a strictly increasing continuous
inverse function x 7! loge x, x2 (0,1). Thus we have
loge exð Þ ¼ x; for x 2 R ;(3)
and
eloge x ¼ x; for x 2 0;1ð Þ:
Now let a¼ ex and b¼ ey. Then, by equation (2), we have
ab ¼ exey ¼ exþy;
so that, by equation (3), we obtain loge(ab)¼ logeaþ logeb, for a, b2 (0,1).
We deduce that, if a> 0 and n2N , then
loge anð Þ ¼ n loge a;
and so
an ¼ en loge a:
With a little more manipulation, we can show that the equation
ax ¼ ex loge a
is true for each rational number x. This suggests that we define ax, for a> 0 and
x irrational, by means of this equation.
Definition If a> 0, then ax ¼ ex loge a; for x 2 R :
For example, 2x ¼ ex loge 2 for x2R , so that the graph y¼ 2x is obtained from
the graph y¼ ex by a scaling in the x-direction with scale factor 1loge 2
.
Sub-section 2.5.1.
Section 3.4.
Sub-section 3.4.3.
162 4: Continuity
This relationship between the graphs y¼ ex and y¼ 2x suggests that the
function x 7! 2x must also be continuous. In fact, we can deduce the continuity
of the function x 7! 2x ¼ ex loge 2; x 2 R , from the continuity of the function
x 7! ex, by using the Multiple Rule and the Composition Rule for continuity.
Remark
Since x 7! 2x is continuous, we deduce that the sequence
21; 21:4; 21:41; 21:414; 21:4142; . . .;
where 1, 1.4, 1.41, 1.414, 1.4142, . . . are truncations offfiffiffi
2p¼ 1:4142 . . .,
does converge to 2ffiffi
2p
, and so both definitions of 2ffiffi
2p
agree.
In general, we have the following result.
Theorem 1 If a> 0, then the function x 7! ax ¼ ex loge a; x 2 R , is
continuous.
Problem 1 Prove that the following functions are continuous:
(a) f(x)¼ x�, where x2 (0,1) and �2R ;
(b) f(x)¼ xx, where x2 (0,1).
4.4.2 Further properties of exponentials
Our definition of ax enables us to give straight-forward proofs of the following
Exponent Laws.
Exponent Laws
� If a, b> 0 and x2R , then axbx ¼ (ab)x.
� If a> 0 and x, y2R , then axay¼ axþ y.
� If a> 0 and x, y2R , then axð Þy ¼ axy.
You first met these laws inSub-section 1.5.3, but therex, y2Q .
4.4 Defining exponential functions 163
For exampl e, to prove the final Exponent Law, notice that, from the defini tion
of ax, we have loge( ax) ¼ x log e a; so that
axð Þy ¼ e y log e a xð Þ
¼ e xy loge a ¼ axy :
Thus manipul ations such as
ffiffiffi
2p ffiffi
2p� �
ffiffi
2p
¼ffiffiffi
2p ffiffi
2p�ffiffi
2pð Þ ¼
ffiffiffi
2p� �2
¼ 2
are inde ed justi fied.
Problem 2 Prove that, if a > 0 and x, y 2R , then axay ¼ a x þ y.
Finally, our defini tion of a x enables us to prove the rule for rearra nging
inequal ities by taking powers.
Rule 5 For any non-ne gative a, b 2R , and any p > 0, a < b , ap < bp.
If a ¼ 0, the result is obvio us. In general , sinc e the functi ons x 7! loge x and
x 7! ex are strictly increas ing, we have
a < b , loge a < log e b
, p loge a < p loge b since p > 0ð Þ
, e p log e a < e p log e b
, ap < bp :
The followi ng exampl e illus trates the use of Rule 5.
Exampl e 1 Determine which of the numbers e p and pe is g reater.
Solut ion We use the inequal ity
e x > 1 þ x ; for x > 0 :
Applying this ineq uality with x ¼ pe
� �
� 1, we obta in
epeð Þ� 1 > 1 þ p
e
� �
� 1� �
¼ pe¼ pe � 1 ;
then, by multipl ying thr ough by the positive factor e , we obtain that epe > p.
Finally, by applying Rule 5 with p¼ e, we obtain ep> pe. &
Problem 3 Prove that ex> xe, for x> e.
4.5 Exercises
Section 4.1
1. Use the appropriate rules, together with the list of basic continuous
functions, to prove that the following functions are continuous:
You met this Rule in Sub-section 1.2.1.
This was one of theExponential Inequalities:part (a) of Theorem 4 inSub-section 4.1.3.
164 4: Continuity
(a) f xð Þ ¼ exp sin x2 þ 1ð Þð Þ; x 2 R ;
(b) f xð Þ ¼ effiffi
xpþ x5; x 2 0;1½ Þ:
2. Determine whether the following functions are continuous at 0:
(a) f xð Þ ¼ sin x sinð1xÞ; x 6¼ 0,
0; x ¼ 0;
�
(b) f xð Þ ¼1xsin 1
x
� �
; x 6¼ 0,
0; x ¼ 0:
�
3. Prove that each of the following sequences is convergent, and determine its
limit:
(a) sin e1n � 1
� �n o
; (b) cos 12n
� �� �12
n o
:
4. Let f be defined on an open interval I, and c2 I. Prove that, if f is continuous
at c and f(c) 6¼ 0, then there is an open interval J� I such that c2 J and
f(x) 6¼ 0, for any x2 J.
5. Determine the points where the function f xð Þ¼ 1; x ¼ 0; 1;xþ 2x½ �; 0 < x < 1;
�
is
(a) continuous on the left; (b) continuous on the right;
(c) continuous.
6. Prove that the following function is continuous on R
f xð Þ ¼�1; x � 1
2p;
sin x; � 12p < x < 1
2p;
1; x � 12p:
8
<
:
7. Determine at which points the following function is continuous
f xð Þ ¼ x; �1 x < 0;ex; 0 x 1:
�
8. Write down examples of functions with the following properties:
(a) f and g are discontinuous on R , but g� f is continuous on R ;
(b) f and g are discontinuous on R , but fþ g and fg are continuous on R ;
(c) f is continuous on R � 1; 12; 1
3; 1
4; . . .
� �
but discontinuous at
1; 12; 1
3; 1
4; . . .
� �
:
Section 4.2
1. The function f is continuous on (0, 1), and takes every real value at most
once. Use the Intermediate Value Theorem to prove that f is strictly mono-
tonic on (0, 1).
2. Give examples of functions f continuous on the half-open interval [0, 1) in
R , to show that f ([0,1)) can be open, closed or half-open.
3. Prove that each of the following polynomials has the stated number of (real)
zeros:
(a) p xð Þ ¼ x4 � 4x3 þ 3x2 þ 2x� 1; 4 zeros;
(b) p xð Þ ¼ 3x3 � 8x2 þ xþ 3; 3 zeros:
4. Prove that the function f xð Þ ¼ x� sin x� 23p; x 2 R , has a zero in
23p; 5
6p
� �
.
We shall use this resultin Chapter 6.
4.5 Exercises 165
5. Using the Zeros Localisation Theorem and the Extreme Values Theorem,
prove that every polynomial of even degree n
p xð Þ ¼ anxn þ an�1xn�1 þ � � � þ a1xþ a0; x 2 R ;
where an 6¼ 0, has a minimum value on R .
6. Write down an example of a function f(x), x2 (0, 1), that is continuous on
(0, 1) and for which f ((0, 1))¼ (0, 1), but such that there is no point c in
(0, 1) for which f (c)¼ c.
Section 4.3
1. For each of the following functions, prove that it has a continuous inverse
function and determine the domain of that inverse function:
(a) f xð Þ ¼ x3 þ 1� 1x2 ; x 2 0;1ð Þ;
(b) f xð Þ ¼ 1
1þx3ð Þ2 ; x 2 �1;1ð Þ:2. Determine whether each of the following statements is true:
(a) sin sin�1 x� �
¼ x, for x 2 ½�1; 1�;(b) sin�1 sin xð Þ ¼ x, for x 2 R .
3. (a) Prove that tan�1 xþ tan�1 y ¼ tan�1 xþy1�xy
� �
, provided that
tan�1 xþ tan�1 y lies in � 12p; 1
2p
� �
.
(b) Use the result in part (a) to evaluate tan�1 12
� �
þ tan�1 13
� �
.
This result is closely related toProblem 2 in Sub-section 4.2.1, so you mightlike to look back at thatproblem and compare the two.
This is known as the AdditionFormula for tan�1.
166 4: Continuity
5 Limits and continuity
In Chapter 4 we made some progress in pinning down the idea of ‘a well-
behaved function’ in precise terms. Using our earlier work on sequences we
defined what was meant by a function being continuous: roughly speaking, its
graph has no jumps or gaps.
However, a number of functions arise quite naturally in mathematics where
we need to handle functions that are already defined near some particular point,
but that are not defined at the point itself.
For example, the function
f xð Þ ¼ sin x
x; x 6¼ 0;
which arises when we examine the question of whether the sine function is
differentiable. Clearly from their graphs, the behaviour of f differs significantly
from the behaviour of the function
g xð Þ ¼ sin1
x
� �
; x 6¼ 0;
that you have already met. It is possible to assign a value (namely, 1) to f at 0 so
that this extension of the domain of f makes f continuous on a domain that is an
interval. On the other hand, it is not possible to assign a value to g at 0 so that this
extension of the domain of g makes g continuous on a domain that is an interval.
In Section 5.1, we discuss limits of functions, and show that the existence of
this helpful value 1 for f can be stated as limx!0
sin xx¼ 1. The concept of a limit of a
function is closely related to that of a continuous function, and many of the
rules for calculating limits are similar to those for continuity. We also discuss
one-sided limits.
In Section 5.2, we discuss the behaviour of functions near asymptotes of
their graphs. In particular, we prove that, if n2Z, then limx!1
xne�x ¼ 0:
Next, in Section 5.3, we introduce a slightly different definition of the limit
limx!c
f xð Þ of a function f at a point c. Instead of using sequences tending to c to
define the limit of f at c, we define limx!c
f xð Þ directly in terms of inequalities
involving x and f (x), and we verify that this new definition is completely
equivalent to the earlier definition. We also illustrate the changes to proofs of
results about limits that the new definition involves.
In Section 5.4, we introduce a definition for the continuity of a function f at a
point c in terms of inequalities involving x and f (x), rather than in terms of the
behaviour of f on sequences that tend to c. We verify that this new definition is
completely equivalent to the earlier definition of continuity, and illustrate the
changes to proofs of results about continuity that the new definition involves.
Finally, in Section 5.5, we introduce a concept that will be extremely
powerful in your further study of Analysis; namely, that of uniform continuity.
1
–1
y = sin 1x
At first sight this newdefinition will seem morecomplicated. Howeverthroughout the rest of thebook we shall see just howpowerful this new definitionturns out to be!
167
Before starting on Sections 5.3 and 5.4, you may find it useful to quickly
revise the material in Sections 1.2 and 1.3 on inequalities.
5.1 Limits of functions
5.1.1 What is a limit of a function?
The graph of the function
f xð Þ ¼ sin x
x; x 6¼ 0;
shows that, if x takes values which are ‘close to’ but distinct from 0, then f(x)
takes values which are ‘close to’ 1. The closer that x gets to 0, the closer f(x)
gets to 1. We now pin down this idea precisely.
First, we need the following fact.
Theorem 1 If {xn} is a null sequence whose terms are non-zero, then
sin xn
xn
!1 as n!1:
Proof First we deduce from the inequality
sin x � x; for 0 < x <p2;
that
sin x
x� 1; for 0 < x <
p2:
Next, we need to use the formula for the area of a sector of a disc of radius 1
in Figure (a), below. Compare the area of this sector with the area of the
triangle in Figure (b), below.
We find that
x � tan x; for 0 < x <p2:
Since tan x ¼ sin xcos x
and cos x > 0 for 0 < x < p2, we obtain
cos x � sin x
x; for 0 < x <
p2:
Combining this inequality with our earlier upper estimate for sin xx
, we find that
cos x � sin x
x� 1; for 0 < x <
p2:
‘may’ means ‘will’!
That is, limn!1
sin xn
xn¼ 1.
Lemma 1, Sub-section 4.1.3.
We discussed the area andperimeter of a disc of radius 1in Sub-section 2.5.4 and inExercise 4 on Section 2.5 inSection 2.6.
168 5: Limits and continuity
In fact, this inequality holds for 0 < jxj < p2, since
cos �xð Þ ¼ cos x andsin �xð Þ�xð Þ ¼
sin x
x:
Now, if {xn} is any null sequence with non-zero terms, then the terms xn
must eventually satisfy the inequality xnj j < p2, and so there is some number X
such that
cos xn �sin xn
xn
� 1; for n > X:
But cos xn! 1 as n!1, since the cosine function is continuous at 0 and
cos 0¼ 1. Hence, by the Squeeze Rule for sequences
sin xn
xn
! 1: &
This behaviour of sin xcos x
near 0 is an example of a function f tending to a limit as x
tends to a point c.
To define this concept, we need to ensure that the function f is defined near
the point c, but not necessarily at the point c itself. We first introduce the idea
of a punctured neighbourhood of c.
Definitions A neighbourhood of a point c of R is an open interval that
contains the point c, and a punctured neighbourhood of a point c of R is a
neighbourhood of c from which the point c itself has been deleted.
For example, the sets (0, 9), (1,1) and R are neighbourhoods of the point 2,
and (1, 2)[ (2, 5) and (�1, 2)[ (2, 4) are punctured neighbourhoods of the
point 2. In general, a neighbourhood of c is an interval of the form (c� r, cþ s),
for some r, s> 0, and a punctured neighbourhood of c is the union of a pair of
intervals (c� r, c)[ (c, cþ s), for some r, s> 0. In practice, we often choose as
a neighbourhood of c an open interval (c� r, cþ r), r > 0, with centre at c; and
as a punctured neighbourhood of c the union (c� r, c)[ (c, cþ r) of two open
intervals of equal length.
We now define the limit of a function in terms of limits of sequences.
Definition Let the function f be defined on a punctured neighbourhood N
of a point c. Then f (x) tends to the limit ł as x tends to c if:
for each sequence xnf g in N such that xn ! c; then f xnð Þ ! ‘: (1)
We write this as either ‘limx!c
f xð Þ ¼ ‘’ or ‘f (x)! ‘ as x! c’.
Let us check that this definition holds for f ðxÞ ¼ sin xx
at 0. This function f is
defined on the domain R � {0}, and so in particular on every punctured
neighbourhood of 0. We have just seen that the statement (1) holds; it follows
then that
limx!0
sin x
x¼ 1:
Since the definition of the limit of a function involves the limit of sequences,
we can use our various Combination Rules for sequences to determine the
limits of many functions.
Take " ¼ p2
in the definition ofnull sequence.
So the ‘puncture’ is at c.
These choices are simplymatters of convenience!
Note that xn 6¼ c, for any n.
For example, f is defined on(�1, 0)[ (0, 1).
5.1 Limits of functions 169
Example 1 Prove that each of the following functions tends to a limit as x
tends to 2, and determine these limits:
(a) f xð Þ¼ x2 � 4x� 2
; (b) f xð Þ¼ x3� 3x� 2x2� 3xþ 2
:
Solution
(a) First, notice that f is defined on every punctured neighbourhood of 2.
Next, notice that
f xð Þ ¼ x2 � 4
x� 2¼ xþ 2; for x 6¼ 2:
Thus, if {xn} is any sequence that lies in some punctured neighbourhood
N of 2 and xn! 2, then
f xnð Þ ¼ xn þ 2! 4 as n!1;
by the Sum Rule for sequences. It follows that
limx!2
x2 � 4
x� 2¼ 4:
(b) The function
f xð Þ ¼ x3 � 3x� 2
x2 � 3xþ 2¼ x� 2ð Þ x2 þ 2xþ 1ð Þ
x� 2ð Þ x� 1ð Þ
has domain R � {1, 2}, and so f is defined on the punctured neighbourhood
N¼ (1, 2)[ (2, 3) of 2.
Next, notice that
f xð Þ ¼ x3 � 3x� 2
x2 � 3xþ 2¼ x2 þ 2xþ 1
x� 1;
for x 2 N ¼ ð1; 2Þ [ ð2; 3Þ:Thus, if {xn} is any sequence that lies in N¼ (1, 2)[ (2, 3) and xn! 2, then
f xnð Þ ¼x2
n þ 2xn þ 1
xn � 1! 4þ 4þ 1
2� 1¼ 9 as n!1;
by the Combination Rules for sequences. It follows that
limx!2
x3 � 3x� 2
x2 � 3xþ 2¼ 9: &
Later in this section we give several further techniques for calculating limits.
Our next example illustrates how to prove that a limit does not exist.
Example 2 Prove that each of the following functions does not tend to a limit
as x tends to 0:
(a) f ðxÞ ¼ 1x; x 6¼ 0; (b) f ðxÞ ¼ sin 1
x
� �
; x 6¼ 0; (c) f xð Þ ¼ffiffiffi
xp; x� 0:
Solution
(a) The function f is defined on the punctured neighbourhood N¼ (�2, 0)[ (0, 2)
of 0. The null sequence 1n
� �
lies in N and tends to 0, but f 1n
� �
¼ n!1.
Hence f does not tend to a limit as x tends to 0.
(b) The function f is defined on the punctured neighbourhood N¼ (�2, 0)[ (0, 2)
of 0.
We can cancel x� 2, sincex 6¼ 2.
For example,N¼ (1, 2)[ (2, 3); any otherpunctured neighbourhood of2 would serve our purposeequally well.
Of course any smallerpunctured neighbourhood of 2would serve our purposeequally well.
Sub-sections 5.1.2 and 5.1.3.
Any other puncturedneighbourhood of 0 wouldserve equally well here, ofcourse.
170 5: Limits and continuity
To prove that f(x) does not tend to a limit as x tends to 0, we choose two null
sequences {xn} and {x0n} that lie in N, such that
f xnð Þ ! 1 and f x0n� �
! �1:
Since sin 2npþ 12p
� �
¼ 1 and sin 2npþ 32p
� �
¼ �1 for n2Z, we can choose
xn ¼1
2npþ 12p
and x0n ¼1
2npþ 32p; for n ¼ 0; 1; 2; . . .:
Hence f does not tend to a limit as x tends to 0.
(c) The function f xð Þ¼ffiffiffi
xp
has domain [0,1), and so f is not defined on any
punctured neighbourhood of zero. Hence f does not tend to a limit as x
tends to zero. &
We collect these techniques together in the form of a strategy.
Strategy To show that limx!c
f xð Þ does not exist:
1. Show that there is no punctured neighbourhood N of c on which f is
defined;
OR
2. Find two sequences {xn} and {x0n} (in some punctured neighbourhood N
of c) which tend to c, such that { f (xn)} and { f (x0n)} have different limits;
OR
3. Find a sequence {xn} (in some punctured neighbourhood N of c) which
tends to c, such that f (xn)!1 or f (xn)!�1.
Problem 1 Determine whether the following limits exist:
(a) limx!0
x2þxx
; (b) limx!0
xj jx:
5.1.2 Limits and continuity
Consider the function
f xð Þ ¼ 1; x 6¼ 0;0; x ¼ 0:
Does this function tend to a limit as x tends to zero; and, if it does, what is the
limit?
Certainly, f is defined on any punctured neighbourhood of zero, since the
domain of f is R . Also, if {xn} is any null sequence with non-zero terms, then
f (xn)¼ 1, for n¼ 1, 2, . . ., so that f (xn)! 1 as n!1. It follows that limx!0
f xð Þ ¼ 1.
This example serves to emphasise that the value of a limit limx!c
f xð Þ has nothing
to do with the value of f (c) – even if f happens to be defined at the point c.
However, if f is defined at c, and f is also continuous at c, then the only
possible value for limx!c
f xð Þ is f (c).
We put these observations together in the following result.
Theorem 2 Let the function f be defined on an open interval I, with c 2 I.
Then
f is continuous at c, limx!c
f xð Þ ¼ f cð Þ:
In particular, the terms xn and
x 0n will be non-zero.
Notice that I is aneighbourhood of c.
5.1 Limits of functions 171
Using Theorem 2 and our knowledge of continuous functions, we can eval-
uate many limits rather easily.
For example, to determine limx!2
3x5 � 5x2 þ 1� �
notice that the function
f (x)¼ 3x5� 5x2þ 1 is defined on R and is continuous at 2, since f is a poly-
nomial. Hence, by Theorem 2
limx!2
3x5 � 5x2 þ 1� �
¼ f 2ð Þ¼ 77:
We saw earlier that the following functions are continuous at 0
f xð Þ ¼ x sin 1x
� �
; x 6¼ 0;0; x ¼ 0;
and f xð Þ ¼ x2 sin 1x
� �
; x 6¼ 0;0; x ¼ 0:
It therefore follows from Theorem 2 that
limx!0
x sin1
x
� �
¼ 0 and limx!0
x2 sin1
x
� �
¼ 0:
On the other hand, we saw in part (b) of Example 2 that
limx!0
sin1
x
� �
¼ 0 does not exist:
It follows from Theorem 2 that, no matter how we try to extend the domain
of f ðxÞ ¼ sin 1x
� �
to include x¼ 0, we can never obtain a continuous function.
Remark
If f is defined on an open interval I, with c 2 I, and limx!c
f xð Þ exists but
limx!c
f xð Þ 6¼ f cð Þ, then f is said to have a removable discontinuity at c. For
example, the function f xð Þ ¼ x sin 1x
� �
; x 6¼ 0;3; x ¼ 0;
has a removable discontinuity
at 0.
Problem 2 Use Theorem 1 to determine the following limits:
(a) limx!2
ffiffiffi
xp
; (b) limx!p
2
ffiffiffiffiffiffiffiffiffi
sin xp
; (c) limx!1
ex
1þx
In the remainder of this chapter we shall frequently use Theorem 2. When the
function f is one of our basic continuous functions, however, we shall not
always refer to the theorem explicitly. For example, f (x)¼ x2þ 1 and g(x)¼sin x are basic continuous functions, and so we can write lim
x!2x2 þ 1ð Þ¼ 5 and
limx!0
sin x ¼ 0 without further explanation.
5.1.3 Rules for limits
As you might expect from your experience with sequences, series and con-
tinuous functions, we often find limits by using various rules. First, we state the
Combination Rules.
Theorem 3 Combination Rules
If limx!c
f xð Þ¼ ‘ and limx!c
g xð Þ¼ m, then:
Sum Rule limx!c
f xð Þ þ g xð Þð Þ ¼ ‘þ m;
Multiple Rule limx!c
lf xð Þ¼l‘; for l 2 R ;
Polynomials are basiccontinuous functions: seeSub-section 4.1.3.
Problem 8 and Example 5,Sub-section 4.1.2,respectively.
Sub-section 5.1.1.
This means that, if weredefine the value of f just at citself, the resulting function isthen continuous at c.
Basic continuous functions:
� polynomials andrational functions;
� modulus function;� nth root function;� trigonometric functions
(sine, cosine and tangent);� the exponential function.
172 5: Limits and continuity
Product Rule limx!c
f xð Þg xð Þ ¼ ‘m;
Quotient Rule limx!c
f xð Þg xð Þ ¼ ‘
m; provided that m 6¼ 0.
For example, since limx!0
sin xx¼ 1 and lim
x!0x2 þ 1ð Þ ¼ 1; we have
limx!0
sin x
xþ 2 x2 þ 1� �
� �
¼ 1þ 2� 1 ¼ 3:
The proofs of these rules are all simple consequences of the corresponding
results for sequences. We illustrate this by proving just one rule.
Proof of the Sum Rule Let the functions f and g be defined on a punctured
neighbourhood N of a point c. We want to show that:
for each sequence xnf g in N such that xn ! c, then f xnð Þ þ g xnð Þ ! ‘þ m:
Now, since limx!c
f xð Þ ¼ ‘, we know that, for any sequence {xn} in N for which
xn! c, then f (xn)! ‘. Also, since limx!c
g xð Þ¼m, we know that, as xn! c, then
g(xn)!m.
It follows by the Sum Rule for sequences that, as xn! c, then
f (xn)þ g(xn)! ‘þm. This completes the proof. &
We can also use the following Composition Rule.
Theorem 4 Composition Rule
If limx!c
f xð Þ¼ ‘ and limx!‘
g xð Þ ¼ L, then limx!c
g f xð Þð Þ ¼ L, provided that:
EITHER f (x) 6¼ ‘ in some punctured neighbourhood of c;
OR g is defined at ‘ and is continuous at ‘.
Remarks on the Composition Rule
(a) In any particular case, the Composition Rule may be FALSE if we do not
ensure that one or other of the two provisos holds! For example, if
f xð Þ ¼ 0; x 2 R ; and g xð Þ ¼sin x
x; x 6¼ 0;
0; x ¼ 0;
then
limx!0
f xð Þ ¼ 0 and limx!0
g xð Þ ¼ 1; but limx!0
g f xð Þð Þ ¼ limx!0
0ð Þ ¼ 0:
(b) Suppose the first proviso ‘f (x) 6¼ ‘ in some punctured neighbourhood of c’
in Theorem 4 holds. Then we know that, for each sequence {xn} in a
punctured neighbourhood N of c for which xn! c, f (xn) does not take the
value ‘ but must lie in a punctured neighbourhood of ‘. In this case, the
desired result will follow from the facts that (i) the sequence {f (xn)}
converges to ‘, (ii) f (xn) lies in a punctured neighbourhood of ‘, and
(iii) limx!‘
g xð Þ ¼ L:
(c) Suppose the second proviso ‘g is defined at ‘ and is continuous at ‘’in Theorem 4 holds. In this case, the desired result will follow from
There are no new ideas in theproof. All that we have to do isto set up things so that we canuse the Sum Rule forsequences.
Note that the second limit is alimit as x! ‘, not as x! c.
Here we take c¼ 0, ‘¼ 0 andL¼ 1.
So, in this case,limx!0
g f xð Þð Þ 6¼ L:
We omit the details of theproof in this case, which arenow straight-forward to writedown.
5.1 Limits of functions 173
the facts that (i) the sequence {f (xn)} converges to ‘, and (ii) g is contin-
uous at ‘.The following example illustrates the use of the Composition Rule.
Example 3 Determine the following limits:
(a) limx!0
sin sin xð Þsin x
; (b) limx!0
1þ sin xx
� �2 �
:
Solution
(a) Let f (x)¼ sin x, x2R , and gðxÞ ¼ sin xx
, x 6¼ 0: Then
limx!0
f xð Þ ¼ limx!0
sin x ¼ 0 and limx!0
g xð Þ ¼ limx!0
sin x
x¼ 1:
Also, f (x)¼ sin x 6¼ 0 in the punctured neighbourhood �p; 0ð Þ [ 0; pð Þ of 0
(for example). It follows, by the Composition Rule, that
limx!0
g f xð Þð Þ ¼ limx!0
sin sin xð Þsin x
¼ 1:
(b) Let f ðxÞ ¼ sin xx
, x 6¼ 0, and g xð Þ ¼ 1þ x2, x 2 R : Then
limx!0
sin x
x¼1 and lim
x!11þ x2� �
¼2:
Also, g is defined and continuous at 1. It follows, by the Composition Rule,
that
limx!0
g f xð Þð Þ ¼ limx!0
1þ sin x
x
� �2 !
¼ 2: &
Problem 3 Use the Combination Rules and the Composition Rule to
determine the following limits:
(a) limx!0
sin x2xþx2 ; (b) lim
x!0
sin x2ð Þx2 ; (c) lim
x!0
xsin x
� �12:
Problem 4 For the functions
f xð Þ ¼0; x ¼ 0;�2 x ¼ 1;2þ x; x 6¼ 0; 1;
8
<
:
and g xð Þ ¼ 0; x ¼ 0;1þ x; x 6¼ 0;
determine f � gð Þ xð Þ; f � gð Þ 0ð Þ; f
limx!0
g xð Þ�
and limx!0
f g xð Þð Þ:
There is also a Squeeze Rule for limits.
Theorem 5 Squeeze Rule
Let the functions f, g and h be defined on a punctured neighbourhood N of a
point c. If:
1. g xð Þ � f xð Þ � h xð Þ; for x 2 N; and
2. limx!c
g xð Þ¼ limx!c
h xð Þ ¼ ‘;then lim
x!cf xð Þ¼‘:
Problem 5 Use the Squeeze Rule for limits to prove that:
(a) limx!0
x2 sin 1x
� �
¼ 0; (b) limx!0
x cos 1x
� �
¼ 0:
Here we have c¼ 0, ‘¼ 0 andL¼ 1.
Here we have c¼ 0, ‘¼ 1 andL¼ 2.
The proof of Theorem 5 is astraight-forward applicationof the Squeeze Rule forsequences.
y
cN
x
y = h(x)y = f (x)y = g(x)
174 5: Limits and continuity
The Limit Inequality Rule is another result for limits of functions that is an
analogue of the corresponding result for sequences.
Theorem 6 Limit Inequality Rule
If limx!c
f xð Þ ¼ ‘ and limx!c
g xð Þ ¼ m, and also
f xð Þ � g xð Þ, on some punctured neighbourhood of c,
then
‘ � m:
5.1.4 One-sided limits
In Example 2(c) above, we saw that limx!0
ffiffiffi
xp
does not exist, because the
function f xð Þ ¼ffiffiffi
xp
, x � 0, is not defined on any punctured neighbourhood
of zero. Nevertheless, if {xn} is any sequence in the domain [0, 1) of f for
which xn! 0, thenffiffiffiffiffi
xnp ! 0. Thus f xð Þ ¼
ffiffiffi
xp
tends to f (0)¼ 0 as x tends to 0
from the right.
Definition Let f be defined on (c, cþ r), for some r > 0. Then f(x) tendsto the limit ł as x tends to c from the right if:
for each sequence {xn} in (c, cþ r) such that xn! c, then f (xn)! ‘.
We write this either as ‘ limx!cþ
f xð Þ ¼ ‘’ or as ‘f (x)! ‘ as x! cþ ’.
There is a corresponding definition for limits as x tends to c from the left.
Definition Let f be defined on (c� r, c), for some r > 0. Then f(x) tends
to the limit ł as x tends to c from the left if:
for each sequence {xn} in (c� r, c) such that xn! c, then f(xn)! ‘.
We write this either as ‘ limx!c�
f xð Þ ¼ ‘’ or as ‘f xð Þ ! ‘ as x! c�’.
Sometimes both left and right limits exist. Even when this happens, the two
values need not be equal, as the following example shows.
Example 4 Prove that the function f xð Þ ¼ xj jx
, x 6¼ 0, tends to different limits
as x tends to 0 from the right and from the left.
Solution The function f is defined on (0, 1), and f (x)¼ 1 on this interval.
Thus, if {xn} is a null sequence in (0, 1), then
limn!1
f xnð Þ ¼ limn!1
1ð Þ ¼ 1:
So limx!0þ
f xð Þ¼1:
Similarly, f is defined on (�1, 0), and f(x)¼�1 on this interval. Thus, if {xn}
is a null sequence in (�1, 0), then
limn!1
f xnð Þ¼ limn!1
�1ð Þ¼�1:
So limx!0�
f xð Þ¼�1: &
Theorem 3, Sub-section 2.3.3.
Note that f need not be definedat the point c.
For example, limx!0þ
ffiffiffi
xp¼ 0:
Again, f need not be defined atthe point c.
5.1 Limits of functions 175
Problem 6 Write down a function f defined on the interval [�1, 3] for
which limx!0�
f xð Þ does not exist but limx!0þ
f xð Þ ¼ 1. Verify that f has the
specified properties.
The relationship between one-sided limits and ‘ordinary’ limits is given by
the following result.
Theorem 7 Let f be defined on a punctured neighbourhood of the point c.
Thenlimx!c
f xð Þ ¼ ‘
, limx!cþ
f xð Þ and
limx!c�
f xð Þ both exist and equal ‘:
Remark
If f is defined on an open interval I that contains the point c, and
limx!cþ
f xð Þ and limx!c�
f xð Þ both exist; but limx!cþ
f xð Þ 6¼ limx!c�
f xð Þ;then f is said to have a jump discontinuity at c.
Analogues of the Combination Rules, Composition Rule and Squeeze Rule can
also be used to determine one-sided limits. In the statements of all these rules,
we simply replace limx!c
by limx!cþ
or limx!c�
, and replace the open interval I contain-
ing c by (c, cþ r) or (c� r, c), as appropriate.
Problem 7 Prove that:
(a) limx!0þ
sin xxþ
ffiffiffi
xp� �
¼ 1; (b) limx!0þ
sinffiffi
xpð Þffiffi
xp ¼ 1:
Problem 8 Use the inequalities 1þ x � ex � 11�x
; for jxj< 1, to
prove that
limx!0þ
ex�1
x¼ lim
x!0�
ex�1
x¼1:
Deduce that
limx!0
ex�1
x¼1:
5.2 Asymptotic behaviour of functions
In this section we define formally a number of statements that you will have
already met in your study of Calculus, such as
1
x!1 as x!0þ and ex!1 as x!1;
and describe the relationship between them.
5.2.1 Functions which tend to infinity
Just as we defined the statement limx!c
f xð Þ ¼ ‘ in terms of the convergence of
sequences, so we can define the statement f (x)!1 as x! c.
We omit a proof of this result.
We verified these inequalitiesas part (c) of Theorem 4 (‘TheExponential Inequalities’) inSub-section 4.1.3.
176 5: Limits and continuity
Definition Let f be defined on a punctured neighbourhood N of the point c.
Then f (x)!1 as x! c if:
for each sequence {xn} in N such that xn! c, then f (xn)!1.
The statement ‘f (x)!�1 as x! c’ is defined similarly, with 1 replaced
by �1.
As with sequences, there is a version of the Reciprocal Rule which re-
lates the behaviour of functions which tend to infinity and functions which
tend to 0.
Theorem 1 Reciprocal Rule
(a) If the function f satisfies the following two conditions:
1. f (x)> 0 for all x in some punctured neighbourhood of c, and
2. f (x)! 0 as x! c,
then 1f xð Þ ! 1 as x! c.
(b) If f (x)!1 as x! c, then 1f xð Þ ! 0 as x! c.
For example, 1x2 !1 as x! 0, since f (x)¼ x2> 0, for x2R � {0}, and
limx!0
x2¼0.
The statements
‘f xð Þ ! 1 or�1 as x! cþ’ and ‘f xð Þ ! 1 or�1 as x! c�’
are defined similarly, with the punctured neighbourhood of c being replaced by
open intervals (c, cþ r) or (c� r, c), as appropriate, for some r> 0.
The Reciprocal Rule can also be applied with ‘x! c’ replaced by ‘x! cþ’
or ‘x! c�’, and with the punctured neighbourhood of c being replaced by open
intervals (c, cþ r) or (c� r, c), as appropriate, for some r > 0.
For example, 1x!1 as x! 0þ, since f(x)¼ x> 0, for x 2 0;1ð Þ, and
limx!0þ
x ¼ 0:
Problem 1 Prove the following:
(a) 1jxj ! 1 as x! 0; (b) 1
1� x3 ! �1 as x! 1þ;
(c) sin xx3 !1 as x! 0:
Remark
There are also versions of the Combination Rules and Squeeze Rule for
functions which tend to 1 or �1 as x tends to c, cþ or c�; here we state
only the Combination Rules for functions which tend to1 as x tends to c.
Theorem 2 Combination Rules
If f(x)!1 and g(x)!1 as x! c, then:
Sum Rule f(x)þ g(x)!1;
Multiple Rule l f(x)!1, for l> 0;
Product Rule f(x) g(x)!1.
Theorem 2, Sub-section 2.4.3.
These are similar to resultsstated for sequences inTheorems 3 and 4,Sub-section 2.4.3.
5.2 Asymptotic behaviour of functions 177
Problem 2 Prove that the following statements are false:
(a) If f(x)!1 and g(x)!1 as x! 0, then f(x)� g(x)!1;
(b) If f(x)!1 and g(x)! 0 as x! 0, then f(x) g(x)! 0 or
f(x) g(x)!1.
5.2.2 Behaviour of f(x) as x tends to1 or �1
Finally, we define various types of behaviour of real functions f (x) as x tends to
1 or �1. To avoid repetition, here we allow ‘ to denote either a real number
or one of the symbols1 and �1.
Definition Let f be defined on an interval (R,1), for some real number R.
Then f (x)! ‘ as x!1 if:
for each sequence {xn} in (R,1) such that xn!1, then f (xn)! ‘.
The statement ‘f (x)! ‘ as x!�1’ is defined similarly, with1 replaced
by�1, and (R,1) by (�1, R). In practice, we usually prove that f (x)! ‘ as
x!�1 by showing that f (�x)! ‘ as x!1.
When ‘ is a real number, we also use the notation
limx!1
f xð Þ ¼ ‘ and limx!�1
f xð Þ ¼ ‘:
Once again, we can use versions of the Combination Rules and Reciprocal
Rule to obtain statements about the behaviour of given functions as x tends to1or �1. In the statements of these rules, we need only replace c by1 or �1,
and the punctured neighbourhood of c by (R,1) or (�1, R), as appropriate.
For example, for any positive integer n
xn !1 as x!1 and limx!1
x�n ¼ 0:
More generally, we have the following result for the behaviour of poly-
nomials as x!1.
Theorem 3 If a0, a1, . . ., an�1 are real numbers and
p xð Þ ¼ xn þ an�1xn�1 þ � � � þ a1xþ a0; x 2 R ;
then
p xð Þ ! 1 as x!1 and1
p xð Þ ! 0 as x!1:
There are also versions of the Squeeze Rule for functions as x tends to1,
which have some important applications.
Theorem 4 Squeeze Rule
Let the functions f, g and h be defined on some interval (R,1).
(a) If f, g and h satisfy the following two conditions:
1. g(x)� f(x)� h(x), for all x in (R,1), and
2. limx!1
g xð Þ ¼ limx!1
h xð Þ ¼ ‘;then lim
x!1f xð Þ ¼ ‘:
We adopt this convention forsimplicity ONLY in thissub-section!
We ask you to prove the firstpart of this result inSection 5.6. The second partthen follows at once.
We omit the proof of thistheorem, which is straight-forward.
178 5: Limits and continuity
(b) If f and g satisfy the following two conditions:
1. f(x)� g(x), for all x in (R,1), and
2. g(x)!1 as x!1,
then f (x)!1 as x!1.
An important application of this Squeeze Rule is to show that ex tends to1faster than any power of x, as x tends to1. We prove this in the next example.
Example 1 Prove that for each n¼ 0, 1, 2, . . .,ex
xn !1 as x!1.
Solution We use the power series expansion
ex ¼ 1þ xþ x2
2!þ � � � þ xn
n!þ xnþ1
nþ 1ð Þ!þ � � �
for x� 0. Since x� 0, all the terms on the right here are non-negative, and so
ex � xnþ1
nþ 1ð Þ! ; for x � 0:
It follows that
ex
xn� x
nþ 1ð Þ! ; for x > 0:
Since xðnþ1Þ!!1 as x!1, it follows from part (b) of the Squeeze Rule that
ex
xn !1 as x!1. &
Remark
We deduce from Example 1 and the Product Rule that, for any integer n,
xnex!1 as x!1. Thus, by the Reciprocal Rule, for any integer n,
xne�x! 0 as x!1.
Problem 3 Determine the behaviour of the following functions as
x!1:
(a) f xð Þ ¼ 2x3þxx3 ; (b) f xð Þ ¼ sin x
x.
In our next example, we describe the behaviour of loge x as x!1.
Example 2 Prove that loge x!1 as x!1.
Solution We prove this from first principles.
First, note that the function f (x)¼ loge x is defined on (0,1).
Next, we have to prove that:
for each sequence {xn} in (0,1) such that xn!1, then loge xn!1.
To prove that loge xn!1, we need to show that:
for each positive number K, there is a number X such that
loge xn > K; for all n > X: (1)
However, since we know that xn!1, we can choose X such that
xn > eK ; for all n > X:
Since the function loge is strictly increasing, the statement (1) is therefore
true. It follows that loge x!1 as x!1, as required. &
y y = h(x)
y = f (x)l
y = g(x)
R
yy = f (x)
y = g(x)
(R, ∞) x
R (R, ∞) x
See Section 3.4.
There is no easy way to provethis result by using theSqueeze Rule, since loge xtends to1 ‘ratherreluctantly’; that is, moreslowly than any function thatwe have considered so far.
5.2 Asymptotic behaviour of functions 179
5.2.3 Composing asymptotic behaviour
Earlier we gave a Composition Rule for limits. We now describe a more
general Composition Rule which permits the composition of the different
types of asymptotic behaviour that we have now met.
For example, since
f ðxÞ ¼ 1
x!1 as x! 0þ and gðxÞ ¼ ex !1 as x!1;
we should expect that
g f xð Þð Þ ¼ e1x !1 as x! 0þ:
This result is true, as the following Composition Rule shows. To avoid
repetition, we again allow ‘ and L to denote either a real number or one of
the symbols1 and �1.
Theorem 5 Composition Rule
If:
1. f (x)! ‘ as x! c (or cþ, c�,1 or �1), and
2. g (x)! L as x! ‘,
then
g f xð Þð Þ ! L as x! c ðor cþ; c�;1 or �1; respectivelyÞ;provided that:
EITHER f (x) 6¼ ‘ in some punctured neighbourhood of c (or in (c, cþ r),
(c� r, c), (R,1) or (�1, R), respectively, for some r> 0)
OR ‘ is finite, and g is defined at ‘ and is continuous at ‘.
Example 3 Prove that
(a) ex2
x!1 as x!1; (b) xe
1x !1 as x! 0þ;
(c) x sin 1x
� �
! 1 as x!1:
Solution
(a) Let f xð Þ ¼ x2
, x 2 R , and g xð Þ ¼ ex
x, x 2 R � 0f g, so that
g f xð Þð Þ ¼ ex2
x2
¼ 2ex2
x, for x 2 R � 0f g, and so in
particular for x 2 0;1ð Þ:Now, by the Multiple Rule, f(x)!1 as x!1; and, by Example 1,
g(x)!1 as x!1. It follows, by the Composition Rule, that
g f xð Þð Þ ¼ 2ex2
x!1 as x!1,
so that; by the Multiple Rule;we have ex2
x!1 as x!1:
(b) Let f xð Þ ¼ 1x, x 2 R � 0f g, and g xð Þ ¼ ex
x, x 2 R � 0f g, so that
g f xð Þð Þ ¼ e1x
1x
¼ xe1x, for x 2 R � 0f g, and so in
particular for x 2 0;1ð Þ:
Sub-section 5.1.3, Theorem 4.
We again adopt thisconvention for simplicityONLY in this sub-section!
We omit the proof ofTheorem 5.
If ‘ denotes1 or �1,then the first proviso isautomatically satisfied. Notethat we must have conditions1 and 2 and the first provisosatisfied or conditions 1 and 2and the second provisosatisfied if we are to make anyapplication of this result.
180 5: Limits and continuity
Now, f (x)!1, as x! 0þ; and, by Example 1, g(x)!1 as x!1. It
follows, by the Composition Rule, that
g f xð Þð Þ ¼ xe1x !1 as x! 0þ:
(c) Let f xð Þ ¼ 1x
, x 2 R � 0f g; and g xð Þ ¼ sin xx; x 2 R � 0f g, so that
g f xð Þð Þ ¼sin 1
x
� �
1x
¼ x sin1
x
� �
; for x 2 R � 0f g, and so
for x 2 0;1ð Þ:
Now
f xð Þ ! 0 as x!1;g xð Þ ! 1 as x! 0; and
f xð Þ 6¼ 0; for x 2 0;1ð Þ:
It follows, by the Composition Rule, that
g f xð Þð Þ ¼ x sin1
x
� �
! 1 as x!1: &
Problem 4 Prove that:
(a) loge (loge x)!1 as x!1; (b) xe1x ! 0 as x! 0�.
Hint for part (b): Use the fact that 1x! �1 as x! 0�.
Problem 5 Give examples of functions f and g (and a specific value for
each of ‘ and m) for which
1. f(x)! ‘ as x!1 and
2. g(x)!m as x! ‘,
but for which g f xð Þð Þ6! m as x!1.
5.3 Limits of functions – using e and d
Earlier we gave definitions of the limit of a sequence, continuity of a function
and limit of a function, and strategies for using these definitions. In each case,
the strategy was in two parts:
GUESS that the definition HOLDS: prove that it holds for all cases.
GUESS that the definition FAILS: find ONE counter-example.
Each definition is particularly convenient if we wish to prove that the
definition FAILS, but it is not always easy to work with it when we wish to
prove that the definition HOLDS.
For example, when we wish to prove that a given function f is discontinuous
at a point c, then we have to find ONE sequence in a punctured neighbourhood of
c such that
xn ! c BUT f xnð Þ 6! f cð Þ:
In Theorem 5, we have ‘¼ 0,L¼ 1.
This is condition 1.
This is condition 2.
So the first proviso issatisfied.
5.3 Limits of functions – using " and � 181
On the other hand, if we wish to use the definition to prove that f is
continuous at c, then we have to show that:
for each sequence {xn} in a punctured neighbourhood of c such thatxn ! c; then f xnð Þ ! f cð Þ :
As it is sometimes tricky to determine the behaviour of every such sequence
{xn}, this definition can be very inconvenient to use.
In this section we introduce a definition of the limit of a function that is
equivalent to our earlier definition, but which does not use sequences.
5.3.1 The e – d definition of limit of a sequence
We motivate our discussion by returning to the definition of a convergent
sequence.
Definition The sequence {an} is convergent with limit ł, or converges to
the limit ł, if {an� ‘} is a null sequence. In other words, if:
for each positive number ", there is a number X such that
an � ‘j j < "; for all n > X: (1)
l + ε{an}
X n
l – ε < an < l + ε
|an – l| < εl
l − ε⇔
The condition (1) means that, from some point on, the terms of the sequence
all lie in the shaded strip between ‘� " and ‘ þ ", and thus lie ‘close to’ ‘. If we
choose a smaller number ", then the shaded strip in the diagram becomes
narrower, and we may need to choose a larger number X in order to ensure that
the inequality (1) still holds. But, whatever positive number " we choose, we
can always find a number X for which (1) holds.
Problem 1 Let an ¼ �1ð Þnn2 , n ¼ 1, 2, . . .. How large must we take X in
order that:
(a) |an� 0|< 0.1, for all n>X?
(b) |an� 0|< 0.01, for all n>X?
(c) |an� 0|<", for all n>X, where " is a given positive number?
Earlier we described a ‘game’, based on the above definition, in which
player A chooses a positive number " and challenges player B to find a number
X such that (1) holds. If the sequence in question converges, then such a
number X always exists, and player B can always win. If the sequence does
not converge, then, for SOME choices of ", NO number X exists such that (1)
holds, and so player A can always win.
In Section 5.4 we shallintroduce a definition ofcontinuity which does notuse sequences.
Sub-section 2.3.1.
Sub-section 2.2.1.
182 5: Limits and continuity
For example, consider the null sequence an ¼ 1ffiffi
np , n ¼ 1, 2, . . .. In this case,
the game might proceed as follows:
I'll playε = 0.1
I'll playX = 100
A B
B wins
Here player B wins, since, for n> 100, we have
1ffiffiffi
np � 0
�
�
�
�
�
�
�
�
¼ 1ffiffiffi
np <
1ffiffiffiffiffiffiffiffi
100p ¼ 0:1:
Player A tries again:
I'll playε = 0.01
I'll playX = 10 000
A B
B wins
Player B again wins, since, for n> 10 000, we have
1ffiffiffi
np � 0
�
�
�
�
�
�
�
�
¼ 1ffiffiffi
np <
1ffiffiffiffiffiffiffiffiffiffiffiffiffi
10 000p ¼ 0:01:
But now player B has figured out a winning strategy, and challenges player A
to do his worst!
I'll play ANY positive
ε
I'll playX = 1/ε
2
A B
B wins
This is indeed a winning strategy, since
for n >1
"2, we have
ffiffiffi
np
>1
", and so
1ffiffiffi
np � 0
�
�
�
�
�
�
�
�
¼ 1ffiffiffi
np < " :
The reason for introducing " in the definition of convergent sequence is to
formalise the idea of ‘closeness’. The statement (1) means that we can make
the terms an of the sequence as close as we please to ‘ by choosing X large
enough.
Here X¼ 100.
Here X¼ 10 000.
Here X ¼ 1"2.
5.3 Limits of functions – using " and � 183
Put another way, we can think of a sequence {an} as a function with domain R
n 7! n; anð Þ:
The number " formalises the idea of closeness of an to ‘ (that is, in the
codomain). The condition ‘for all n>X’ restricts the values n in the domain
to values for which condition (1) holds.
an close to l {
{an}
for all n > X
X n
l + ε
l – εl
|an – l| < ε for all n > X
5.3.2 The e – d definition of limit of a function
The concept of the limit of a function as x! c also involves the idea of
closeness. To define
f xð Þ ! ‘ as x! c; (2)
we require the statement
f xð Þ � ‘j j < ";
this means that the values of f (x) and ‘ are within a distance " of each other.
Earlier, we formalised the statement (2) by defining the limit of a function in
terms of limits of sequences.
Definition Let the function f be defined on a punctured neighbourhood N
of a point c. Then f (x) tends to the limit ł as x tends to c if:
for each sequence xnf g in N such that xn ! c, then f xnð Þ ! ‘: (3)
Intuitively, this means that, as xn gets close to c, so f (xn) gets close to ‘. More
precisely, we can make f (xn) as close as we please to ‘ by choosing xn
sufficiently close to c; we can ensure this by considering only xns for suffi-
ciently large n. But the sequence {xn} can be any sequence of points in N that
converges to c, so what we are really saying is that we can make f (x) as close as
we please to ‘ by choosing x sufficiently close to c (but not equal to c).
Thus we now need to specify not only closeness in the codomain
f xð Þ � ‘j j < ";
but also closeness in the domain. To do this, we introduce another small
positive number � (which depends on ") and specify closeness in the domain
by a statement of the form
0 < x� cj j < �:
In general, the smaller thevalue of ", the larger the X thatwe have to choose.
Sub-section 5.1.1.
Note that xn 6¼ c, for any n.
Actually it is the requirementj x� c j<� that specifies thecloseness in the domain. Therequirement 0< j x� c j ismade simply to ensure thatx 6¼ c.
184 5: Limits and continuity
f (xn) close to l
y = f (x) y = f (x)
{ f (xn)}
{xn}
c x
xn close to c
c x
x close to c
y y
l{ f (x) close to l
l{
Problem 2 Let f(x)¼ 2xþ 3, x 2 R . How small must we choose � in
order that:
(a) j f (x)� 5j< 0.1, for 0< jx� 1j<�?(b) j f (x)� 5j< 0.01, for 0< jx� 1j<�?(c) j f (x)� 5j<", for 0< jx� 1j<�, where " is a given positive
number?
In general, the statement ‘f (x)! ‘ as x! c’ means the following:
for EACH positive number ", there is a CORRESPONDING positive number� such that
f xð Þ � ‘j j < "; for all x satisfying 0 < x� cj j < �: (4)
The following diagram illustrates how the numbers " and � are used to
formalise the idea of closeness.
f (x) close to l
y = f (x) y = f (x)
c y
x close to c
c – δ
l + εl
l – ε
c + δc x
y y
l{
This leads us to the following definition of the limit of a function.
Definition Let the function f be defined on a punctured neighbourhood N
of a point c. Then f (x) tends to the limit ł as x tends to c if:
for each positive number ", there is a positive number � such that
f xð Þ � ‘j j < "; for all x satisfying 0 < x� cj j < � : (5)
The same number �may servefor various different values of"; but, in general, the value of� depends on the value of ".
Condition (4) means that, forall x with 0< jx� cj<�, thepoints (x, f (x)) of the graphlie in the heavily shadedrectangle, and thus lie ‘closeto’ (c, ‘).
Thus
f ðxÞ is ‘close to’ ‘
wheneverx is ‘close to’ c:
5.3 Limits of functions – using " and � 185
Remarks
1. We assume that � is chosen sufficiently small in (5), so that the interval
(c� �, cþ �) lies within N. We also require that 0< jx� cj in (5), since we
are not concerned with the value of f at c, or even whether f is defined at c.
2. There are similar definitions of limits from the left and limits from the right
at c, where we simply replace ‘0< jx� cj<�’ by ‘c� � < x< c’ and
‘c< x< cþ �’, respectively; and similar definitions of limx!1
f xð Þ and
limx!�1
f xð Þ.
We now formally state the equivalence of the two definitions of ‘limit of a
function’.
Theorem 1 The ‘"� � definition’ and the ‘sequential definition’ of the
statement limx!c
f xð Þ ¼ ‘ are equivalent.
Proof Let the function f be defined on a punctured neighbourhood N of a
point c.
We have to show that:
for each sequence {xn} in N such that xn ! c, then f xnð Þ ! ‘ (6)
, for each positive number ", there is a positive number � such that
f xð Þ � ‘j j < "; for all x satisfying 0 < x� cj j < � : (7)
First, let us assume that (6) holds. Then, for each sequence {xn} in N and
each positive number ", there is a number X such that
f xnð Þ � ‘j j < "; for all n > X: ()
We will now use a ‘proof by contradiction’. So, suppose that (7) does not
hold. Then there must exist some positive number " for which there is no
positive number � such that
f xð Þ � ‘j j < "; for all x satisfying 0 < x� cj j < �: (8)
In particular, we can always assume that � < r, so that {x : 0< jx� cj<�}N.
It follows that, for each positive integer n, there is some number xn (say) in
N such that
f xnð Þ � ‘j j � "; where 0 < xn � cj j < 1
n: (9)
But, from our assumption (6), the sequence {xn} must satisfy (*). Hence, in
addition, in particular, for the positive number " that we have chosen for (*),
there is a number X such that j f (xn)� ‘j<", for all n>X. But this is incon-
sistent with (9), so that we have a contradiction. It follows, therefore, that (7)
must hold after all!
Next, let us assume that (7) holds. That is, for each positive number ", there
is a positive number � such that
f xð Þ � ‘j j < "; for all x satisfying 0 < x� cj j < �: (10)
For otherwise the values f (x)may be undefined.The existence and value of alimit is a local property, butthe behaviour of the functionAT the point in question isirrelevant.
You may omit this proof at afirst reading.
We shall assume, forconvenience, that
N ¼ c� r; cð Þ [ c; cþ rð Þ¼ x : 0 < x� cj j < rf g:
Here we simply restate (6).
Here we write down what ismeant by ‘(7) does not hold’.
Recall that
N ¼ x : 0 < x� cj j < rf g:This is possible since (8) doesnot hold for any �, and since wemay take as successive choicesof � the values 1; 1
2; 1
3; . . ..
For all n>X, f (xn) now has tosatisfy the two inconsistentinequalities
f xnð Þ � ‘j j � " and
f xnð Þ � ‘j j < ":
Here we simply restate (7).
186 5: Limits and continuity
Now let {xn} be any sequence in N such that xn! c. In particular, for the
positive number � that we have chosen for (10), there is a number X such
that (0< )jxn� cj<� for all n>X. It follows from (10) that jf(xn)� ‘j<"for all n>X.
Since this holds for each positive number ", we deduce that f (xn)! ‘, so that
(6) holds, as required. &
Remark
Depending on the particular circumstances, we can therefore use whichever
definition of limit is the more convenient for our purposes.
We can think of the choice of � in terms of " in the definition of limit of a
function as a game with players A and B, just as we did with the choice of X in
the definition of convergent sequence.
Thus, player A chooses some value for ", and then player B has to try to find a
value of � such that
f xð Þ � ‘j j < "; for all x satisfying 0 < x� cj j < �:
For example, consider how we might show that f (x)¼ x2! 0 as x! 0.
Here, for the function f (x)¼ x2, x 2 R , and c¼ 0, player B has to find a number
� such that
f xð Þ � 0j j ¼ 4x2�
�
�
� < "; for all x satisfying 0 < xj j < �:
The game might proceed as follows:
I'll playε = 0.2
I'll playδ = 0.1
A B
B wins
Here player B wins, since
for 0 < xj j < 0:1; we have that 4x2�
�
�
� � 0:04 < 0:2:
So player A tries again:
I'll playε = 0.02
I'll playδ = 0.1
A B
A wins
Here player A wins, since
x ¼ 0:09 satisfies 0 < xj j < 0:1 BUT 4� 0:092�
�
�
� ¼ 0:0324 6< 0:02:
Player B lost because he did not choose � sufficiently small; so he tries again.
We have 0< jxn� cj since xn
lies in a puncturedneighbourhood of c.
5.3 Limits of functions – using " and � 187
I'll playε = 0.02
I'll playδ = 0.01
A B
B wins
This time player B wins, since
for 0 < xj j < 0:01;we have that 4x2�
�
�
� � 0:0004 < 0:02:
But now player B has figured out a winning strategy! He challenges player A to
do his worst!
I'll play ANY positive
ε
I'll playδ =
12√ε
A B
B wins
This is indeed a winning strategy, since
for 0 < xj j < 1
2
ffiffiffi
"p
;we have that 4x2�
�
�
�< 4� 1
2
ffiffiffi
"p� �2
¼ ":
In general, if f (x)! ‘ as x! c, then such a number � always exists, and
player B can always win. On the other hand, if f ðxÞ 6! ‘ as x! c, then for SOME
choices of " no corresponding positive number � exists, and so player A can
always win.
Our strategy for using the ‘"� � definition’ of limx!c
f xð Þ in particular cases is
as follows.
Strategy for using the ‘e� d definition’ of limit of a function
1. To show that f (x)! ‘ as x! c, find an expression for � in terms of " such
that:
for each positive number "
f xð Þ � ‘j j < "; for all x satisfying 05 x� cj j5�: (11)
2. To show that f ðxÞ 6! ‘ as x! c, find:
ONE positive number " for which there is NO positive number �such that
f xð Þ � ‘j j < "; for all x satisfying 0 < x� cj j < �:
Whatever value of " is chosenby player A, player B needonly specify � ¼ 1
2
ffiffiffi
"p
.
Of course the choice of � maydepend on c too.
188 5: Limits and continuity
Example 1 Using the strategy, prove that
(a) limx!1
2xþ 3ð Þ ¼ 5; (b) limx!0
2x3ð Þ ¼ 0;
(c) limx!0
x sin 1x
� �
¼ 0; (d) limx!0
sin 1x
� �
does not exist.
Solution
(a) The function f (x)¼ 2xþ 3 is defined on every punctured neighbour-
hood of 1.
We must prove that:
for each positive number ", there is a positive number � such that
ð2xþ 3Þ � 5j j < "; for all x satisfying 0 < x� 1j j < � ;
that is
2 x� 1ð Þj j < "; for all x satisfying 0 < x� 1j j < �: (12)
We choose � ¼ 12"; then the statement (12) holds, since
for 0 < x� 1j j < 12", we have 2 x� 1ð Þj j < ":
Hence, limx!1
2xþ 3ð Þ ¼ 5.
(b) The function f (x)¼ 2x3 is defined on every punctured neighbourhood of 0.
We must prove that:
for each positive number ", there is a positive number � such that
2x3 � 0�
�
�
� < "; for all x satisfying 0 < x� 0j j < � ;
that is
2x3�
�
�
� < "; for all x satisfying 0 < xj j < � : (13)
We choose � ¼ 12
ffiffiffi
"3p
; then the statement (13) holds, since:
for 0 < xj j < 12
ffiffiffi
"3p
, we have that
2x3�
�
�
� ¼ 2 xj j3< 2� 1
2
ffiffiffi
"3p� �3
¼ 1
4" < ":
Hence, limx!0
2x3ð Þ ¼ 0.
(c) The function f xð Þ ¼ x sin 1x
is not defined at 0, but it is defined on every
punctured neighbourhood of 0.
We must prove that:
for each positive number ", there is a positive number � such that
x sin1
x� 0
�
�
�
�
�
�
�
�
< "; for all x satisfying 0< x� 0j j< � ;
that is
x sin1
x
�
�
�
�
�
�
�
�
< "; for all x satisfying 0 < xj j < �: (14)
Now, sin 1x
�
�
�
� � 1, for all non-zero x, so that
We could equally well choose� to be any positive numberless than 1
2", since the
statement (12) would stillhold.
Notice that many otherchoices of � would also haveserved our purpose.
5.3 Limits of functions – using " and � 189
x sin1
x
�
�
�
�
�
�
�
�
� xj j; for all non-zero x:
We therefore choose �¼ "; then the statement (14) holds.
Hence, limx!0
x sin 1x
� �
¼ 0.
(d) Suppose that the limit exists; call it ‘. Take " ¼ 12, say. Then there is some
positive number �, and points xn and yn of the form
xn ¼1
2nþ 12
� �
pand yn ¼
1
2nþ 32
� �
p; where n 2 N ;
such that
0 < xnj j < � and f xnð Þ � ‘j j < 1
2; and
0 < ynj j < � and f ynð Þ � ‘j j < 1
2:
:
Hence, for sufficiently large n
f xnð Þ � f ynð Þj j ¼ f xnð Þ � ‘ð Þ � f ynð Þ � ‘ð Þj j
� f xnð Þ � ‘j j þ f ynð Þ � ‘j j <1
2þ 1
2¼ 1:
But f (xn)� f (yn)¼ 1� (�1)¼ 2, which contradicts the previous inequal-
ity. It follows that no such numbers ‘ and � exist, so that in fact limx!0
sin 1x
� �
does not exist. &
Remarks
1. In Example 1, parts (a) and (b), we can find an appropriate choice of � by
‘working backwards’.
For example, in part (b) we have
2x3�
�
�
� < ", x3�
�
�
� <1
2"
, xj j < 1
213
ffiffiffi
"3p
;
so we can choose � ¼ 1
213
ffiffiffi
"3p
. The choice that we made in Example 1 was to
take � ¼ 12
ffiffiffi
"3p
, which was a somewhat smaller (but equally valid) value for �.Both possibilities are valid, since we only require to show that
0 < x� 0j j < � ) 2x3 � 0�
�
�
� < ";
rather than
0 < x� 0j j < � , 2x3 � 0�
�
�
� < ":
On the other hand, in part (c) it is not possible to ‘solve for �’ in this way –
that is, to solve the inequality x sin 1x
�
�
�
� < " to obtain a solution such that
x sin1
x
�
�
�
�
�
�
�
�
< "; for 0 < xj j < �:
For, if " is sufficiently small, the solution set of the inequality x sin 1x
�
�
�
� < " is
not even an interval but a union of disjoint intervals.
Of course any choice for �smaller than this particularvalue will also serve ourpurposes – such as 1
7" or 1
625".
190 5: Limits and continuity
2. In Example 1, part (d), the point is that, whatever the value of �, there are
always points in any punctured neighbourhood of 0 where f takes the values
1 and �1. Thus there cannot exist numbers ‘ and � such that
f xð Þ � ‘j j < 1
2; for all x satisfying 0 < xj j < �:
Problem 3 Using the Strategy (except in part (d)), prove that:
(a) limx!3
5x� 2ð Þ ¼ 13; (b) limx!0
1� 7x3ð Þ ¼ 1;
(c) limx!0
x2 cos 1x3
� �
¼ 0; (d) limx!0
xþ xj jx
does not exist.
Sometimes it is a little tricky to find an expression for � in terms of " such that
f xð Þ � ‘j j < "; for all x satisfying 0 < x� cj j < �:
The following example illustrates one approach that is sometimes useful.
Example 2 Use the "� � definition to prove that limx!2
x2 ¼ 4.
Solution According to the definition, we must show that for x ‘close to 2’ the
following difference is small
x2 � 4�
�
�
� ¼ xþ 2j j � x� 2j j:For x close to 2, the term jxþ 2j is approximately 4, so that the product
jxþ 2j � jx� 2j is approximately equal to 4jx� 2j. Certainly, if we restrict x
to the punctured neighbourhood (1, 2)[ (2, 3) in which jxþ 2j< 5, we know that
x2 � 4�
�
�
� < 5 x� 2j j:
So, we proceed as follows.
The function f (x)¼ x2 is defined on every punctured neighbourhood of 2.
We must prove that
for each positive number ", there is a positive number � such that
x2 � 4�
�
�
� < "; for all x satisfying 0 < x� 2j j < � ;
that is
xþ 2j j � x� 2j j < "; for all x satisfying 0 < x� 2j j < �:
We choose � ¼ min 1; 15"
� �
, so that for 0< jx� 2j<� we have
xþ 2j j < 5 and x� 2j j < 1
5" :
It follows that
xþ 2j j � x� 2j j < 5� 1
5" ¼ "; for all x satisfying 0 < x� 2j j < �:
Hence limx!2
x2 ¼ 4: &
Problem 4 Use the "� � definition to prove that limx!1
x3 ¼ 1.
Hint: Try � ¼ min 1, 17"
� �
.
The discussion so far has beenour motivation.
The formal proof now starts!
5.3 Limits of functions – using " and � 191
Proofs
Since the definition of limit of a function introduced in this section is equiva-
lent to the earlier sequential definition, the rules for combining limits can also
be proved using the "� � approach. To give you a flavour of what this
involves, we prove here just one of the Combination Rules.
Example 3 Prove the Sum Rule:
If limx!c
f xð Þ ¼ ‘ and limx!c
g xð Þ ¼ m; then limx!c
f xð Þ þ g xð Þð Þ ¼ ‘þ m:
Solution Since the limits for f and g exist, the functions f and g must
be defined on some punctured neighbourhoods (c� r1, c)[ (c, cþ r1) and
(c� r2, c)[ (c, cþ r2) of c. It follows that the function fþ g is certainly defined
on the punctured neighbourhood (c� r, c)[ (c, cþ r), where r¼min{r1, r2}.
We want to prove that:
for each positive number ", there is a positive number � such that
f xð Þ þ g xð Þð Þ � ‘þ mð Þj j < "; for all x satisfying 0 < x� cj j < �: (15)
We know that, since limx!c
f xð Þ ¼ ‘, there is a positive number �1 such that
f xð Þ � ‘j j < 1
2"; for all x satisfying 0 < x� cj j < �1; (16)
similarly, since limx!c
g xð Þ ¼ m, there is a positive number �2 such that
g xð Þ � mj j < 1
2"; for all x satisfying 0 < x� cj j < �2: (17)
We now choose �¼min{�1, �2}. Then both statements (16) and (17) hold
for all x satisfying 0< |x� c|<�, so that
f xð Þ þ g xð Þð Þ � ‘þ mð Þj j ¼ f xð Þ � ‘ð Þ þ g xð Þ � mð Þj j� f xð Þ � ‘j j þ g xð Þ � mj j
<1
2"þ 1
2" ¼ ";
so that the statement (15) holds.
Hence limx!c
f xð Þ þ g xð Þð Þ ¼ ‘þ m. &
We now introduce an analogue of a useful tool for writing out proofs that
you met earlier in your work on sequences, the so-called ‘K" Lemma’.
In order to prove that a function f has a limit as x! c, using the definition of
limit, you need to prove that:
for each positive number ", there is a positive number � such that
f xð Þ � ‘j j < "; for all x satisfying 0 < x� cj j < �:
It can be tedious in the process of the proof to make a complicated choice for � in
order to end up with a final inequality that says that some expression is ‘<"’. While
this means that we end up with an inequality that shows at once that the desired
result holds, in fact it is not strictly necessary to end up with precisely ‘< "’.
Of course f and g may bedefined on some larger setstoo!
Sub-section 2.2.2.
192 5: Limits and continuity
Lemma The ‘Ke Lemma’ For the function f, suppose that:
for each positive number ", there is a positive number � such that
f xð Þ � ‘j j < K"; for all x satisfying 0 < x� cj j < �;
where K is a positive real number that does not depend on " or X.
Then f (x)! ‘ as x! c.
We omit a proof of the Lemma, as it is essentially the same as the previous
proof of the K" Lemma for sequences. There are analogues of the K" Lemma
for the definition of continuity (in terms of " and �), differentiability and
integrability, but we shall not always mention them explicitly every time an
analogue could be stated.
From time to time we shall use this Lemma in order to avoid arithmetic
complexity in proofs.
Problem 5 Use the "� � definition to prove the Product Rule:
If limx!c
f xð Þ ¼ ‘ and limx!c
g xð Þ ¼ m; then limx!c
f xð Þg xð Þ ¼ ‘m:
Problem 6 Use the "� � definition to prove that, if limx!c
f xð Þ ¼ ‘ and
‘ 6¼ 0, then there exists some punctured neighbourhood of c on which f (x)
has the same sign as ‘.
Hint: Take " ¼ 12‘j j in the definition of limit.
5.4 Continuity – using e and d
In this section we introduce a definition of continuity of a function that is
equivalent to our earlier definition, but which does not use sequences.
5.4.1 The e–d definition of continuity
Recall our earlier definitions of continuity. The first version was for continuity
of a function at an interior point of an interval.
Definition A function f defined on an interval I that contains c as an
interior point is continuous at c if
for each sequence {xn} in I such that xn! c, then f (xn)! f (c).
We then saw how to extend this definition to include continuity of a function at
an end-point of an interval.
Definition A function f defined on a set S in R that contains a point c is
continuous at c if:
for each sequence {xn} in S such that xn! c, then f (xn)! f (c).
Loosely speaking, we mayexpress this result as ‘K" isjust as good as "’ in thedefinition of limit.
For example, K might be 2 or p7
or 259, but it could not be 2xor 1
".
For example, in Problem 5below.
You may omit Problem 5 ifyou are short of time.
Sub-section 4.1.1.
Sub-section 4.1.1.
Here c may be an interiorpoint of an interval, or it maybe an end-point of an interval;both possibilities are coveredby the phrase ‘each sequence{xn} in S’.
5.4 Continuity – using " and � 193
We start by reformulating the first version of this definition in terms of " and �.So, if we replace ‘ by f (c) in the definition of limit of a function, we
immediately obtain the following definition of continuity of a function at an
interior point of an interval.
Definition A function f defined on an interval I that contains c as an inte-
rior point is continuous at c if:
for each positive number ", there is a positive number � such that
f xð Þ � f cð Þj j < "; for all x satisfying x� cj j < �: (1)
Remarks
1. Notice that in (1) we write jx� cj<� rather than 0< jx� cj<�, since in
fact (1) is always satisfied when x¼ c.
2. We may assume that � has been chosen sufficiently small so that (c� �,cþ �) lies in I.
3. It is sufficient in (1) to have
f xð Þ � f cð Þj j < K"; for all x satisfying x� cj j < �;
where K is some positive constant. This fact is often referred to as the ‘K"Lemma’ for continuity.
4. The condition (1) means that, for all x ‘sufficiently close to’ c, the values of f(x)
lie ‘close to’ f(c). If we choose a smaller number", then we may need to choose
a smaller number � in order to ensure that (1) still holds. But, whatever positive
number " we choose, we can always find a number � for which (1) holds.
5. We can describe the definition of continuity in terms of an ‘"� � game’ in
much the same way as we described the definition of limit. WHATEVER choice
of " player A makes, then player B can ALWAYS choose a value of � such that
the required condition (1) holds.
6. For a given value of the positive number ", we may need to choose different
values of � for different points c.
7. It follows immediately from our definition of continuity in terms of limits
that, if a function f is defined on an open interval containing c as an interior
point, then
f is continuous at c, limx!c
f xð Þ ¼ f cð Þ.
8. We can define one-sided continuity in a similar fashion. Thus a function f is
continuous on the left at c if:
for each positive number ", there is a positive number � such that
f xð Þ � f cð Þj j < "; for all x satisfying c� � < x < c;
and is continuous on the right at c if:
for each positive number ", there is a positive number � such that
f xð Þ � f cð Þj j < "; for all x satisfying c < x < cþ �:It follows from these definitions and the previous remark that, if a function f
is defined on an open interval containing c as an interior point, then
f is continuous at c, f is continuous on the left and on the right at c.
Sub-section 5.3.2.
Note that the inequalityj f (x)� f (c)j<" specifiescloseness in the codomain,whereas the inequalityjx� cj<� specifies closenessin the domain.
c – δ c + δ xc
y = f (x)
f (c)f (c) + ε
f (c) – ε
y
We shall return to this point inSection 5.5.
We shall not spend muchtime looking at one-sidedcontinuity; however there is awide range of results similarto the results for ‘ordinary’continuity.
194 5: Limits and continuity
We can use the definitions in the last Remark to extend the definition of
continuity of a function f from points of an open interval to all points of a
general interval, in a natural way. We say that f is continuous on an interval I if
f is continuous at all interior points of I, continuous on the right at the left end-
point of I if it belongs to I, and continuous on the left at the right end-point of
I if it belongs to I.
This leads us to extend our initial definition of continuity, as follows.
Definition A function f defined on an interval I that contains a point c is
continuous at c if:
for each positive number ", there is a positive number � such that
f xð Þ � f cð Þj j < "; for all x in I satisfying x� cj j < �:
f is said to be continuous on I if it is continuous at each point c of I.
Since the ‘"� � definition’ and the ‘sequential definition’ of limits are
equivalent, and the two definitions of continuity are simply reformulations
of those definitions, it is obvious that the two definitions of continuity must
also be equivalent. Depending on the particular circumstances, we can there-
fore use whichever definition of continuity is the more convenient for our
purposes.
Theorem 1 The ‘"� � definition’ and the ‘sequential definition’ of the
statement ‘f is continuous at c’ are equivalent.
Our strategy for using the ‘"� � definition’ of continuity in particular cases
is also similar to the strategy for using the ‘"� � definition’ of limits.
Strategy for using the ‘e� d definition’ of continuity
1. To show that f is continuous at an interior point c of an interval, find an
expression for � in terms of " such that:
for each positive number "
f xð Þ � f cð Þj j5 "; for all x satisfying x� cj j < �: (2)
2. To show that f is discontinuous at an interior point c of an interval, find
ONE positive number " for which there is NO positive number � such that
f xð Þ � f cð Þj j < "; for all x satisfying x� cj j < �: (3)
Example 1 Use the Strategy to prove that f (x)¼ x2 is continuous at 2.
Solution The function f is defined on R .
We must prove that:
for each positive number ", there is a positive number � such that
x2 � 4�
�
�
� < "; for all x satisfying x� 2j j < �;
that is
xþ 2j j x� 2j j < "; for all x satisfying x� 2j j < �:
This was Theorem 1 in Sub-section 5.3.2.
Note that, for x near to 2, xþ 2is near to 4 and x� 2 near to 0.
5.4 Continuity – using " and � 195
We choose � ¼ min 1; 15"
� �
; then, for jx� 2j<�, we have
xþ 2j j < 5 and x� 2j j < 1
5":
Hence
x2 � 4�
�
�
� < 5� 1
5" ¼ "; for all x satisfying x� 2j j < �:
It follows that f is continuous at 2. &
Problem 1 Use the Strategy to prove that f (x)¼ x3 is continuous at 1.
Hint: Try � ¼ min 1; 17"
� �
.
Problem 2 Use the Strategy to prove that f xð Þ ¼ffiffiffi
xp
, x� 0, is contin-
uous at 4.
Example 2 Use the Strategy to prove that the following function is discon-
tinuous at 0
f xð Þ ¼ x; if x � 0;1; if x < 0:
Solution The function f is defined on R .
The graph of f suggests that, while f is ‘well behaved’ to the right of 0, we
should examine the behaviour of f to the left of 0. If we choose " to be any
positive number less than 1, then there will always be points x (with x< 0) as
close as we please to 0, where f (x) is not within a distance " of f(0)¼ 0.
So, take " ¼ 12, say; and let xn ¼ � 1
n, for n¼ 1, 2, . . .. Then, for any positive
number �
xn ¼ �1
n
� �
2 ��; 0ð Þ; for all n > X; where X ¼ 1
�;
but
f xnð Þ � f 0ð Þj j ¼ f � 1
n
� �
� f 0ð Þ�
�
�
�
�
�
�
�
¼ 1� 0
¼ 1 6< 1
2¼ ":
Thus, with this choice of ", no value of � exists such that
f xð Þ � f 0ð Þj j < "; for all x satisfying x� 0j j < �:
This proves that f is discontinuous at 0. &
Problem 3 Use the Strategy to prove that the following function is
discontinuous at 2
f xð Þ ¼x� 1
2; if x > 2;
1; if x ¼ 2;x� 1; if x < 2:
8
<
:
Finally, we demonstrate a nice application of the "� � approach to continuity
in order to obtain a result that will be useful later on.
We could equally well choose� to be any positive numbersmaller than � ¼ min 1; 1
5"
� �
,since the rest of the argumentwould still apply. We choose� < 1 to get a bound on theterm xþ 2, and � < 1
5" as a
small multiple of " such thatthe final product is at most ".
y
0
1
x
For example, the points ofthe sequence {xn}, wherexn ¼ � 1
n, for all n� 1, since
then f (xn)¼ 1.
We make a suitable choice of" such that no corresponding �exists that satisfies therequirement (3).
This was requirement (3).
196 5: Limits and continuity
Theorem 2 Let f be continuous at an interior point c of an interval I, and
f cð Þ 6¼ 0. Then there exists a neighbourhood N¼ (c� r, cþ r) of c on
which:
(a) f(x) has the same sign as f(c), and
(b) f xð Þj j > 12
f cð Þj j.
Proof Since f(c) 6¼ 0, we may take 12
f cð Þj j as the positive number " in the
definition of continuity at c. It follows that there exists some positive number r
such that
f xð Þ � f cð Þj j < 1
2f cð Þj j; for all x satisfying x� cj j < r: (4)
We now rewrite (4) in the following equivalent form
� 1
2f cð Þj j < f xð Þ � f cð Þ < 1
2f cð Þj j; for all x 2 c� r; cþ rð Þ: (5)
Suppose, first, that f (c)> 0. It follows from the left inequality in (5) that, for
x2 (c� r, cþ r), we have �12
f cð Þ < f xð Þ � f cð Þ – in other words, that12
f cð Þ < f xð Þ. Thus, in this case, both (a) and (b) hold for x2 (c� r, cþ r).
Suppose, next, that f (c)< 0. It follows from the right inequality in (5) that,
for x2 (c� r, cþ r), we have f xð Þ � f cð Þ < 12
f cð Þj j ¼ �12
f cð Þ – in other
words, that f xð Þ < 12
f cð Þ. Thus, in this case too, both (a) and (b) hold for
x2 (c� r, cþ r).
This completes the proof of the theorem. &
5.4.2 The Dirichlet and Riemann functions
We can use our new "� � approach to continuity to tackle two very strange
functions that were devised in the second half of the nineteenth century. We
start with the Dirichlet function.
Definition Dirichlet’s function is defined on R by the formula
f xð Þ ¼ 1; if x is rational;0; if x is irrational:
The graph of f looks rather like two parallel lines, but each line has
‘infinitely many gaps in it’.
Theorem 3 The Dirichlet function is discontinuous at each point of R .
Proof Let c be any point of R . Then, for each n = 1, 2, . . ., by the Density
Property of R , the open interval c� 1n
, cþ 1n
� �
contains a rational number, xn
say, and an irrational number, yn say. Then xn! c and yn! c as n!1, but
f xnð Þ ¼ 1 and f ynð Þ ¼ 0:
Thus f (x) does not tend to a limit as x tends to c, whether c is rational or
irrational.
It follows that f is discontinuous at each point c of R . &
Here r> 0.
For convenience, here we usethe symbol r rather than � inthe definition of continuity –this makes no real difference.
y
1f (xn)
f (yn)
c xxn yn c + 1n
Recall that the DensityProperty of R asserts that,between any two unequal realnumbers, there exists at leastone rational and one irrationalnumber – Sub-section 1.1.4.
5.4 Continuity – using " and � 197
Our next function possesses even stranger behaviour.
Definition Riemann’s function is defined on R by the formula
f xð Þ ¼1q;
0;
if x is a rational numberpq; in lowest terms; with q > 0;
if x is irrational:
It is not clear from the graph of f whether f is continuous at any point of R .
Theorem 4 Riemann’s function is discontinuous at each rational point of
R , and continuous at each irrational point of R .
Proof First, let c ¼ pq
be any rational point of R , where q> 0 andpq
is in its
lowest terms.
Then, by the Density Property of R , for each n = 1, 2, . . ., the open inter-
val c� 1n
, cþ 1n
� �
contains an irrational number xn, for which
f xnð Þ ¼ 0 6¼ 1q¼ f cð Þ.
Hence as n!1xn ! c but f xnð Þ 6! f cð Þ:
Thus f is discontinuous at any rational point c.
Next, suppose that c is any irrational point of R . We must prove that:
for each positive number ", there is a positive number � such that
f xð Þ � f cð Þj j < "; for all x satisfying x� cj j < �: (6)
Since c is irrational, f (c)¼ 0; also f (x)� 0 for all x. It follows that condition
(6) becomes
f xð Þ < "; for all x satisfying x� cj j < �: (7)
Our task is therefore to find a value of � such that (7) holds.
Now, let N be any positive integer such that N > 1", and let SN be the set of
rational numbers pq
(in lowest terms) in the interval (c� 1, cþ 1) for which
0< q<N – that is
SN ¼p
q: c� p
q
�
�
�
�
�
�
�
�
� 1; 0 < q < N
:
Since there are only a finite number of points in SN and c =2 SN, we can define a
positive number � as follows
The expression ‘in lowestterms’ means that p and qhave no common factor.
In this part of the proof, weuse a sequential approach.
Note that xn 6¼ c.
In this part of the proof, weuse an "� � approach.
Thus, in particular, 1N< ".
We choose (c� 1, cþ 1)simply so that we are onlylooking at points ‘near to c’.
c =2 SN since c is irrational.
198 5: Limits and continuity
� ¼ min x� cj j : x 2 SNf g:Thus there are NO rational numbers p
q, with 0< q<N, that lie in the interval
(c� �, cþ � ).
It follows that, if jx� cj<�, then:
EITHER x is irrational, so that f(x) (¼ 0)<";OR x is rational, so that x ¼ p
q, with q�N, and f xð Þð¼ 1
qÞ � 1
N< ".
In either case, f(x)<".Hence we have proved that (7) holds, and so that f is continuous at c. &
5.4.3 Proofs
Since the definition of continuity introduced in this section is equivalent to the
earlier sequential definition, the rules for continuity can also be proved using
the "� � approach. To give you a flavour of what this involves, we first prove
one of the Combination Rules.
Example 3 (Sum Rule) Let f and g be defined on an open interval I contain-
ing the point c. Then, if f and g are continuous at c, so is the sum function fþ g.
Solution The functions f, g and fþ g are certainly defined on some neigh-
bourhood (c� r, cþ r) I of c, where r> 0.
We want to prove that:
for each positive number ", there is a positive number � such that
f xð Þþg xð Þð Þ� f cð Þþg cð Þð Þj j<"; for all x satisfying x� cj j<� (8)
We know that, since f is continuous at c, there is a positive number �1 (which
we may assume is� r) such that
f xð Þ � f cð Þj j < 1
2"; for all x satisfying x� cj j < �1; (9)
similarly, since g is continuous at c, there is a positive number �2 (which we
may assume is also� r) such that
g xð Þ � g cð Þj j < 1
2"; for all x satisfying x� cj j < �2: (10)
We now choose �¼min{�1, �2}. Then both statements (9) and (10) hold for all
x satisfying jx� cj<�, so that
f xð Þ þ g xð Þð Þ � f cð Þ þ g cð Þð Þj j ¼ f xð Þ � f cð Þð Þ þ g xð Þ � g cð Þð Þj j� f xð Þ � f cð Þj j þ g xð Þ � g cð Þj j
<1
2"þ 1
2" ¼ ";
so that the statement (8) holds.
Hence fþ g is continuous at c. &
Problem 4 Prove that if f is defined on an open interval I containing
the point c at which it is continuous, and f (c) 6¼ 0, then 1f
is defined on
some neighbourhood of c and is also continuous at c.
Hint: Use the result of Theorem 2 in Sub-section 5.4.1.
� is the distance from c to thenearest point of SN.
You might like to comparethis solution with that ofExample 3 in Sub-section 5.3.2.
5.4 Continuity – using " and � 199
Finally, we give a proof of the Composition Rule for continuity; this is a
particularly pleasing example of the "� � approach.
Example 4 (Composition Rule) Prove that if f is continuous at c and g is
continuous at f (c), then g� f is continuous at c.
Solution We must prove that:
for each positive number ", there is a positive number � such that
g f xð Þð Þ � g f cð Þð Þj j < "; for all x satisfying x� cj j < �: (11)
We know that, since g is continuous at f(c), there is a positive number �1
such that:
g f xð Þð Þ � g f cð Þð Þj j < "; for all f xð Þ satisfying f xð Þ � f cð Þj j < �1: (12)
f (c) – δ1 f (c) + δ1 g(f (c)) – ε
g
g( f (c)) + εg( f (c)) g( f (x))f (c) f (x)
Also, since f is continuous at c, there is a positive number � such that
f xð Þ � f cð Þj j < �1; for all x satisfying x� cj j < �: (13)
c – δ c + δc x f (c) – δ1 f (c) + δ1
f
f (c) f (x)
Combining (12) and (13), we deduce that:
for all x satisfying jx� cj<�, then f xð Þ � f cð Þj j < �1,
so that
g f xð Þð Þ � g f cð Þð Þj j < ":
Hence g� f is continuous at c. &
5.5 Uniform continuity
We start by reminding you of the "� � definition of continuity at points of an
interval I.
Definition A function f defined on an interval I that contains a point c is
continuous at c if:
for each positive number ", there is a positive number � such that
f xð Þ � f cð Þj j < "; for all x in I satisfying x� cj j < �:
f is said to be continuous on I if it is continuous at each point c of I.
We do not bother to mentionappropriate neighbourhoodsof c and f (c) on which f and gare defined, respectively, justto simplify the statement ofthe Rule.
By (13).
By (12).
Here the choice of � depends,in general, on both " and c.
200 5: Limits and continuity
Earlier we pointed out that, for a given value of the positive number ", we may
need to choose different values of � for different points c. This suggests the
following related definition.
Definition A function f defined on an interval I is uniformly continuouson I if:
for each positive number ", there is a positive number � such that
f xð Þ � f cð Þj j < "; for all x and c in I satisfying x� cj j < �: (1)
Since the definition of uniform continuity is more restrictive than that of
continuity, it follows that uniform continuity implies continuity.
Theorem 1 If a function f is uniformly continuous on an interval I, then it
is continuous on I.
However the converse result is not true in general.
Example 1 Prove that the function f xð Þ ¼ 1x, 0 < x � 1; is not uniformly
continuous on (0, 1].
Solution To prove that f is NOT uniformly continuous on (0,1], we need to
find ONE positive number " for which there is NO positive number � such that
f xð Þ� f cð Þj j< "; for all x and c in ð0;1� satisfying x� cj j5�: (2)
We will take as our choice of " the number 1, and prove that there is NO
corresponding positive number � such that
f xð Þ� f cð Þj j< 1; for all x and c in ð0;1� satisfying x� cj j<�: (3)
Suppose on the contrary that some number � did exist such that (3) was valid.
We will show that this assumption leads us to a contradiction.
Choose a positive integer N with
1
N< �:
Now each point of (0,1] belongs to at least one interval of the form
0
2N;
2
2N
� �
;1
2N;
3
2N
� �
; . . .;k
2N;k þ 2
2N
� �
; . . .;2N � 2
2N;2N
2N
� �
;2N � 1
2N; 1
� �
;
(4)and, in addition, all points x and c in any given such interval satisfy the
inequality
x� cj j < 2
2N¼
� �
1
N< �;
Hence, by (3)
all points x and c in ð0; 1� in any interval in ð4Þ satisfy f xð Þ � f cð Þj j < 1:
(5)Now, from the definition of f, we have
f xð Þ < 2N
2N � 1¼ K; say; for all x in
2N � 1
2N; 1
� �
:
Sub-section 5.4.1, Remark 6.
Here the choice of � dependsONLY on ", the same choicewhatever c may be.
Since f is a rational function, itis a basic continuous function,and so is continuous on (0, 1].
For the statement (2) assertsthat statement (1) is not valid.
This is what we now set outto do.
For, successive intervals in (4)overlap.
In particular, f (x)< f (c)þ 1.
For, f is strictly decreasingand f 2N�1
2N
� �
¼ 2N2N�1
.
5.5 Uniform continuity 201
Then, since the interv als 2 N � 12N
; 1� �
and 2 N � 22 N
; 2 N2 N
� �
overlap, ther e is a point c in2N � 2
2 N; 2 N
2 N
� �
for which f ( c ) < K. It follow s from ( 5) that in the second last
interv al 2N � 22 N
; 2 N2 N
� �
in ( 4) we have
f xð Þ < K þ 1; for all x in2N � 2
2N;
2N
2N
� �
:
Work ing backwar ds through the 2 N intervals, we can check that
f xð Þ < K þ 2N � 1ð Þ; for all x in0
2N ;
2
2N
� �
:
It follow s then, that in fact f( x ) < K þ (2N � 1), for all x in (0,1]. In othe r words,
f is bounded in (0,1]. But this is not the case, since f ( x ) !1 as x ! 0þ .
We have thus reac hed the desired cont radiction . &
Problem 1 Let f be the funct ion f (x ) ¼ x 2, x 2 [2, 3]. Fo r any give n
positive number ", find a formula for � in terms of " such that (1) holds.
This proves directly from the defi nition that f is uniforml y cont inuous on
[2, 3].
Hint: Use the fact that x 2 � c 2 ¼ (x þ c)( x � c ).
Notice, thoug h, that the function in Probl em 1 is a continuo us funct ion on a
closed interv al, and reca ll that we saw earlier that continuo us functi ons on
closed intervals have ‘nice properties ’. It tur ns out that clos ed interv als are
important too for unif orm continuit y.
Theorem 2 If a funct ion f is cont inuou s on a closed interv al [ a, b], then it
is uniforml y continuo us on [a , b].
Proof of The orem 2 We use the Bolza no–We ierstras s Th eorem, and a proof
by contradict ion.
So, suppos e that, in fact, f is n ot uniform ly continuo us on [ a, b]. Then for
some choi ce of ", there is NO corr espond ing positiv e numb er � such that
f xð Þ � f yð Þj j < "; for all x and y in a, b½ � satisfying x � yj j < �: (6)
We can reform ulate this statemen t (6) in the followi ng conven ient way :
for any posi tive number � , there are two points x , y in [ a, b] such that
f xð Þ � f yð Þj j � " and x � yj j < �: (7)
By applying ( 7) for the choi ces �n ¼ 1n , n ¼ 1, 2, . . ., in turn, we can then define
two seque nces {xn} and { y n} in [a,b] such that
f xnð Þ � f ynð Þj j � " and xn � y nj j < 1
n ; for n 2 N : (8)
Now, the seque nce {xn} is bo unded, so that b y the Bolza no–We ierstras s
Theore m it cont ains a subsequenc e xnkf g that conver ges to some point c of
[ a, b] as k !1. Then, since xnk� ynk
j j < 1nk
, it follows that the correspo nding
subsequence ynkf g must also converge to c as k!1.
Since f is continuous at c, we must have
f xnkð Þ ! f cð Þ and f ynk
ð Þ ! f cð Þ; as k!1: (9)
For f is strictly decreasing.
Section 4.2.
Example 1 shows that, ingeneral, f may NOT beuniformly continuous if itsinterval of continuity is a half-closed interval.
You may omit this proof on afirst reading.
" is now fixed from here on inthis proof.
We use y rather than c in (6),since it fits in better with thenotation that we will use laterin the proof.
The choice of x and y willdepend, in general, on thechoice of �.
Sub-section 2.5.1, Theorem 3.
From (8).
202 5: Limits and continuity
Now, it follows from (8) that
f xnkð Þ � f ynk
ð Þj j � "; for each k 2 N : (10)
So, if we now let k!1 in (10) and use the limits in (9), we deduce that
0 � ";which is absurd.
This contradiction shows that (7) and so (6) cannot be valid. It follows
that f must be uniformly continuous on [a, b] after all. &
We shall use Theorem 2 later, to prove that a function continuous on a closed
interval [a, b] is integrable on [a, b].
Theorem 2 is one of the important theorems in Analysis beyond the scope of
this book.
5.6 Exercises
Section 5.1
1. Determine whether the following limits exist:
(a) limx!1
x3�1x�1
; (b) limx!1
x3�1x�1j j; (c) lim
x!0esin x.
2. Determine the following limits:
(a) limx!0
sin xþ ex�1x
� �
; (b) limx!0
ex�1sin x
;
(c) limx!0
e xj j�1xj j ; (d) lim
x!1þx3�1x�1j j.
3. Write out the proof of Theorem 4 (the Composition Rule) in Sub-
section 5.1.3.
4. For the functions
f xð Þ ¼0; x ¼ 0;1; x ¼ 1;2; x 6¼ 0; 1;
8
<
:
and g xð Þ ¼ 0; x ¼ 0;1þ xj j; x 6¼ 0;
determine f g 0ð Þð Þ; f
limx!0
g xð Þ�
and limx!0
f g xð Þð Þ.
Section 5.2
1. Prove that
(a) 1x4 !1 as x! 0; (b) cot x!1 as x! 0þ;
(c) exp ex � xð Þ ! 1 as x!1; (d) loge x! �1 as x! 0þ;
(e) xþ sin x!1 as x!1; (f) xx !1 as x!1.
2. Let p xð Þ ¼ xn þ an�1xn�1 þ � � � þ a1xþ a0, where a0; a1; . . .; an�1 2 R .
Prove that p xð Þ ! 1 as x!1.
Hint: Write p xð Þ ¼ xn 1þ r 1x
� �� �
, for a suitable polynomial r.
3. For a> 0 and n2Z, prove that:
(a) eaxxn !1 as x!1; (b) eaxxn ! 0 as x! �1.
Here we use the LimitInequality Rule, Theorem 6of Sub-section 5.1.3.
In Theorem 3, Sub-section 7.2.2.
5.6 Exercises 203
4. For a< 0 and n2Z, prove that:
(a) eaxxn ! 0 as x!1;
(b) eaxxn ! �1; as x! �1; if n is odd;1; as x! �1; if n is even:
Section 5.3
1. Use the strategy based on the "� � definition of limit to prove the following
statements:
(a) limx!3
3x� 4ð Þ ¼ 5; (b) limx!2þ
ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � 4p
¼ 0.
2. Use the strategy based on the "� � definition of limit to prove the following
statements:
(a) limx!0
sin xxj j does not exist; (b) lim
x!0cos 1
x
� �
does not exist.
3. Use the strategy based on the "� � definition of limit to prove that, if
limx!c
f xð Þ ¼ ‘, where ‘ 6¼ 0, then limx!c
1f xð Þ exists and equals 1
‘.
Section 5.4
1. Use the strategy based on the "� � definition of continuity to prove that:
(a) f xð Þ ¼ x5 is continuous at 0; (b) f xð Þ ¼ 1x
is continuous at 2.
2. Use the strategy based on the "� � definition of continuity to prove that the
function
f xð Þ ¼ sin 1x; if x 6¼ 0;
0; if x ¼ 0;
is discontinuous at 0.
3. Use the "� � definition of continuity to prove that, if f and g are continuous
at c, then so is the product fg.
Section 5.5
1. A function f is defined on a bounded interval I, on which it satisfies the
following inequality for some given number K2R :
f xð Þ � f yð Þj j � K x� yj j; for all x; y in I:
Prove that f is (a) bounded on I, and (b) uniformly continuous on I.
2. Give an example of a function f that is continuous and bounded on R but is
not uniformly continuous on R , and verify your assertions.
3. Prove that, if a function f is continuous on a half-closed interval (a, b], then
it is uniformly continuous on (a, b] if and only if limx!aþ
f xð Þ exists (and is
finite).
204 5: Limits and continuity
6 Differentiation
The family of all functions is so large that there is really no possibility of
finding many interesting properties that they all possess. In the last two
chapters we concentrated our attention on the class of all continuous
functions, and we found that continuous functions share some important
properties – for example, they satisfy the Intermediate Value Theorem, the
Extreme Values Theorem and the Boundedness Theorem. However, many
of the most interesting and powerful properties of functions are obtained
only when we further restrict our attention to the class of all differentiable
functions.
You will have already met the idea of differentiating a given function f; that
is, finding the slope of the tangent to the graph y¼ f(x) at those points of the
graph where a tangent exists. The slope of the tangent at the point (c, f(c)) is
called the derivative of f at c, and is written as f 0(c). But when does a function
have a derivative? Geometrically, the answer is: whenever the slope of the
chord through the point (c, f(c)) and an arbitrary point (x, f(x)) of the graph
approaches a limit as x! c. In this chapter we investigate which functions are
differentiable, and we discuss some of the important properties that all differ-
entiable functions possess.
In Section 6.1 we give a strategy for determining whether a given function
f is differentiable at a given point c. In particular, we prove that, if f is
differentiable at c, then it is continuous at c; whereas a function f may be
continuous at c, but not differentiable at c. We also consider functions that
possess higher derivatives; that is, functions which can be differentiated more
than once.
In Section 6.2 we obtain various standard derivatives and rules for differ-
entiation, including the Inverse Function Rule.
In Sections 6.3 and 6.4 we study the properties of functions that are differ-
entiable on an interval, and establish some powerful results about derivatives
which are easy to describe geometrically.
In Section 6.5 we meet an important and useful result, called l’Hopital’s
Rule, which enables us to find limits of the form
limx!c
f xð Þg xð Þ ;
in some of the awkward cases when f (c)¼ g(c)¼ 0.
In Section 6.6 we construct the Blancmange function, a function that is
continuous everywhere on R , but differentiable nowhere on R . It is related to
certain types of fractals.
In fact there exist functionsthat are continuouseverywhere on R , butdifferentiable nowhere on R .This discovery by KarlWeierstrass in 1872 caused asensation in Analysis.
In this case, the Quotient Rulefor limits of functions fails.
205
6.1 Differentiable functions
6.1.1 What is differentiability?
Differentiability arises from the geometric concept of the tangent to a graph.
The tangent to the graph y¼ f (x) at the point (c, f (c)) is the straight line through
the point (c, f (c)) whose direction is the limiting direction of chords joining the
points (c, f (c)) and (x, f (x)) on the graph, as x! c.
The following three examples illustrate some of the possibilities that can
occur when we try to find tangents in particular instances.
The function f (x)¼ x2, x2R , is continuous on R , and its graph has a tangent
at each point; for example, the line y¼ 2x� 1 is the tangent to the graph at the
point (1, 1).
On the other hand, although the function g(x)¼ jx� 1j, x2R , is continuous
on R , its graph does not have a tangent at the point (1, 0); no line through the
point (1, 0) is a tangent to the graph.
Finally, the signum function
k xð Þ ¼�1; x < 0,
0; x ¼ 0,
1; x > 0,
(
is discontinuous at 0, and no line through the point (0, 0) is a tangent to the
graph.
We now make these ideas more precise, using the concept of limit to pin
down what we meant above by ‘limiting direction’. We define the slope of the
graph at (c, f (c)) to be the limit, as x tends to c, of the slope of the chord through
the points (c, f (c)) and (x, f (x)). The slope of this chord is
f xð Þ � f cð Þx� c
;
and this expression is called the difference quotient for f at c. Thus the slope
of the graph of f at (c, f (c)) is
limx!c
f xð Þ � f cð Þx� c
; (1)
provided that this limit exists.
Sometimes it is more convenient to use an equivalent form of the differ-
ence quotient, particularly when we are examining a specific function for
206 6: Differentiation
differentiability at a specific point. If we replace x by cþ h, then ‘x! c’ in (1)
is equivalent to ‘h! 0’. Thus we can rewrite the difference quotient for f at c as
Q hð Þ ¼ f cþ hð Þ � f cð Þh
;
and the slope of the graph of f at (c, f (c)) is then
limh!0
Q hð Þ; (2)
provided that this limit exists.
We say that f is differentiable at c if the graph y¼ f (x) has a tangent at the
point (c, f (c)), and that the derivative of f at c is the limit of the difference
quotient given by expression (1) or expression (2). To formalise this concept,
we need to ensure that f is defined near the point c, and so we assume that c
belongs to some open interval I in the domain of f.
Definitions Let f be defined on an open interval I, and c2 I. Then the
derivative of f at c is
limx!c
f xð Þ � f cð Þx� c
or limh!0
Q hð Þ;
provided that this limit exists. In this case, we say that f is differentiable at c.
The derivative of f at c is denoted by f 0(c), and the function f 0: x 7! f 0 xð Þis called the derived function. The operation of obtaining f 0(x) from f (x)
is called differentiation.
For future reference, we reformulate the definition in terms of " and � as
follows.
Definition Let f be defined on an open interval I, and c2 I. Then f is
differentiable at c with derivative f 0(c) if:
for each positive number ", there is a positive number � such thatf xð Þ�f cð Þ
x�c� f 0 cð Þ
�
�
�
�
�
�< ", for all x satisfying 0< jx� cj<�.
Remarks
1. In ‘Leibniz notation’, f 0(x) is written as dydx
, where y¼ f (x). As we shall see
later, this notation is sometimes useful; however, it is important to recall
that the symbol dydx
is purely notation and does NOT mean some quantity dy
‘divided by’ another quantity dx.
2. The existence of the derivative f 0(c) is not quite equivalent to the existence
of a tangent to the graph y¼ f (x) at the point (c, f (c)). For, if limx!c
f xð Þ�f cð Þx�c
exists (and so is some real number), then the graph has a tangent at the point
(c, f (c)), and the slope of the tangent is the value of this limit. On the other
hand, the graph may have a vertical tangent at the point (c, f (c)). In this
case,f xð Þ�f cð Þ
x�c!1 or �1 as x! c, so that lim
x!c
f xð Þ�f cð Þx�c
does not exist; thus
f is not differentiable at c.
3. Since the concept of a derivative is defined in terms of a limit, we shall use
many results obtained for limits of functions to prove analogous results for
derivatives.
Note that the differencequotient depends on thechoice of c; for differencechoices of c, the differencequotient is generally different.
Sometimes f 0 is denoted by Dfand f 0(x) by Df (x).
This is a simple reformulation,with nothing else taking place.We shall use the definitionoccasionally in this form,especially when provinggeneral results as distinct fromlooking at specific functionsand specific points.
For instance, in the statementof the Composition Rule inSub-section 6.2.2.
6.1 Differentiable functions 207
Example 1 Prove that the function f (x)¼ x3, x2R , is differentiable at any
point c, and determine f 0(c).
Solution At the point c
Q hð Þ ¼ f cþ hð Þ � f cð Þh
¼ cþ hð Þ3� c3
h
¼ 3c2hþ 3ch2 þ h3
h
¼ 3c2 þ 3chþ h2:
Thus, Q(h)! 3c2 as h! 0. It follows that f is differentiable at c, and that
f 0(c)¼ 3c2. &
For comparison, we now prove the same result using the "� � definition of
differentiability; for simplicity, though, we shall assume that c¼ 2 and we shall
simply check that f 0(2)¼ 12. We think that you will see why the Q(h) method is
often preferred!
Solution We have to show that for each positive number ", there is a positive
number � such that
x3 � 8
x� 2� 12
�
�
�
�
�
�
�
�
< "; for all x satisfying 0 < x� 2j j < �;
that is
x2 þ 2x� 8�
�
�
� < "; for all x satisfying 0 < x� 2j j < �:
Now, x2þ 2x� 8¼ (xþ 4)(x� 2); so, for 0< jx� 2j< 1, we have x2 (1, 3)
and jxþ 4j< 7.
Next, choose � ¼ min 1; 17"
� �
. With this choice of �, it follows that, for all x
satisfying 0< jx� 2j<�, we have
x2 þ 2x� 8�
�
�
� ¼ xþ 4j j � x� 2j j
< 7� 1
7" ¼ ":
It follows that f is differentiable at 2, with derivative f 0(2)¼ 12. &
To prove that a function is not differentiable at a point, we use the strategy for
limits that applies in these situations.
Strategy To show that limh!0
Q hð Þ does not exist:
1. Show that there is no punctured neighbourhood of c on which Q is defined;
OR
2. Find two non-zero null sequences {hn} and h0n� �
such that {Q(hn)} and
Q h0n� �� �
have different limits;
OR
3. Find a null sequence {hn} such that Q(hn)!1 or Q(hn)!�1.
Example 2 Prove that the modulus function f(x)¼ jxj, x2R , is not differ-
entiable at 0.
Recall that h 6¼ 0.
Thus, the derived function off is f 0(x)¼ 3x2, x2R .
Here we use the fact thatx3�8¼ x�2ð Þ x2þ2xþ4ð Þ:
We arrange for x2 (1, 3), inorder to concentrate simply onvalues of x near to 2. Thebound 7 for jxþ 4j is simplyused in order to keep somegiven bound (not necessarily asmall bound) for this term.
It is now clear why we madethat particular choice for �; itarose from trying variousvalues and then adjusting thechoice appropriately in orderto end up with a neat ‘"’.Alternatively, we could haveused a different value for � interms of ", and the K " Lemma.
For convenience, we rephrasethis strategy in terms oflimh!0
Q hð Þ rather than limx!c
f xð Þ.
208 6: Differentiation
Solution The graph suggests that the slopes of chords joining the origin to
points on the graph of f to the left and to the right do not tend to the same limit
as these points approach the origin. This suggests that we investigate whether
case 2 of the above strategy may be useful.
At the point 0
Q hð Þ ¼ f 0þ hð Þ � f 0ð Þh
¼ hj j � 0j jh
¼ hj jh:
Now let {hn} and h0n� �
be two null sequences, where hn ¼ 1n
and h0n ¼ � 1n.
Then
Q hnð Þ ¼ Q1
n
� �
¼1n
�
�
�
�
1n
¼1n1n
! 1 as n!1;
and
Q h0n� �
¼ Q � 1
n
� �
¼�1
n
�
�
�
�
�1n
¼1n
�1n
! �1 as n!1:
It follows that f is not differentiable at 0. &
Problem 1
(a) Prove that the function f (x)¼ xn, x2R , n2N , is differentiable at
any point c, and determine f 0(c).
(b) Prove that the constant function on R is differentiable at any point c,
with derivative zero.
Problem 2 Prove that the function f xð Þ ¼ 1x; x 2 R � 0f g, is differen-
tiable at any point c 6¼ 0, and determine f 0(c).
Problem 3 Use the "� � definition of differentiability to prove that the
function f (x)¼ x4 is differentiable at 1, with derivative f 0(1)¼ 4.
Problem 4 Prove that the following functions f are not differentiable
at the given point c:
(a) f xð Þ ¼ xj j12; x 2 R , c¼ 0; (b) f(x)¼ [x], x2R , c¼ 1.
Looking back at Example 2, it looks as though chords joining the origin to
points (h, f (h)) have slopes that tend to a limit 1 if h! 0þ, whereas they have
slopes that tend to a limit�1 if h! 0�. This suggests the concept of one-sided
derivatives that will be useful in our work later on.
This is motivation for theapproach to the solution ratherthan part of the solution itself.
This is the start of thesolution.
y
y = ⏐x⏐
x01n
1n
–
In particular, the derivative ofthe identity function f (x)¼ x,x2R , is the constant functionf 0(x)¼ 1.
Here [x] denotes the integerpart of x.
6.1 Differentiable functions 209
c
slopef ′(c)
slopef ′(c)
y
x
y = f (x)
c
y
x
y = f (x)L
R
Definitions Let f be defined on an interval I, and c2 I. Then the leftderivative of f at c is
f 0L cð Þ ¼ limx!c�
f xð Þ � f cð Þx� c
or f 0L cð Þ ¼ limh!0�
Q hð Þ;
provided that this limit exists. In this case, we say that f is differentiable on
the left at c.
Similarly, the right derivative of f at c is
f 0R cð Þ ¼ limx!cþ
f xð Þ � f cð Þx� c
or f 0R cð Þ ¼ limh!0þ
Q hð Þ;
provided that this limit exists. In this case, we say that f is differentiable on
the right at c.
A function f whose domain contains an interval I is differentiable on I if
it is differentiable at each interior point of I, differentiable on the right at the
left end-point of I (if this belongs to I ) and differentiable on the left at the
right end-point of I (if this belongs to I ).
With this notation, the function f in Example 2 is differentiable on the left at
0 and f 0L 0ð Þ ¼ �1; it is also differentiable on the right at 0, and f 0R 0ð Þ ¼ 1.
The connection between the definitions of differentiability and of one-sided
differentiability is rather obvious.
Theorem 1 A function f whose domain contains an interval I that contains
c as an interior point is differentiable at c if and only if f is both differenti-
able on the left at c and differentiable on the right at c AND
f 0L cð Þ ¼ f 0R cð Þ:
Example 3 Determine whether the function
f xð Þ ¼ xþ x2; �1 � x < 0,
sin x; 0 � x � 2p,
is differentiable at the points c¼�1, 0 and 2p, and determine the correspond-
ing derivatives when they exist.
Solution We investigate the behaviour of the difference quotient Q(h) at
each point in turn.
At �1, the function is not defined to the left of �1; and, for 0< h< 1,
we have
Q hð Þ ¼ f �1þ hð Þ � f �1ð Þh
All these definitions areanalogous to definitions forcontinuity that you met inSub-section 4.1.1.
We omit a proof of thisstraight-forward result.
The common value issimply f 0(c).
y
–1 π2π
x
y = f (x)
210 6: Differentiation
¼�1þ hð Þ þ �1þ hð Þ2
n o
� �1ð Þ þ �1ð Þ2n o
h
¼ �hþ h2
h
¼ �1þ h! �1 as h! 0þ:
It follows that f is differentiable on the right at �1, and f 0R �1ð Þ ¼ �1.
At 0, the function is defined on either side of 0, but by a different formula;
we therefore examine each side separately.
At 0, for 0< h< 2p, we have
Q hð Þ ¼ f hð Þ � f 0ð Þh
¼ sin h� sin 0
h
¼ sin h
h! 1 as h! 0þ:
Also at 0, for �1< h< 0, we have
Q hð Þ ¼ f hð Þ � f 0ð Þh
¼hþ h2� �
� sin 0f gh
¼ hþ h2
h
¼ 1þ h! 1 as h! 0�:
It follows that f is differentiable on the right at 0 and f 0R 0ð Þ ¼ 1, and that f
is differentiable on the left at 0 and f 0L 0ð Þ ¼ 1. Since the left and right deriva-
tives at 0 are equal, it follows that f is differentiable at 0 and f 0 0ð Þ ¼ 1.
Finally, at 2p, the function is not defined to the right of 2p; and,
for �2p< h< 0, we have
Q hð Þ ¼ f 2pþ hð Þ � f 2pð Þh
¼ sin 2pþ hð Þ � sin 2ph
¼ sin h
h! 1 as h! 0�:
It follows that f is differentiable on the left at 2p, and f 0L 2pð Þ ¼ 1. &
Problem 5 Determine whether the function
f xð Þ ¼�x2; �2 � x <0,
x4; 0 � x < 1,
x3; 1 � x � 2,
0; x > 2,
8
>
>
<
>
>
:
is differentiable at the points c¼�2, 0, 1 and 2, and determine the
corresponding derivatives when they exist.
Hint: Sketch the graph y¼ f(x) first.
Remarks
Just as with continuity, the definition of differentiability involves a function
f defined on a set in R , the domain A (say), that maps A to another set in R ,
the codomain B (say).
These remarks are analogousto similar remarks forcontinuity inSub-section 4.1.1.
6.1 Differentiable functions 211
1. Let f and g be functions defined on open intervals I and J, respectively,
where I� J; and let f (x)¼ g(x) on J. Technically g is a different func-
tion from f. However, if f is differentiable at an interior point c of J, it
is a simple matter of some definition checking to verify that g too is
differentiable at c. Similarly, if f is non-differentiable at c, g too is non-
differentiable at c.
2. The underlying point here is that differentiability at a point is a local
property. It is only the behaviour of the function near that point that
determines whether it is differentiable at the point.
6.1.2 Differentiability and continuity
All the functions that we have met so far that are differentiable at a particular
point c are also continuous at that point. In fact, this property holds in general.
Theorem 2 Let f be defined on an open interval I, and c2 I. If f is differ-
entiable at c, then f is also continuous at c.
Proof If f is differentiable at c, then there is some number f 0(c) such that
limx!c
f xð Þ � fðcÞx� c
¼ f 0 cð Þ:
It follows that
limx!c
f xð Þ � f cð Þf g ¼ limx!c
f xð Þ � f cð Þx� c
� x� cð Þ
¼ limx!c
f xð Þ � f cð Þx� c
� limx!c
x� cð Þn o
¼ f 0 cð Þ � 0 ¼ 0:
Hence, by the Sum and Multiple Rules for limits
limx!c
f xð Þ ¼ f cð Þ:
Thus f is continuous at c. &
In fact this gives us a useful test for non-differentiability.
Corollary 1 If f is discontinuous at c, then f is not differentiable at c.
For example, the signum function
f xð Þ ¼�1; x < 0,
0; x ¼ 0,
1; x > 0,
(
is discontinuous at c, and so cannot be differentiable at c.
It is often worth checking whether a function is even continuous at a point
before setting out on a complicated investigation of difference quotients to
determine whether it is differentiable at that point.
differentiable) continuous
There is a similar result forone-sided derivatives.
By the Product Rule forLimits, Sub-section 5.1.3.
Sub-section 5.1.3
Problem 5, Sub-section 4.1.1.
212 6: Differentiation
Problem 6 Prove that the function
f xð Þ ¼ sin 1x; x 6¼ 0,
0; x ¼ 0,
is not differentiable at 0.
Example 4 Show that the function
f xð Þ ¼ x sin 1x; x 6¼ 0,
0; x ¼ 0,
is continuous at 0, but not differentiable at 0.
Solution We proved earlier that f is continuous at 0.
At 0
Q hð Þ ¼ f 0þ hð Þ � f 0ð Þh
¼h sin 1
h
� �
� 0
h
¼ sin1
h
� �
:
Thus, if we define two null sequences xnf g ¼ 1np
� �
and x0n� �
¼n
1
2nþ12ð Þpo
;
n 2 N , we have
Q xnð Þ ¼ sin npð Þ ¼ 0 and Q x0n� �
¼ sin 2nþ 1
2
� �
p� �
¼ sin1
2p
� �
¼ 1;
so that {Q (xn)} and Q x0n� �� �
tend to different limits as n!1. It follows that
f cannot be differentiable at 0. &
Problem 7 Prove that the function
f xð Þ ¼ x2 sin 1x; x 6¼ 0,
0; x ¼ 0,
is differentiable at 0. Given that, for x 6¼ 0; f 0 xð Þ ¼ 2x sin 1x� cos 1
x,
is f 0 continuous at 0?
6.1.3 The sine, cosine and exponential functions
In order to study the differentiability of each of the sine, cosine and exponential
functions, we need to use the following three standard limits.
Theorem 3 Three standard limits
(a) limx!0
sin xx¼ 1; (b) lim
x!0
1�cos xx¼ 0; (c) lim
x!0
ex�1x¼ 1:
Proof The limits (a) and (c) have been verified earlier, so we have only to
verify (b) now.
Using the half-angle formula for cosine, we have
limx!0
1� cos x
x¼ lim
x!0
2 sin2 12
x� �
x
Sub-section 4.1.2,Problem 8 (a).
y
x
y = x sin 1x
y = x2 sin 1x
y
x
(a) was proved in Sub-section 5.1.1; (c) inSub-section 5.1.4, Problem 8.
For cos x ¼ 1� 2 sin2 12
x� �
.
6.1 Differentiable functions 213
¼ limx!0
sin 12
x� �
12
x� sin
1
2x
� �� �
!
¼ limx!0
sin 12
x� �
12
x
!
� limx!0
sin1
2x
� �� �
¼ 1� 0 ¼ 0: &
With this tool, we can now tackle the various functions.
Theorem 4 The function f (x)¼ sin x is differentiable on R , and f 0(x)¼ cos x.
Proof Let c be any point in R . At c, the difference quotient for f is
Q hð Þ ¼ sin cþ hð Þ � sin c
h
¼ sin c cos hþ cos c sin h� sin c
h
¼ cos c� sin h
h� sin c� 1� cos h
h;
so that
limh!0
Q hð Þ ¼ cos c� limh!0
sin h
h� sin c� lim
h!0
1� cos h
h
¼ cos c� 1� sin c� 0
¼ cos c:
It follows that f is differentiable at c, and f 0(c)¼ cos c. &
Problem 8 Prove that the function f (x)¼ cos x is differentiable on R ,
and f 0(x)¼�sin x.
Finally we find the derivative of the exponential function. This is an
extremely important result, as the function f (x)¼ lex, l an arbitrary constant,
is the only function f that satisfies the differential equation f 0(x)¼ f (x) on R .
Theorem 5 The function f (x)¼ ex is differentiable on R , and f 0(x)¼ ex.
Proof Let c be any point in R . At c, the difference quotient for f is
Q hð Þ ¼ ecþh � ec
h
¼ ec � eh � 1
h;
so that
limh!0
Q hð Þ ¼ ec � limh!0
eh � 1
h
¼ ec � 1
¼ ec:
It follows that f is differentiable at c, and f 0(c)¼ ec. &
By the Product Rule forlimits.
Since sine is continuous at 0.
That is, f is its own derivedfunction.
214 6: Differentiation
6.1.4 Higher-order derivatives
In the previous sub-section, we saw that sin0 ¼ cos and cos0 ¼�sin. An exactly
similar argument shows that (�sin)0 ¼�cos and (�cos)0 ¼ sin. Thus �sin
and �cos are all differentiable on R , and so clearly we can differentiate the
functions sin and cos as many times as we please on R .
In general, when we differentiate any given differentiable function f, we
obtain a new function f 0 (whose domain may be smaller than that of f ). The
notion of differentiability can then be applied to the function f 0, just as before,
yielding another function f 00 ¼ ( f 0)0, whose domain consists of those points
where f 0 is differentiable.
Definitions Let f be differentiable on an open interval I, and c2 I. If f 0 is
differentiable at c, then f is called twice differentiable at c, and the number
f 00(c) is called the second derivative of f at c. The function f 00 is called the
second derived function of f.
Provided that the derivatives exist, we can define f 000 or f (3), f (4), . . .,f (n), . . .. The functions f 00, f (3), f (4), . . ., f (n), . . . are called the higher-order
derived functions of f, whose values f 00(x), f (3)(x), f (4)(x), . . ., f (n)(x), . . .are called the higher-order derivatives of f.
However, not all derived functions are themselves differentiable. You have
already seen, for example, that the function
f xð Þ ¼ x2 sin 1x; x 6¼ 0,
0; x ¼ 0,
is differentiable at 0. In fact, it has as its derived function
f 0 xð Þ ¼ 2x sin 1x� cos 1
x; x 6¼ 0,
0; x ¼ 0,
which is not even continuous at 0.
Example 5 Prove that the function
f xð Þ ¼ �12
x2; x < 0,12
x2; x � 0,
is differentiable on R , but that its derived function is not differentiable at 0.
Solution For c> 0, the difference quotient for f at c is
Q hð Þ ¼12
cþ hð Þ2� 12
c2
h
¼ 1
22cþ hð Þ ! c as h! 0;
so that f is differentiable at c and f 0(c)¼ c.
When c¼ 0 and h> 0, a similar argument shows that f is differentiable on
the right at 0 and f 0R 0ð Þ ¼ 0.
For c< 0, the difference quotient for f at c is
Q hð Þ ¼�1
2cþ hð Þ2þ 1
2c2
h
¼ �1
22cþ hð Þ ! �c as h! 0;
f 0 0 ¼ (f 0)0 is sometimes writtenas f (2).
We assume that h issufficiently small that jhj< c,so that cþ h> 0; and hencethat we are using the correctvalue for f at cþ h.
We assume that h issufficiently small thatjhj<�c, so that cþ h< 0;and hence that we are usingthe correct value for f at cþ h.
6.1 Differentiable functions 215
so that f is differentiable at c and f 0(c)¼�c.
When c¼ 0 and h< 0, a similar argument shows that f is differentiable on
the left at 0 and f 0L 0ð Þ ¼ 0.
Since f has f 0L 0ð Þ ¼ f 0R 0ð Þ ¼ 0, it follows that f is differentiable at 0 and
f (0)¼ 0.
Thus the derived function for f is given by
f 0 xð Þ ¼ xj j; x 2 R :
This is the modulus function, which as we showed earlier is not differentiable
at 0. It follows that f 0 is not differentiable at 0. &
Problem 9 Prove that the function
f xð Þ ¼ x2; x < 0;x3; x � 0;
is differentiable on R . Is f 0 differentiable at 0?
6.2 Rules for differentiation
In the last section we showed that the functions sin, cos and exp are differenti-
able on R , by appealing directly to the definition of differentiability.
However, it would be very tedious if, every time that we wished to prove that
a given function is differentiable and to determine its derived function, we had
to use the difference quotient method described in Section 6.1. Sometimes we
do need to use that method, but usually we can avoid the algebra involved in
the difference quotient method by using various rules for differentiation. In this
section, we introduce the Combination Rules, the Composition Rule and the
Inverse Function Rule for differentiable functions. These are similar to the
rules for continuous functions that you met earlier.
Note that each of the rules for differentiation supplies two pieces of
information:
1. a function of a certain type is differentiable;
2. an expression for the derivative.
6.2.1 The Combination Rules
The Combination Rules for differentiable functions are a consequence of the
Combination Rules for limits.
Theorem 1 Combination Rules
Let f and g be defined on an open interval I, and c2 I. Then, if f and g are
differentiable at c, so are:
Sum Rule fþ g, and ( fþ g)0(c)¼ f 0(c)þ g0(c);
Multiple Rule l f, for l2R , and (l f )0(c)¼ l f 0(c);
Product Rule fg, and ( fg)0(c)¼ f 0(c)g (c)þ f (c)g0(c);
Example 2, Sub-section 6.1.1.
In Sections 4.1 and 4.3.
We differentiate each termin turn.
We differentiate one term at atime.
216 6: Differentiation
Quotient Rule fg, provided that g(c) 6¼ 0; and
fg
� �0cð Þ ¼ g cð Þf 0 cð Þ�f cð Þg0 cð Þ
g cð Þð Þ2 :
Now, we saw in Section 6.1 that, for any n2N , the function
x 7! xn; x 2 R ;
is differentiable on R , and that its derived function is
x 7! nxn�1; x 2 R :
We can use this fact, together with the Combination Rules, to prove that any
polynomial function is differentiable on R , and that its derivative can be
obtained by differentiating the polynomial term-by-term.
Corollary 1 Let p(x)¼ a0þ a1xþ a2x2þ � � � þ anxn, x2R , where a0, a1,
a2, . . ., an2R . Then p is differentiable on R , and its derivative is
p0 xð Þ ¼ a1 þ 2a2xþ � � � þ nanxn�1; x 2 R :
Since a rational function is a quotient of two polynomials, it follows from
Corollary 1 and the Quotient Rule that a rational function is differentiable at
all points where the denominator does not vanish (that is, the denominator
does not take the value 0).
Example 1 Prove that the function f xð Þ ¼ x3
x2�1; x 2 R � �1f g, is differenti-
able on its domain, and find its derivative.
Solution The function f is a rational function whose denominator does not
vanish on R � {�1}; hence f is differentiable on this set (the whole of its
domain).
By Corollary 1, the derived function of x 7! x3 is x 7! 3x2, and the derived
function of x 7! x2 � 1 is x 7! 2x. It follows from the Quotient Rule that the
derivative of f is
f 0 xð Þ ¼ x2 � 1ð Þ � 3x2 � x3 � 2xð Þx2 � 1ð Þ2
¼ x4 � 3x2
x2 � 1ð Þ2: &
Problem 1 Find the derivative of each of the following functions:
(a) f (x)¼ x7� 2x4þ 3x3� 5xþ 1, x2R ;
(b) f xð Þ ¼ x2þ1x3�1
; x 2 R � 1f g;(c) f (x)¼ 2 sin x cos x, x2R ;
(d) f xð Þ ¼ ex
3þsin x�2 cos x; x 2 R :
Problem 2 Find the third order derivative of the function f (x)¼ xe2x,
x2R .
In the last section we found the derived functions for sin, cos and exp. We
now ask you to find the derived functions for the remaining trigonometric
functions and the three most common hyperbolic functions.
In particular,
1g
� �0cð Þ ¼ � g0 cð Þ
g cð Þð Þ2 :
6.2 Rules for differentiation 217
Problem 3 Find the derivative of each of the following functions:
(a) f xð Þ ¼ tan x; x 2 R � � 12p;� 3
2p;� 5
2p; . . .
� �
;
(b) f xð Þ ¼ cosec x; x 2 R � 0;�p;�2p; . . .f g;(c) f xð Þ ¼ sec x; x 2 R � � 1
2p;� 3
2p;� 5
2p; . . .
� �
;
(d) f xð Þ ¼ cot x; x 2 R � 0;�p;�2p; . . .f g:
Problem 4 Find the derivative of each of the following functions
(a) f (x)¼ sinh x, x2R ; (b) f(x)¼ cosh x, x2R ;
(c) f (x)¼ tanh x, x2R .
6.2.2 The Composition Rule
In the last sub-section we extended our stock of differentiable functions to
include all rational, trigonometric and hyperbolic functions. However, to differ-
entiate many other functions we need to differentiate composite functions,
such as the function f (x)¼ sin(cos x), x2R , which is the composite of two
differentiable functions – namely, f ¼ sin cos. The Composition Rule tells us
that the composite of two differentiable functions is itself differentiable.
Theorem 2 Composition Rule
Let g and f be defined on open intervals I and J, respectively, and let c2 I and
g Ið Þ J. If g is differentiable at c and f is differentiable at g(c), then f g is
differentiable at c and
ð f gÞ0 cð Þ ¼ f 0 g cð Þð Þg0 cð Þ:
Remarks
1. When written in Leibniz notation, the Composition Rule has a form that is
easy to remember: if we put
u ¼ g xð Þ and y ¼ f uð Þ ¼ f g xð Þð Þthen
dy
dx¼ dy
du� du
dx:
2. We can extend the Composition Rule to a composite of three (or more)
functions; for example
f g hð Þ0 xð Þ ¼ f 0 g h xð Þð Þ½ �g0 h xð Þð Þh0 xð Þ:In Leibniz notation, if we put
v ¼ h xð Þ; u ¼ g vð Þ and y ¼ f uð Þ ¼ f g h xð Þð Þ½ �;
This rule is sometimes knownas the Chain Rule.
We give the proof of the Rulein Sub-section 6.2.4.
We frequently use thisextension of the CompositionRule without mentioning itexplicitly.
218 6: Differentiation
then we obtain the chain
dy
dx¼ dy
du� du
dv� dv
dx:
Example 2 Prove that each of the following composite functions is differ-
entiable on its domain, and find its derivative:
(a) k(x)¼ sin (cos x), x2R ; (b) k(x)¼ cosh (e2x), x2R ;
(c) k xð Þ ¼ tan 14
ex2 þ sin px� �
; x 2 �1; 1ð Þ:
Solution
(a) Here k(x)¼ sin (cos x), so let
g xð Þ ¼ cos x and f xð Þ ¼ sin x; for x 2 R ;
then
g0 xð Þ ¼ �sin x and f 0 xð Þ ¼ cos x; for x 2 R :
By the Composition Rule, k ¼ f g is differentiable on R , and
k0 xð Þ ¼ f 0 g xð Þð Þg0 xð Þ¼ cos cos xð Þ � � sin xð Þ¼ �cos cos xð Þ � sin x:
(b) Here k(x)¼ cosh (e2x), so let
h xð Þ ¼ 2x; g xð Þ ¼ ex and f xð Þ ¼ cosh x; for x 2 R ;
then h, g and f are differentiable on R , and
h0 xð Þ ¼ 2; g0 xð Þ ¼ ex and f 0 xð Þ ¼ sinh x; for x 2 R :
By the (extended form of the) Composition Rule, k ¼ f g h is differ-
entiable on R , and
k0 xð Þ ¼ f 0 g h xð Þð Þ½ �g0 h xð Þð Þh0 xð Þ¼ sinh e2x� �
� e2x � 2
¼ 2e2x sinh e2x:
(c) Here k xð Þ ¼ tan 14
ex2 þ sin px� �
; x 2 �1; 1ð Þ, so let
g xð Þ¼ 1
4ex2 þ sin px� �
; x 2 �1; 1ð Þ; and
f xð Þ¼ tan x; x 2 � 1
2p;
1
2p
� �
:
Now, when x2 (�1, 1), we have
g xð Þj j � 1
4ex2 þ sin pxj j� �
� 1
4eþ 1ð Þ
< 1
<1
2p;
so that g(x) lies in � 12p; 1
2p
� �
, the domain of the differentiable func-
tion tan.
6.2 Rules for differentiation 219
Now, by the Composition and Combination Rules, g is differenti-
able on (�1, 1), and
g0 xð Þ ¼ 1
42xex2 þ p cos px� �
; x 2 �1; 1ð Þ:
Also
f 0 xð Þ ¼ sec2 x; x 2 � 1
2p;
1
2p
� �
:
Finally, it follows, by the Composition Rule, that k ¼ f g is differentiable
on (�1, 1), and
k0 xð Þ¼ f 0 g xð Þð Þg0 xð Þ
¼ sec2 1
4ex2 þ sinpx� �
� �
�1
42xex2 þpcospx� �
; x2 �1;1ð Þ: &
Problem 5 Find the derivative of each of the following functions:
(a) f(x)¼ sinh (x2), x2R ; (b) f (x)¼ sin (sinh 2x), x2R ;
(c) f xð Þ ¼ sin cos 2xx2
� �
; x 2 0;1ð Þ:
6.2.3 The Inverse Function Rule
Earlier we showed that, if a function f with domain some interval I and
image J is strictly monotonic on I, then f possesses a strictly monotonic and
continuous inverse function f�1 on J. In particular, the power function, the
trigonometric functions, the exponential function and the hyperbolic func-
tions all have inverse functions, provided that we restrict the domains, where
necessary.
These standard functions are all differentiable on their domains. Do their
inverse functions also have this property, as their graphs suggest?
As you saw earlier, we obtain the graph y¼ f�1(x) by reflecting the graph
y¼ f(x) in the line y¼ x. This reflection maps a typical point P(c, d) to the
point Q(d, c).
It follows that, if the slope of the tangent to the graph y¼ f (x) at the point P
is f 0(c)¼m, then the slope of the tangent to the graph y¼ f� 1(x) at Q is
f�1ð Þ0 dð Þ ¼ 1m:
However, if the graph of f has a horizontal tangent (m¼ 0) at a point P, then
the graph of f�1 has a vertical tangent at the corresponding point Q; in this case,
f�1 is not differentiable at Q, since 1m
is not defined when m¼ 0. We therefore
require the condition ‘f 0(x) is non-zero’ in our statement of the Rule.
Thus, in the Inverse Function Rule we require:
(a) f is strictly monotonic and continuous on an open interval I, so that f has
a strictly monotonic and continuous inverse function on the open interval
J¼ f(I);
(b) f is differentiable on I and f 0(x) 6¼ 0 on I, so that 1f 0ðxÞ is defined.
The Rule tells us that, under these conditions, f �1 is also differentiable, and
there is a simple formula for ( f �1)0.
Section 4.3.
Recall that strictly monotonicmeans that f is either strictlyincreasing or strictlydecreasing.
y
PQ
x
y = x
y = f –1(x)
y = f (x)
y
P
Q
x
y = x
y = f –1(x)y = f (x)
220 6: Differentiation
Theorem 3 Inverse Function Rule
Let f : I! J, where I is an interval and J is the image f(I), be a function such
that:
1. f is strictly monotonic on I;
2. f is differentiable on I;
3. f 0(x) 6¼ 0 on I.
Then f�1 is differentiable on J. Further, if c2 I and d¼ f (c), then
f�1� �0
dð Þ ¼ 1
f 0 cð Þ :
Remark
The Leibniz notation for derivatives can be used to express the Inverse
Function Rule in a form that is easy to remember: if
y ¼ f xð Þ and x ¼ f�1 yð Þ;
and we write dydx
for f 0 xð Þ, and dxdy
for ( f�1)0(y), then
dx
dy¼ 1
dydx
:
Proof Let F¼ f�1, and let y¼ f (x), where x2 I, so that x¼F(y), where y2 J.
Then the difference quotient for F at d is
F yð Þ � F dð Þy� d
¼ x� c
f xð Þ � f cð Þ
¼ 1f xð Þ�f cð Þ
x�c:
Since f is one–one and differentiable on I, it is necessarily one–one and
continuous on I. Thus F is one–one and continuous on J. It follows that, for
y 6¼ d, we must have x 6¼ c, since x¼ f�1(y) and c¼ f�1(d).
Also, since f�1 is continuous, x! c as y! d. It follows that
limy!d
F yð Þ � F dð Þy� d
¼ limx!c
1f xð Þ�f cð Þ
x�c
!
¼ 1
limx!c
f xð Þ�f cð Þx�c
� �
¼ 1
f 0 cð Þ :
Thus F is differentiable at c, with derivative 1f 0 cð Þ. &
Example 3 For each of the following functions, show that f�1 is differenti-
able and determine its derivative:
(a) f (x)¼ xn, x> 0, n2N; (b) f xð Þ ¼ tan x; x 2 � 12p; 1
2p
� �
;
(c) f (x)¼ ex, x2R .
y
J
dy
x c I x
6.2 Rules for differentiation 221
Solution
(a) The function
f xð Þ ¼ xn; x > 0;
is continuous and strictly increasing on (0,1), and f ((0, 1))¼ (0, 1).
Also, f is differentiable on (0,1), and its derivative f 0(x)¼ nxn�1 is non-
zero there. So f satisfies the conditions of the Inverse Function Rule.
Hence f�1 is differentiable on (0,1); and, if y¼ f (x), then
f�1� �0
yð Þ ¼ 1
f 0 xð Þ ¼1
nxn�1
¼ 1
nx1�n ¼ 1
ny
1�nn :
If we now replace the domain variable y by x, we obtain
f�1� �0
xð Þ ¼ 1
nx
1n�1; x 2 0;1ð Þ:
(b) The function
f xð Þ ¼ tan x; x 2 � 1
2p;
1
2p
� �
;
is continuous and strictly increasing on �12p; 1
2p
� �
, and f � 12p; 1
2p
� �� �
¼ R .
Also, f is differentiable on �12p; 1
2p
� �
, and its derivative f 0 xð Þ ¼ sec2 x is
non-zero there. So f satisfies the conditions of the Inverse Function Rule.
Hence f�1¼ tan�1 is differentiable on R ; and, if y¼ f (x), then
f�1� �0
yð Þ ¼ 1
f 0 xð Þ ¼1
sec2 x
¼ 1
1þ tan2 x¼ 1
1þ y2:
If we now replace the domain variable y by x, we obtain
tan�1� �0
xð Þ ¼ 1
1þ x2; x 2 R :
(c) The function
f xð Þ ¼ ex; x 2 R ;
is continuous and strictly increasing on R , and f (R)¼ (0, 1). Also, f is
differentiable on R , and its derivative f 0(x)¼ ex is non-zero there. So
f satisfies the conditions of the Inverse Function Rule.
Hence f� 1¼ loge is differentiable on (0,1); and, if y¼ f (x), then
f�1� �0
yð Þ ¼ 1
f 0 xð Þ ¼1
ex
¼ 1
y:
If we now replace the domain variable y by x, we obtain
logeð Þ0 xð Þ ¼ 1
x; x 2 0;1ð Þ: &
Problem 6 For each of the following functions f, show that f�1 is
differentiable and determine its derivative:
(a) f(x)¼ cos x, x2 (0, p); (b) f(x)¼ sinh x, x2R .
y¼ xn, so that x ¼ y1n:
222 6: Differentiation
Most equations of the form y¼ f (x) cannot be solved explicitly to give x as
some formula involving y alone. The Inverse Function Rule is often used in
such situations to solve problems that would otherwise be intractable. The
following problem illustrates this type of application.
Problem 7 Prove that the function f (x)¼ x5þ x� 1, x2R , has an
inverse function f�1 which is differentiable on R . Find the values of
( f�1)0(d ) at those points d corresponding to the points c¼ 0, 1 and �1,
where d¼ f (c).
Exponential functions
Earlier we defined the number ax, for a> 0, by the formula
ax ¼ exp x loge að Þ:Since the functions exp and log are differentiable on R and (0,1), respectively,
it follows that we can use this formula to determine the derivatives of functions
such as x 7! x�; x 7! ax and x 7! xx.
Example 4 Prove that, for �2R , the power function f (x)¼ x�, x2 (0,1), is
differentiable on its domain, and that f 0(x)¼�x��1.
Solution By definition
f xð Þ ¼ exp � loge xð Þ:The function x 7!� loge x is differentiable on (0,1), with derivative �
x. By the
Composition Rule, f is differentiable on (0,1) with derivative
f 0 xð Þ ¼ exp � loge xð Þ � �
x
� �
¼ x� � �
x
� �
¼ �x��1; x 2 0;1ð Þ: &
Remark
In the case when � is a rational number, � ¼ mn
say, the result of Example 4 can
also be proved by applying the Composition Rule to the functions x 7! xm and
x 7! x1n. The function x 7! x
1n is differentiable on (0,1), by the Inverse Function
Rule, as it is the inverse of the function x 7! xn.
Example 5 Prove that, for a> 0, the function f (x)¼ ax, x2R , is differenti-
able on its domain, and that f 0(x)¼ ax loge a.
Solution By definition
f xð Þ ¼ exp x loge að Þ:The function x 7! x loge a is differentiable on R , with derivative loge a. By the
Composition Rule, f is differentiable on R , with derivative
f 0 xð Þ ¼ exp x loge að Þ � loge a
¼ ax loge a; x 2 R : &
Problem 8 Prove that the function f (x)¼ xx, x2 (0,1), is differenti-
able on its domain, and find its derivative.
Section 4.4.
Section 6.1.
6.2 Rules for differentiation 223
We end this sub-section with a list of basic differentiable functions.
Basic differentiable functions The following functions are differentiable
on their domains:
� polynomials and rational functions
� nth root function
� trigonometric functions (sine, cosine and tangent)
� exponential function
� hyperbolic functions (sinh, cosh and tanh).
6.2.4 Proofs
We now supply the proofs of the Combination Rules and the Composition
Rule, which we omitted earlier. We also illustrate how to prove such results
using the "� � method.
Theorem 1 Combination Rules
Let f and g be defined on an open interval I, and c2 I. Then, if f and g are
differentiable at c, so are:
Sum Rule fþ g, and ( fþ g)0(c)¼ f 0(c)þ g0(c);
Multiple Rule l f, for l2R , and (l f )0(c)¼ l f 0(c);
Product Rule fg, and ( fg)0(c)¼ f 0(c) g(c)þ f (c) g0(c);
Quotient Rule fg, provided that g (c) 6¼ 0; and f
g
� �0cð Þ¼ g cð Þf 0 cð Þ�f cð Þg0 cð Þ
g cð Þð Þ2 :
Proof
Sum Rule
Let F¼ fþ g. Then
limx!c
F xð Þ � F cð Þx� c
¼ limx!c
f xð Þ þ g xð Þf g � f cð Þ þ g cð Þf gx� c
¼ limx!c
f xð Þ � f cð Þx� c
þ limx!c
g xð Þ � g cð Þx� c
¼ f 0 cð Þ þ g0 cð Þ:Thus F is differentiable at c, with derivative f 0(c)þ g0(c).
Multiple Rule
This is just a special case of the Product Rule, with g(x)¼ l.
Product Rule
Let F¼ fg. Then
limx!c
F xð Þ�F cð Þx� c
¼ limx!c
f xð Þg xð Þ� f cð Þg cð Þx� c
You may omit these proofs ata first reading.
In this proof we use theCombination Rules for limitsgiven in Sub-section 5.1.
224 6: Differentiation
¼ limx!c
f xð Þ � f cð Þf gg xð Þ þ f cð Þ g xð Þ � g cð Þf gx� c
¼ limx!c
f xð Þ � f cð Þx� c
g xð Þ þ limx!c
f cð Þ g xð Þ � g cð Þx� c
¼ limx!c
f xð Þ � f cð Þx� c
� limx!c
g xð Þ þ f cð Þ � limx!c
g xð Þ � g cð Þx� c
¼ f 0 cð Þg cð Þ þ f cð Þg0 cð Þ;since f and g are differentiable at c, and since g is continuous at c, so that
g(x)! g(c) as x! c.
Thus F is differentiable at c, with derivative f 0(c)g(c)þ f (c)g0(c).
Quotient Rule
Let F ¼ fg.
Recall, first, that, since g is continuous at c (as it is differentiable there) and
g (c) 6¼ 0, there exists an open interval J containing c on which g(x) 6¼ 0. Thus
the domain of F contains the open interval J, and c2 J.
Then
limx!c
F xð Þ�F cð Þx� c
¼ limx!c
f xð Þg xð Þ�
f cð Þg cð Þ
x� c
¼ limx!c
f xð Þg cð Þ� f cð Þg xð Þx� cð Þ�g xð Þg cð Þ
¼ limx!c
f xð Þ� f cð Þf gg cð Þ� f cð Þ g xð Þ�g cð Þf gx� cð Þ�g xð Þg cð Þ
¼ limx!c
f xð Þ� f cð Þx� cð Þ�g xð Þ� lim
x!c
f cð Þg cð Þ�
g xð Þ�g cð Þx� cð Þ�g xð Þ
¼ limx!c
f xð Þ� f cð Þx� c
� limx!c
1
g xð Þ�f cð Þg cð Þ� lim
x!c
g xð Þ�g cð Þx� cð Þ � lim
x!c
1
g xð Þ
¼ f 0 cð Þg cð Þ �
f cð Þg0 cð Þg2 cð Þ ¼
f 0 cð Þg cð Þ� f cð Þg0 cð Þg2 cð Þ ;
since f and g are differentiable at c, and g is continuous at c.
Thus F is differentiable at c, with derivativef 0 cð Þg cð Þ�f cð Þg0 cð Þ
g2 cð Þ : &
As an illustration of how such results may be proved using the "� � method
that we introduced in Section 5.4, we prove just the Sum Rule. Notice that in
our proofs we avoid many complications by judicious use of the ‘K" Lemma’.
Proof of the Sum Rule Let F¼ fþ g.
In view of the K" Lemma, we must prove that:
for each positive number ", there is a positive number � such that
F xð Þ � F cð Þx� c
� f 0 cð Þ þ g0 cð Þf g�
�
�
�
�
�
�
�
< K" for all x satisfying
0 < x� cj j < �: (1)
First, we write the expression on the left-hand side of (1) in a convenient
form as
For, differentiable)continuous.
The ‘K" Lemma’ firstappeared in Sub-section 2.2.2.
As you work through thisproof, compare it with theearlier proof using limits.
Recall that K must NOT dependon x or ".
6.2 Rules for differentiation 225
f xð Þ þ g xð Þf g � f cð Þ þ g cð Þf gx� c
� f 0 cð Þ þ g0 cð Þf g
¼ f xð Þ � f cð Þf g þ g xð Þ � g cð Þf gx� c
� f 0 cð Þ þ g0 cð Þf g
¼ f xð Þ � f cð Þx� c
� f 0 cð Þ
þ g xð Þ � g cð Þx� c
� g0 cð Þ
: (2)
We now use the information that we already have to tackle each of the terms
in (2) in turn.
Since f is differentiable at c, there exists some positive number �1 such that
f xð Þ� f cð Þx� c
� f 0 cð Þ�
�
�
�
�
�
�
�
<"; for all x satisfying 0< x� cj j<�1: (3)
Also, since g is differentiable at c, there exists some positive number �2
such that
g xð Þ�g cð Þx� c
�g0 cð Þ�
�
�
�
�
�
�
�
<"; for all x satisfying 0< x� cj j<�2: (4)
Now let �¼min{�1, �2}. It follows that, for all x satisfying 0< jx� cj<�,both (3) and (4) hold. Hence, if we apply the Triangle Inequality to (2) and then
use both (3) and (4), we find that, for all x satisfying 0< jx� cj<�f xð Þ þ g xð Þf g � f cð Þ þ g cð Þf g
x� c� f 0 cð Þ þ g0 cð Þf g
�
�
�
�
�
�
�
�
� f xð Þ � f cð Þx� c
� f 0 cð Þ�
�
�
�
�
�
�
�
þ g xð Þ � g cð Þx� c
� g0 cð Þ�
�
�
�
�
�
�
�
< "þ " ¼ 2":
But this is just the statement (1) that we set out to prove, with K¼ 2. &
Remark
Note that our use of the ‘K" Lemma’ meant that we did not need to know at the
steps (3) and (4) to use expressions like 12" rather than " in order to end up with
a final conclusion that ‘some expression is<"’! For ‘some expression is <K"’is sufficient.
We end with the proof of the Composition Rule.
Theorem 2 Composition Rule
Let g and f be defined on open intervals I and J, respectively, and let c2 I
and g Ið Þ J. If g is differentiable at c and f is differentiable at g(c), then
f g is differentiable at c and
f gð Þ0 cð Þ ¼ f 0 g cð Þð Þg0 cð Þ:
This expression appears in (2).
By the Triangle Inequality.
By (3) and (4).
We would only discover theneed for such a choice oncewe had made a first attemptat the proof.
226 6: Differentiation
Proof Let F ¼ f g, and let y¼ g(x) and d¼ g(c).
The difference quotient for F at c is
F xð Þ � F cð Þx� c
¼ f g xð Þð Þ � f g cð Þð Þx� c
: (5)
Now, the right-hand side of (5) is equal to
f yð Þ � f dð Þy� d
� g xð Þ � g cð Þx� c
; (6)
provided that y 6¼ d.
Unfortunately, for some choices of g and c it is possible for y¼ d to occur
when y¼ g(x) and x is arbitrarily close to c but not actually equal to c. In such
situations the expression (6) will not exist.
To get round this problem, we introduce a carefully chosen auxiliary
function
h yð Þ ¼f yð Þ�f dð Þ
y�d; y 6¼ d,
f 0 dð Þ; y ¼ d.
Since f is differentiable at d, h(y)! f 0(d) as y! d; and, since h(d )¼ f 0(d ), it
follows that h is continuous at d.
We then apply the Composition Rule for continuous functions, to deduce
that the composite function
h g xð Þ ¼f g xð Þð Þ�f g cð Þð Þ
g xð Þ�g cð Þ ; g xð Þ 6¼ g cð Þ;f 0 g cð Þð Þ; g xð Þ ¼ g cð Þ;
(
is continuous at c.
Next, notice that, if g(x) 6¼ g(c), it follows from (5) that
F xð Þ � F cð Þx� c
¼ h g xð Þ � g xð Þ � g cð Þx� c
; (7)
and also that this last statement (7) is also valid when g(x)¼ g(c), since then
both sides of (7) are zero.
Hence, if we let x! c in (7) and use the continuity of the function h g, we
obtain
limx!c
F xð Þ � F cð Þx� c
¼ h g cð Þ � g0 cð Þ
¼ f 0 g cð Þð Þg0 cð Þ:
Thus F is differentiable at c, with derivative f 0(g(c))g0(c). &
Remark
Let
g xð Þ ¼ x2 sin 1x
� �
; x 6¼ 0,
0; x ¼ 0,
c ¼ 0; and d ¼ g 0ð Þ ¼ 0:
Some texts ignore thispossibility, and so giveincomplete proofs ofTheorem 2! In the Remarkbelow, we describe one suchsituation.
6.2 Rules for differentiation 227
Since sin(np)¼ 0 for all n2N , it follows that y¼ g(x) takes the value d¼ 0
at the points 1np that are arbitrarily close to c¼ 0 but not equal to 0.
6.3 Rolle’s Theorem
In the next two sections we describe some of the fundamental properties of
functions that are differentiable, not just at a particular point but on an
interval. Our results are motivated by the geometric significance of differ-
entiability in terms of tangents, and explain why the graphs of differentiable
functions possess certain geometric properties.
6.3.1 The Local Extremum Theorem
Earlier we described some of the fundamental properties of functions which
are continuous on a closed interval. In particular, we proved the Extreme
Values Theorem, which states that, if a function ƒ is continuous on a closed
interval [a, b], then there are points c, d in [a, b] such that
f cð Þ � f xð Þ � f dð Þ; for all x 2 ½a; b�;
we called f(c) the minimum of ƒ on [a, b], and f(d) the maximum of ƒ on [a, b].
We used the term extremum to denote either a maximum or a minimum.
But how do we determine the points c, d where these extrema occur? In
general, unfortunately, this is not easy. However, if the function ƒ is differenti-
able, then we can, in principle, determine c and d by first finding any local
maxima or local minima of the function ƒ on the interval [a, b].
Roughly speaking, for a point c in (a, b), the value f(c) is a local maximum of
ƒ on [a, b] if
f ðxÞ � f ðcÞ; for all x in ½a; b� sufficiently close to c;
and a local minimum of ƒ on [a, b] if
f ðxÞ � f ðcÞ; for all x in ½a; b� sufficiently close to c:
To make this idea more precise, we use the concept of neighbourhood that you
met earlier.
Definitions Let ƒ be defined on an interval I, and let c2 I. Then:
� ƒ has a local maximum f(c) at c if there exists a neighbourhood N of c
such that
f ðxÞ � f ðcÞ; for x 2 N \ I;
� ƒ has a local minimum f(c) at c if there exists a neighbourhood N of c
such that
f ðxÞ � f ðcÞ; for x 2 N \ I;
� ƒ has a local extremum at c if f(c) is either a local maximum or a local
minimum.
This example shows that theparagraph ‘Unfortunately . . .’after equation (6) above doesdescribe a situation that canoccur.
Section 4.2.
Sub-section 4.2.3.
From this point onwards, weshall commonly use the letterc to denote an extremum – thatis EITHER a maximum OR aminimum.
Sub-section 5.1.1.
Notice that N \ I necessarilyincludes an interval of theform (c� r, c] or [c, cþ s),where r, s> 0.
228 6: Differentiation
When we wish to locate local extrema of a differentiable function ƒ, instead of
using the above definition we usually use the following result, which gives a
connection between local extrema of a function ƒ and the points where f 0 vanishes.
Theorem 1 Local Extremum Theorem
Let ƒ be defined on an interval [a, b]. If ƒ has a local extremum at c, where
a< c< b, and if ƒ is differentiable at c, then
f 0ðcÞ ¼ 0:
Proof Suppose that ƒ has a local maximum at c.
Since a< c< b, it follows from the definition of local maximum that there
exists a neighbourhood N of c with N ½a, b� such that
f ðxÞ � f ðcÞ; for x 2 N:
We now choose numbers r, s> 0 such that N¼ (c� r, cþ s).
First, looking to the left of c, we have
f ðxÞ � f ðcÞ � 0 and x� c < 0; for c� r < x < c;
so that
f ðxÞ � f ðcÞx� c
� 0; for c� r < x < c:
Thus
fL0ðcÞ ¼ lim
x!c�
f ðxÞ � f ðcÞx� c
� 0: (1)
Next, looking to the right of c, we have
f ðxÞ � f ðcÞ � 0 and x� c > 0; for c < x < cþ s;
so that
f ðxÞ � f ðcÞx� c
� 0; for c < x < cþ s:
Thus
fR0ðcÞ ¼ lim
x!cþ
f ðxÞ � f ðcÞx� c
� 0: (2)
Since ƒ is differentiable at c, the left and right derivatives at c exist and are
equal. Hence, in view of inequalities (1) and (2), their common value, f 0ðcÞ,must be 0. &
We prove only the localmaximum version: the proofof the local minimum versionis similar.
6.3 Rolle’s Theorem 229
Remarks
1. The Local Extremum Theorem applies only if the function is differentiable
at a local extremum. For example, the function
f ðxÞ ¼ jxj; x 2 ½�1; 1�;has a local minimum 0 at 0, but ƒ is not differentiable at 0.
2. The Local Extremum Theorem does not assert that a point where the deriva-
tive vanishes is necessarily a local extremum. For example, the function
f ðxÞ ¼ x3; x 2 ½�1; 1�;does not have a local extremum at 0, although f 0(0)¼ 0.
3. The Local Extremum Theorem does not make any assertion about a
local extremum that occurs at a point c that is one of the end-points of the
interval [a, b].
Problem 1 Find the local extrema of the function
f xð Þ ¼ 1
4x4 � 1
3x3; x 2 �1; 2½ �:
Clearly any extremum of ƒ on [a, b] that occurs at a point other than a or b
must be a local extremum. It follows from Theorem 1 that such a point c must
be a point where f 0(c)¼ 0. Thus an immediate consequence of the Local
Extremum Theorem is the following criterion for finding all the extrema of
‘well-behaved functions’ on closed intervals.
Corollary 1 Let ƒ be continuous on the closed interval [a, b] and differ-
entiable on the open interval (a, b). Then the extrema of ƒ on [a, b] can occur
only at a, at b, or at points c in (a, b) where f 0(c)¼ 0.
We now reformulate Corollary 1 as a strategy for locating minima and
maxima.
Strategy To determine the maximum and the minimum of a function ƒwhich is continuous on [a, b] and differentiable on (a, b):
1. Determine the points c1, c2, . . . in (a, b) where f 0 vanishes;
2. Compare the values of f (a), f (b), f (c1), f (c2), . . .;
the least is the minimum, and the greatest is the maximum.
Problem 2 Use the above Strategy to determine the minimum and the
maximum of the function f xð Þ ¼ sin2 xþ cos x, for x 2 0; 12p
�
:
6.3.2 Rolle’s Theorem
In the previous sub-section we saw that, if a function ƒ is continuous on
the closed interval [a, b] and differentiable on the open interval (a, b), then
the extrema of ƒ can occur only at a, at b, or at points c in (a, b) where
f 0(c)¼ 0.
230 6: Differentiation
Now the function
f ðxÞ ¼ sin1
2px
� �
; x2 � 2
3;
2
3
� �
;
shows that it can happen that no interior point of (a, b) corresponds to a
maximum or a minimum of ƒ, and that f 0 need not vanish at any interior
point at all.
However, the situation is quite different for the function
f ðxÞ ¼ sin1
2px
� �
; x2½�2; 2�:
Here f(�2)¼ f(2)¼ 0; on [�2, 2] the function f has a maximum at 1 and a
minimum at �1, and at both these interior points f 0 vanishes. This is a special
case of Rolle’s Theorem which asserts that, if f(a)¼ f(b), then there is at least
one point c strictly between a and b at which f 0 vanishes.
Theorem 2 Rolle’s Theorem
Let ƒ be defined on the closed interval [a, b] and differentiable on the open
interval (a, b). If f(a)¼ f(b), then there exists some point c, with a< c< b,
for which f 0(c)¼ 0.
Remarks
1. This apparently simple result is one of the most important results in
Analysis.
2. In geometric terms, Rolle’s Theorem means that, if the line joining the
points (a, f(a)) and (b, f(b)) on the graph of ƒ is horizontal, then so is the
tangent to the graph at some point c in (a, b).
3. There may be more than one point c in (a, b) at which f 0 vanishes (as in the
diagram in the margin). Rolle’s Theorem simply asserts that at least one
such point c exists.
Proof If ƒ is constant on [a, b], then f 0(x)¼ 0 everywhere in (a, b); in this
case, we may take c to be any point of (a, b).
If ƒ is non-constant on [a, b], then either the maximum or the minimum (or
both) of ƒ on [a, b] is different from the common value f (a)¼ f (b). Since one
of the extrema occurs at some point c with a< c< b, the Local Extremum
Theorem applied to the point c shows that f 0(c) must be zero. &
Example 1 Verify that the conditions of Rolle’s Theorem are satisfied by the
function
f xð Þ ¼ 3x4 � 2x3 � 2x2 þ 2x; x 2 �1; 1½ �;and determine a value of c in (�1, 1) for which f 0(c)¼ 0.
Solution Since f is a polynomial function, f is continuous on [�1, 1] and
differentiable on (�1, 1). Also, f (�1)¼ f (1)¼ 1. Thus f satisfies the condi-
tions of Rolle’s Theorem on [�1, 1].
It follows that there exists a number c2 (�1, 1) for which f 0(c)¼ 0. Now
That is, the extrema maynot occur at interior pointsof (a, b).
This is an existence theorem.Often it is difficult to evaluatec explicitly.
Since ƒ is continuous on[a, b], ƒ must have both amaximum and a minimum on[a, b], by the Extreme ValuesTheorem.
6.3 Rolle’s Theorem 231
f 0 xð Þ ¼ 12x3 � 6x2 � 4xþ 2
¼ 12 x2 � 1
3
� �
x� 1
2
� �
;
so that f 0 vanishes at the points 1ffiffi
3p , � 1
ffiffi
3p and 1
2in (�1, 1). Any of these three
numbers will serve for c. &
Problem 3 Verify that the conditions of Rolle’s Theorem are satisfied
by the function
f xð Þ ¼ x4 � 4x3 þ 3x2 þ 2; x 2 1; 3½ �;and determine a value of c in (1,3) for which f 0(c)¼ 0.
Problem 4 For each of the following functions, state whether Rolle’s
Theorem applies for the given interval:
(a) f xð Þ ¼ tan x; x 2 0; p½ �;(b) f xð Þ ¼ xþ 3 x� 1j j; x 2 0; 2½ �;(c) f xð Þ ¼ x� 9x17 þ 8x18; x 2 0; 1½ �;(d) f xð Þ ¼ sin xþ tan�1 x; x 2 0; 1
2p
�
:
6.4 The Mean Value Theorem
Here we continue to study the geometric properties of functions that are
differentiable on intervals, and describe some of their applications.
6.4.1 The Mean Value Theorem
First, recall the geometric interpretation of Rolle’s Theorem: Under suitable
conditions, if the chord joining the points (a, f (a)) and (b, f (b)) of the graph of f
is horizontal, then so is the tangent at some point c of (a, b).
If you imagine pushing the chord (as shown in the margin), always parallel
to itself, until it is just about to lose contact with the graph of f, it looks as
though at this point the chord becomes a tangent to the graph. Similarly, the
‘chord-pushing’ approach suggests that, even if the original chord is not
horizontal (that is, if f(a) 6¼ f(b)), there must still be some point c of (a, b) at
which the tangent is parallel to the chord.
Example 1 Consider the function
f xð Þ ¼ 3� 3xþ x3; x 2 1; 2½ �:Find a point c of (1, 2) such that the tangent to the graph of f is parallel to the
chord joining (1, f (1)) to (2, f (2)).
Solution Since f(1)¼ 3� 3þ 1¼ 1 and f(2)¼ 3� 6þ 8¼ 5, the slope of the
chord joining the endpoints of the graph is
f 2ð Þ � f 1ð Þ2� 1
¼ 5� 1
2� 1¼ 4:
232 6: Differentiation
Now, since f is a polynomial, it is differentiable on (1, 2) and its derivative is
f 0(x)¼�3þ 3x2; hence f 0(c)¼ 4 when 3c2¼ 7, or c ¼ffiffi
73
q
’ 1:53. Thus, at
the point (c, f(c)) the tangent to the curve is parallel to the chord joining the
end-points. &
We now generalise Rolle’s Theorem and assert that there is always a point
where the tangent to the graph is parallel to the chord joining the end-points.
This result is known as the Mean Value Theorem, so-called since
f bð Þ � f að Þb� a
can be thought of as the mean value of the derivative between a and b.
Theorem 1 Mean Value Theorem
Let f be continuous on the closed interval [a, b] and differentiable on the
open interval (a, b). Then there exists a point c in (a, b) such that
f 0 cð Þ ¼ f bð Þ � f að Þb� a
:
The idea of the proof is as follows. We define h(x) to be the vertical distance
from the chord to the curve; then h(a) and h(b) are both 0; in fact, h satisfies all
the conditions of Rolle’s Theorem. Applying Rolle’s Theorem to h, we obtain
the desired result.
Proof The slope of the chord joining the points (a, f (a)) and (b, f (b)) is
m ¼ f bð Þ � f að Þb� a
;
and so the equation of the chord is
y ¼ m x� að Þ þ f að Þ:It follows that the vertical height, h(x), between points with ordinate x on the
graph and those on the chord is given by
h xð Þ ¼ f xð Þ � m x� að Þ þ f að Þ½ �:Now h(a)¼ h(b)¼ 0, and h is continuous on [a, b] and differentiable on (a, b).
Thus h satisfies all the conditions of Rolle’s Theorem.
It follows from Rolle’s Theorem that there exists some point c in (a, b) for
which h0(c)¼ 0. But, since h0(c)¼ f 0(c)�m, it follows that
f 0 cð Þ ¼ m ¼ f bð Þ � f að Þb� a
: &
Example 2 Verify that the conditions of the Mean Value Theorem are
satisfied by the function f xð Þ ¼ x�1xþ1
; x 2
1; 72
�
; and find a value for c that
satisfies the conclusion of the theorem.
Solution The function f is a rational function whose denominator is non-zero
on
1; 72
�
; so f is continuous on
1; 72
�
and differentiable on�
1; 72
�
: Thus
f satisfies the conditions of the Mean Value Theorem.
Nowf 7
2
� �
� f 1ð Þ72� 1
¼59� 052
¼ 2
9;
Again, this is an existencetheorem.
Note that when f(a)¼ f(b), theMean Value Theorem simplyreduces to Rolle’s Theorem.
y = f (x)
1
y
x72
59
6.4 The Mean Value Theorem 233
and
f 0 xð Þ ¼ 2
xþ 1ð Þ2;
so the Mean Value Theorem asserts that there exists some point c in 1; 72
� �
for
which f 0(c)¼ 29. Thus
2
cþ 1ð Þ2¼ 2
9;
and so cþ 1ð Þ2¼ 9. It follows that c¼ 2. &
Problem 1 For each of the following functions, verify that the condi-
tions of the Mean Value Theorem are satisfied, and find a value for c that
satisfies the conclusion of the theorem:
(a) f xð Þ ¼ x3 þ 2x; x 2 �2; 2½ �; (b) f xð Þ ¼ ex; x 2 0; 3½ �:
6.4.2 Positive, negative and zero derivatives
We now study some consequences of the Mean Value Theorem for functions
whose derivatives are always positive, always negative, or always zero.
First, we prove a crucial result about monotonic functions which you use
regularly to sketch the graph of a function f. It concerns the behaviour of f on a
general interval I, so here we denote the interior of I (the set of all interiorpoints of I ) by Int I.
Theorem 2 Increasing-Decreasing Theorem
Let f be continuous on an interval I and differentiable on Int I.
(a) If f 0(x)� 0 on Int I, then f is increasing on I.
(b) If f 0(x)� 0 on Int I, then f is decreasing on I.
Proof Choose any two points x1 and x2 in I, with x1< x2. The function f
satisfies the conditions of the Mean Value Theorem on the interval [x1, x2],
so there exists a point c in (x1, x2) such that
f x2ð Þ � f x1ð Þx2 � x1
¼ f 0 cð Þ:
It follows that f (x2)� f(x1) must have the same sign as f 0(c).
(a) If f 0(x)� 0 on Int I, then f(x2)� f(x1)� 0, so that f(x2)� f(x1). Thus f is
increasing on I.
(b) If f 0(x)� 0 on Int I, then f(x2)� f(x1) � 0, so that f(x2)� f (x1). Thus f is
decreasing on I. &
Remark
If the inequalities in the statement of Theorem 2 are replaced by strict inequal-
ities, the conclusions of the Theorem become the following:
(a) If f 0(x)> 0 on Int I, then f is strictly increasing on I;
(b) If f 0(x)< 0 on Int I, then f is strictly decreasing on I.
This is a quadratic equationwith roots c¼ 2 and �4; weignore the solution c¼�4,since it lies outside
�
1; 72
�
.
For example, if I¼ [0, 1), thenInt I¼ (0, 1).
The proofs of these assertionsare similar to the proofs inTheorem 2.
234 6: Differentiation
Problem 2 For each of the following functions f, determine whether
f is increasing, strictly increasing, decreasing or strictly decreasing:
(a) f xð Þ ¼ 3x43 � 4x; x 2 1;1½ Þ; (b) f xð Þ ¼ x� loge x; x 2 0; 1ð �:
Two useful consequences of Theorem 2 are the following corollaries.
Corollary 1 Zero Derivative Theorem
Let f be continuous on an interval I and differentiable on Int I. If f 0(x)¼ 0
for all x in Int I, then f is constant on I.
Proof Cases (a) and (b) of Theorem 2 both apply, so that f is both increasing
and decreasing on I.
Hence f is constant on I. &
As an illustration of the use of this important result, we can now prove the
claim made earlier that the function f (x)¼ lex, l an arbitrary constant, is the
only function f that satisfies the differential equation f 0(x)¼ f (x) on R . For, if
f 0(x)¼ f (x), thend
dxe�xf xð Þð Þ ¼ e�xf 0 xð Þ � e�xf xð Þ
¼ e�x f 0 xð Þ � f xð Þð Þ ¼ 0;
it then follows from Corollary 1 that e�x f(x) is just some constant l, say, so
that f (x)¼ lex.
Corollary 2 Let f and g be continuous on an interval I and differentiable
on Int I. If f 0(x)¼ g0(x) for all x in Int I, then
f(x)¼ g(x)þ c for all x in I, for some constant c.
Proof This follows immediately by applying Corollary 1 to the function
h¼ f� g, since h0(x)¼ 0 for all x in the interior of I. &
Example 3 Prove that sinh�1 x ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 1p
� �
, for all x2R .
Solution Let
f xð Þ ¼ sinh�1 x; x 2 R ;
g xð Þ ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 1p
� �
; x 2 R :
Then f and g are continuous on R and differentiable on R .
Now
f 0 xð Þ ¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 1p ; x 2 R ;
also, by the Composition Rule for Derivatives
g0 xð Þ ¼ 1
xþffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 1p � 1þ 1
2� 2x x2 þ 1
� ��12
� �
¼1þ x
ffiffiffiffiffiffiffiffiffi
x2 þ 1p
xþffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 1p
¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 1p ; x 2 R :
Sub-section 6.1.3.
For h is continuous on I anddifferentiable on Int I.
6.4 The Mean Value Theorem 235
Hence f 0(x)¼ g0(x) for all x in R .
It follows from Corollary 2 that
sinh�1 x ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 1p
� �
þ c;
for some constant c. Putting x¼ 0 in the above identity, we obtain 0¼ loge(1)þ c,
so that c¼ 0. &
Problem 3 Use Corollary 1 to prove the following identities:
(a) sin�1 xþ cos�1 x ¼ 12p, for x 2 �1; 1½ �;
(b) tan�1 xþ tan�1 1x
� �
¼ 12p, for x> 0.
Next, note that second derivatives can often be used to identify whether a
point c where f 0(c)¼ 0 is a local maximum or a local minimum of a function f.
Suppose that f is defined on a neighbourhood of c, and that
f 0 cð Þ ¼ 0 and f 00 cð Þ > 0:
It can be shown that there is some punctured neighbourhood N ¼ c� r; cð Þ [c; cþ sð Þ of c such that
f 0 xð Þ � f 0 cð Þx� c
> 0; for x 2 N;
that is
f 0 xð Þx� c
> 0; for x 2 N:
We can rewrite this inequality in the form
f 0 xð Þ < 0; for x 2 c� r; cð Þ;f 0 xð Þ > 0; for x 2 c; cþ sð Þ:
It follows from the Increasing–Decreasing Theorem that f has a local
minimum at c.
A similar argument shows that, if
f 0 cð Þ ¼ 0 and f 00 cð Þ < 0;
then f has a local maximum at c.
Thus we have proved the following result.
Theorem 3 Second Derivative Test
Let f be defined on a neighbourhood of c, and f 0(c)¼ 0.
(a) If f 00(c)> 0, then f (c) is a local minimum of f.
(b) If f 00(c)< 0, then f (c) is a local maximum of f.
Remark
The theorem gives us no information in the case that f 00(c)¼ 0.
Problem 4 For the function f(x)¼ x3� 3x2þ 1, x2R , determine those
points c where f 0(c)¼ 0. Using the Second Derivative Test, determine
whether these correspond to local maxima, local minima or neither.
In Exercise 6 on this sub-section, in Section 6.7, we askyou to verify a result thatimplies this assertion.
The following diagrams arehelpful in remembering thisresult.
236 6: Differentiation
Inequalities
We now demonstrate how the Increasing–Decreasing Theorem can be used to
prove certain inequalities involving differentiable functions.
Example 4 Prove that, for �> 1,
1þ xð Þ�� 1þ �x; for x � �1:
Solution Let
f xð Þ ¼ 1þ xð Þ�� 1þ �xð Þ; for x 2 �1;1½ Þ:The function f is continuous on [�1,1) and differentiable on (�1,1), and
f 0 xð Þ ¼ � 1þ xð Þ��1��
¼ � 1þ xð Þ��1�1h i
:
Firstly, if �1< x< 0, then
0 < 1þ x < 1:
Since �> 1, we can then take the (�� 1)th power of each side of the inequality
1þ x< 1 to obtain
1þ xð Þ��1< 1; for �1 < x < 0;
so that
f 0 xð Þ < 0; for �1 < x < 0:
Next, if x> 0, then
1þ x > 1;
so that
1þ xð Þ��1> 1; for x > 0:
Hence
f 0 xð Þ > 0; for x > 0:
Bringing together these two arguments, we have
f 0 xð Þ < 0; for �1 < x < 0;
f 0 xð Þ > 0; for x > 0:
Also, f (0)¼ 0.
Hence, by the Increasing–Decreasing Theorem
f is decreasing on �1; 0½ �;f is increasing on ½0;1Þ;
so we have
f xð Þ � f 0ð Þ ¼ 0; for x 2 �1; 0½ �;f xð Þ � f 0ð Þ ¼ 0; for x 2 0;1½ Þ:
It follows that
1þ xð Þ�� 1þ �xð Þ � 0; for x 2 �1;1½ Þ;as required. &
This is a generalisation ofBernoulli’s Inequality
1þ xð Þn � 1þ nx;
for x�� 1, n2N , that youmet in Sub-section 1.3.3.
Since �� 1> 0, we can useRule 5 for inequalities, whichyou met in Sub-section 1.2.1.
Since �� 1> 0, we can againuse Rule 5 for inequalities.
6.4 The Mean Value Theorem 237
Example 4 illustrates the following general strategy.
Strategy To prove that g(x)� h(x) on [a, b]:
1. Let
f xð Þ ¼ g xð Þ � h xð Þ;and show that f is continuous on [a, b] and differentiable on (a, b);
2. Prove that:
EITHER f að Þ � 0; and f 0 xð Þ � 0 on a; bð Þ;OR f bð Þ � 0; and f 0 xð Þ � 0 on a; bð Þ;
Problem 5 Prove the following inequalities:
(a) x� sin x, for x 2 0; 12p
�
;
(b) 23
xþ 13� x
23, for x2 [0; 1].
6.5 L’Hopital’s Rule
In order to differentiate the functions sin and exp, we needed to use the
following results
limx!0
sin x
x¼ 1 and lim
x!0
ex � 1
x¼ 1:
Each of these limits is the limit as x! 0 of a quotient in which the numerator
and denominator take the value 0 when x is 0. The evaluation of these limits
was not trivial and required considerable care.
In Analysis and in Mathematical Physics we often need to evaluate limits of
the form
limx!c
f xð Þg xð Þ ; where f cð Þ ¼ g cð Þ ¼ 0:
Such limits cannot be evaluated by the Quotient Rule for limits of functions,
because it does not apply in this situation.
For example, do the limits
limx!p
2
cos 3x
sin x� ecos xand lim
x!0
x2
cosh x� 1
exist? If they do, what are their values?
Do we have to evaluate all such limits by a direct argument, or is there a handy
rule which we can apply? We shall find that there is, called l’Hopital’s Rule.
6.5.1 Cauchy’s Mean Value Theorem
Recall that the Mean Value Theorem asserts that, under certain conditions, the
graph of a function f(x), for x2 [a, b], has the property that at some intermedi-
ate point c the tangent to the graph is parallel to the chord joining the end-
points of the graph. In other words, there exists a point c in (a, b) such that
There is a correspondingversion of the Strategy inwhich the weak inequalitiesare replaced by strictinequalities.
This also works if b is1.This also works if a is �1.
Sub-section 6.1.3.
See Sub-section 5.1.1 andProblem 8 in Sub-section 5.1.4,respectively.
The Quotient Rule for limitsof functions states that
limx!c
f xð Þg xð Þ ¼
limx!c
f xð Þlimx!c
g xð Þ ;
provided that these last twolimits exist.
Theorem 1, Sub-section 6.4.1.
238 6: Differentiation
f 0 cð Þ ¼ f bð Þ � f að Þb� a
: (1)
The key tool that we shall need in Sub-section 6.5.2 in the proof of
l’Hopital’s Rule is the following result.
Theorem 1 Cauchy’s Mean Value Theorem
Let f and g be continuous on [a, b] and differentiable on (a, b). Then there is
some point c in (a, b) for which
f 0 cð Þ � g bð Þ � g að Þf g ¼ g0 cð Þ � f bð Þ � f að Þf g: (2)
In particular, if g(b) 6¼ g(a) and g0(c) 6¼ 0, this equation can be written in the
form
f 0 cð Þg0 cð Þ ¼
f bð Þ � f að Þg bð Þ � g að Þ : (3)
Notice that the Mean Value Theorem is simply the special case when
g(x)¼ x. For then g0(c)¼ 1 and g(b)� g(a)¼ b� a, and equation (3) reduces
to equation (1).
However, Theorem 1 is NOT a simple consequence of the Mean Value
Theorem. For, if we apply the Mean Value Theorem separately to the functions
f and g, we establish the existence of two points c1 and c2 in (a, b) for which
f 0 c1ð Þ ¼f bð Þ � f að Þ
b� aand g0 c2ð Þ ¼
g bð Þ � g að Þb� a
:
However, since c1 and c2 are usually unequal, we cannot deduce the exist-
ence of a single point c satisfying the statement (2).
Our proof of Theorem 1 is similar to that of the Mean Value Theorem: we
choose a suitable ‘auxiliary’ function h on [a, b] for which the conditions of
Rolle’s Theorem are satisfied.
Proof Consider the function
h xð Þ ¼ f xð Þ � g bð Þ�g að Þf g�g xð Þ � f bð Þ� f að Þf g; for x2 a;b½ �:
By the Combination Rules for continuity and differentiability, h is continuous
on [a, b] and differentiable on (a, b). Also
h að Þ ¼ f að Þ � g bð Þ � g að Þf g � g að Þ � f bð Þ � f að Þf g¼ f að Þg bð Þ � g að Þ f bð Þ
This form is easier toremember.
Here two terms cancel.
6.5 L’Hopital’s Rule 239
and
h bð Þ ¼ f bð Þ � g bð Þ � g að Þf g � g bð Þ � f bð Þ � f að Þf g¼ f að Þg bð Þ � g að Þf bð Þ;
so that h(a)¼ h(b).
Thus h satisfies the conditions of Rolle’s Theorem on [a, b]. Therefore there
exists a point c in (a, b) for which
h0ðcÞ ¼ 0 ;
that is, these exists a point c in (a, b) for which
f 0 cð Þ � g bð Þ � g að Þf g ¼ g0 cð Þ � f bð Þ � f að Þf g:This is precisely equation (2).
Equation (3) follows immediately from equation (2). &
Example 1 By applying Cauchy’s Mean Value Theorem to the functions
f xð Þ ¼ x6 � 1 and g xð Þ ¼ 1
2x4 þ 3x3 þ 3x� 3 on ½1; 2�;
prove that the equation 3x5� 2x3� 9x2¼ 3 has at least one root in (1, 2).
Solution Since f and g are both polynomials, they are continuous on [1, 2]
and differentiable on (1, 2). It follows that Cauchy’s Mean Value Theorem
applies to the functions f and g on [1, 2].
Now, g 2ð Þ ¼ 12� 16þ 3� 8þ 3� 2� 3 ¼ 8þ 24þ 6� 3 ¼ 35 and
g 1ð Þ ¼ 12þ 3þ 3� 3 ¼ 3 1
2, so that g(2) 6¼ g(1). Also, f 0(x)¼ 6x5 and
g0(x)¼ 2x3þ 9x2þ 3 6¼ 0 on (1, 2). It therefore follows from Cauchy’s Mean
Value Theorem that there exists at least one point c in (1, 2) for which
6c5
2c3 þ 9c2 þ 3¼ 63� 0
35� 3 12
¼ 63
31 12
¼ 2:
By cross-multiplying, we see that c satisfies the equation 6c5¼ 4c3þ 18c2þ 6,
or
3c5 � 2c3 � 9c2 ¼ 3:
In other words, the equation 3x5� 2x3� 9x2¼ 3 has at least one root in
(1, 2). &
Problem 1 By applying Cauchy’s Mean Value Theorem to the functions
f xð Þ¼ x3þ x2 sinx and g xð Þ¼ xcosx� sinx on 0;p½ �;prove that the equation 3x¼ (p2� 2) sin x� xcos x has at least one root
in (0, p).
6.5.2 l’Hopital’s Rule
We are now in a position to state the key result that we need to evaluate certain
types of limits in a fairly routine way.
Again two terms cancel.
The common value of h(a)and h(b) does not matter; theimportant thing is that the twoare equal.
Here we use equation (3).
240 6: Differentiation
Theorem 2 l’Hopital’s Rule
Let f and g be differentiable on a neighbourhood of the point c, at which
f(c)¼ g(c)¼ 0. Then
limx!c
f xð Þg xð Þ exists and equals lim
x!c
f 0 xð Þg0 xð Þ ;
provided that this last limit exists.
Proof We assume that
limx!c
f 0 xð Þg0 xð Þ exists and equals ‘: (4)
Hence there is some punctured neighbourhood N¼ (c� r, c)[ (c, cþ s) of c
on which f 0
g0 is defined and g0(x) 6¼ 0.
We now prove that our assumption (4) implies that
limx!c
f xð ÞgðxÞ exists and equals ‘: (5)
Let y be any specific point in N for which y> c. The functions f and g are
continuous on [c, y] and differentiable on (c, y), so they satisfy the conditions of
Cauchy’s Mean Value Theorem on [c, y].
Now g(y)� g(c) 6¼ 0; for otherwise we would have g(c)¼ 0 (from our
assumptions) and so g(y)¼ 0; this would imply, by Rolle’s Theorem, that g0
would vanish somewhere in (c, y), which it does not.
It follows from the conclusion (3) of Cauchy’s Mean Value Theorem that
there exists some point z in (c, y) for which
f 0 zð Þg0 zð Þ ¼
f yð Þ � f cð Þg yð Þ � g cð Þ
¼ f yð Þg yð Þ ; since f cð Þ ¼ g cð Þ ¼ 0:
Now let y! cþ. Since c< z< y, it follows that z! cþ too.
But we know that
limz!cþ
f 0 zð Þg0 zð Þ exists and has the value ‘:
It follows that
limy!cþ
f yð Þg yð Þ exists and has the value ‘:
This is exactly the same as the statement that
limx!cþ
f xð Þg xð Þ exists and has the value ‘:
A similar argument shows that
limx!c�
f xð Þg xð Þ exists and has the value ‘:
Combining the last two statements, we obtain the desired result (5). &
We now show how we can use l’Hopital’s Rule to evaluate the two limits that
we mentioned at the start of this section.
You may omit this proof at afirst reading.
Theorem 1, Sub-section 6.5.1.
This is just a special case of(4), with z in place of x andz! cþ in place of x! c.
Here x is simply a ‘dummyvariable’: it does not matterwhat letter we assign to thevariable in the limit.
We simply repeat the wholeargument, starting with aspecific point y in N forwhich y< c.
6.5 L’Hopital’s Rule 241
Example 2 Prove that limx!p
2
cos 3x
sin x � e cos x ( 6)
exists and determine its value.
Solution Let
f xð Þ ¼ cos 3x and g xð Þ ¼ sin x� ecos x; for x 2 R :
Then f and g are differentiable on R , and
fp2
� �
¼ gp2
� �
¼ 0;
so that f and g satisfy the conditions of l’Hopital’s Rule at p2.
Now
f 0 xð Þg0 xð Þ ¼
�3 sin 3x
cos xþ sin x� ecos x:
Then, by l’Hopital’s Rule, the limit (6) exists and equals
limx!p
2
f 0 xð Þg0 xð Þ ¼ lim
x!p2
�3 sin 3x
cos xþ sin x� ecos x;
provided that this last limit exists.
By the Quotient Rule for continuous functions, we know that f 0
g0 is continuous
at p2, so that
limx!p
2
f 0 xð Þg0 xð Þ ¼
f 0 p2
� �
g0 p2
� �
¼ 3
1¼ 3:
It follows, from l’Hopital’s Rule, that the original limit (6) must also exist,
and that its value is 3. &
Example 3 Prove that limx!0
x2
cosh x � 1 ( 7)
exists and determine its value.
Solution Let
f xð Þ ¼ x2 and g xð Þ ¼ cosh x� 1; for x 2 R :
Then f and g are differentiable on R , and
f 0ð Þ ¼ g 0ð Þ ¼ 0;
so that f and g satisfy the conditions of l’Hopital’s Rule at 0.
Now, the derivatives of f and g are
f 0 xð Þ ¼ 2x and g0 xð Þ ¼ sinh x; for x 2 R :
It follows from l’Hopital’s Rule that the limit (7) exists and equals
limx!0
2x
sinh x(8)
provided that this last limit exists.
Now, both f 0 and g0 are differentiable on R , and f 0(0)¼ g0(0)¼ 0; thus f 0 and
g0 satisfy the conditions of l’Hopital’s Rule at 0.
Since
f 00 xð Þ ¼ 2 and g00 xð Þ ¼ cosh x; for x 2 R ;
Here we are using the fact (seeRemark 7, Sub-section 5.4.1)that, if f is continuous at c, then
limx!c
f xð Þ ¼ f cð Þ:
(In future, we shall notmention this fact explicitly inthis particular connection.)
We cannot assert that
limx!0
f 0 xð Þg0 xð Þ ¼
f 0 0ð Þg0 0ð Þ ;
since f 0(0)¼ g0(0)¼ 0.
242 6: Differentiation
it follows from l’Hopital’s Rule that the limit (8) exists and equals
limx!0
2
cosh x; (9)
provided that this last limit exists.
The function cosh is continuous on R , and cosh 0¼ 1, so that limx!0
cosh x ¼ 1.
It follows, from the Quotient Rule for limits, that the limit (9) does exist and
that its value is
2
cosh 0¼ 2
1¼ 2:
Working backwards, we conclude that the limit (8) exists, and equals 2.
Working further backwards, we conclude that the limit (7) also exists and
equals 2. &
Before applying a theorem, it is important to check that its conditions are
satisfied; for if they are not satisfied you cannot use the theorem! For instance,
a thoughtless application of l’Hopital’s Rule can give an incorrect answer!
For example, consider the problem of evaluating
limx!1
2x2 � x� 1
x2 � x: (10)
If we put f(x)¼ 2x2� x� 1 and g(x)¼ x2� x, we might be tempted to evaluate
(10) as follows
limx!1
f xð Þg xð Þ ¼ lim
x!1
f 0 xð Þg0 xð Þ ¼ lim
x!1
4x� 1
2x� 1
¼ limx!1
f 00 xð Þg00 xð Þ ¼ lim
x!1
4
2
¼ 2:
(11)
In fact, the value of the limit (10) is 3; so let us review the above argument
carefully!
The line (11) in the calculation is valid, since f and g are differentiable on R
and f(1)¼ g(1)¼ 0; so the conditions of l’Hopital’s Rule are satisfied for the
first application of the Rule. However, f 0(1)¼ 3 and g0(1)¼ 1, so the condi-
tions are not satisfied for the second application of l’Hopital’s Rule!
In fact, we should have concluded directly from line (11) that
limx!1
f 0 xð Þg0 xð Þ ¼ lim
x!1
4x� 1
2x� 1¼ 3:
The moral is that you must apply l’Hopital’s Rule with care, particularly to
make the proviso ‘provided that this last limit exists’ at the appropriate points. At
the end, you then work backwards through the chain of applications of the Rule
to reach your conclusion about the limit that you set out originally to examine.
Remark
Notice that we cannot apply l’Hopital’s Rule to evaluate the limits
limx!0
sin x
x¼ 1 and lim
x!0
ex � 1
x¼ 1;
because we used these limits to find the derivatives of sin x and ex, respectively!
Note that the carefully set outlogic of the argument here isessential, since it is aconsequence of what weknow from Theorem 2.
Note that f(1)¼ g(1)¼ 0.
Here the careful argumentsthat we gave in Examples 2and 3 have been abandoned infavour of thoughtless‘formula-pushing’!
6.5 L’Hopital’s Rule 243
Problem 2 Prove that the following limits exist, and evaluate them.
(a) limx!0
sinh 2xsin 3x
; (b) limx!0
1þxð Þ15� 1�xð Þ
15
1þ2xð Þ25� 1�2xð Þ
25
;
(c) limx!0
sin x2þsin x2ð Þ1�cos 4x
; (d) limx!0
sin x�x cos xx3 .
6.6 The Blancmange function
We now meet a function, called the Blancmange function, which is continuous
at each point of R but is differentiable nowhere on R . The construction of the
first function with these properties by Karl Weierstrass caused a huge excite-
ment among mathematicians!
You saw earlier that, if a function is differentiable at a point c, then it is also
continuous at c. The Blancmange function shows, in a very striking way, that
the converse result is false!
6.6.1 What is the Blancmange function?
We give the formal definition first, and then look at the underlying geometry
of the graph of the function.
Definition Let f be the function
f ðxÞ ¼ x� ½x�; 0 � x � [x] � 12,
1� ðx� ½x�Þ; 12< x � [x] < 1,
where [x] denotes the integer part of x. Then the Blancmange function is
defined on R by the formula
BðxÞ ¼ f ðxÞ þ 1
2f ð2xÞ þ 1
4f ð4xÞ þ 1
8f ð8xÞ þ � � �
¼X
1
n¼0
1
2nf ð2nxÞ:
So, what does the graph of B look like? And can we obtain any idea why the
function has its strange properties?
The function f is continuous on R , but is not differentiable at the points 12
n,
for any integer n.
You may omit this section,apart from the next twoparagraphs, at a first reading.
Sub-section 6.1.2.
This series converges by theComparison Test, sincef 2nxð Þj j � 1
2; for all x 2 R :
244 6: Differentiation
Next, we construct the graphs
y ¼ 12
f 2xð Þ; y ¼ 14
f 4xð Þ; y ¼ 18
f 8xð Þ; . . . ;
we obtain each graph by scaling the previous graph by a factor of 12
in both the
x-direction and the y-direction.
Each graph has twice as many peaks in any interval as its predecessor, and
each of these peaks is half the height of the peaks in its predecessor. Thus the
number of points in the interval where the function is not differentiable doubles
(approximately) at each stage.
We obtain the Blancmange function B by adding together all these functions
to form an infinite series B xð Þ ¼ f xð Þ þ 12
f 2xð Þ þ 14
f 4xð Þ þ 18
f 8xð Þ þ � � �.Thus, for example
B 12
� �
¼ f 12
� �
þ 12
f 1ð Þ þ 14
f 2ð Þ þ 18
f 4ð Þ þ � � �¼ 1
2þ 1
2� 0
� �
þ 14� 0
� �
þ 18� 0
� �
þ � � �¼ 1
2;
and
B 14
� �
¼ f 14
� �
þ 12
f 12
� �
þ 14
f 1ð Þ þ 18
f 2ð Þ þ � � �¼ 1
4þ 1
2� 1
2
� �
þ 14� 0
� �
þ 18� 0
� �
þ � � �¼ 1
4þ 1
4¼ 1
2:
To get an idea of the shape of the graph of B, we look at the graphs of
successive partial sum functions of B. To help you follow the construction, we
also draw in the graph at the previous stage (in light dashes) and the function
being added to it (in heavy dashes).
The first few of these graphsare included in the diagramsbelow.
6.6 The Blancmange function 245
Eventually, we obtain the following graph of B:
It looks as though B is continuous. However, it is NOT true in general that the
sum of infinitely many continuous function is itself continuous, so a proof of
the continuity of B is necessary.
Similarly, it looks as though B might not be differentiable.
For example, we have marked the points on the above graphs corresponding to
x ¼ 13. At the first stage, 1
3lies in the interval 0; 1
2
�
, and so the point 13; f 1
3
� �� �
lies on a line segment of slope 1. At the next stage, 13
lies in the interval 14; 1
2
�
,
and so the point 13; f 1
3
� �
þ 12
f 23
� �� �
lies on a line segment of slope 0. And so on.
At successive stages in the construction of the graph of B, the point corres-
ponding to 13
lies alternately on line segments of slope 0 and 1. Thus it seems
plausible that the slopes of chords joining the points 13; B 1
3
� �� �
and (x, B(x))
do not tend to any fixed value as x tends to 13, so that B would not be
differentiable at 13.
Notice in the last diagram above that however closely we look at the graph
of the Blancmange function B, it seems to have small blancmanges growing on
it everywhere. At the nth stage, each horizontal line segment has one or two
mini-blancmanges growing on it, and each sloping line segment has one or two
‘sheared’ mini-blancmanges growing on it.
Finally, notice that the Blancmange function B is periodic, with period 1.
This occurs because f is periodic, with period 1.
Here the dashes indicate thegraphs at the early stages andalso the graphs of functionsoccurring in the summation.
We give this inSub-section 6.6.2.
We prove this inSub-section 6.6.3.
246 6: Differentiation
6.6.2 Continuity of the Blancmange function
To verify that B is continuous on R , we must use the power of the "� �definition of continuity.
Theorem 1 The Blancmange function B is continuous at each point c2R .
Proof We must show that
for each positive number ", there is a positive number � such that
BðxÞ � BðcÞj j < "; for all x satisfying x� cj j < �: (1)
Now, it follows, from the definition of B, that B xð Þ � B cð Þ ¼P
1
n¼0
12n f 2nxð Þ � f 2ncð Þð Þ; hence, by the infinite form of the Triangle Inequality
B xð Þ � B cð Þj j �X
1
n¼0
1
2nf 2nxð Þ � f 2ncð Þj j: (2)
For all x and c, both of the numbers f(2nx) and f(2nc) lie in 0; 12
�
, so that the
modulus of their difference is at most 12; that is
f 2nxð Þ � f 2ncð Þj j � 1
2:
Now we choose an integer N such that 12N <
12". (This choice is possible, since
the sequence 12n
� �
is null.) It follows that
X
1
n¼N
1
2nf 2nxð Þ � f 2ncð Þj j �
X
1
n¼N
1
2n� 1
2
¼ 1
2N
<1
2": (3)
Next, since each of the functions x 7! f(2nx), n¼ 0, 1, . . ., N� 1, is continuous,
it follows that
for each n¼ 0, 1, . . ., N� 1, there is a positive number �n such that
f 2nxð Þ � f 2ncð Þj j < 14"; for all x satisfying x� cj j < �n:
Now we choose �¼min{�0, �1, . . ., �N�1}. Thus, for each n¼ 0, 1, . . .,N� 1, we must have
f 2nxð Þ � f 2ncð Þj j < 14"; for all x satisfying x� cj j < �:
It follows that
X
N�1
n¼0
1
2nf 2nxð Þ � f 2ncð Þj j <
X
N�1
n¼0
1
2n� 1
4"
<X
1
n¼0
1
2nþ2� "
¼ 1
2": (4)
Sub-section 3.3.1.
We apply the definition ofcontinuity to each functionin turn.
We use 14" here, in order
to obtain an " in our finalresult (1).
6.6 The Blancmange function 247
Substituting the upper bounds from (3) and (4) into inequality (2), we deduce
that, for all x satisfying jx� cj<�, we have
B xð Þ � B cð Þj j �X
N�1
n¼0
1
2nf 2nxð Þ � f 2ncð Þj j þ
X
1
n¼N
1
2nf 2nxð Þ � f 2ncð Þj j
<1
2"þ 1
2"
¼ ":This is the desired result (1). Hence B is continuous at c. &
6.6.3 The Blancmange function is differentiablenowhere
The proof that B is nowhere differentiable on R is more tricky, and the
following lemma about difference quotients in general (which is of interest
in its own right) plays a crucial role.
Lemma Let f be defined on an open interval I and differentiable at the
point c2 I. Let {xn} and {yn} be sequences in I converging to c such that
xn � c � yn and xn < yn; for n ¼ 0; 1; 2; . . .:
Then
limn!1
f ynð Þ � f xnð Þyn � xn
exists and equals f 0 cð Þ:
Proof We must prove that
for each positive number ", there exists some number X such that
f ynð Þ � f xnð Þyn � xn
� f 0 cð Þ�
�
�
�
�
�
�
�
< "; for all n>X: (5)
Since f is differentiable at c, we know that, for each positive number ", there
exists some positive number � such that
f xð Þ � f cð Þx� c
� f 0 cð Þ�
�
�
�
�
�
�
�
<1
4"; for all x satisfying 0 < x� cj j < �;
so that
f xð Þ� f cð Þ� f 0 cð Þ x� cð Þj j � 1
4" x� cj j; for all x satisfying x� cj j<�: (6)
Now, since xn! c and yn! c as n!1, it follows that there are numbers X1
and X2 such that
xn � cj j < � for all n>X1; and yn � cj j < � for all n>X2;
so, if we set X¼max{X1, X2}, we certainly have
xn � cj j < � and yn � cj j < � for all n>X:
It now follows from (6) that, for all n>X, we have
f xnð Þ � f cð Þ � f 0 cð Þ xn � cð Þj j � 1
4" xn � cj j and
f ynð Þ � f cð Þ � f 0 cð Þ yn � cð Þj j � 1
4" yn � cj j:
xn ync
yy = f (x)
x
This is the definition ofdifferentiability at c, butwith 1
4" in place of ", in order
to obtain an " in our finalresult (5).
In fact, this is a strictinequality if x 6¼ c. Howeverlater in the argument we willneed to allow the possibilitythat x¼ c, so we must write(6) with a weak inequalitysign.
We put x¼ xn and then x¼ yn
into (6).
248 6: Differentiation
Hence, we can apply the Triangle Inequality to obtain
f ynð Þ� f xnð Þ� f 0 cð Þ yn� xnð Þj j¼ f ynð Þ� f cð Þ� f 0 cð Þ yn� cð Þf g� f xnð Þ� f cð Þ� f 0 cð Þ xn� cð Þf gj j� f ynð Þ� f cð Þ� f 0 cð Þ yn� cð Þj jþ f xnð Þ� f cð Þ� f 0 cð Þ xn� cð Þj j
� 1
4" yn� cj jþ 1
4" xn� cj j
� 1
4" yn� xnj jþ 1
4" yn� xnj j
¼ 1
2" yn� xnj j; for all n> X:
Since we know that xn 6¼ yn, we can divide this inequality by the non-zero
term yn� xn to obtain, for n>X
f ynð Þ � f xnð Þyn � xn
� f 0 cð Þ�
�
�
�
�
�
�
�
� 1
2" < ":
This is the result (5) that we set out to prove. &
Remark
The hypothesis that xn and yn must lie on opposite sides of c cannot generally
be omitted. For example, consider the function
f ðxÞ ¼ x2 sin 1x
� �
; x 6¼ 0,
0; x ¼ 0,
that you saw earlier is differentiable at 0, with f 0(0)¼ 0. If we set
xn ¼1
2nþ 1ð Þp and yn ¼1
2nþ 12
� �
p; for n ¼ 0; 1; 2; . . .;
then
f ynð Þ � f xnð Þyn � xn
¼1
2nþ12ð Þ2p2
� �
� 0ð Þ
1
2nþ12ð Þp
� �
� 12nþ1ð Þp
� �
¼ 2 2nþ 1ð Þ2nþ 1
2
� �
p
! 2
pas n!1:
However, the value of this limit is not 0, the value of f 0(0).
We are now ready to prove the principal result in this sub-section.
Theorem 2 The Blancmange function B is not differentiable at any point
c2R .
Proof In order to apply the lemma, we construct two sequences {xn} and
{yn} converging to c such that the corresponding sequence of difference
quotients is not convergent. We use the method of repeated bisection to
construct {xn} and {yn}.
We first insert some terms andtake them away again.
We use the two inequalities thatwe have just verified for n>X.
Since xn� c� yn, bothjyn� cj and jxn� cj are� |yn� xn|.
Problem 7, Sub-section 6.1.2.
y = x2
y =f (x)
y = – x2
y
xn xyn
6.6 The Blancmange function 249
Since B is periodic with period 1, we assume for simplicity that c2 [0, 1].
We start by defining [x0, y0]¼ [0, 1]. Then, since c lies in one of the intervals
0; 12
�
and 12; 1
� �
, we can define
x1; y1½ � ¼ 0; 12
�
;12; 1
�
;
if c 2 0; 12
�
;
if c 2 12; 1
� �
:
With this definition of [x1, y1], we have that:
1. x1; y1½ � x0; y0½ �;2. y1 � x1 ¼ 1
2;
3. x1� c� y1;
4. x1 ¼ 12
p1 and y1 ¼ 12
p1 þ 1ð Þ, for some integer p1.
We can then repeat this process, bisecting the interval [x1, y1] to obtain [x2, y2],
and so on. In this way, we obtain a sequence of closed intervals [xn, yn], for
n¼ 1, 2, . . ., such that:
1. xnþ1; ynþ1½ � xn; yn½ �;2. yn � xn ¼ 1
2
� �n;
3. xn� c� yn;
4. xn ¼ 12n pn and yn ¼ 1
2n pn þ 1ð Þ, for some integer pn.
Properties 2 and 3 imply that both the sequences {xn} and {yn} converge to c,
but we shall show that the sequence of difference quotients {Qn}, where
Qn ¼B ynð Þ � B xnð Þ
yn � xn
¼ 2n B ynð Þ � B xnð Þð Þ;
is not convergent. It will then follow, from the lemma, that B is not differenti-
able at c.
To prove that the sequence {Qn} is divergent, it is sufficient to prove that
Qnþ1 ¼ Qn � 1; for n ¼ 0; 1; 2; . . .; (7)
since it then follows that {Qn} cannot converge.
To prove (7), we put
zn ¼1
2xn þ ynð Þ ¼ 1
2nþ12pn þ 1ð Þ:
Note that
xnþ1; ynþ1½ � ¼ xn; zn½ �;zn; yn½ �;
if c 2 xn; zn½ �;if c 2 zn; ynð �:
We now claim that, for n¼ 0, 1, 2, . . .
B znð Þ ¼1
2B xnð Þ þ B ynð Þð Þ þ 1
2nþ1: (8)
It would then follow from (8) that, if [xnþ1, ynþ1]¼ [xn, zn], then
Qnþ1 ¼ 2nþ1 B znð Þ � B xnð Þð Þ¼ 2n B ynð Þ � B xnð Þð Þ þ 1 ¼ Qn þ 1;
We use Property 2 for thevalue of yn� xn.
The structure of the proof isas follows(8)) (7)) Theorem 2.
250 6: Differentiation
whereas, if [xnþ1, ynþ1]¼ [zn, yn], then
Qnþ1 ¼ 2nþ1 B ynð Þ � B znð Þð Þ¼ 2n B ynð Þ � B xnð Þð Þ � 1 ¼ Qn � 1:
In either case, the required result (7) would then hold. Thus, in order to
complete the proof that (7) holds, it is sufficient to prove that (8) holds.
To verify (8), note that, for k¼ 0, 1, 2, . . ., we have
2kxn ¼1
2n�kpn;
2kyn ¼1
2n�kpn þ 1ð Þ;
2kzn ¼1
2n�kþ12pn þ 1ð Þ;
9
>
>
>
>
>
=
>
>
>
>
>
;
(9)
and f ( p)¼ 0 for any integer p. It then follows from the definition of B as an
infinite series that
B xnð Þ þ B ynð Þ ¼X
1
k¼0
1
2kf 2kxn
� �
þ f 2kyn
� �� �
¼X
n�1
k¼0
1
2kf 2kxn
� �
þ f 2kyn
� �� �
; and
B znð Þ ¼X
1
k¼0
1
2kf 2kzn
� �
¼X
n
k¼0
1
2kf 2kzn
� �
:
We deduce from (9) that, for k¼ 0, 1, 2, . . ., n� 1, the terms 2kxn, 2kyn and
2kzn all lie in the interval
2kxn; 2kyn
�
¼ 1
2n�kpn;
1
2n�kpn þ 1ð Þ
� �
;
and so in some interval of the form 12
qk;12
qk þ 1ð Þ �
, where qk is an integer.
Now, the restriction of f to such an interval is linear, so that we have
f 2kzn
� �
¼ 12
f 2kxn
� �
þ f 2kyn
� �� �
; k ¼ 0; 1; 2; . . .; n� 1;
hence
X
n�1
k¼0
1
2kf 2kzn
� �
¼ 1
2
X
n�1
k¼0
1
2kf 2kxn
� �
þ f 2kyn
� �� �
: (10)
Finally
1
2nf 2nznð Þ ¼ 1
2nf pn þ
1
2
� �
¼ 1
2nþ1: (11)
If we then add (10) and (11), we obtain (8), as required.
This completes the proof of Theorem 2. &
We shall shortly use the
expressionP
1
k¼0
12k f 2kx� �
for
B(x), hence some ks nowappear.
For, f (2kxn)¼ f (2kyn)¼ 0 ifk> n� 1.
For, f (2kzn)¼ 0 if k> n.
6.6 The Blancmange function 251
6.7 Exercises
Section 6.1
1. Determine whether each of the following functions f is differentiable at the
specified point c; if it is, evaluate the derivative f 0(c).
(a) f xð Þ ¼ �x2; x � 0,
x2; x > 0,
c ¼ 0;
(b) f xð Þ ¼ 1; x < 1,
x2; x � 1,
c ¼ 1;
(c) f xð Þ ¼ tan x; � 12p < x < 1
3p;
x2; x � 13p;
c ¼ 13p;
(d) f ðxÞ ¼ 1� x; x < 1,
x� x2; x � 1,
c ¼ 1;
(e) f xð Þ ¼ x sin 1x2
� �
; x 6¼ 0;0; x ¼ 0;
c ¼ 0;
(f) f xð Þ ¼ sin x sin 1x
� �
; x 6¼ 0;0; x ¼ 0;
c ¼ 0:
2. Write down an expression for a function f with domain (�1, 2] and the
following properties:
(a) fL0 exists at 1, and fL
0(1)¼ 1;
(b) fR0 exists at 1, and fR
0(1)¼ 2.
Verify that f has the properties (a) and (b).
Section 6.2
1. Use the rules for differentiation to verify that the function
f xð Þ ¼ loge 1þ xð Þ þ ex2
; x 2 �1;1ð Þ, is differentiable on its domain,
and determine its derivative.
2. Write down (without justification) the derivatives of the following
functions:
(a) f xð Þ ¼ x2þ1x�1
; x 2 1;1ð Þ;(b) f xð Þ ¼ loge sin xð Þ; x 2 0; pð Þ;(c) f xð Þ ¼ loge sec xþ tan xð Þ; x 2 �1
2p; 1
2p
� �
;
(d) f xð Þ ¼ cos xþsin xcos x�sin x
; x 2 0; 14p
� �
:
3. Prove that the function f(x)¼ tanh x, x2R , has an inverse function f�1 that
is differentiable on (�1, 1), and find an expression for (f�1)0(x).
4. Prove that the function f(x)¼ tan xþ 3x, x 2 �12p; 1
2p
� �
, has an inverse
function f�1 that is differentiable on R , and find the value of ( f�1)0(0).
5. Determine the derivatives of the following functions, assuming that they are
differentiable on their domains:
(a) f xð Þ ¼ coth x; x 2 R � 0f g;(b) f xð Þ ¼ loge xð Þloge x; x 2 1;1ð Þ:
252 6: Differentiation
Section 6.3
1. The function f has domain [�2, 2] and its graph consists of four line
segments, as shown.
Identify (a) the local minima of f, (b) the minima of f, (c) the local maxima
of f, and (d) the maxima of f, and state where these occur,
2. Let f be the function
f xð Þ ¼ 2x3 � 3x2; x 2 0; 2½ �:Find (a) the local minima and minima of f, (b) the local maxima and
maxima of f and state where these occur.
3. Prove that, of all rectangles with given perimeter, the square has the greatest
area.
4. Verify that the conditions of Rolle’s Theorem are satisfied by the function
f xð Þ ¼ 1þ 2x� x2; x 2 0; 2½ �;and determine a value of c in (0,2) for which f 0(c)¼ 0.
5. Use Rolle’s Theorem to prove that, if p is a polynomial and l1, l2, . . ., ln are
distinct zeros of p, then p0 has at least (n�1) zeros.
6. Use Rolle’s Theorem to prove that, for any real number l, the function
f xð Þ ¼ x3 � 3
2x2 þ l; x 2 R ;
never has two distinct zeros in [0, 1].
Hint: Assume that f has two distinct zeros in [0, 1], and obtain a
contradiction.
7. The function f is twice differentiable on an interval [a, b]; and, for some
point c in (a, b), f(a)¼ f(b)¼ f(c). Prove that there exists some point d in
(a, b) with f 00(d)¼ 0.
8. The function f is differentiable on a neighbourhood N of a point c; f 0 is
continuous at c and f 0(c)> 0. Prove that f is increasing on some neighbour-
hood of c.
Harder: Give an example of a function f that is differentiable on R with the
properties that (a) f 0(0)> 0 and (b) there is NO open neighbourhood of 0 on
which f is increasing.
9. Let the function f : 0; 1½ � 7! 0; 1½ � be continuous on [0, 1] and differentiable
on (0, 1). We know that there is at least one point c in [0, 1] for which
f (c)¼ c. Use Rolle’s Theorem to prove that, if f 0 xð Þ 6¼ 1 in (0, 1), then there
is exactly one such point c.
Hint: Consider the function h xð Þ ¼ f xð Þ � x; x 2 0; 1½ �; assume that two
such points c exist, and obtain a contradiction.
You saw this in Problem 2of Sub-section 4.2.1.
6.7 Exercises 253
Section 6.4
1. For each of the following functions, verify that the conditions of the Mean
Value Theorem are satisfied, and determine a value of c that satisfies the
conclusion of the theorem:
(a) f xð Þ ¼ 2x�1x�2
; x 2 �1; 1½ �; (b) f xð Þ ¼ x3 þ 2x2 þ x; x 2 0; 1½ �:2. Use the Mean Value Theorem to prove that
sin b� sin aj j � b� aj j; for a; b 2 R :
3. Use the Zero Derivative Theorem to prove that
cosh�1 x ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � 1p� �
; for x � 1:
4. For the function f(x)¼ x3� 2x2þ x, x2R , determine those points c where
f 0(c)¼ 0. Using the Second Derivative Test, determine whether these
correspond to local maxima, local minima, or neither.
5. Prove the following inequalities:
(a) loge x � 1� 1x; for x 2 1;1½ Þ;
(b) 4x14 � xþ 3; for x 2 0; 1½ �;
(c) loge 1þ xð Þ > x� 12
x2; for x 2 0;1ð Þ:6. Let f be defined on a neighbourhood of c, with f 0(c)> 0. Prove that there is
some punctured neighbourhood N of c such that
f xð Þ � f cð Þx� c
> 0; for x 2 N:
7. By applying the Mean Value Theorem to the function f xð Þ ¼ffiffiffi
xp
on the
interval [100, 102], prove that 10 111<
ffiffiffiffiffiffiffiffi
102p
< 10 110:
8. Let f be differentiable on a closed interval [a, b].
(a) Prove that, if the minimum of f on [a, b] occurs at a, then fR0(a)� 0; and
that, if the minimum of f on [a, b] occurs at b, then fL0(b)� 0.
(b) Prove that, if fR0(a)< 0 and fL
0(b)> 0, then f 0(x)¼ 0 for some x2 (a, b).
Hint: Check that the minimum of f on [a, b] occurs at some point in (a, b).
(c) By considering the function g (x)¼ f(x)� kx, x2 [a, b], prove that, if
fR0(a)< k< fL
0(b), then f 0(x)¼ k for some x2 (a, b).
Section 6.5
1. Verify that the following functions satisfy the conditions of Cauchy’s Mean
Value Theorem on [0, 2], and determine a value of c that satisfies the
conclusion of the theorem
f xð Þ ¼ x4 þ 2x2 and g xð Þ ¼ x3 þ 3x; for x 2 0; 2½ �:2. Use l’Hopital’s Rule to prove that the following limits exist, and evaluate
the limits:
(a) limx!0
sinh xþ sin xð Þsin x
; (b) limx!1
5xþ3ð Þ13� xþ3ð Þ
12
x�1;
(c) limx!0
1� cos xx2 ; (d) lim
x!0
sinh x�xsin 3x3ð Þ :
This result is known asDarboux’s Theorem, and isan intermediate value theoremfor f 0.
254 6: Differentiation
7 Integration
We have already used the idea of the area of a set in the plane; for example, to
define the number p and to prove that sin xx! 1 as x! 0. However, our earlier
discussions begged the question of what exactly we mean by ‘area’.
For simplicity, we shall restrict our attention in this book to defining only the
area between a graph and the x-axis.
In Section 7.1 we use the idea of lower and upper estimates to give a rigorous
definition of such an area. This is done by trapping the desired area between
underestimates and overestimates, each of which is the sum of the areas of
suitably chosen rectangles.
Then the area between the graph y¼ f (x), x2 [a, b], and the segment [a, b] of
the x-axis is defined to be A if:
the least upper bound of the underestimates¼A,
and:
the greatest lower bound of the overestimates¼A.
We call A the integral of f over [a, b], and denote it byZ b
a
f or
Z b
a
f xð Þdx:
In practice, it would be inconvenient if we had to revert to the definition in
order to tell whether a given function was integrable or not, and we shall verify
several criteria for integrability.
In Section 7.2, we identify some large classes of integrable functions, and
verify the standard rules for integrable functions.
In Section 7.3 we meet the Fundamental Theorem of Calculus which enables
us to avoid use of the definition of integrability in order to evaluate many
integrals. For instance, we verify the usual methods for integration by parts and
integration by substitution.
It follows from theFundamental Theorem ofCalculus that we can think ofintegration as the operationinverse to differentiation.
255
Often it is not possible to evaluate an integral explicitly, and the best that we
can do is to obtain upper and lower estimates for its value. In Section 7.4, we
meet a range of inequalities for integrals, including the Triangle Inequality for
integralsZ b
a
f
�
�
�
�
�
�
�
�
�Z b
a
fj j; where a < b:
We use these inequalities to prove Wallis’s Formula for p, namely that
limn!1
2
1:2
3:4
3:4
5:6
5:6
7: � � � : 2n
2n� 1:
2n
2nþ 1
� �
¼ p2;
and to establish the Maclaurin Integral Test, which enables us to determine the
convergence or divergence of series such asP
1
n¼1
1np, for p> 0, and
P
1
n¼2
1n loge n
.
Finally, in Section 7.5, we discuss the evaluation of n!. It is easy to evaluate
n! for values of n up to 10, say, by direct multiplication; but for n¼ 100 or
n¼ 200, the number n! cannot be evaluated by a standard scientific calculator.
Surprisingly, we can use integration techniques to obtain an excellent estimate
for n!, called Stirling’s Formula
n! �ffiffiffiffiffiffiffiffi
2pnp n
e
� �n
as n!1:
Before starting to read the chapter, we recommend that you refresh your
memory of greatest lower bounds and least upper bounds of functions.
Recall that, if ƒ is a function defined on an interval I�R , then:
� A real number m is the greatest lower bound, or infimum, of ƒ on I if:
1. m is a lower bound of ƒ(I);
2. if m0>m, then m0 is not a lower bound of ƒ(I).
� M is the least upper bound, or supremum, of ƒ on I if:
1. M is an upper bound of ƒ(I);
2. if M 0<M, then M 0 is not an upper bound of ƒ(I).
In Sections 7.1 and 7.2 we shall discuss some new properties of infimum and
supremum that are needed for our work on integrability.
Also, we shall recommend that you omit working your way through quite a
number of detailed proofs during your first reading of this chapter. This is not
necessarily because they are particularly difficult, but simply in order to guide
you through the key ideas first before you return to study some of the ‘gory
details’ later on once you have grasped the overall picture.
7.1 The Riemann integral
Earlier we defined the number p to be the area of the unit disc. We obtained
lower and upper estimates for p by trapping the area of the disc between the
area of a 3� 2n-sided inner polygon and the area of a 3� 2n-sided outer
polygon, and letting n!1.
At least not in 2005, whenthese words are being written.
We shall also explain theprecise meaning of thesymbol tilda: ‘�’.
You met sup and inf in Sub-section 1.4.2.
We often denote m byinf { f(x) : x2 I}, inf
x2If xð Þ,
infI
f or simply by inf f.
We often denote M bysup { f(x) : x2 I}, sup
x2I
f xð Þ,
supI
f or simply by sup f.
Sub-section 2.5.4.
256 7: Integration
For simplicity, in this book we do not consider general areas in the plane, but
restrict our attention to the area between a graph y¼ f(x), x2 [a, b], and the
interval [a, b] of the x-axis. For a continuous function f, we certainly require
such a definition to agree with our intuitive notion of area!
However, it is not obvious that we can always say that the region between a
graph and the x-axis has an area. For example, how can we define such an area
for the following functions?
(a) f(x) = x2, 0 ≤ x ≤ 1,2, 1 < x ≤ 2.{ {(c) f(x) =
1, 0 ≤ x ≤ 1, x rational,x irrational.0, 0 ≤ x ≤ 1.{(b) f(x) =
–2, 0 ≤ x < 1,3, x = 1.
Our approach is to find lower and upper estimates for the area, if such a
thing exists, that we can assign to each region by splitting it up into smaller
regions with areas which we can approximate by rectangles, and use the fact
that
Area of a rectangle¼ base� height.
We now illustrate the underlying idea of these estimates by considering the
situation when the function f is positive on [a, b] (see the diagrams below).
First, we divide up the interval [a, b] into a family of smaller intervals, called a
partition of [a, b]. Then we approximate the area by finding two sequences of
rectangles each with one of the subintervals as base. In one sequence, we
choose rectangles as large as possible so that the sum of their individual areas
forms an underestimate for the ‘area’ of the region; in the other, we choose
rectangles as small as possible so that the sum of their individual areas forms an
overestimate for the ‘area’.
Then, if there is a real number A with the properties:
the least upper bound of the underestimates¼A,
When defining p, we usedtriangles rather thanrectangles.
7.1 The Riemann integral 257
and:
the greatest lower bound of the overestimates¼A,
we define A to be the area between the graph and the x-axis.
We find that we can define such an area for the graphs (a) and (b) above, but
not for the graph (c).
7.1.1 The Riemann integral and integrability
We now start on a process leading to the formal definition of the integral.
Definitions A partition P of an interval [a, b] is a family of a finite
number of subintervals of [a, b]
P ¼ x0; x1½ ; x1; x2½ ; . . . ; xi�1; xi½ ; . . . ; xn�1; xn½ f g; (1)
where
a ¼ x0 < x1 < x2 < � � � < xi�1 < xi < � � � < xn�1 < xn ¼ b:
The points xi, 0� i� n, are called the partition points in P.
The length of the ith subinterval is denoted by �xi¼ xi� xi�1, and the
mesh of P is the quantity Pk k ¼ max1�i�n
�xif g.A standard partition is a partition with equal subintervals.
For example, consider the partition P of [0, 1], where
P ¼ 0;1
2
�
;1
2;3
5
�
;3
5;3
4
�
;3
4; 1
� �
:
Here
�x1 ¼1
2� 0 ¼ 1
2; �x2 ¼
3
5� 1
2¼ 1
10; �x3 ¼
3
4� 3
5¼ 3
20and
�x4 ¼ 1� 3
4¼ 1
4;
and the mesh of P is
Pk k ¼ max1
2;
1
10;
3
20;1
4
�
¼ 1
2:
P is not a standard partition of [0, 1], since not all its subintervals are of equal
length.
For brevity, we sometimesshorten this expression for Psimply to xi�1; xi½ gn
i¼1 or�
xi�1; xi½ : 1 � i � nf g:
a = x0 x1 xi–1 xn = bxi.... ....
Equivalently
P ¼ 0; 0:5½ ; 0:5; 0:6½ ;f0:6; 0:75½ ; 0:75; 1½ g:
258 7: Integration
Next, we introduce some notation, mi and Mi, associated with the height of
the graph of a bounded function f on its subintervals. Since we are intending to
develop a definition of integral that will apply to discontinuous functions as
well as to continuous functions, we use the concepts of greatest lower bound
and least upper bound of f on subintervals rather than minimum and maximum
of f on the subintervals.
We also introduce some quantities, L( f, P) and U( f, P), that correspond to
underestimates and overestimates of the possible area.
Definitions Let f be a bounded function on [a, b], and P a partition of [a, b]
given by P¼ {[xi�1, xi] : 1� i� n}. We denote by mi and Mi the quantities
mi¼ inf f xð Þ : x2 xi�1; xi½ f g and Mi¼ sup f xð Þ : x2 xi�1; xi½ f g:Then the corresponding lower and upper Riemann sums for f on [a, b] are
L f , Pð Þ ¼X
n
i¼1
mi�xi and U f , Pð Þ ¼X
n
i¼1
Mi�xi:
m1
x0
y
x x
y
x1 x2 x3 xn–1 xn xn–1 xnx0 x1 x2 x3 ......
m2 m3
mny = f (x)
y = f (x)M1
M2 M3
Mn
Example 1 Let f(x)¼ x, x2 [0, 1], and let P ¼ 0; 15
�
; 15; 1
2
�
; 12; 1
�� �
be a
partition of [0, 1]. Evaluate L( f, P) and U( f, P).
Solution In this case, the function f is increasing and continuous. Thus, on
each subinterval in [0, 1], the infimum of f is the value of f at the left end-point
of the subinterval and the supremum of f is the value of f at the right end-point
of the subinterval.
Hence, on the three subintervals in P, we have
m1 ¼ f 0ð Þ ¼ 0; M1 ¼ f1
5
� �
¼ 1
5; �x1 ¼
1
5� 0 ¼ 1
5;
m2 ¼ f1
5
� �
¼ 1
5; M2 ¼ f
1
2
� �
¼ 1
2; �x2 ¼
1
2� 1
5¼ 3
10;
m3 ¼ f1
2
� �
¼ 1
2; M3 ¼ f 1ð Þ ¼ 1; �x3 ¼ 1� 1
2¼ 1
2:
It then follows, from the definitions of L( f, P) and U( f, P), that
L f , Pð Þ ¼X
3
i¼1
mi�xi ¼ m1�x1 þ m2�x2 þ m3�x3
¼ 0� 1
5þ 1
5� 3
10þ 1
2� 1
2
¼ 0þ 3
50þ 1
4¼ 31
100;
We need to do this, since, if fis not continuous on a typicalsubinterval [xi�1, xi], it maynot possess a minimum or amaximum on the subinterval.
You met infimum andsupremum earlier, in Sub-section 1.4.2. The quantitiesmi and Mi exist, since f isbounded.
y
1
0
y = x
15
12
1 x
We use the fact that, if afunction is continuous on aclosed interval, then it attainsits infimum and supremumthere, by the Extreme ValuesTheorem in Sub-section 4.2.3.
7.1 The Riemann integral 259
U f , Pð Þ ¼X
3
i¼1
Mi�xi ¼ M1�x1 þM2�x2 þM3�x3
¼ 1
5� 1
5þ 1
2� 3
10þ 1� 1
2
¼ 1
25þ 3
20þ 1
2¼ 69
100: &
Problem 1 Evaluate L ( f, P) and U ( f, P) for the following function
and partition of [0, 1]:
f xð Þ ¼2x; 0 � x <
1
2;
1
2< x < 1;
0; x ¼ 1
2;
1; x ¼ 1;
8
>
>
<
>
>
:
x 2 0; 1½ ; and P ¼ 0; 13
�
; 13; 3
4
�
; 34; 1
�� �
:
Hint: Be careful over the values of m2 and M3; you may find it helpful
to sketch the graph of f.
Example 2 Evaluate L( f, Pn) and U( f, Pn) for the following function and
standard partition of [0, 1]
f xð Þ ¼ x; x 2 0; 1½ ; and
Pn ¼ 0;1
n
�
;1
n;2
n
�
; . . . ;i� 1
n;
i
n
�
; . . . ; 1� 1
n; 1
� �
;
and determine limn!1
L f ;Pnð Þ and limn!1
U f ;Pnð Þ, if these exist.
Solution In this case, the function f is increasing and continuous. Thus, on
each subinterval in [0, 1], the infimum of f is the value of f at the left end-point
of the subinterval and the supremum of f is the value of f at the right end-point
of the subinterval.
Hence, on the ith subinterval i�1n; i
n
�
in Pn, for 1� i� n, we have
mi ¼ fi� 1
n
� �
¼ i� 1
n; Mi ¼ f
i
n
� �
¼ i
n; and
�xi ¼i
n� i� 1
n¼ 1
n:
It then follows, from the definitions of L( f, Pn) and U( f, Pn), that
L f ;Pnð Þ ¼X
n
i¼1
mi�xi ¼X
n
i¼1
i� 1
n� 1
n
¼ 1
n2
X
n
i¼1
i�X
n
i¼1
1
( )
¼ 1
n2
n nþ 1ð Þ2
� n
�
¼ n� 1
2n;
Notice how a careful layingout of the calculation for thevarious terms makes itstraight-forward to calculatethese two sums.
It is quite important that youtackle this problem, to makesure that you understand thenotation being used. Youshould also read its solutioncarefully.
y
1
0
y = x
1n
2n
n –1n
1 x. . .
We follow the generalstructure of the solution toExample 1.
260 7: Integration
U f ;Pnð Þ ¼X
n
i¼1
Mi�xi ¼X
n
i¼1
i
n� 1
n
¼ 1
n2
X
n
i¼1
i
¼ 1
n2� n nþ 1ð Þ
2¼ nþ 1
2n:
:
It follows that
limn!1
L f ;Pnð Þ ¼ limn!1
n� 1
2n¼ 1
2and
limn!1
U f ;Pnð Þ ¼ limn!1
nþ 1
2n¼ 1
2: &
Problem 2 Evaluate L( f, Pn) and U( f, Pn) for the following function
and standard partition of [0, 1]
f xð Þ ¼ x2; x 2 0; 1½ ; and
Pn ¼ 0;1
n
�
;1
n;2
n
�
; . . .;i� 1
n;
i
n
�
; . . .; 1� 1
n; 1
� �
;
and determine limn!1
L f , Pnð Þ and limn!1
U f , Pnð Þ, if these exist.
Properties of Riemann sums
In all the examples that we have seen so far, the lower Riemann sum for f over
an interval has been less than or equal to the corresponding upper Riemann
sum. Also, in Example 2 we found that the value that ‘we hope for’ forR 1
0xdx,
namely 12, is greater than all the lower Riemann sums and less than all the upper
Riemann sums; it seems reasonable to ask whether such a property holds in
general.
So we now examine some of the key properties of Riemann sums. We shall
need to use some properties of least upper bounds and greatest lower bounds.
Theorem 1 For any function f bounded on an interval [a, b] and any
partition P of [a, b], L( f, P)�U( f, P).
Proof Let P¼ {[xi�1, xi] : 1� i� n}. Then, on each subinterval [xi�1, xi],
1� i� n, we have inf{ f(x) : x2 [xi�1, xi]}� sup { f(x):x2 [xi�1, xi]} – in other
words, mi�Mi.
It follows that
X
n
i¼1
mi�xi �X
n
i¼1
Mi�xi;
in other words, L( f, P)�U( f, P), as required. &
Next, we need some techniques for comparing the Riemann sums for
different partitions on the interval [a, b]; this will enable us shortly to verify
that, for two partitions P and P0 of [a, b], we must have L( f, P)�U( f, P0).
Here we are assuming that ourrigorous treatment ofintegration will give the sameanswers as those that youobtained in your initialCalculus course!
We will set these out carefullyas we need them.
For each term in the left-handsum is less than or equal to thecorresponding term in theright-hand sum.
Theorem 3, below.
7.1 The Riemann integral 261
Notice that this fact does NOT follow from Theorem 1, which deals only with
one partition at a time.
Definition Let P¼ {[x0, x1], [x1, x2], . . ., [xi�1, xi], . . ., [xn�1, xn]} be a
partition of an interval [a, b]. Then a partition P0 of [a, b] is a refinement of
P if the partition points of P0 include the partition points of P. A partition Q
of [a, b] is the common refinement of two partitions P and P0 of [a, b] if the
partition points of Q comprise the partition points of P together with the
partition points of P0.
For example, the partition P0 ¼ 0, 12
�
, 12
, 35
�
, 35
, 34
�
, 34
, 1 �� �
of [0, 1] is
a refinement of the partition P ¼ 0, 12
�
, 12
, 34
�
, 34
, 1 �� �
, since it simply has
one additional partition point 35
as compared with P. Similarly,
P00 ¼ 0, 13
�
, 13
, 12
�
, 12
, 35
�
, 35
, 34
�
, 34
, 1 �� �
is also a refinement of P. However the
partition Q ¼ 0, 15
�
, 15
, 12
�
, 12
, 1 �� �
of [0, 1] is not a refinement of P, since its
partition points do not include all the partition points of P.
Also, the common refinement of the partitions P ¼ 0, 12
�
, 12
, 34
�
, 34
, 1 �� �
and P0 ¼ 0, 13
�
, 13
, 23
�
, 23
, 1 �� �
is
0;1
3
�
;1
3;1
2
�
;1
2;2
3
�
;2
3;3
4
�
;3
4; 1
� �
:
We shall need the following crucial result in our work on refinements.
Lemma 1 For any bounded function f defined on intervals I and J, with
I� J, we have
infx2J
f � infx2I
f and supx2I
f � supx2J
f :
Proof By definition of greatest lower bound, we know that infx2J
f � f xð Þ, for
x2 J. It follows, from the fact that I� J, that
infx2J
f � f xð Þ; for x 2 I:
Hence infx2J
f is a lower bound for f on I, so that infx2J
f � infx2I
f .
The proof that supx2I
f � supx2J
f is similar, so we omit it. &
Problem 3 Prove that, for any bounded function f defined on intervals
I and J where I � J; supx2I
f � supx2J
f .
Lemma 2 Let f be a bounded function on an interval [a, b]. Let P and P0 be
partitions of [a, b], where P0 is a refinement of P that contains just one
additional partition point. Then
L f , Pð Þ � L f , P0ð Þ and U f , P0ð Þ � U f , Pð Þ:
Proof Let P be the partition
P ¼ x0; x1½ ; x1; x2½ ; . . . ; xi�1; xi½ ; . . . ; xn�1; xn½ f gof [a, b], and suppose that P0 contains an additional partition point c in the
particular subinterval [�, �] of P.
The point 34
is missing from Q.
I
J
Loosely speaking, the largerinterval gives the functionmore space to get smaller andmore space to get larger.
For infx2I
f is the greatest lower
bound of f on I.
Loosely speaking, theaddition of one partition pointincreases L and decreases U.
a = x0 x1 xi–1 xn = bxi... ...
262 7: Integration
Note that c must be an interior point of [�, �]. For we are assuming that all
the partition points of partitions are distinct; and, if c were an end-point � or �,
then P0 would contain that point twice in its set of partition points.
Since [�, c] [�, �], it follows from Lemma 1 that
infx2 �;�½
f � infx2 �;c½
f and infx2 �;�½
f � infx2 c;�½
f : (2)
Now, the terms in the lower sums L ( f, P) and L ( f, P0) are the same, except that
the contribution to L ( f, P) associated with the interval [�, �] is
infx2 �;�½
f � � � �ð Þ;
whereas the contribution to L( f, P0) associated with the intervals [�, c] and
[c, �] is
infx2 �; c½
f � c� �ð Þ þ infx2 c; �½
f � � � cð Þ:
It then follows, from the inequalities (2), that
infx2 �; c½
f � c� �ð Þ þ infx2 c; �½
f � � � cð Þ � infx2 �; �½
f � c� �ð Þ þ infx2 �; �½
f � � � cð Þ
¼ infx2 �; �½
f � c� �ð Þ þ � � cð Þf g
¼ infx2 �; �½
f � � � �ð Þ:
Since the lower sums L ( f, P) and L ( f, P0) are the same, apart from these
contributions to each, it follows that L ( f, P0)� L ( f, P), as required.
The proof that U ( f, P0)�U ( f, P) is similar, so we omit it. &
Lemma 2 shows that the addition of just one point to a partition increases the
lower Riemann sum and decreases the upper Riemann sum. By applying this
fact a finite number of times, we deduce the following general result.
Theorem 2 Let f be a bounded function on an interval [a,b], and let P and
P0 be partitions of [a,b], where P0 is a refinement of P. then
L f ;Pð Þ � L f ;P0ð Þ and U f ;P0ð Þ � U f ;Pð Þ:
We now compare the Riemann sums for two different partitions, and dis-
cover that all the lower Riemann sums of a bounded function on an interval are
less than or equal to all the upper Riemann sums. This is a significant
improvement on the result of Theorem 1 that, for a given partition P of [a, b],
L( f, P)�U( f, P).
Theorem 3 Let f be a bounded function on an interval [a, b], and let P and
P0 be partitions of [a, b]. Then
L f , Pð Þ � U f , P0ð Þ:
Proof Let Q be the common partition of P and P0. Then, since Q is a
refinement of both P and P0, we have:
L( f, P)� L( f, Q), by Theorem 2,
L( f, Q)�U( f, Q), by Theorem 1,
U( f, Q)�U( f, P0), by Theorem 2.
We use � and � here ratherthan xi�1 and xi simply inorder to avoid sub-subscripts.
αc
β
Loosely speaking, refining apartition increases L anddecreases U.
7.1 The Riemann integral 263
It follows from this chain of inequalities that L( f, P)�U( f, P0), as
required. &
This result is exactly what makes our definitions of lower and upper
Riemann sums useful – whatever the integral of f over an interval [a, b]
might be, if it even exists, we are certainly using lower Riemann sums, all of
which provide underestimates, and upper Riemann sums, all of which provide
overestimates.
Definitions Let f be a bounded function on an interval [a, b]. Then we
define:
� the lower integral of f on [a, b] to beR
�b
af ¼ sup
P
L f , Pð Þ,
� the upper integral of f on [a, b] to be �R b
af ¼ inf
PU f , Pð Þ,
where P denotes partitions of [a, b].
Further, if the lower and upper integrals are equal, we define the integral
of f on [a, b],R b
af , to be their common value; that is
Z b
a
f ¼Z
�
b
a
f ¼�Z b
a
f :
It is all too easy to be lulled into a sense of false security by a firmly stated
definition! So we now prove the following result that assures us that the above
definitions make sense.
Theorem 4 Let f be a bounded function on an interval [a, b]. Then:
(a) The lower integralR
�b
af and the upper integral
R
�b
a f both exist;
(b)R
�b
af �
R
�ba f .
Proof
(a) Since f is bounded on [a, b], there is some number M such that j f (x)j �M
on [a, b]; in particular,
f xð Þ � M; for x 2 a; b½ :Thus for any partition P¼ {[xi�1, xi]: 1� i� n} of [a, b], we have
f xð Þ � M; for x 2 xi�1; xi½ and 1 � i � n;
in particular, we have
mi ¼ infxi�1; xi½
f xð Þ � M:
It follows that
L f ;Pð Þ ¼X
n
i¼1
mi�xi �X
n
i¼1
M�xi
¼ MX
n
i¼1
�xi ¼ M b� að Þ:
Since all the lower sums L ( f, P) are bounded above by M (b� a), it follows
that the greatest lower bound of the lower sums, supP
L f ;Pð Þ, must exist.
This greatest lower bound is precisely the lower integralR
�b
af .
Sometimes written asR
�
b
af xð Þdx.
Sometimes written asR
�b
af xð Þdx.
Often written asR b
af xð Þdx.
For, infxi�1 ; xi½
f xð Þ � f xið Þ � M:
Also,R
�b
af � M b� að Þ.
264 7: Integration
The proof of the existence of the upper integralR
�b
af is very similar, so
we omit it here.
(b) Let P and P0 be any two partitions of [a, b]. We know, from Theorem 3,
that
L f , Pð Þ � U f , P0ð Þ:
If we fix P0 for the moment, then we know that U( f, P0) serves as an upper
bound for all the lower Riemann sums L( f, P), whichever partition P may
be. It follows, then, that the least upper bound of the L( f, P), the lower
integralR
�
b
af , must satisfy the inequalityZ
�
b
a
f � U f , P0ð Þ: (3)
It follows from the inequality (3) thatR
�
b
af serves as a lower bound for all
the upper Riemann sums U( f, P0), whichever partition P0 may be. It
follows, then, that the greatest lower bound of the U( f, P0), the upper
integralR
�b
af , must satisfy the inequality
Z
�
b
a
f �Z
�b
a
f : &
Remark
It follows, from the definition of the integral (when it exists) asZ b
a
f ¼ supP
L f , Pð Þ ¼ infP
U f , Pð Þ;
that, for any partition P of [a, b]
L f ;Pð Þ �Z b
a
f � U f ;Pð Þ:
Now, we have already seen that, if f(x)¼ x, x2 [0, 1], and
Pn ¼ 0; 1n
�
; 1n; 2
n
�
; . . . ; 1� 1n; 1
�� �
is a standard partition of [0, 1], then
limn!1
L f , Pnð Þ ¼ 1
2and lim
n!1U f , Pnð Þ ¼ 1
2: (4)
Since f is bounded on [0, 1], we know that the lower integralR
�1
0f exists. But
L f , Pnð Þ �R
�1
0f , so that, by the Limit Inequality Rule for sequences, it follows
from (4) that
1
2�Z
�
1
0
f : (5)
Similarly, since f is bounded on [0, 1], we know that the upper integralR
�1
0f
exists. ButR
�1
0f � U f , Pnð Þ, so that, by the Limit Inequality Rule for
sequences, it follows from (4) that
Z
�1
0
f � 1
2: (6)
Example 2, above.
7.1 The Riemann integral 265
We then observe thatZ
�1
0
f � 1
2�Z
�
1
0
f , in light of (5) and (6), and
Z
�
1
0
f �Z
�1
0
f , by part (b) of Theorem 4.
It follows that we must haveR
�1
0f ¼
R
�1
0f ¼ 1
2, so that f is integrable on [0, 1]
andR 1
0f ¼ 1
2.
Problem 4 Use the result of Problem 2 to prove that the function
f(x)¼ x2 is integrable on [0, 1], and evaluateR 1
0f .
Problem 5 Let f be the constant function x 7! k on [0, 1], and
Pn ¼ 0; 1n
�
; 1n; 2
n
�
; . . . ; 1� 1n; 1
�� �
a standard partition of [0, 1]. Cal-
culate the Riemann sums L( f, Pn) and U( f, Pn). Hence prove that f is
integrable on [0, 1], and evaluateR 1
0f .
However, not all bounded functions defined on closed intervals are integrable!
Example 3 Prove that the function
f xð Þ ¼ 1; 0 � x � 1; x rational,
0; 0 � x � 1; x irrational,
is not integrable on [0, 1].
Solution Let P¼ {[x0, x1], [x1, x2], . . . , [xi�1, xi], . . . , [xn�1, xn]} be any
partition of [0, 1].
Then
L f , Pð Þ ¼X
n
i¼1
mi�xi ¼X
n
i¼1
0� �xi
¼ 0;and
U f , Pð Þ ¼X
n
i¼1
Mi�xi ¼X
n
i¼1
1� �xi
¼X
n
i¼1
�xi ¼ 1:
Since all the lower Riemann sums are 0, their least upper boundR
�1
0f is also
zero. Similarly, since all the upper Riemann sums are 1, their greatest lower
boundR
�1
0f is also 1.
SinceR
�1
0f 6¼
R
�1
0f , it follows that f is not integrable on [0, 1]. &
7.1.2 Criteria for integrability
It would be tedious to have to go through the ab initio discussion of integr-
ability on each occasion that we wished to determine whether a given function
was integrable on a given closed interval. Therefore we now meet three criteria
that we often use to avoid that process.
In other words,R 1
0xdx ¼ 1
2.
xi–1 xi
1
y
10 x
y = 1, x ∈
y = 0, x ∉
For, each subinterval [xi�1, xi]contains both rational andirrational points.
We suggest that at a firstreading you omit ALL theproofs in this sub-section butread the rest of the text.
266 7: Integration
The first criterion is of particular interest in that its statement says nothing
about the value of the integral itself: it mentions only the difference between
the upper and the lower Riemann sums.
Theorem 5 Riemann’s Criterion for integrability
Let f be a bounded function on an interval [a, b]. Then
f is integrable on a; b½
if and only if:
for each positive number ", there is a partition P of [a, b] for which
U f ;Pð Þ � L f ;Pð Þ < ": (7)
Proof Suppose, first, that f is integrable on [a, b]. Then
supP
L f ;Pð Þ ¼ infP
U f ;Pð Þ; over all partitions P of a; b½ ;
denote by I the common value of these two quantities.
It follows that, for any given positive number ", there are partitions Q and Q0
of [a, b] for which
L f ;Qð Þ> I � 1
2" and U f ;Q0ð Þ< I þ 1
2": (8)
Now let P be the common refinement of Q and Q0. It follows from (8) and
Theorem 2 that
L f ;Pð Þ � L f ;Qð Þ > I � 1
2";
U f ;Pð Þ � U f ;Q0ð Þ < I þ 1
2":
Hence, we may deduce from subtracting these inequalities that
U f ;Pð Þ � L f ;Pð Þ < I þ 1
2"
�
� I � 1
2"
�
¼ ":This is precisely the assertion (7).
Suppose next that the statement (7) holds; that is, that for each positive
number ", there is some partition P of [a, b] for which
U f ;Pð Þ � L f ;Pð Þ < ": (9)
SinceR
�b
af � U f ;Pð Þ and
R
�b
af � L f ;Pð Þ, whatever partition P is, it follows
from (9) that
Z
�b
a
f �Z
�
b
a
f � U f ;Pð Þ � L f ;Pð Þ
< ": (10)
Since the left-hand side of (10) is some non-negative number independent of ",
it follows thatR
�b
af �
R
�b
af ¼ 0: In other words, f is integrable on [a, b], as
required. &
y
area = U( f, P) – L( f, P)
y = f (x)
xba
Recall that L( f, P)�U( f, P),for any partition P.
That is, that refining apartition increases L anddecreases U.
By (9).
7.1 The Riemann integral 267
Problem 6 Use Riemann’s Criterion to determine the integrability of
the following functions on [0, 1]:
(a) f xð Þ ¼ �2; 0 � x < 1,
3; x ¼ 1;
(b) f xð Þ ¼ 1; 0 � x � 1, x rational,
0; 0 � x � 1, x irrational.
Our next criterion for integrability is phrased in terms of a sequence of
partitions and its statement involves a value for the integral.
Theorem 6 Common Limit Criterion
Let f be a bounded function on an interval [a, b].
(a) If f is integrable on [a, b], then there is a sequence {Pn} of partitions of
[a, b] such that
limn!1
L f ;Pnð Þ ¼Z b
a
f and limn!1
U f ;Pnð Þ ¼Z b
a
f : (11)
(b) If there is a sequence {Pn} of partitions of [a, b] such that
limn!1
L f ;Pnð Þ and limn!1
U f ;Pnð Þ both exist and are equal; (12)
then f is integrable on [a, b] and the common value of these two limits
isR b
af .
Proof
(a) Suppose first that f is integrable on [a, b], and let I denoteR b
af .
Then, corresponding to each integer n� 1, there exist some partitions
of [a, b], Qn and Qn0 say, for which
L f ;Qnð Þ > I � 1
nand U f ;Q0n
� �
< I þ 1
n: (13)
Now let Pn be the common refinement of Qn and Qn0. It follows that
I � 1
n< L f ;Qnð Þ ðby ð13ÞÞ
� L f ;Pnð Þ ðsince Pn is a refinement of QnÞ� U f ;Pnð Þ� U f ;Q0n
� �
ðsince Pn is a refinement of Q0nÞ
< I þ 1
n: ðby ð13ÞÞ
Since the sequences I � 1n
� �
and I þ 1n
� �
both converge to I as n!1,
it follows from the Squeeze Rule for sequences and the above inequal-
ities that
L f ;Pnð Þ ! I and U f ;Pnð Þ ! I as n!1:This is precisely the statement (11).
(b) Suppose next that the statement (12) holds. Let I denote the common value
of the two limits.
268 7: Integration
It follows from (12) that, for any given positive number ", there are
partitions Q and Q0 of [a, b] for which
L f ;Qð Þ> I � 1
2" and U f ;Q0ð Þ< I þ 1
2":
So, if we let P be the common refinement of Q and Q0, it follows that
L f ;Pð Þ � L f ;Qð Þ > I � 1
2";
U f ;Pð Þ � U f ;Q0ð Þ < I þ 1
2":
Hence, we may deduce from subtracting these inequalities that
U f ;Pð Þ � L f ;Pð Þ < I þ 1
2"
�
� I � 1
2"
�
¼ ":It follows from Theorem 5 that the function f is therefore integrable on [a, b].
Finally, we use the sequence {Pn} of partitions mentioned in the state-
ment of part (b). For them, if we let n!1 in the inequality
L f ;Pnð Þ �Z b
a
f � U f ;Pnð Þ;
and use the assumption (12), it follows that limn!1
L f ;Pnð Þ ¼R b
af and
limn!1
U f ;Pnð Þ ¼R b
af . This concludes the proof. &
Now the statement of Theorem 6 can be loosely paraphrased as saying that a
function f is integrable on [a, b] if and only if there is a sequence of partitions
{Pn} such that the corresponding lower and upper Riemann sums tend to a
common value. This gives us no information on what sort of sequences we
should examine to verify that f is integrable. Our next result says that we only
need to examine any sequence of partitions that we please whose mesh tends to
zero. For instance, a function f is integrable on [a, b] if and only if the lower and
upper Riemann sums for the sequence of standard partitions {Pn} of [a, b] tend
to a common value.
Theorem 7 Null Partitions Criterion
Let f be a bounded function on an interval [a, b], and {Pn} any sequence of
partitions of [a, b] with jjPnjj! 0.
(a) If f is integrable on [a, b], then
limn!1
L f ;Pnð Þ ¼Z b
a
f and limn!1
U f ;Pnð Þ ¼Z b
a
f :
(b) If there is a number I such that limn!1
L f ;Pnð Þ and limn!1
U f ;Pnð Þ both
exist and equal I, then f is integrable on [a, b] and I ¼R b
af :
Part (b) is simply a special case of part (b) of Theorem 6, for it is true even
without the additional assumption that jjPnjj! 0.
Part (a) must be less straight-forward, for, if we drop the assumption
that jjPnjj! 0, then it is may not be the case that limn!1
L f ;Pnð Þ ¼R b
af or
limn!1
U f ;Pnð Þ ¼R b
af . To see this, consider the function f(x)¼ x, x2 [0, 1].
We have just proved that f isintegrable on [a, b]. Here wedetermine the value of itsintegral over [a, b].
This is a wonderfulsimplication!
Often the partitions Pn will bestandard partitions of [a, b].
7.1 The Riemann integral 269
For inf0;1½
f ¼ 0 and sup0;1½
f ¼ 1.
There is an analogous resultfor lower Riemann sums, butwe do not need it.
Theorem 2, Sub-section 7.1.1.
xi–1 xi
a b
xi–1 xi
a b
c
P
Q
Lemma 1 (in Sub-section 7.1.1)said that, for intervals I and Jwhere I� J
infx2J
f � infx2I
f and
supx2I
f � supx2J
f :
For 0< xi� xi�1� �.
We have already seen thatR 1
0f ¼ 1
2. However, if we choose for each partition
Pn simply the trivial partition Pn¼ {[0,1]}, then L( f, Pn)¼ 0 and U( f, Pn)¼ 1
for each n.
We have not used a condition such as jjPnjj! 0 before, so we need to
establish the following preliminary result before we can tackle the proof of
part (a) of Theorem 7.
Lemma 3 Let f be a bounded function on an interval [a, b], and let � be
any positive number. Let P be a partition of [a, b] with jjPjj<�, and P0 a
partition of [a, b] with the same partition points as P together with at most N
additional partition points. Then
U f ;P0ð Þ � U f ;Pð Þ � N M � mð Þ�;where m ¼ inf
x2 a;b½ f xð Þ and M ¼ sup
x2 a;b½ f xð Þ.
Remark
We already know, since P0 is a refinement of P, that U( f, P0)�U( f, P).
Lemma 3 gives us some lower bound to how much smaller U( f, P0) can be
than U( f, P).
Proof Let c be the first partition point of P0 that is not a partition point of P.
Then, if P¼ {[x0, x1], [x1, x2], . . . , [xn�1, xn]}, there is some integer i for which
xi�1< c< xi. Denote by Q the partition of [a, b] whose partition points are
those of P together with the additional point c.
Now the upper Riemann sums U( f, P) and U( f, Q) are the same, apart from
the contributions
supx2 xi�1;xi½
f � xi � xi�1ð Þ to U f ;Pð Þ
and
supx2 xi�1; c½
f � c� xi�1ð Þ þ supx2 c; xi½
f � xi � cð Þ to U f ;Qð Þ:
Now
supx2 xi�1;xi½
f � M; by Lemma 1;
and
supx2 xi�1; c½
f � m and supx2 c; xi½
f � m;
since m is a lower bound for f on all of a; b½ :It follows that
U f ;Qð Þ � U f ;Pð Þ ¼ supx2 xi�1; c½
f � c� xi�1ð Þ þ supx2 c; xi½
f � xi � cð Þ
� supx2 xi�1; xi½
f � xi � xi�1ð Þ
� m c� xi�1ð Þ þ m xi � cð Þ �M xi � xi�1ð Þ¼ m xi � xi�1ð Þ �M xi � xi�1ð Þ¼ � M � mð Þ xi � xi�1ð Þ� � M � mð Þ�;
270 7: Integration
so that
U f ;Qð Þ � U f ;Pð Þ � M � mð Þ�:We can interpret this inequality as follows: the insertion of one additional point
into the partition P (to obtain a new partition, Q) does not decrease the upper
Riemann sum by more than (M�m)�.We can repeat this insertion process up to a further N� 1 times to end up
with the final partition P0; the effect of this is that the upper Riemann sum of P
does not decrease by more than N(M�m)�. In other words
U f ;P0ð Þ � U f ;Pð Þ � N M � mð Þ�: &
We now use this result to prove part (a) of Theorem 7.
Theorem 7, part (a) Let f be a bounded function on an interval [a, b], and
{Pn} any sequence of partitions of [a, b] with jjPnjj! 0. If f is integrable on
[a, b], then
limn!1
L f ;Pnð Þ ¼Z b
a
f and limn!1
U f ;Pnð Þ ¼Z b
a
f :
Proof We are considering a bounded function f integrable on an interval
[a,b], and {Pn} any sequence of partitions of [a, b] with jjPnjj! 0. We will use
Lemma 3 to prove that
limn!1
U f ;Pnð Þ ¼Z b
a
f :
(The proof that limn!1
L f ;Pnð Þ ¼R b
af is similar, so we omit it.)
First, note that we can assume that f is non-constant on [a, b], since the result
is clearly immediately true in that situation.
Denote by I the value of the integralR b
af , and let m¼ inf f(x) and M¼
sup f(x).
For each n� 1, we can choose a partition Qn of [a, b] for which
U f ;Qnð Þ < I þ 1
n:
Let Nn denote the number of partition points in Qn, and let
� ¼ 1
Nnn M � mð Þ :
Next, choose any partition Pn of [a, b] for which jjPnjj<�.We then apply Lemma 3, with Pn in place of P, and with Pn
0 as the common
refinement of Pn and Qn in place of P0. It follows that
U f ;P0n� �
� U f ;Pnð Þ � Nn M � mð Þ�
¼ U f ;Pnð Þ � 1
n:
We can now obtain our crucial estimate for U( f, Pn) from this inequality, as
follows
Since f is integrable on [a, b].
Such a number � exist sinceM 6¼m, for we are assumingthat f is non-constant.
Notice that Pn0 has at most
Nn extra points as comparedwith Pn.
By Lemma 3.
From the definition of �.
7.1 The Riemann integral 271
U f ;Pnð Þ � U f ;P0n� �
þ 1
n
� U f ;Qnð Þ þ 1
n
< I þ 2
n:
We can therefore deduce that, for each n� 1
I � U f ;Pnð Þ< I þ 2
n:
It follows, by the Squeeze Rule for sequences, that U( f, Pn)! I as n!1.&
We shall use these various criteria in the next section.
7.2 Properties of integrals
In Section 7.1 you met the definition of integrability, and saw that the constant
function and the identity function were integrable on any interval [a, b]. But we
want to know that many more functions than that are integrable! In this section
we extend our family of integrable functions to a very large family indeed –
much wider than, for example, just the continuous functions.
7.2.1 Infimum and Supremum of functions (revisited)
In order to capitalise on the definition of integrability via lower and upper
Riemann sums, we need some further properties of the infimum and supremum
of a function. So, first, we remind you of their definitions.
Definitions Let ƒ be a function defined on an interval I�R . Then:
� A real number m is the greatest lower bound, or infimum, of ƒ on I if:
1. m is a lower bound of ƒ(I );
2. if m0>m, then m0 is not a lower bound of ƒ(I ).
� A real number M is the least upper bound, or supremum, of ƒ on I if:
1. M is an upper bound of ƒ(I );
2. if M0<M, then M0 is not an upper bound of ƒ(I ).
Remark Notice that, if m is a lower bound of ƒ on I, then infI
f � m;
and, if M is an upper bound of ƒ on I, then supI
f � M.
We will also sometimes need a strategy for verifying that a particular
number is the infimum or supremum of a function on a particular interval.
This is the last inequality,rewritten.
For Pn0 is a refinement of Qn.
By our choice of Qn.
You met these earlier, in Sub-section 1.4.2 and again inSection 7.1.
These are the definitions forthe least upper bound and thegreatest lower bound of afunction on an interval I; thereis a similar definition of thesebounds on a general set S in R .
272 7: Integration
When using these strategies,we shall often use thenumbers mþ " and M� " inplace of m0 and M0, to fit inwith our increased use of " asa positive number that can beas small as we please.
Lemma 1, Sub-section 7.1.1.
Loosely speaking, the largerinterval gives the functionmore space to get smaller andmore space to get larger.
Note that K must beindependent of the choiceof x and y in I.
For sup f xð Þ : x 2 If g issimply the least upper bound.(See also the Remark above.)
For inf f yð Þ:y 2 If g is simplythe greatest lower bound. (Seealso the Remark above.)
Strategies Let ƒ be a function defined on an interval I�R . Then:
� To show that m is the greatest lower bound, or infimum, of f on I, check that:
1. f (x)�m, for all x2 I;
2. if m0>m, then there is some x2 I such that f (x)<m0.
� To show that M is the least upper bound, or supremum, of f on I, check that:
1. f (x)�M, for all x2 I;
2. if M0<M, then there is some x2 I such that f (x)>M0.
Problem 1 Let f be a bounded function on an interval I. Prove that, for
any constant k
infx2I
k þ f xð Þf g ¼ k þ infx2I
f xð Þf g and
supx2I
k þ f xð Þf g ¼ k þ supx2I
f xð Þf g:
In fact you have already met one property of inf f and sup f in your study of
integration.
Lemma 1 For any bounded function f defined on intervals I and J where
I� J, we have
infx2J
f � infx2I
f and supx2I
f � supx2J
f :
A key tool in our work will be the following innocuous-looking result.
Lemma 2 Let f be a bounded function on an interval I. If, for some
number K, f (x)� f (y)�K for all x and y in I, then supI
f � infI
f � K.
Proof Since f (x)� f (y)�K for all x and y in I, we have
f xð Þ � K þ f yð Þ; for all x and y in I: (1)
It follows from (1) that, for any choice whatsoever of y in I, then Kþ f (y) serves
as an upper bound for the set of all possible values of f (x) for x in I. It follows that
sup f xð Þ : x 2 If g � K þ f yð Þ;which we may rewrite in the form
sup f xð Þ : x 2 If g � K � f yð Þ (2)
In a similar way, it follows from (2) that sup f xð Þ : x 2 If g � K serves as a
lower bound for the set of all possible values of f(y) for y in I. It follows that
sup f xð Þ : x 2 If g � K � inf f yð Þ :y 2 If g: (3)
We may rearrange the inequality (3) in the form sup f xð Þ : x 2 If g�inf f yð Þ : y 2 If g � K. This is precisely the required result, since the letters
x and y in this last inequality are simply ‘dummy variables’. &
Remarks
1. If f is bounded on I and f (x)� f (y)<K for all x and y in I, then it may not be
true that supI
f � infI
f <K. For example, if f is the identity function on
I¼ (0, 1), then
7.2 Properties of integrals 273
f xð Þ � f yð Þ < 1, for all x and y in ð0,1Þ, but sup0,1½
f � inf0,1½
f ¼ 1� 0 ¼ 1:
2. The conclusion of Lemma 2 also holds if we know that, for some number K,
j f (x)� f (y)j �K, for all x and y in I, since
f xð Þ � f yð Þ � f xð Þ � f yð Þj j:
In some situations, the following result related to Lemma 2 is also useful.
Lemma 3 Let f be a bounded function on an interval I. Then
supx; y2I
f xð Þ � f yð Þf g ¼ supI
f � infI
f :
Proof We use the strategy for supremum mentioned earlier.
Let x and y be any numbers in I. Then, in particular
f xð Þ � supI
f ; for all x 2 I ;
and, since f yð Þ � infI
f , for all y2 I
� f yð Þ ��infI
f ; for all y 2 I:
If we add these inequalities, we obtain
f xð Þ � f yð Þ � supI
f � infI
f ; for all x; y 2 I;
so that
supx;y2I
f xð Þ � f yð Þf g� supI
f � infI
f :
To prove the desired result, we now need to prove that
for each positive number ", there are X and Y in I for which
f Xð Þ � f Yð Þ >�
supI
f � infI
f�
� ": (4)
Now, by the definition of infimum and supremum on I, we know that, since12" > 0, there exist X and Y in I such that
f Xð Þ> supI
f � 1
2" and f Yð Þ < inf
If þ 1
2":
It follows from these two inequalities that
f Xð Þ � f Yð Þ >�
supI
f � 1
2"�
��
infI
f þ 1
2"�
¼ supI
f � infI
f � ":
This completes the proof. &
Problem 2 Let f (x)¼ x2 on the interval I¼ (�2, 3]. Determine
infI
f; supI
f; infx;y2I
f xð Þ � f yð Þf g and supx; y2I
f xð Þ � f yð Þf g:
By the earlier Remark.
By the strategy for supremum.
274 7: Integration
Theorem 1 Combination Rules
Let f and g be bounded functions on an interval I. Then:
Sum Rule infI
f þ gð Þ� infI
f þ infI
g and supI
f þ gð Þ � supI
f þ supI
g;
Multiple Rule infI
lfð Þ ¼ l�
infI
f�
and supI
lfð Þ ¼ l supI
f
� �
; for l> 0;
Negative Rule infI�fð Þ ¼ �
�
supI
f�
and supI
�fð Þ ¼ ��
infI
f�
.
Proof We will prove only the first part of each Rule, for the proofs of the
second parts are similar.
Sum Rule
For any x and y in I
f xð Þ � infI
f and g xð Þ � infI
g;
so that, by adding these two inequalities, we obtain
f xð Þ þ g xð Þ� infI
f þ infI
g:
Since infI
f þ infI
g is a lower bound for f(x)þ g(x) on I, it follows that
infI
f þ gð Þ � infI
f þ infI
g:
Multiple Rule
For any x in I, f xð Þ � infI
f so that
lf xð Þ � l infI
f ; for all x 2 I:
Thus
infx2I
lf xð Þð Þ � l infI
f :
To prove that infx2I
lf xð Þð Þ ¼ l�
infI
f�
, we now need to prove that:
for each positive number ", there is some X in I for which
lf Xð Þ< l�
infI
f�
þ ": (5)
Now, since "> 0 and l> 0, we have "l > 0. It follows, from the definition of
infimum of f on I, that there exists an X in I for which
f Xð Þ< infI
f þ "l:
Multiplying both sides by the positive number l, we obtain the desired
result (5).
Negative Rule
For any x in I, f xð Þ � supI
f so that
� f xð Þ� � supI
f; for all x 2 I :
Thus
infx2I�f xð Þð Þ� �sup
I
f :
Since l> 0.
7.2 Properties of integrals 275
To prove that infx2I�f xð Þð Þ ¼ �sup
I
f , we now need to prove that:
for each positive number ", there is some X in I for which
� f Xð Þ<� supI
f þ ": (6)
Now, since "> 0 it follows, from the definition of supremum of f on I, that
there exists an X in I for which
f Xð Þ > supI
f � "so that
� f Xð Þ<�supI
f þ ":
This is precisely the inequality (6). &
Problem 3 Prove that if f is a bounded function on an interval I, then:
(a) supI
lfð Þ ¼ l supI
f ; for l > 0; (b) supI
�fð Þ ¼ �infI
f .
We will now use these results on greatest lower bounds and least upper bounds
to address the main topic of this section: integration!
7.2.2 Monotonic and continuous functions
We now prove that bounded functions on an interval [a, b] are integrable if they
are either monotonic on [a, b] or continuous on [a, b]. This provides us with
very many integrable functions!
Neither of these two classes of functions includes the other. For example, the
function
f xð Þ ¼ x� x2; x 2 0; 1½ ;is continuous on [0,1] but is not monotonic on [0, 1]; while the function
f xð Þ ¼ x; 0 � x < 1;2; x ¼ 1;
; x 2 0; 1½ ;
is monotonic on [0,1] but is not continuous on [0, 1].
Theorem 2 Let f be a bounded function on an interval [a, b]. If f is mon-
otonic on [a, b], then it is integrable on [a, b].
Proof We shall assume that f is increasing on [a, b]; if it is decreasing, the
proof is similar. If f is constant on [a, b], it is certainly integrable on [a, b]. So,
we shall assume, in addition, that f is also non-constant on [a, b]; it follows, in
particular, that f (a) 6¼ f (b).
We will prove that:
for each positive number ", there is a partition P of [a, b] for which
U f ;Pð Þ � L f ;Pð Þ < ":
It then follows from the Riemann Criterion for integrability that f is integrable
on [a, b].
For any given "> 0, let P be any partition of [a, b] with mesh jjPjj such that
Pk k < "f bð Þ�f að Þ. Then, with the usual notation for Riemann sums
You should sketch the graphsof these two functions toappreciate these statements.
Theorem 5, Sub-section 7.1.2.
276 7: Integration
U f ;Pð Þ � L f ;Pð Þ ¼X
n
i¼1
Mi � mið Þ�xi ¼X
n
i¼1
f xið Þ � f xi�1ð Þð Þ�xi
� Pk k �X
n
i¼1
f xið Þ � f xi�1ð Þð Þ
¼ Pk k � f bð Þ � f að Þð Þ< ":
This completes the proof. &
Notice that the brevity of the above proof hides the fact that within the proof
we are using quite a lot of work needed to prove the Riemann Criterion!
Theorem 3 Let f be a bounded function on an interval [a, b]. If f is cont-
inuous on [a, b], then it is integrable on [a, b].
Proof We will prove that:
for each positive number ", there is a partition P of [a, b] for which
U f ;Pð Þ � L f ;Pð Þ < ":
It then follows from the Riemann Criterion for integrability that f is integrable
on [a, b].
Our key new tool here is the fact that, since f is continuous on an interval
[a, b], which is a closed interval, it is therefore uniformly continuous on [a, b].
It follows that, for any given " > 0; "2 b�að Þ > 0, so that there is a positive
number � for which
f xð Þ � f yð Þj j < "
2 b� að Þ ; for all x and y in a, b½ satisfying x� yj j < � : (7)
Now, let P ¼ xi�1, xi½ f gni¼1 be any partition of [a, b] with mesh jjPjj such
that jjPjj<�. Then, for each i we have, for all x and y in [xi�1, xi]
x� yj j � xi � xi�1
� Pk k < �:
It follows, from (7), that
f xð Þ � f yð Þj j < "
2 b� að Þ ; for all x and y in xi�1; xi½ ;
so that, by Remark 2 following Lemma 2
supx2 xi�1;xi½
f xð Þ � infx2 xi�1;xi½
f xð Þ � "
2 b� að Þ :
Then, with the usual notation for Riemann sums
U f ;Pð Þ � L f ;Pð Þ ¼X
n
i¼1
Mi � mið Þ�xi
� "
2 b� að Þ �X
n
i¼1
�xi
¼ "
2 b� að Þ � b� að Þ
¼ 1
2" < ":
This completes the proof. &
Since f is increasing on [a, b].
You met uniform continuityin Section 5.5; this assertion isTheorem 2 there.
Here the choice of � dependsONLY on ", the same choicewhatever x and y may be. Wechoose "
2 b�að Þ rather than ", forconvenience later on.
In Sub-section 7.2.1. Recallthat the conclusion ofLemma 2 is a weak inequality,not a strong inequality.
7.2 Properties of integrals 277
7.2.3 Rules for integration
Our first task is to prove the Combination Rules for integrable functions.
Theorem 4 Combination Rules
Let f and g be integrable on [a, b]. Then so are:
Sum Rule fþ g, andR b
af þ gð Þ ¼
R b
af þ
R b
ag;
Multiple Rule lf, andR b
alfð Þ ¼ l
R b
af ;
Product Rule fg;
Modulus Rule j f j.
Proof We use the usual notation for Riemann sums and integrals.
Sum Rule
Let {Pn} be any sequence of partitions of [a, b], where jjPnjj! 0 as n!1.
Since f and g are integrable on [a, b], it follows, from the Null Partitions
Criterion for integrability, that
L f ;Pnð Þ and U f ;Pnð Þ !Z b
a
f as n!1;and
L g;Pnð Þ and U g;Pnð Þ !Z b
a
g as n!1:
Then
L f ;Pnð Þ þ L g;Pnð Þ ¼X
n
i¼1
infxi�1; xi½
f � �xi þX
n
i¼1
infxi�1; xi½
g� �xi
¼X
n
i¼1
infxi�1; xi½
f þ infxi�1; xi½
g
�
� �xi
�X
n
i¼1
infxi�1; xi½
f þ gð Þ � �xi ¼ L f þ g;Pnð Þð Þ
�X
n
i¼1
supxi�1; xi½
f þ gð Þ � �xi ¼ U f þ g;Pnð Þð Þ
�X
n
i¼1
supxi�1; xi½
f þ supxi�1; xi½
g
( )
� �xi
¼ U f ;Pnð Þ þ U g;Pnð Þ:
We now let n!1 in this last set of inequalities. Then, since
L f ;Pnð Þ þ L g;Pnð Þ !Z b
a
f þZ b
a
g and U f ;Pnð Þ þ U g;Pnð Þ !Z b
a
f þZ b
a
g ;
it follows, from the Limit Inequality Rule for sequences, that
L f þ g;Pnð Þ !Z b
a
f þZ b
a
g and U f þ g;Pnð Þ !Z b
a
f þZ b
a
g; as n!1:
Hence, by part (b) of the Null Partitions Criterion for integrability, we deduce
that fþ g is integrable on [a, b] andR b
af þ gð Þ ¼
R b
af þ
R b
ag:
You may omit the detailsof all the proofs in thissub-section at a first reading.
Pn ¼ f½xi�1; xigni¼1:
Theorem 7, Sub-section 7.1.2.
By the definition of lowerRiemann sum.
By the Combination Rules forinf and sup, Theorem 1,Sub-section 7.2.1.
Theorem 7, Sub-section 7.1.2.
By the Combination Rules forinf and sup, again.
278 7: Integration
Multiple Rule
Let {Pn} be any sequence of partitions of [a, b], where jjPnjj! 0 as n!1.
Since f is integrable on [a, b], it follows from the Null Partitions Criterion for
integrability that
L f ;Pnð Þ and U f ;Pnð Þ !Z b
a
f as n!1:
Then, if l> 0
L l f ;Pnð Þ ¼X
n
i¼1
infxi�1; xi½
lfð Þ � �xi
¼ lX
n
i¼1
infxi�1; xi½
f � �xi
¼ lL f ;Pnð Þ ! lZ b
a
f as n!1;
while, if l< 0
L lf ;Pnð Þ ¼X
n
i¼1
infxi�1; xi½
lfð Þ � �xi
¼ lX
n
i¼1
supxi�1; xi½
f � �xi
¼ lU f ;Pnð Þ ! lZ b
a
f as n!1:
It follows that, for all real l, L lf ;Pnð Þ ! lR b
af as n!1.
A similar argument shows that, for all real l, U lf ;Pnð Þ ! lR b
af as n!1.
Since the limits of the two sequences of Riemann sums are equal, it follows
from part (b) of the Null Partitions Criterion for integrability, that lf is
integrable on [a, b] and
Z b
a
lfð Þ ¼ lZ b
a
f :
Product Rule
Since f and g are bounded on [a, b], there is some number M, say, such that
f xð Þj j � M and g xð Þj j � M; for all x 2 a; b½ :
Now let {Pn} be any sequence of partitions of [a, b], where jjPnjj! 0 as
n!1. It follows that, if x, y2 [xi�1, xi] for some i, then
f xð Þg xð Þ � f yð Þg yð Þj j ¼ f xð Þg xð Þ � f xð Þg yð Þf g þ f xð Þg yð Þ � f yð Þg yð Þf gj j¼ f xð Þ g xð Þ � g yð Þf g þ g yð Þ f xð Þ � f yð Þf gj j� M g xð Þ � g yð Þj j þM f xð Þ � f yð Þj j
� M supxi�1;xi½
g� infxi�1;xi½
g
( )
þM supxi�1;xi½
f � infxi�1;xi½
f
( )
;
Pn ¼ f½xi�1; xigni¼1:
Theorem 7, Sub-section 7.1.2.
By the Combination Rules forinf and sup, Theorem 1,Sub-section 7.2.1.
By the Combination Rules forinf and sup.
For, if l¼ 0, the result istrivial.
Theorem 7, Sub-section 7.1.2.
Pn ¼ f½xi�1; xigni¼1:
7.2 Properties of integrals 279
so that, by Lemma 2 in Sub-section 7.2.1 and Remark 2 following that Lemma
supxi�1;xi½
fgð Þ � infxi�1;xi½
fgð Þ � M supxi�1;xi½
g� infxi�1;xi½
g
( )
þM supxi�1;xi½
f � infxi�1;xi½
f
( )
: (8)
In terms of the Riemann sums for fg, it follows that
U fg;Pnð Þ�L fg;Pnð Þ ¼X
n
i¼1
supxi�1;xi½
fgð Þ� infxi�1;xi½
fgð Þ !
�xi
�MX
n
i¼1
supxi�1;xi½
g� infxi�1;xi½
g
( )
�xiþMX
n
i¼1
supxi�1;xi½
f � infxi�1;xi½
f
( )
�xi
¼M U g;Pnð Þ�L g;Pnð Þf gþM U f ;Pnð Þ�L f ;Pnð Þf g! 0 as n!1:
It then follows, from part (b) of the Null Partitions Criterion for integrability,
that fg is integrable on [a, b].
Modulus Rule
Let {Pn} be any sequence of partitions of [a, b], where jjPnjj! 0 as n!1.
Then, if x, y2 [xi� 1, xi] for some i, we have�
� f xð Þj j � f yð Þj j�
� � f xð Þ � f yð Þj j;so that
�
� f xð Þj j � f yð Þj j�
� � supxi�1;xi½
f � infxi�1;xi½
f
and therefore
supxi�1;xi½
fj j � infxi�1;xi½
fj j � supxi�1;xi½
f � infxi�1;xi½
f : (9)
In terms of the Riemann sums for j f j, it follows that
U fj j;Pnð Þ � L fj j;Pnð Þ ¼X
n
i¼1
supxi�1;xi½
fj j � infxi�1;xi½
fj j !
�xi
�X
n
i¼1
supxi�1;xi½
f � infxi�1;xi½
f
!
�xi
¼ U f ;Pnð Þ � L f ;Pnð Þ
! 0 as n!1:
It then follows, from part (b) of the Null Partitions Criterion for integrability,
that jf j is integrable on [a, b]. &
Our next result is often overlooked as obvious – which it is not!
Theorem 5 The Sub-interval Theorem
(a) Let f be integrable on [a, b], and let a< c< b. Then f is integrable on
both [a, c] and [c, b], andR b
af ¼
R c
af þ
R b
cf :
(b) Let f be integrable on [a, c] and [c, b], where a< c< b. Then f is
integrable on [a, b].
By definition of U and L.
By (8).
By part (a) of the NullPartitions Criterion forintegrability (Theorem 7,Sub-section 7.1.2), since f andg are integrable on [a, b].
Pn ¼ xi�1; xi½ f gni¼1:
This follows from the ‘reverseform’ of the TriangleInequality, that you met inSub-section 1.3.1.
By Remark 2 followingLemma 2, Sub-section 7.2.1.
By definition of U and L.
By (9).
By part (a) of the NullPartitions Criterion forintegrability (Theorem 7,Sub-section 7.1.2), since f isintegrable on [a, b].
a c b
280 7: Integration
Proof
(a) Let {Pn}, where Pn ¼ xi�1, xi½ f gni¼1, be any sequence of partitions of [a, b]
such that c is a partition point of each Pn, for n� 2, and such that jjPnjj! 0
as n!1.
For n� 2, let Qn consist of those subintervals in Pn that lie in [a, c], and
Qn0 consist of those subintervals in Pn that lie in [c, b]. Thus {Qn} is a
partition of [a, c] with jjQnjj! 0 as n!1, and {Qn0} is a partition of [c, b]
with jjQn0jj! 0 as n!1.
Now, with the usual notation for Riemann sums, we have
U f ;Qnð Þ � L f ;Qnð Þ ¼ the sum of those terms inP
n
i¼1
Mi � mið Þ�xi
that correspond to subintervals xi�1; xi½ that lie in a; c½
�X
n
i¼1
Mi � mið Þ�xi
¼ U f ;Pnð Þ � L f ;Pnð Þ ! 0 as n!1:It follows from the Null Partitions Criterion that f is integrable on [a, c].
A similar argument proves that f is integrable on [c, b].
In particular, it follows that
L f ;Qnð Þ !Z c
a
f and L f ;Q0n� �
!Z b
c
f as n!1:
So, if we let n!1 in the identity
L f ;Pnð Þ ¼ L f ;Qnð Þ þ L f ;Q0n� �
;
we deduce thatZ b
a
f ¼Z c
a
f þZ b
c
f :
(b) In the opposite direction, suppose that f is integrable on [a, c] and [c, b].
Let {Pn}, where Pn ¼ xi�1; xi½ f gni¼1, be any sequence of partitions of
[a, b] such that c is a partition point of each Pn, for n� 2, and jjPnjj! 0 as
n!1.
For n� 2, let Qn consist of those subintervals in Pn that lie in [a, c], and
Qn0 consist of those subintervals in Pn that lie in [c, b]. Thus {Qn} is a
partition of [a, c] with jjQnjj! 0 as n!1, and {Qn0} is a partition of [c, b]
with jjQn0jj! 0 as n!1.
Then, with the usual notation for Riemann sums, we have
U f ;Pnð Þ � L f ;Pnð Þ ¼ the sum of those terms inP
n
i¼1
Mi � mið Þ�xi that
correspond to subintervals xi�1; xi½ that lie in a; c½
þ the sum of those terms inX
n
i¼1
Mi � mið Þ�xi that
correspond to subintervals xi�1; xi½ that lie in c; b½ ¼ U f ;Qnð Þ � L f ;Qnð Þf g þ U f ;Q0n
� �
� L f ;Q0n� �� �
! 0 as n!1:
It follows, from the Null Partitions Criterion, that f is integrable on
[a, b]. &
This identity follows from thedefinitions of Pn, Qn and Qn
0.
By the definition of integral.
7.2 Properties of integrals 281
In light of Theorem 5 we now introduce some notational conventions.
Conventions Let f be integrable on [a, b], where a< b. Then we make the
following definitionsZ a
a
f ¼ 0 and
Z a
b
f ¼ �Z b
a
f :
With these conventions, we can assert that if f is integrable on any interval that
contains the points a, b and c, thenR b
af ¼
R c
af þ
R b
cf .
7.3 Fundamental Theorem of Calculus
7.3.1 Fundamental Theorem of Calculus
It would be very tedious if, each time that we wished to evaluate an integral, we
had to calculate upper and lower sums and find their greatest lower bound and
least upper bound, respectively. Fortunately, this is usually unnecessary, as
there is a short-cut using the idea of a primitive.
Definition Let f be a function defined on an interval I. Then the function F
is a primitive of f on I if F is differentiable on I and
F0 ¼ f :
For example, if
f ðxÞ ¼ 1
4� x2; x 2 ½�1; 1;
then
FðxÞ ¼ 1
4loge
2þ x
2� x
� �
; x 2 ½�1; 1;
is a primitive of f on [�1, 1], since
F0ðxÞ ¼ 1
4� 2� x
2þ x� ð2� xÞ þ ð2þ xÞ
ð2� xÞ2
¼ 1
4� 4
ð2þ xÞð2� xÞ ¼1
4� x2:
Problem 1
(a) Prove that the function f ðxÞ ¼ 1ffiffiffiffiffiffiffiffi
x2�4p , x2 (2, 1), has a primitive
F xð Þ ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � 4p� �
.
(b) Prove that the function f xð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffi
4� x2p
, x2 (�2, 2), has a primitive
F xð Þ ¼ 12
xffiffiffiffiffiffiffiffiffiffiffiffiffi
4� x2p
þ 2 sin�1 x2
� �
.
The connection between primitives of a function f and the integral of f on an
interval I is given by the following result, which is one of the most important
theorems in Analysis.
These conventions applywhere the upper and lowerlimits of the integral are thesame, or are in reverse orderon the real line.
To prove this assertion, weneed to look at all the possibleorderings of the three pointson R , which is rather tedious;so we omit it!
Sometimes we denote aprimitive of f by
R
f , and wedenote F(x) by
R
f ðxÞdx:
282 7: Integration
Theorem 1 Fundamental Theorem of Calculus
Let f be integrable on [a, b], and F a primitive of f on [a, b]. ThenZ b
a
f ¼ F bð Þ � F að Þ:
Proof Since f is integrable on [a, b], there is a sequence
Pn ¼ x0; x1½ ; x1; x2½ ; . . .; xn�1; xn½ f gof partitions of [a, b], with jjPnjj! 0 as n!1, such that
limn!1
L f ;Pnð Þ ¼Z b
a
f and limn!1
U f ;Pnð Þ ¼Z b
a
f : (1)
Now, the function F satisfies the conditions of the Mean Value Theorem on
each subinterval [xi�1, xi], for i¼ 1, 2, . . ., n. It follows that there exists some
point ci2 (xi�1, xi) such that
F xið Þ � F xi�1ð Þ ¼ F0 cið Þ xi � xi�1ð Þ¼ f cið Þ�xi: (2)
Next, since mi� f(ci)�Mi, for i¼ 1, 2 . . ., n, it follows that
X
n
i¼1
mi�xi �X
n
i¼1
f cið Þ�xi �X
n
i¼1
Mi�xi: (3)
We may now apply (2) to the general term in the sumP
n
i¼1
f cið Þ�xi; thus it
follows from (3) that
L f ;Pnð Þ �X
n
i¼1
F xið Þ � F xi�1ð Þf g � U f ;Pnð Þ:
Since the middle term is a ‘telescoping’ sum, we deduce that
L f ;Pnð Þ � F bð Þ � F að Þ � U f ;Pnð Þ: (4)
Now let n!1 in (4), using the facts in (1); this thus givesZ b
a
f � F bð Þ � F að Þ �Z b
a
f :
In other words, FðbÞ � FðaÞ ¼R b
af ; as required. &
Example 1 EvaluateR 1
02xdx.
Solution The function f(x)¼ 2x is continuous on [0, 1] and so is integrable
on [0, 1]; from the Table of Standard Primitives, it has a primitive
F xð Þ ¼ 2x
loge 2on ½0; 1:
It follows from the Fundamental Theorem of Calculus that
Z 1
0
2xdx ¼ 2x
loge 2
� 1
0
¼ 1
loge 2: &
Problem 2 Using the Table of Standard Primitives, evaluate the
following integrals:
(a)R 4
0x2 þ 9ð Þ
12dx; (b)
R e
1loge x dx:
Often F(b)�F(a) is writtenas
F xð Þ½ ba or F xð Þjba:
We use the standard notationfor lower and upper sums andpartitions.
For F 0 ¼ f and �xi¼ xi� xi�1.
Recall that
mi ¼ inf f xð Þ : x 2 xi�1; xi½ f g;Mi ¼ sup f xð Þ : x 2 xi�1; xi½ f g:
By the Limit Inequality Rulefor sequences.
The Table of StandardPrimitives appears inAppendix 2.
7.3 Fundamental Theorem of Calculus 283
7.3.2 Finding primitives
The Fundamental Theorem of Calculus asserts that, if f is integrable on [a, b],
and F a primitive of f on [a, b], thenR b
af ¼ F bð Þ � F að Þ. This does not make it
clear whether a primitive of f is unique.
In fact, for any given function f, a primitive of f is NOT unique. For example,
the functionsx 7! sin�1 x and x 7! �cos�1 x; x 2 �1; 1ð Þ;
are both primitives of the function x 7! 1ffiffiffiffiffiffiffiffi
1�x2p ; so that, for instance
Z 1ffiffi
2p
0
1ffiffiffiffiffiffiffiffiffiffiffiffiffi
1� x2p dx ¼ sin�1 x
�
1ffiffi
2p
0 ¼ � cos�1 x �
1ffiffi
2p
0
¼ 1
4p:
However, two primitives of a given function f are related; they can only
differ by a constant function.
Theorem 2 Uniqueness Theorem for Primitives
Let F1 and F2 be primitives of a function f on an interval I. Then there exists
some constant c such that
F2 xð Þ ¼ F1 xð Þ þ c; for x 2 I:
Proof Since F1 and F2 are primitives of f on I
F01 xð Þ ¼ f xð Þ and F02 xð Þ ¼ f xð Þ; for x 2 I;
so that
F02 xð Þ ¼ F01 xð Þ; for x 2 I:
It follows that there exists some constant c such that
F2 xð Þ ¼ F1 xð Þ þ c; for x 2 I: &
Our stock of primitives can be considerably extended by use of the following
Combination Rules. For convenience, we include here an extra rule which
applies to a composite function f �g for which the function g is just an x-scaling.
Theorem 3 Combination Rules
Let F and G be primitives of f and g, respectively, on an interval I, and l2R .
Then, on I:
Sum Rule fþ g has primitive FþG;
Multiple Rule lf has primitive lF;
Scaling Rule x j! f(lx) has primitive x 7! 1l F lxð Þ.
For example, it follows from the Table of Standard Primitives and the
Combination Rules that the function x 7!3 sinh 2xð Þ þ 1x, x2 (0, 1), has a
primitive x 7! 32
cosh 2xð Þ þ loge x.
Problem 3 Using the Table of Standard Primitives and the Combina-
tion Rules, find a primitive of each of the following functions:
(a) f xð Þ ¼ 4 loge x� 24þx2 ; x 2 0;1ð Þ;
(b) f xð Þ ¼ 2 tan 3xð Þ þ e2x sin x; x 2 � 16p; 1
6p
� �
:
For example
sin�1 x� �cos�1 x� �
¼ 1
2p:
By Corollary 2, Sub-section 6.4.2.
These rules are easily provedusing the corresponding rulesfor derivatives.
In applications of Theorem 3we do not usually bother tomention the theoremexplicitly.
284 7: Integration
7.3.3 Techniques of integration
We are now in a position to use the Fundamental Theorem of Calculus to give
rigorous proofs of some standard techniques for integration.
Theorem 4 Integration by Parts
If f and g are differentiable on an interval [a, b], and f 0 and g0 are continuous
on [a, b], thenZ b
a
fg0 ¼ fg½ ba �Z b
a
f 0g:
Proof We may reformulate the Product Rule for derivatives in the form
fgð Þ0 ¼ f 0gþ fg0;
in other words, that fg is a primitive on [a,b] of f 0gþ fg0.It follows from the Fundamental Theorem of Calculus that
Z b
a
f 0gþ fg0ð Þ ¼ fg½ ba;
so thatR b
afg0 ¼ fg½ ba�
R b
af 0g, as required. &
Strategy to evaluate an integral using integration by parts
1. Write the original function in the form fg0, where f is a function that you
can differentiate and g0 a function that you can integrate.
2. Use the formulaR b
afg0 ¼ fg½ ba�
R b
af 0g:
Example 2 Evaluate the integralR 1
0tan�1 xdx:
Solution Here we use a common trick: we consider g0(x) to be the factor 1,
and use integration by parts. ThusZ 1
0
tan�1 xdx ¼Z 1
0
1� tan�1 xdx
¼ x tan�1 x �1
0�Z 1
0
xdx
1þ x2
¼ tan�1 1� 1
2loge 1þ x2
� �
� 1
0
¼ 1
4p� 1
2loge 2þ 1
2loge 1 ¼ 1
4p� 1
2loge 2: &
Problem 4
(a) Find a primitive of the function f xð Þ ¼ x13 loge x; x 2 0;1ð Þ:
(b) Evaluate the integralR p
2
0x2 cos xdx:
Hint: Use integration by parts twice.
You will meet further instances of integration by parts in the next section.
Theorem 1, Sub-section 6.2.1.
Recall the notation that
fg½ ba¼ f bð Þg bð Þ � f að Þg að Þ:
There is a similar strategy forfinding a primitive byintegration by parts.
Here
f xð Þ ¼ tan�1 x; g0 xð Þ ¼ 1;
so that
f 0 xð Þ ¼ 1
1þ x2; g xð Þ ¼ x:
7.3 Fundamental Theorem of Calculus 285
Note that, in this situation, forg to possess an inversefunction, g must be strictlyincreasing.
There is an analogous result inthe situation that g is strictlydecreasing.
Theorem 2, Sub-section 6.2.2.
Recall that, by hypothesis
g �; �½ ð Þ � a; b½ :
Next we look at the method of integration by substitution, and discover that
one needs to be just a little careful in making substitutions.
Theorem 5 Integration by Substitution
If f is continuous on [a, b], g differentiable on [�, �], g0 continuous on [�, �],
and g([�, �]) � [a, b], thenZ g �ð Þ
g �ð Þf xð Þdx ¼
Z �
�
f g tð Þð Þg0 tð Þdt:
If, in addition, g(�)¼ a, g(�)¼ b and g possesses an inverse function g�1 on
[a, b], thenZ b
a
f xð Þdx ¼Z g�1 bð Þ
g�1 að Þf g tð Þð Þg0 tð Þdt:
Proof Let F be a primitive of f on [a, b], and define the function h by
h tð Þ ¼ F g tð Þð Þ; t 2 �; �½ : (5)
By the Composition Rule for derivatives, applied to the composite
h(t)¼F(g(t)), we have that
h0 tð Þ ¼ F0 g tð Þð Þg0 tð Þ: (6)
By the Fundamental Theorem of Calculus applied to h on [�, �], we haveZ �
�
h0 tð Þdt ¼ h �ð Þ � h �ð Þ: (7)
If we now substitute for h and h0 from (5) and (6) into the two sides of equation
(7), we obtainZ �
�
F0 g tð Þð Þg0 tð Þdt ¼ F g �ð Þð Þ � F g �ð Þð Þ: (8)
Since F is a primitive of f on [a, b], and so a primitive of f on g([�, �]), we can
rewrite the left-hand side of equation (8) asR �
� f g tð Þð Þg0 tð Þdt:
Further, since F is a primitive of f on [a, b], and so a primitive of f on
g([�, �]), we can apply the Fundamental Theorem of Calculus to the function F
on g([�, �]) to express the right-hand side of equation (8) as
F g �ð Þð Þ � F g �ð Þð Þ ¼Z g �ð Þ
g �ð ÞF0 xð Þdx
¼Z g �ð Þ
g �ð Þf xð Þdx:
So, if we make these substitutions on both sides of equation (8), we deduce thatZ �
�
f g tð Þð Þg0 tð Þdt ¼Z g �ð Þ
g �ð Þf xð Þdx: (9)
This is the first part of the theorem.
Suppose next that g possesses an inverse function g�1 on [�, �], and
a ¼ g �ð Þ and b ¼ g �ð Þ;then
� ¼ g�1 að Þ and � ¼ g�1 bð Þ:Making these substitutions into equation (9), we obtain the second part of the
theorem. &
286 7: Integration
There is a similar strategy forfinding a primitive byintegration by substitution.
Here
when t ¼ 0; x ¼ 1;
when t ¼ 32; x ¼ 4:
Also, t ¼ 12
x� 1ð Þ.
For, if we make thesubstitution x¼ t2, then
t ¼ �1) x ¼ 1;
t ¼ 2) x ¼ 4:
Strategy to evaluate an integral �b
a f ððgððtÞÞÞÞg0ððtÞÞdt using integration by
substitution:
1. Choose a function x¼ g(t) such that dxdt¼ g0 tð Þ, and express dt in terms of
x and dx.
2. Make the necessary substitutions to give an integral in terms of x and dx.
3. Calculate this integral.
Example 3 EvaluateR 3
2
0tþ1
2tþ1ð Þ12
dt,
Solution Let x¼ g(t)¼ 2tþ 1, t2R . The function g is one–one on R .
Then dxdt¼ 2, so that dx¼ 2dt. Making the various substitutions into the
integral, we get
Z 32
0
t þ 1
2t þ 1ð Þ12
dt ¼Z 4
1
12
x� 1ð Þ þ 1
x12
1
2dx
¼Z 4
1
12
xþ 1ð Þx
12
1
2dx
¼ 1
4
Z 4
1
x12 þ x�
12
� �
dx
¼ 1
4
2
3x
32 þ 2x
12
� �4
1
¼ 1
4
16
3þ 4
� �
� 1
4
2
3þ 2
� �
¼ 5
3: &
Problem 5
(a) Find a primitive of the function f (x)¼ sin(2 sin 3x) cos 3x, x2R .
(b) Evaluate the integralR 1
0ex
1þexð Þ2 dx.
Remark
If you use the second assertion of Theorem 5, you must verify that the function
g does have an inverse; you may end up with a contradiction if you simply
calculate thoughtlessly.
For example, take�¼�1, �¼ 2, g(t)¼ t2, and f xð Þ ¼ x12; then a¼ g(�1)¼ 1
and b¼ g(2)¼ 4. ThenZ b
a
f xð Þdx ¼Z 4
1
x12dx ¼ 2
3x
32
� �4
1
¼ 16
3
� �
� 2
3
� �
¼ 14
3;
butZ 2
�1
f g tð Þð Þg0 tð Þdt ¼Z 2
�1
t2�
12 2tdt ¼
Z 2
�1
2t2dt
¼ 2
3t3
� �2
�1
¼ 16
3
� �
� � 2
3
� �
¼ 6:
This contradiction arises since the function g does NOT have an inverse on [1, 4]
that maps [1, 4] to [�1, 2].
Our second substitution technique applies when we let t¼ g(x), where the
function g has an inverse, so that x¼ g�1(t), and we express x in terms of t, and
dx in terms of t and dt.
7.3 Fundamental Theorem of Calculus 287
There is a similar strategy forfinding a primitive byintegration by substitution.
Here
when x ¼ 0; t ¼ 0;when x ¼ loge 5; t ¼ 2:
One of the goals of NumericalAnalysis is to obtain goodestimates for integrals.
By the Null PartitionsCriterion, Theorem 7,Sub-section 7.1.2.
Strategy to evaluate an integral �b
a fððxÞÞdx using integration bysubstitution
1. Choose a function t¼ g(x), where g has an inverse, so that x¼ g�1(t);
express dx in terms of t and dt.
2. Make the necessary substitutions to give an integral in terms of t and dt.
3. Calculate this integral.
Example 4 EvaluateR loge 5
0e2x
ex�1ð Þ12
dx.
Solution Let t ¼ g xð Þ ¼ ex � 1ð Þ12, x2 [0, 1). The function g is one–one
on [0,1).
Then t2¼ ex� 1, so that ex¼ t2þ 1 and x¼ loge(t2þ 1). It follows that
dx
dt¼ 2t
t2 þ 1; so that dx ¼ 2t
t2 þ 1dt:
HenceZ loge 5
0
e2x
ex � 1ð Þ12
dx ¼Z 2
0
t2 þ 1ð Þ2
t� 2t
t2 þ 1dt
¼Z 2
0
2 t2 þ 1� �
dt
¼ 2
3t3 þ 2t
� 2
0
¼ 16
3þ 4 ¼ 28
3: &
Problem 6
(a) Find a primitive of the function f xð Þ ¼ 1
3 x�1ð Þ32 þ x x�1ð Þ
12
; x 2 1;1ð Þ:
(b) Evaluate the integralR loge 3
0ex
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ exp
dx:
7.4 Inequalities for integrals and their applications
Often it is not possible to evaluate an integral explicitly, and a numerical estimate
for its value is sufficient for our purposes. This can occur both in applications
of mathematics and in the proofs of theorems that involve integration.
7.4.1 The key inequalities
Our principal tool connecting inequalities and integrals is the following.
Theorem 1 Fundamental Inequality for Integrals
If f is integrable on [a, b], and f(x)� 0 on [a, b], thenR b
af xð Þdx � 0.
Proof Since f is integrable on [a, b], the sequence {Pn} of standard partitions
of [a, b] has the property that
limn!1
L f ;Pnð Þ ¼Z b
a
f :
Now, for each value of n, we have
288 7: Integration
We use the usual notation forpartitions and lower and upperRiemann sums.
Theorem 3, Sub-section 2.3.3.
Section 7.1.1.
L f ;Pnð Þ ¼X
n
i¼1
mi�xi; where mi ¼ inf f xð Þ : x 2 xi�1; xi½ f g:
But, since f (x)� 0 on [a, b], it follows that f (x)� 0 on each subinterval [xi�1, xi],
so that 0 is a lower bound for f on [xi�1, xi], for each i. Since mi is the greatest
lower bound for f on [xi�1, xi], we must therefore have mi� 0, for each i.
Since each term mi�xi in the sum for L(f,Pn) is non-negative, it follows that
L f ;Pnð Þ ¼X
n
i¼1
mi�xi � 0: (1)
Letting n!1 in (1) and using the Limit Inequality Rule for sequences, we
deduce that
Z b
a
f ¼ limn!1
L f ;Pnð Þ � 0: &
We can now use the Fundamental Inequality for Integrals to prove the most
commonly used inequalities for integrals.
Theorem 2 Inequality Rule for Integrals
Let f and g be integrable on [a, b]. Then:
(a) If f(x) � g(x) on [a, b], thenR b
af xð Þdx �
R b
ag xð Þdx:
(b) If m � f(x) � M on [a, b], then m b� að Þ �R b
af xð Þdx � M b� að Þ:
Recall that earlier we motivated our discussion of integrals in terms of areas.
The following diagrams illustrate the results of Theorem 1 in the special case
that the functions are positive so that we can interpret the integrals as areas
between the curves and the x-axis.
7.4 Inequalities for integrals 289
Proof of Theorem 2
(a) Since f and g are integrable on [a, b], so is the function g – f.
Since f (x) � g(x) on [a, b], it follows that
g xð Þ � f xð Þ � 0; x 2 a; b½ ;so, by Theorem 1, we deduce that
Z b
a
g� fð Þ � 0;
from which it follows at once thatR b
af �
R b
ag.
(b) First, the constant function x j!M, x2 [a, b], is continuous on [a, b] and so
is integrable on [a, b].
We can therefore apply part (a) of the Theorem to the inequality f (x)�M
on [a, b] to obtainZ b
a
f �Z b
a
Mdx ¼ M b� að Þ:
Next, the constant function x j!m, x2 [a, b], is continuous on [a, b] and so
is integrable on [a, b].
We can therefore apply part (a) of the Theorem to the inequality
f (x)�m on [a, b] to obtainZ b
a
f �Z b
a
mdx ¼ m b� að Þ: &
Example 1 Prove thatR 1
0x3
2�sin4 xdx � 1
4loge 2.
Solution Since, by the Sine Inequality, jsin xj � jxj, for x2R , it follows that
sin4 x � x4, for x2R . Hence
x3
2� sin4 x� x3
2� x4; for x 2 0; 1½ :
If we apply part (a) of Theorem 2 to this inequality, we obtainZ 1
0
x3
2� sin4 xdx �
Z 1
0
x3
2� x4dx
¼ �1
4loge 2� x4
� �
� 1
0
¼ �1
4loge 1� loge 2ð Þ ¼ 1
4loge 2: &
Example 2 Prove that 3ffiffiffiffi
34p �
R 2
�1dxffiffiffiffiffiffiffiffiffi
2þ x5p � 3:
Solution Since the function x 7!ffiffiffiffiffiffiffiffiffiffiffiffiffi
2þ x5p
is increasing on [�1, 2], we have
1ffiffiffiffiffi
34p � 1
ffiffiffiffiffiffiffiffiffiffiffiffiffi
2þ x5p � 1; for x 2 �1; 2½ :
If we apply part (b) of Theorem 2 to these inequalities, we obtain
3ffiffiffiffiffi
34p �
Z 2
�1
dxffiffiffiffiffiffiffiffiffiffiffiffiffi
2þ x5p � 3:
&
By the Combination Rules forintegrals, Theorem 4 ofSection 7.2.3.
Theorem 2, Sub-section 4.1.3.
For the length of [�1, 2] is 3.
290 7: Integration
Problem 1 Use the Inequality Rule for integrals to prove thatR 3
1x sin 1
x10
� �
dx � 4:
Problem 2 Use the Inequality Rule for integrals to prove that12�R 1
2
0ex2
dx � 23:
Hint: Use the fact that 1þ x � ex � 11�x
, for x2 [0, 1).
Our final result is of great value in many applications. We saw earlier that, if the
function f is integrable on [a, b], so is the function j f j. We can then use
Theorem 2 to obtain a connection between the values of the integrals of f and j f j.
Theorem 3 Triangle Inequality
If f is integrable on [a, b], thenZ b
a
f
�
�
�
�
�
�
�
�
�Z b
a
fj j:
Furthermore, if jf(x)j � M on [a, b], thenZ b
a
f
�
�
�
�
�
�
�
�
� M b� að Þ:
Proof From the definition of the modulus function, we have
� f xð Þj j � f xð Þ � f xð Þj j:Thus, since jfj is integrable on [a, b], it follows from part (a) of Theorem 2 that
�Z b
a
fj j �Z b
a
f �Z b
a
fj j:
We can then rewrite this pair of inequalities in the desired formZ b
a
f
�
�
�
�
�
�
�
�
�Z b
a
fj j:
Next, we assume that jf(x)j � M on [a,b]. Then, from the definition of the
modulus function, it follows that
�M � f xð Þ � M; for x 2 a; b½ :It follows from part (b) of Theorem 2 that
�M b� að Þ �Z b
a
f � M b� að Þ:
We can then rewrite this pair of inequalities in the desired formZ b
a
f
�
�
�
�
�
�
�
�
� M b� að Þ: &
Example 3 Prove thatR p
2
0
x� p2
2þ cos xdx
�
�
�
�
�
�� p2
16.
Solution By the Triangle Inequality for IntegralsZ p
2
0
x� p2
2þ cos xdx
�
�
�
�
�
�
�
�
�Z p
2
0
x� p2
�
�
�
�
2þ cos xj j dx
¼Z p
2
0
p2� x
2þ cos xdx:
By the Modulus Rule,Theorem 4, Sub-section 7.2.3.
The name arises because ofthe similarity between thisinequality and the TriangleInequality for numbers
X
n
i¼1
ai
�
�
�
�
�
�
�
�
�
�
�X
n
i¼1
aij j:
So we must now examine thislast integral.
7.4 Inequalities for integrals 291
But 2þ cos x� 2, for x 2 0; p2
�
, so that
1
2þ cos x� 1
2; for x 2 0;
p2
h i
:
By applying this inequality to the integrand in the last integral, we obtain from
the Inequality Rule for Integrals thatZ p
2
0
x� p2
2þ cos xdx
�
�
�
�
�
�
�
�
�Z p
2
0
1
2
p2� x
� �
dx
¼ 1
2
p2
x� 1
2x2
� p2
0
¼ 1
2
p2
4� p2
8
� �
¼ p2
16: &
Problem 3 Prove the following inequalities:
(a)R 4
1
sin 1xð Þ
2þ cos 1xð Þ
dx
�
�
�
�
�
�
�
�
� 3; (b)R p
4
0tan x
3� sin x2ð Þ dx�
�
�
�
�
�� 1
4loge 2:
We end this sub-section with a nice application of the Triangle Inequality,
which is closely related to the Fundamental Theorem of Calculus.
Theorem 4 Let f be continuous on [a, b], and F be defined on [a, b] by the
formula
F xð Þ ¼Z x
a
f tð Þdt:
Then F is differentiable on [a, b], and its derivative is
F0 xð Þ ¼ f xð Þ:
Proof We will prove that F is differentiable at c and F0(c)¼ f (c) in the case
that a< c< b. (The cases c¼ a and c¼ b are similar, with the appropriate one-
sided derivatives being used.)
We must prove that:
for each positive number ", there is a positive number � such that
F xð Þ � F cð Þx� c
� f cð Þ�
�
�
�
�
�
�
�
< "; for all x satisfying 0 < x� cj j < �:
Now
F xð Þ � F cð Þx� c
¼ 1
x� c
Z x
a
f tð Þdt �Z c
a
f tð Þdt
�
¼ 1
x� c
Z x
c
f tð Þdt
and
f cð Þ ¼ 1
x� c
Z x
c
f cð Þdt:
Theorem 1, Sub-section 7.3.1.
You may omit this proof at afirst reading.
We will assume that � ischosen small enough that(c� �, cþ �) [a, b].
292 7: Integration
It follows that
F xð Þ � F cð Þx� c
� f cð Þ ¼ 1
x� c
Z x
c
f tð Þ � f cð Þf gdt
so that
F xð Þ � F cð Þx� c
� f cð Þ�
�
�
�
�
�
�
�
¼ 1
x� cj j
Z x
c
f tð Þ � f cð Þf gdt
�
�
�
�
�
�
�
�
:
We now suppose that x> c; it then follows from the above equation, using the
Triangle Inequality for integrals, that
F xð Þ � F cð Þx� c
� f cð Þ�
�
�
�
�
�
�
�
¼ 1
x� c
Z x
c
f tð Þ � f cð Þf gdt
�
�
�
�
�
�
�
�
� 1
x� c
Z x
c
f tð Þ � f cð Þj jdt ðsince x > cÞ:
But, since f is continuous at c, we know that there must exist some positive
number � such that
f tð Þ � f cð Þj j < 1
2"; for all x satisfying x� cj j < �:
It follows that, for x satisfying c< x< cþ �, we have
F xð Þ � F cð Þx� c
� f cð Þ�
�
�
�
�
�
�
�
� 1
x� c
Z x
c
1
2"dt
¼ 1
2"
< ":
A similar argument applies in the case that x< c; so that, for all x satisfying
0< jx� cj<�, we have
F xð Þ � F cð Þx� c
� f cð Þ�
�
�
�
�
�
�
�
< ":
This completes the proof. &
7.4.2 Wallis’s Formula
In this sub-section we use the method of Reduction of Order together with
various inequalities between integrals to establish a remarkable formula for the
number p.
Reduction of Order method
Quite often we need to evaluate an integral In that involves a non-negative
integer n. A common approach to such integrals is to relate the value of In to the
value of In�1 or In�2 by a reduction formula, using integration by parts.
Example 4 Let In ¼R p
2
0sinn xdx, n¼ 0, 1, 2, . . ..
(a) Evaluate I0 and I1.
(b) Prove that In ¼ n�1n
In�2, for n� 2.
We omit the details.
This is just another name for arecurrence formula.
7.4 Inequalities for integrals 293
(c) Deduce from part (b) that, for n� 1
I2n ¼1:3: . . . : 2n� 3ð Þ 2n� 1ð Þ
2:4: . . . : 2n� 2ð Þ 2nð Þ �p2
and
I2nþ1 ¼2:4: . . . : 2n� 2ð Þ 2nð Þ
3:5: . . . : 2n� 1ð Þ 2nþ 1ð Þ :
Solution
(a) We can evaluate the first two integrals easily
I0 ¼Z p
2
0
1dx ¼ p2; and
I1 ¼Z p
2
0
sin xdx ¼ � cos x½ p2
0¼ 1:
(b) We first write In in the form In ¼R p
2
0sin x sinn�1 xdx. Using integration by
parts, we find that, for n� 2
In ¼ �cos xð Þ sinn�1 x �
p2
0�Z p
2
0
�cos xð Þ n� 1ð Þ sinn�2 x cos xdx
¼ 0þ n� 1ð ÞZ p
2
0
cos2 x sinn�2 xdx
¼ n� 1ð ÞZ p
2
0
1� sin2 x� �
sinn�2 xdx
¼ n� 1ð Þ In�2 � Inf g:
We can then rewrite this result in the form nIn¼ (n� 1)In�2, so that
In ¼n� 1
nIn�2:
(c) If we replace n in the formula for In in part (b) by 2n, 2n� 2, 2n� 4, . . ., in
turn, we obtain
I2n ¼2n� 1
2nI2n�2; I2n�2 ¼
2n� 3
2n� 2I2n�4; I2n�4 ¼
2n� 5
2n� 4I2n�6; . . .:
Hence
I2n ¼2n� 1
2nI2n�2
¼ 2n� 3ð Þ 2n� 1ð Þ2n� 2ð Þ 2nð Þ I2n�4
¼ 2n� 5ð Þ 2n� 3ð Þ 2n� 1ð Þ2n� 4ð Þ 2n� 2ð Þ 2nð Þ I2n�6;
We will integrate sin x anddifferentiate sinn�1 x.
294 7: Integration
and so on. Continuing this process, we obtain
I2n ¼1:3: . . . : 2n� 3ð Þ 2n� 1ð Þ
2:4: . . . : 2n� 2ð Þ 2nð Þ I0
¼ 1:3: . . . : 2n� 3ð Þ 2n� 1ð Þ2:4: . . . : 2n� 2ð Þ 2nð Þ �
p2:
Similarly
I2nþ1 ¼2n
2nþ 1I2n�1
¼ 2n� 2ð Þ 2nð Þ2n� 1ð Þ 2nþ 1ð Þ I2n�3
..
.
¼ 2:4: . . . : 2n� 2ð Þ 2nð Þ3:5: . . . : 2n� 1ð Þ 2nþ 1ð Þ I1
¼ 2:4: . . . : 2n� 2ð Þ 2nð Þ3:5: . . . : 2n� 1ð Þ 2nþ 1ð Þ : &
Problem 4 Let In ¼R 1
0exxndx, n¼ 0, 1, 2, . . ..
(a) Evaluate I0.
(b) Prove that In¼ e � nIn�1, for n� 1.
(c) Deduce the values of I1, I2, I3 and I4.
Wallis’s Formula
We now use the various results that we have just proved for the integral
In ¼R p
2
0sinn xdx to verify some surprising results.
Theorem 5 Wallis’s Formula
(a) limn!1
21: 2
3: 4
3: 4
5: 6
5: 6
7: . . . : 2n
2n�1: 2n
2nþ1
� �
¼ p2:
(b) limn!1
n!ð Þ222n
2nð Þ!ffiffi
np ¼
ffiffiffi
pp
:
In the next problem, we ask you to establish a number of relationships
between the terms of the two sequences in the statement of Theorem 5.
Problem 5 Let an ¼ 21: 2
3: 4
3: 4
5: 6
5: 6
7: . . . : 2n
2n� 1: 2n
2nþ 1and bn ¼ n!ð Þ222n
2nð Þ!ffiffi
np ,
n� 1.
(a) Evaluate an and bn, for n¼ 1, 2 and 3.
(b) Verify that b2n ¼ 2nþ1
nan, for n¼ 1, 2 and 3.
(c) Prove that b2n ¼ 2nþ1
nan, for all n� 1.
For I0 ¼ p2:
For I1¼ 1.
Each of the two limits iscalled Wallis’s Formula.
You may omit tackling thisproblem and the followingproof at a first reading.
7.4 Inequalities for integrals 295
Proof of Theorem 5 Let In ¼R p
2
0sinn xdx, for all n� 0, and let the sequences
{an} and {bn} be as given in Problem 5.
(a) Using the formulas for I2n and I2nþ1 in part (c) of Example 4 above, we
obtain
I2n
I2nþ1
¼ 1:3:3:5:5: . . .: 2n� 1ð Þ 2n� 1ð Þ 2nþ 1ð Þ2:2:4:4:6: . . .: 2n� 2ð Þ 2nð Þ 2nð Þ � p
2;
which we arrange in the form
an ¼I2nþ1
I2n
� p2:
Notice that, in order to complete the proof of part (a), it is sufficient to
show that
I2nþ1
I2n
! 1 as n!1: (2)
We do this as follows.
Since 0 � sin x � 1 for x 2 0; p2
�
, we have
sin2n x � sin2nþ1 x � sin2nþ2 x; for x 2 0;p2
h i
:
It follows, from the Inequality Rule for integrals, that
I2n � I2nþ1 � I2nþ2:
Thus
1 � I2nþ1
I2n
� I2nþ2
I2n
¼ 2nþ 1
2nþ 2: (3)
By letting n!1 in (3) and using the Squeeze Rule for sequences, we
deduce that the limit (2) holds, as desired.
(b) We know, from part (c) of Problem 5, that
b2n ¼
2nþ 1
nan; for all n � 1:
We also know, from the proof of part (a) of this theorem, that
an !p2
as n!1:
It follows, by applying the Product Rule for sequences to the above
formula for bn2, that
b2n ! 2� p
2¼ p as n!1:
Hence, by the continuity of the square root function, we conclude that
bn !ffiffiffi
pp
as n!1. &
Unfortunately the sequences in Theorem 5 converge rather slowly as n
increases, so they are of little value in determining p orffiffiffi
pp
to any reasonable
degree of appro ximat ion. In the next chapter , we shal l mee t prac tical way s of
estimating p.
The numerator of I2nþ1 isthe same as the denominatorof I2n.
Note that
I2nþ2 ¼2nþ 1
2nþ 2I2n;
by the reduction formulafor In.
Section 8.5.
296 7: Integration
7.4.3 Maclaurin Integral Test
We now introduce a method for determining the convergence or divergence of
certain series of the formP
1
n¼1
f nð Þ, where the terms are positive and decrease to
zero. In particular, we show that the seriesP
1
n¼1
1np converges if p> 1 and
diverges if 0< p� 1.
We use the fact that each term in such a series can be regarded as a
contribution to a lower Riemann sum or upper Riemann sum for a suitable
integral.
Theorem 6 Maclaurin Integral Test
Let f be positive and decreasing on [1, 1), and let f(x)! 0 as x!1.
Then:
(a)P
1
n¼1
f nð Þ converges if the sequenceR n
1f :n 2 N
� �
is bounded above;
(b)P
1
n¼1
f nð Þ diverges if the sequenceR n
1f :n 2 N
� �
tends to1 as n!1.
Before proving the theorem, we illustrate the underlying ideas.
Example 5 Let Pn�1 be the standard partition of [1, n] with n� 1 subintervals
½1; 2; ½2; 3; . . .; ½i; iþ 1; . . .; ½n� 1; nf g:
(a) Determine the lower and upper Riemann sums, L( f,Pn�1) and U( f,Pn�1),
for the function f xð Þ ¼ 1x2, x 2 1;1½ Þ.
(b) Deduce that the seriesP
1
n¼1
1n2 converges, and that 1�
P
1
n¼1
1n2 � 2.
Solution
(a) Since f is decreasing on [1,1), it follows that, for i¼ 1, 2, . . ., n� 1,
mi ¼ f iþ 1ð Þ ¼ 1
iþ 1ð Þ2;
Mi ¼ f ið Þ ¼ 1
i2:
Also, each subinterval in the partition has length 1.
You saw in Sub-section 3.2.1that this series converges ifp� 2.
In these results, the number 1may be replaced by anyconvenient positive integer.
In fact, the sum of this series isp2
6, but to prove this goes
outside the scope of this book.
7.4 Inequalities for integrals 297
Hence the lower and upper Riemann sums for f are
L f ;Pn�1ð Þ ¼X
n�1
i¼1
mi � 1 ¼ 1
22þ 1
32þ � � � þ 1
n2;
U f ;Pn�1ð Þ ¼X
n�1
i¼1
Mi � 1 ¼ 1
12þ 1
22þ � � � þ 1
n� 1ð Þ2:
(b) Let sn denote the nth partial sum of the seriesP
1
n¼1
1n2; thus
sn ¼1
12þ 1
22þ � � � þ 1
n� 1ð Þ2þ 1
n2:
It follows that
L f ;Pn�1ð Þ ¼ sn � 1 and U f ;Pn�1ð Þ ¼ sn �1
n2:
Since f is monotonic on [1, n], it is integrable on [1, n]; hence we have
L f ;Pn�1ð Þ �Z n
1
f � U f ;Pn�1ð Þ;
so that
sn � 1 �Z n
1
dx
x2� sn �
1
n2: (4)
Now, the sequence {sn} is increasing, since the series has positive terms.
Also, from (4), we have
sn �Z n
1
dx
x2þ 1
¼ � 1
x
� n
1
þ1
¼ 1� 1
n
� �
þ 1 ¼ 2� 1
n� 2: (5)
Thus {sn} is bounded above.
Hence, by the Monotone Convergence Theorem for sequences, {sn} is
convergent, so that
X
1
n¼1
1
n2is convergent:
Since s1¼ 1 and {sn} is increasing, the sum of the series is at least 1.
Finally, we deduce from (5), using the Limit Inequality Rule for
sequences, that limn!1
sn � 2; so the sum of the series is at most 2. Hence
1 �X
1
n¼1
1
n2� 2: &
Example 6 Let Pn�1 be the standard partition of [1, n] with n� 1 subintervals
½1; 2; ½2; 3; . . .; ½i; iþ 1; . . .; ½n� 1; nf g:
298 7: Integration
(a) Determine the lower and upper Riemann sums, L( f, Pn�1) and U( f, Pn�1),
for the function f xð Þ ¼ 1x; x 2 1;1½ Þ.
(b) Deduce that the seriesP
1
n¼1
1n
diverges.
Solution
(a) Since f is decreasing on [1,1), it follows that, for i¼ 1, 2, . . . , n� 1
mi ¼ f iþ 1ð Þ ¼ 1
iþ 1;
Mi ¼ f ið Þ ¼ 1
i:
Also, each subinterval in the partition has length 1.
Hence the lower and upper Riemann sums for f are
L f ;Pn�1ð Þ ¼X
n�1
i¼1
mi � 1 ¼ 1
2þ 1
3þ � � � þ 1
n;
U f ;Pn�1ð Þ ¼X
n�1
i¼1
Mi � 1 ¼ 1
1þ 1
2þ � � � þ 1
n� 1:
(b) Let sn denote the nth partial sum of the seriesP
1
n¼1
1n; thus
sn ¼1
1þ 1
2þ � � � þ 1
n� 1ð Þ þ1
n:
It follows that
L f ;Pn�1ð Þ ¼ sn � 1 and U f ;Pn�1ð Þ ¼ sn �1
n:
Since f is monotonic on [1, n], it is integrable on [1, n]; hence we have
L f ;Pn�1ð Þ �Z n
1
f � U f ;Pn�1ð Þ;
so that
sn � 1 �Z n
1
dx
x� sn �
1
n;
or
sn � 1 � loge n � sn �1
n: (6)
In particular, sn� loge n. Then, since loge n!1 as n!1, it follows, from
the Squeeze Rule for sequences that tend to infinity, that sn!1 as n!1.
Hence, the seriesP
1
n¼1
1n
diverges. &
Remarks
We can in fact get more information from the above argument than just the
result of the example.
1. If we define the sequence {�n} by the expression
�n ¼ 1þ 1
2þ 1
3þ � � � þ 1
n� loge n
7.4 Inequalities for integrals 299
then
�nþ1 � �n ¼1
nþ 1� loge
nþ 1
n
¼ 1
nþ 1�Z nþ1
n
dx
x
¼Z nþ1
n
1
nþ 1� 1
x
� �
dx � 0;
so that {�n} is decreasing. Also, it follows from (6), using the fact that
�n¼ sn� loge n, that
�n �1
n;
thus the sequence {�n} is bounded below by 0.
Hence the sequence {�n} tends to a limit, � say, as n!1. The value of �clearly lies between 1 (for �1¼ 1) and 0.
This number is called Euler’s constant, and occurs often in Analysis.
In fact,
� ¼ 0:57721 56649 01532 86060 65120 90082 40243 . . .:
2. Although the sequence sn ¼ 1þ 12þ � � � þ 1
n�1þ 1
ntends to infinity, we can
make more precise statements than that, arising from (6). We can rewrite
the inequalities in (6) in the following form
1
nþ loge n � sn � 1þ loge n;
if we then divide throughout by logen and let n!1, we obtain, by the
Limit Inequality Rule for sequences, that
sn
loge n! 1 as n!1:
We can write this last limit in an equivalent form as ‘sn� logen as n!1’ –
in other words, ‘the behaviour of sn and logen are essentially the same for
large n’.
Problem 6 Use the Maclaurin Integral Test to determine the behaviour
of the seriesP
1
n¼1
1np, for p> 0, p 6¼ 1.
Problem 7 Show thatR
dx
x loge xð Þ2 ¼ �1
loge x, for x> 1, and hence prove
that the seriesP
1
n¼2
1
n loge nð Þ2 converges.
Problem 8 Show thatR
dxx loge x
¼ loge loge xð Þ, for x> 1, and hence
prove that the seriesP
1
n¼2
1n loge n
diverges.
Sequences revisited
We can apply many ideas similar to those in Example 5 to obtain useful
information about certain types of convergent sequences.
By definition of the sequence{�n}.
For loge n � sn � 1n:
We give here just the first 35decimal places.
We shall address this �notation more carefully inthe next section.
300 7: Integration
Example 7 Prove that 1nþ 1þ 1
nþ 2þ � � � þ 1
2n! loge 2 as n!1.
Solution Let f xð Þ ¼ 11þx
, x2 [0, 1]; and let Pn ¼ 0; 1n
�
; 1n; 2
n
�
; . . .; n�1n; 1
�� �
be the standard partition of [0, 1] into n sub-intervals of equal length 1n.
Since f is decreasing on [0, 1], it follows that on the ith sub-interval i�1n; i
n
�
we have
mi ¼ fi
n
� �
¼ 1
1þ in
:
Thus the lower Riemann sum for f on Pn is
L f ;Pnð Þ ¼X
n
i¼1
1
1þ in
� 1
n
¼X
n
i¼1
1
nþ i¼ 1
nþ 1þ 1
nþ 2þ � � � þ 1
2n:
Since f is decreasing on [0, 1], it is integrable on [0, 1]. Hence, as n!1, we
have
1
nþ 1þ 1
nþ 2þ �� �þ 1
2n¼ L f ;Pnð Þ!
Z 1
0
f ¼Z 1
0
dx
1þ x
¼ loge 1þ xð Þ½ 10 ¼ loge 2:&
Example 7 illustrates a general technique that is often useful.
Strategy If f is positive and decreasing on [0, 1], then: 1n
P
n
i¼1
f in
� �
!R 1
0f as
n!1.
Problem 9 Prove that n 1n2þ12 þ 1
n2þ22 þ � � � þ 1n2þn2
� �
! p4
as n!1.
You have already met the corresponding ideas for divergent sequences in
Example 6 and Remark 2 after that example, so we state without comment
the general strategy illustrated there.
Strategy If f is positive and decreasing on [1,1), f(x)! 0 as x!1, andR n
1f !1 as n!1, then
P
n
i¼1
f ið Þ �R n
1f as n!1.
Remark
In fact under the hypotheses of this strategyP
n
i¼1
f ið Þ � ðR n
1f Þ þ c, for any fixed
number c; in particular situations we may choose c in any convenient way.
Problem 10 Prove that 1þ 1ffiffi
2p þ 1
ffiffi
3p þ � � � þ 1
ffiffi
np � 2
ffiffiffi
np
as n!1.
Recall that
mi ¼ inf f xð Þ : x 2 i�1n; i
n
�� �
:
The trick in applying thisstrategy is to make a goodchoice of f.
Recall that this means that
P
n
i¼1
f ið ÞR n
1f! 1 as n!1:
The reason for this will beexplained in Section 7.5.1.
Here you will use both thestrategy and the subsequentremark.
7.4 Inequalities for integrals 301
Proof of the Maclaurin Integral Test
Theorem 6 Maclaurin Integral Test
Let f be positive and decreasing on [1,1), and let f(x)! 0 as x!1. Then:
(a)P
1
n¼1
f nð Þ converges if the sequenceR n
1f : n 2 N
� �
is bounded above;
(b)P
1
n¼1
f nð Þ diverges if the sequenceR n
1f : n 2 N
� �
tends to1 as n!1.
Proof Let In denote the integralR n
1f , let sn¼ f(1)þ f(2)þ � � � þ f(n) denote
the nth partial sum of the series, and let Pn�1 be the standard partition of [1, n]
with n� 1 subintervals
½1; 2; ½2; 3; . . .; ½i; iþ 1; . . .; ½n� 1; nf g:
Since f is decreasing on [1,1), it follows that, for i¼ 1, 2, . . ., n� 1
mi ¼ f iþ 1ð Þ and Mi ¼ f ið Þ:Also, each subinterval in the partition has length 1.
Hence the lower and upper Riemann sums for f are
L f ;Pn�1ð Þ ¼X
n�1
i¼1
mi � 1 ¼ f 2ð Þ þ f 3ð Þ þ � � � þ f nð Þ
¼ sn � f 1ð Þ;
U f ;Pn�1ð Þ ¼X
n�1
i¼1
Mi � 1 ¼ f 1ð Þ þ f 2ð Þ þ � � � þ f n� 1ð Þ
¼ sn � f nð Þ:
Since f is monotonic on [1, n], it is integrable on [1, n]; hence we have
L f ;Pn�1ð Þ � In � U f ;Pn�1ð Þ;
so that
sn � f 1ð Þ � In � sn � f nð Þ: (7)
Case 1: {In} is bounded
We are now assuming that, for some M, In�M, for all n. It follows from
(7) that
sn � f 1ð Þ þM; for all n:
302 7: Integration
Thus the increasing sequence {sn} is bounded above, and so, by the Monotone
Convergence Theorem, it is convergent.
Hence the seriesP
1
n¼1
f nð Þ is convergent.
Case 2: {In} is not bounded
The sequence {In} is increasing, since Inþ1 � In ¼R nþ1
nf � 0. Since we are
now assuming that {In} is not bounded, it follows that In!1 as n!1.
Now, from (7), sn� In; so, by the Squeeze Rule for sequences which tend to
infinity,
sn !1 as n!1:
Hence the seriesP
1
n¼1
f nð Þ is divergent. &
7.5 Stirling’s Formula for n!
For small values of n, we can evaluate n! directly by multiplication or by using
a scientific calculator.
Problem 1 Complete the following table of values of n!
n n! n n! n n!
1 1 6 720 20 2 2 7 5040 30 3 6 8 40 320 40 4 24 9 362 880 50 5 120 10 3 628 800 60
As n increases, n! grows very quickly; for instance:
� around 14! seconds have elapsed since the birth of Christ;
� around 18! seconds have elapsed since the formation of the Earth.
A calculator soon becomes useless for evaluating n!; thus the author’s
current scientific calculator (2005) gives that 69!’ 1.7� 1098, but an error
message when asked the value of 70!.
Stirling’s Formula gives us a way of estimating n! for large values of n. In order
to state the formula, though, we must first introduce some notation that enables us
to compare the behaviour of positive functions of n for large values of n.
7.5.1 The tilda notation
For many numbers that arise in the normal way, we often use the symbol ‘’’ to
denote ‘is approximately equal to’. For example,ffiffiffi
2p¼ 1:4142135623730950
4880 . . . (where we have given only the first 20 decimal places); for many
purposes it is sufficient to use estimates such asffiffiffi
2p’ 1:414 or even
ffiffiffi
2p’ 1:4,
14! ’ 8:7� 1010
18! ’ 6:4� 1015
Often an estimate is sufficientfor our purposes.
7.5 Stirling’s Formula for n! 303
since the error involved is small. But what do we mean by ‘the error involved is
small’? Sometimes we mean that the error, the difference between the exact
value and the approximation, is a small number; sometimes we mean that the
percentage error, namely
percentage error ¼ actual error
exact value� 100;
is a small number.
To handle the behaviour of positive functions of n for large values of n, we
introduce a very precise mathematical notation.
Notation For positive functions f and g with domain N , we write
f nð Þ � g nð Þ as n!1;
to mean thatf nð Þg nð Þ ! 1 as n!1.
For example, n2þ 1000nþ 10� n2 as n!1, since n2þ1000nþ 10n2 !1 as
n!1; and sin 1n
� �
� 1n
as n!1, sincesin 1
nð Þ1n
! 1 as n!1.
Notice that, if f(n)� g(n) as n!1, then g(n)� f(n) as n!1.
Also, if f(n)� g(n) as n!1, g(n)!1 and c is a given number, then
f(n)þ c� g(n) as n!1, since, as n!1f nð Þ þ c
g nð Þ ¼f nð Þg nð Þ þ
c
g nð Þ ! 1þ 0 ¼ 1:
Notice, however, that the statement f(n)� g(n) as n!1 does NOT mean
that f(n)� g(n)! 0 or even that f(n)� g(n) is bounded. For example,
n2þ 1000nþ 10� n2 as n!1, yet
n2 þ 1000nþ 10� �
� n2� �
¼ 1000nþ 10!1 as n!1:
In situations like this example, ‘�’ compares the sizes of the dominant terms
in f and g.
Problem 2 Find two pairs from the following functions such that
fi(n)� fj(n) as n!1f1 nð Þ ¼ sin n2ð Þ; f2 nð Þ ¼ sin 1
n2
� �
; f3 nð Þ ¼ 1� cos 1n
� �
;f4 nð Þ ¼ 2
n2 ; f5 nð Þ ¼ 1n2 ; f6 nð Þ ¼ 1� 1
n; f7 nð Þ ¼ 1
2n2 :
Remark
If ‘ is finite and positive, then the statements
‘f nð Þ ! ‘ as n!1’ and ‘f nð Þ � ‘ as n!1’
are equivalent, since each is equivalent to the statement
‘f nð Þ‘! 1 as n!1’:
We can also, in this situation, legitimately say that f(n)� ‘ for large n.
The Combination Rules for handling tilda are direct consequences of the
corresponding Combination Rules for sequences.
‘For example, the error is lessthan 1’.
For example, ‘the percentageerror is less than 1%’.
The notation is ONLY usedfor positive functions.
We say ‘f tilda g’ or ‘ftwiddles g’.
You can think of this as sayingthat the percentage errorinvolved by replacing f(n) byg(n) is small for large n.
For, limx!0
sin xx¼ 1:
For,f nð Þg nð Þ ! 1, g nð Þ
f nð Þ ! 1.
For example, we may write
1þ 1
n
� �n
� e as n!1;
since
1þ 1
n
� �n
! e as n!1:
Note that we only use � forpositive funtions.
304 7: Integration
Combination Rules
Let f1(n)� g1(n) and f2(n)� g2(n) as n!1. Then:
Sum Rule f1(n)þ f2(n)� g1(n)þ g2(n);
Multiple Rule lf1(n)� lg1(n), for any number l> 0;
Product Rule f1(n)� f2(n)� g1(n)� g2(n);
Reciprocal Rule 1f1 nð Þ � 1
g1 nð Þ.
Example 1 Prove that, if f(n)� g(n) as n!1, then f nð Þð Þ1n � g nð Þð Þ
1n
as n!1.
Solution Since f(n)� g(n) as n!1, we have
f nð Þg nð Þ ! 1 as n!1;
so that
loge
f nð Þg nð Þ ! 0 as n!1;
and therefore
1
nloge
f nð Þg nð Þ ! 0 as n!1:
But
1
nloge
f nð Þg nð Þ ¼ loge
f nð Þg nð Þ
� �1n
¼ loge
f nð Þð Þ1n
g nð Þð Þ1n
;
so that
loge
f nð Þð Þ1n
g nð Þð Þ1n
! 0 as n!1:
We now use the fact that the exponential function is continuous at 0; it follows
from the previous limit that
f nð Þð Þ1n
g nð Þð Þ1n
! 1 as n!1;
which is precisely the statement that f nð Þð Þ1n � g nð Þð Þ
1n as n!1. &
Problem 3 Give specific examples of functions f and g to show that
f nð Þð Þ1n � g nð Þð Þ
1n as n!1 does not imply that f(n)� g(n) as n!1.
Hint: Try f(n)¼ n2 and g(n)¼ n.
7.5.2 Stirling’s Formula
Stirling’s Formula was discovered in the eighteenth century, in an analysis of
a gambling problem!
For example, if
f1 nð Þ ¼ n2 þ n; g1 nð Þ ¼ n2;
f2 nð Þ ¼ n3 þ n; g2 nð Þ ¼ n3;
then
n2 þ n� �
þ n3 þ n� �
� n2 þ n3;
5 n2 þ n� �
� 5n2 l ¼ 5ð Þ;n2 þ n� �
� n3 þ n� �
� n2 � n3 ¼ n5;
1
n2 þ n� 1
n2:
Here we use the fact thatthe function loge is continuousat 1.
7.5 Stirling’s Formula for n! 305
Theorem 1 Stirling’s Formula
n! �ffiffiffiffiffiffiffiffi
2pnp n
e
� �n
as n!1:
Problem 4 Use your calculator to evaluateffiffiffiffiffiffiffiffi
2pnp
ne
� �nfor n¼ 5. For
this value of n does the expression approximate n! to within 1%?
For small values of n, Stirling’s Formula gives reasonable approximations to
n!, and the percentage error quickly decreases as n increases:
n n! Stirling’s approximation Error
10 3 628 800 3 598 696 0.83% (1 in 120)
20 2.433� 1018 2.423� 1018 0.42% (1 in 240)
52 8.066� 1067 8.053� 1067 0.16% (1 in 620)
100 9.333� 10157 9.325� 10157 0.09% (1 in 1170)
We illustrate the use of Stirling’s Formula as follows.
If 200 coins are tossed, then the probability of there being exactly 100 heads
and 100 tails is
200
100
� �
� 1
2
� �200
¼ 200!
100!ð Þ2�2200:
It follows from Stirling’s Formula that this probability is
200!
100!ð Þ2�2200’
ffiffiffiffiffiffiffiffiffiffi
400pp
� 200e
� �200
ffiffiffiffiffiffiffiffiffiffi
200pp
� 100e
� �100� �2
� 2200
¼ 1
10ffiffiffi
pp
¼ 1
17:724 . . .:
In other words, the probability of there being exactly 100 heads and 100 tails
is about 1 in 18 – rather higher than you might expect.
Problem 5 Prove that limn!1
nn
n!
� �1n ¼ e.
Hint: Use the result of Example 1.
Problem 6 Use Stirling’s Formula to estimate the following numbers
to two significant figures:
(a)300
150
� �
� 1
2
� �300
; (b)300!
100!ð Þ3� 1
3
� �300
:
Problem 7 Use Stirling’s Formula to determine a number l such that
4n
2n
� ��
2n
n
� �
� l22n as n!1:
This fact is easily provedusing Probability Theory.
Here we substitute n¼ 200and n¼ 100 into Stirling’sFormula.
306 7: Integration
7.5.3 Proof of Stirling’s Formula
Theorem 1 Stirling’s Formula
n! �ffiffiffiffiffiffiffiffi
2pnp n
e
� �n
as n!1:
Proof We divide up the proof into a number of steps, for clarity.
Step 1: Setting things up
We consider the function
f ðxÞ ¼ loge x; x 2 1; n½ ;and the standard partition Pn�1 of the interval [1, n] with n� 1 subintervals
1; 2½ ; 2; 3½ ; . . .; i; iþ 1½ ; . . .; n� 1; n½ f g:We consider also the sequence of numbers cnf g1n¼2, where cn is the total area
between the concave curve y¼ loge x, for x2 [1, n], and the polygonal line with
vertices (1, 0), (2, loge 2), (3, loge 3), . . ., (n, loge n), as illustrated below. This
consists of n� 1 small slivers.
y
x1 2 3 ... ...i
total area ofsilvers = cn
y = loge x y = loge x
(i, logei)
(i + 1,loge(i + 1))
contributionto cn
i + 1 n
Step 2: Calculating areas
The area between the curve y¼ loge x and the x-axis, for x2 [1, n], isZ n
1
loge xdx ¼ x loge x� x½ n1
¼ n loge n� n� 1ð Þ: (1)
Next, the area between the polygonal line and the x-axis is
12
L f , Pn�1ð Þ þ U f ;Pn�1ð Þf g; (2)
and, since f is increasing, we have
L f , Pn�1ð Þ ¼ loge 1þ loge 2þ � � � þ loge n� 1ð Þ¼ loge n� 1ð Þ!¼ loge n!� loge n (3)
and
U f , Pn�1ð Þ ¼ loge 2þ � � � þ loge n
¼ loge n!: (4)
Substituting from (3) and (4) into (2), we find that the area between the
polygonal line and the x-axis is
7.5 Stirling’s Formula for n! 307
loge n!� 12
loge n: (5)
It follows from (1) and (5) that
cn ¼ n loge n� n� 1ð Þ � loge n!þ 12
loge n
¼ loge
nnþ12
en�1n!: (6)
Step 3: Behaviour of the sequence {cn}
It is obvious from the definition of cn that the sequence {cn} is positive and
increasing. In order to apply the Monotone Convergence Theorem to the
sequence, we must prove next that {cn} is bounded above.
So, let i be an integer with 1� i� n� 1, and let
A ¼ i; loge ið Þ;B ¼ iþ 1; logeðiþ 1Þð Þ; and C ¼ iþ 1; loge ið Þ:Then, as illustrated in the diagram in the margin
BC ¼ loge iþ 1ð Þ � loge i
¼ loge 1þ 1
i
� �
:
Next, let the tangent at A to the curve meet the line BC at D. Since AC¼ 1
and the slope of the line AD is 1i, it follows that CD ¼ 1
i. Hence
BD ¼ CD� BC
¼ 1
i� loge 1þ 1
i
� �
� 1
2i2:
Hence the contribution to the area cn of this sliver between the lines x¼ i and
x¼ iþ 1 is at most
area of � ABD ¼ 1
2AC � BD
� 1
2� 1� 1
2i2¼ 1
4i2:
If we now sum such areas over i¼ 1, 2, . . ., n� 1, we obtain
cn �1
4
X
n�1
i¼1
1
i2
� 1
4
X
1
i¼1
1
i2:
Since this infinite series converges, it follows that the sequence {cn} is
bounded above.
It follows from the Monotone Convergence Theorem that the sequence {cn}
is convergent.
Step 4: Properties of the sequence {an}, where an ¼ ecn
Since {cn} is convergent and the exponential function is continuous, it follows
that, if we set an ¼ ecn , for n¼ 2, 3, . . ., then the sequence {an} is also
convergent. Thus, using the expression (6) for cn, we have
an ¼nnþ1
2
en�1n!! L; as n!1; (7)
for some non-zero number L.
Here we use the inequality
loge 1þ xð Þ > x� 12x2;
for x > 0;
that you met in Section 6.7,Exercise 5(c) on Section 6.4.
The actual value of the bounddoes not matter here.
We now introduce {an} sothat we can use Wallis’sFormula in the next Step.
L 6¼ 0 since the exponentialdoes not take the value 0.
308 7: Integration
It follows from the formula for an in (7) that
a2n
a2n
¼ n2nþ1
e2n�2 n!ð Þ2� e2n�1 � 2nð Þ!
2nð Þ2nþ12
¼ 2nð Þ!n12e
n!ð Þ222nþ12
; (8)
after some cancellation.
Step 5: Proving Stirling’s Formula
We can rewrite (8) in the form
a2n
a2n
¼ 2nð Þ!n12
n!ð Þ222n� e
ffiffiffi
2p : (9)
But, by Wallis’s Formula, the first quotient on the right-hand side of (9) tends
to 1ffiffi
pp as n!1. It follows that, if we let n!1 on both sides of (9), we obtain
L2
L¼ 1
ffiffiffi
pp � e
ffiffiffi
2p ;
so that
L ¼ effiffiffiffiffiffi
2pp :
Knowing the value of L, we can then rewrite (7) in the form
nnþ12
en�1n!! e
ffiffiffiffiffiffi
2pp ;
by the Reciprocal Rule for sequences, it follows that
en�1n!
nnffiffiffi
np !
ffiffiffiffiffiffi
2pp
e:
. By the definition of tilda, this limit is exactly equivalent to the relation
n! �ffiffiffiffiffiffiffiffi
2pnp n
e
� �n
as n!1;
that we set out to prove. &
Remark
It is quite remarkable how many of the techniques that we have met so far in the
book are needed in order to prove this apparently straight-forward result.
7.6 Exercises
Section 7.1
1. Sketch the graph of the function
f xð Þ ¼ 1� xj j; �1 < x < 1;1; x ¼ �1:
Determine the minimum, maximum, infimum and supremum of f on [�1, 1].
You met Wallis’s Formulain Theorem 5 of Sub-section 7.4.2.
limn!1
n!ð Þ222n
2nð Þ!ffiffiffi
np ¼
ffiffiffi
pp
:
7.6 Exercises 309
2. Let f be the function
f xð Þ ¼ xj j; � 1 < x < 1;12; x ¼ �1:
Evaluate L( f, P) and U( f, P) for each of the following partitions P of [�1, 1]:
(a) P ¼ �1;� 12
�
; � 12; 0
�
; 0; 12
�
; 12; 1
�� �
;
(b) P ¼ �1;� 14
�
; � 14; 1
3
�
; 13; 1
�� �
:
3. Let f be the function
f xð Þ ¼ 1� x; 0 � x < 1;2; x ¼ 1:
(a) Using the standard partition Pn of [0, 1] with n equal subintervals,
evaluate L( f, Pn) and U( f, Pn).
(b) Deduce that f is integrable on [0, 1], and evaluateR 1
0f :
4. Let f be the function f(x)¼ x3, x2 [0, 1].
(a) Using the standard partition Pn of [0, 1] with n equal subintervals,
evaluate L( f, Pn) and U( f, Pn).
(b) Deduce that f is integrable on [0, 1], and evaluateR 1
0f :
5. Let f be the function f (x)¼ sin x, x2 0; p2
�
:
(a) Using the standard partition Pn of 0; p2
�
with n equal subintervals,
evaluate L ( f, Pn) and U ( f, Pn).
(b) Deduce that f is integrable on 0; p2
�
, and evaluateR p
2
0f :
Hint: Use the formula
sin Aþ sin Aþ Bð Þ þ � � � þ sin Aþ n� 1ð ÞBð Þ
¼sin 1
2nB
� �
sin 12
B� � sin Aþ n� 1
2B
� �
; B 6¼ 0:
6. Prove that the function
f xð Þ ¼ 1þ x; 0 � x � 1; x rational;1� x; 0 � x � 1; x irrational;
is not integrable on [0, 1].
7. Prove that Dirichlet’s function
f xð Þ ¼1q; if x is a rational number p
q, in lowest terms, with q > 0,
0; if x is irrational.
is integrable on [0, 1].
Hint: Verify that there are at most 12
n nþ 1ð Þ points x in [0, 1] for which
f xð Þ > 1n:
Section 7.2
1. For bounded functions f and g on an interval I, prove that
supI
f þ gð Þ � sup fI
þ supI
g:
For the interval I¼ [�1, 2], write down functions f and g for which strict
inequality holds in this result.
This formula can be proved byMathematical Induction.
You met this function in Sub-section 5.4.2.
310 7: Integration
2. Write down functions f and g and an interval I for which it is not true that
supI
fgð Þ ¼ sup fI
� supI
g:
3. Write down functions f and g on [0, 1] that are not integrable on [0, 1] but
such that fþ g is integrable on [0, 1].
4. (a) Write down functions f and g on [0, 1] such that f is integrable, g is not
integrable, and fg is integrable.
(b) Write down a function f on [0, 1] such that jfj is integrable but f is not
integrable on [0, 1].
5. Prove that, if the functions f and g are integrable on [a, b], then so is the
function max{f, g}.
Hint: Use the formula max a; bf g ¼ 12
aþ bþ a� bj jf g, for a, b2R .
6. Prove that, if f and g are integrable on [a, b], then
Z b
a
fg
� �2
�Z b
a
f 2
� �
Z b
a
g2
� �
:
Hint: Use the Cauchy-Schwarz InequalityP
n
i¼1
aibi
� �2
�P
n
i¼1
a2i
� �
P
n
i¼1
b2i
� �
:
Section 7.3
1. Write down a primitive of each of the following functions:
(a) f xð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � 9p
; x 2 3;1ð Þ;(b) f xð Þ ¼ sin 2xþ 3ð Þ � 4 cos 3x� 2ð Þ; x 2 R ;
(c) f xð Þ ¼ e2x sin 3xð Þ; x 2 R :
2. Using the result of part (c) of Exercise 1, write down a primitive F of the
function f xð Þ ¼ e2x sin 3xð Þ, x2R , for which F(p)¼ 0.
3. Show that the following functions are all primitives of the function
f(x)¼ sech x, x 2 � p2; p
2
� �
:
(a) F1 xð Þ ¼ tan�1 sinh xð Þ; (b) F2 xð Þ ¼ 2 tan�1 exð Þ;(c) F3 xð Þ ¼ sin�1 tanh xð Þ; (d) F4 xð Þ ¼ 2 tan�1 tanh 1
2x
� �� �
:
4. Let In ¼R e
1x loge xð Þndx, for n � 0:
(a) Prove that In ¼ 12
e2 � 12
nIn�1, for n� 1.
(b) Evaluate I0, I1, I2 and I3.
5. Evaluate each of the following integrals, using the suggested substitution
where given:
(a)R p
2
0tan sin xð Þ cos xdx; (b)
R 1
0
tan�1 xð Þ21þx2 dx ðtry u ¼ tan�1 xÞ;
(c)R 1
0x5 þ 2x2� �
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x6 þ 4x3 þ 4p
dx;
(d)R p
2
0dx
2þcos xdx try u¼ tan 1
2x
� �
and use the identity cos x¼ 1� tan2 12xð Þ
1þ tan2 12xð Þ
� �
;
(e)R e
18x7 loge xdx; (f)
R e2
eloge loge xð Þ
xdx;
(g)R p
2
0sin 2xð Þ
1þ3 cos2 xdx; (h)
R 4
1dx
1þxð Þffiffi
xp ðtry u¼
ffiffiffi
xpÞ:
This is known as theCauchy–Schwarz Inequalityfor integrals.
You met this in Sub-section 1.3.3.
7.6 Exercises 311
Section 7.4
1. Prove the following inequalities:
(a)R 1
0x3
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 1þ x99ð Þp
dx � 12; (b)
R 1
0x4ffiffiffiffiffiffiffiffiffiffiffi
1þ3x97p dx � 1
10:
2. Prove that 12�R 1
01þx30
2�x99 dx � 2:
3. Prove thatR 2
0
x2 x�3ð Þ sin 99xð Þ1þx20 dx
�
�
�
�
�
�� 4:
4. Show thatR
dx
x loge xð Þ32
¼ �2 loge xð Þ�12, and deduce that
P
1
n¼2
1
n loge nð Þ32
is
convergent.
5. Show thatR
dx
x loge xð Þ12
¼ 2 loge xð Þ12, and deduce that
P
1
n¼2
1
n loge nð Þ12
is divergent.
6. Prove that limn!1
n 1
nþ1ð Þ2 þ1
nþ2ð Þ2 þ � � � þ1
2nð Þ2� �
¼ 12:
7. Determine the convergence or divergence of the following series:
(a)P
1
n¼2
1
loge nð Þloge n; (b)P
1
n¼2
1
loge nð Þloge loge nð Þ:
8. Prove thatP
n
k¼2
loge kk� 1
2loge nð Þ2, as n!1.
9. Prove that, if f is integrable on [a, b], where a< b, and f(x)> 0, for all x2 [a, b],
thenR b
af > 0:
Hint: Use the following steps:
(a) Assume that the result is false, so that in factR b
af ¼ 0;
(b) Verify that, if c; d½ � a; b½ , thenR d
cf ¼ 0;
(c) Prove that there is a partition P of [a, b] with U( f, P)� b� a, and
deduce that there is some subinterval [a1, b1] of [a, b] with a1< b1 and
supa1,b1½
f � 1;
(d) Using the fact thatR b1
a1f ¼ 0, prove that there is a sequence
an, bn½ f g1n¼ 1 of intervals with anþ1, bnþ1½ an, bn½ and supan,bn½
f � 1n;
(e) Prove that there is some point c in [a, b] such that an! c as n!1, and
that c2 [an, bn], for all n� 1;
(f) Verify that f(c)¼ 0.
Section 7.5
1. Prove that limn!1
3nð Þ!n3n
� �1n¼ 27
e3 :
2. Use Stirling’s Formula to estimate each of the following numbers to two
significant figures:
(a)�
400
200
�
� 1
2
� �400
; (b)400!
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
800pð Þp
100!ð Þ44400:
3. Use Stirling’s Formula to determine the number l such that
8nð Þ!2nð Þ!ð Þ4
� l216n
n32
as n!1:
312 7: Integration
8 Power series
The evaluation of given functions at given points in their domains is of great
importance. If we are dealing with a polynomial function, the calculation of the
function’s values presents no problem: it is simply a matter of arithmetic. For
example, if
f xð Þ ¼ 1þ 1
2x� 1
2x2 � 1
6x3þ 1
4x4;
then
f 1ð Þ ¼ 1þ 1
2� 1
2� 1
6þ 1
4¼ 13
12:
On the other hand, the sine function is rather different: there is no way of
calculating its values precisely merely by the use of arithmetic.
It is important to be able to estimate values of functions that cannot be
evaluated exactly, and to know how close the estimates are to the actual values
of the function.
In this chapter we are primarily concerned with a procedure for calculating
approximate values of functions, like the sine function, whose values cannot be
calculated easily at all points in their domains. We see how, in principle, we
can use a certain sequence of polynomials to calculate the values of the sine
function, for example, to any desired degree of accuracy; and how we can
represent sin x, for any x, as the sum of a series. We shall see, for instance, that
the polynomial p xð Þ ¼ x� 16
x3 approximates f (x)¼ sin x to within 10�5 for all
x in the interval [0, 0.1], and that, in general
sin x ¼ x� x3
3!þ x5
5!� x7
7!þ � � �; for x 2 R :
In Section 8.1 we define the Taylor polynomial Tn(x) and discuss some
particular examples of functions for which the Taylor polynomials appear to
provide useful approximations.
In Section 8.2 we investigate how closely Taylor polynomials approximate a
given function in the neighbourhood of a point a in its domain, and we establish
a criterion for when we can say that
f xð Þ ¼ limn!1
Tn xð Þ ¼X
1
n¼0
an x� að Þn:
In this case, we say that f is the sum function of the power seriesP
1
n¼0
an x� að Þn.
In Sections 8.3 and 8.4 we look at the behaviour of power series in their own
right; that is, we consider functions which are defined by power series. In
particular, in Section 8.3 we see that a power seriesP
1
n¼0
an x� að Þn behaves in
one of three ways: it converges only for x¼ a, or it converges for all x, or it
Section 8.2, Example 1, andTheorem 3.
313
converges in an interval (a�R, aþR), where R is a positive number called the
radius of convergence.
In Section 8.4 we discuss various rules for power series, including the Sum,
Multiple and Product Rules. We also find that it is valid to differentiate or
integrate a given power series term-by-term, and that this does not affect the
radius of convergence.
Finally, in Section 8.5 we look briefly at various methods for estimating the
number p, and prove that p is irrational.
Some of the proofs in this chapter are not particularly illuminating, and you
may wish to leave them till a second reading of the chapter. You will need a
calculator handy while you work through Sections 8.1 and 8.2.
8.1 Taylor polynomials
8.1.1 What are Taylor polynomials?
Let f be a continuous function defined on an open interval I containing the
point a. It follows from the definition of continuity that
limx!a
f xð Þ ¼ f að Þ:
Thus we can write
f xð Þ ’ f að Þ; for x near a:
In geometrical terms, this means that we can approximate the graph y¼ f(x)
near a by the horizontal line y¼ f(a) through the point (a, f(a)). For most
continuous functions, this does not give a very good approximation.
However, if the function is differentiable on I, then we can obtain a better
approximation by using the tangent line through (a, f(a)) instead of the
horizontal line. The tangent to the graph at (a, f(a)) has equation
y� f að Þx� a
¼ f 0 að Þ
or
y ¼ f að Þ þ f 0 að Þ x� að Þ;so, for x near a, we can write
f xð Þ ’ f að Þ þ f 0 að Þ x� að Þ:This approximation is called the tangent approximation to f at a.
Notice that the function f and the approximating polynomial
f að Þ þ f 0 að Þ x� að Þhave the same value at a and the same first derivative at a.
Example 1 Determine the tangent approximation to the function f(x)¼ ex at 0.
Solution Here
f xð Þ ¼ ex; f 0ð Þ ¼ 1;
f 0 xð Þ ¼ ex; f 0 0ð Þ ¼ 1:
That is, I is a neighbourhoodof a.
We can think of the tangent at(a, f(a)) as the line of bestapproximation to the graphnear a.
314 8: Power series
Hence the tangent approximation to f at 0 is
ex’ f 0ð Þ þ f 0 0ð Þ x� 0ð Þ ¼ 1þ x: &
Problem 1 Determine the tangent approximation to each of the
following functions f at the given point a:
(a) f xð Þ ¼ ex; a ¼ 2; (b) f xð Þ ¼ cos x; a ¼ 0:
So far we have seen two approximations to f(x) for x near a:
f xð Þ ’ f að Þ ða constant functionÞ;f xð Þ ’ f að Þ þ f 0 að Þ x� að Þ ða linear functionÞ:
If the function f is twice differentiable on a neighbourhood I of a, then we can
find an even better approximation to f(x) by considering the expression
f xð Þ � f að Þ þ f 0 að Þ x� að Þf gx� að Þ2
:
Let
F xð Þ ¼ f xð Þ � f að Þ þ f 0 að Þ x� að Þf g; for x 2 I;
G xð Þ ¼ x� að Þ2; for x 2 I:
Then F and G are differentiable on I, and
F0 xð Þ ¼ f 0 xð Þ � f 0 að Þ;G0 xð Þ ¼ 2 x� að Þ:
Also, F að Þ ¼ G að Þ ¼ 0:Hence, by l’Hopital’s Rule, it follows that
limx!a
F xð ÞG xð Þ exists and equals lim
x!a
F0 xð ÞG0 xð Þ ;
provided that this latter limit exists.
Now
limx!a
F0 xð ÞG0 xð Þ ¼ lim
x!a
f 0 xð Þ � f 0 að Þ2 x� að Þ ;
and, since the function f is twice differentiable at a, this limit exists and equals12
f 00 að Þ. Hence
limx!a
f xð Þ � f að Þ þ f 0 að Þ x� að Þf gx� að Þ2
exists and equals1
2f 00 að Þ:
We can reformulate this result as follows: for x near a
f xð Þ ’ f að Þ þ f 0 að Þ x� að Þ þ 1
2f 00 að Þ x� að Þ2 ða quadratic functionÞ:
Notice that the function f and the approximating polynomial f að Þþf 0 að Þ x� að Þ þ 1
2f 00 að Þ x� að Þ2 have the same value at a and the same
first and second derivatives at a.
Theorem 2, Sub-section 6.5.2.
8.1 Taylor polynomials 315
This suggests that, if the function is n-times differentiable on I, then we may
be able to find a better approximating polynomial of degree n whose value at a
and whose first n derivatives at a are equal to those of f. This leads to the
following definition.
Definition Let f be n-times differentiable on an open interval containing
the point a. Then the Taylor polynomial of degree n for f at a is the
polynomial
Tn xð Þ ¼ f að Þ þ f 0 að Þ1!
x� að Þ
þ f 00 að Þ2!
x� að Þ2þ � � � þ f ðnÞ að Þn!
x� að Þn:
Notice that
Tn að Þ ¼ f að Þ; T 0n að Þ ¼ f 0 að Þ; . . .; T nð Þn að Þ ¼ f nð Þ að Þ;
that is, f and Tn have the same value at a and have equal derivatives at a of all
orders up to and including n.
Example 2 Determine the Taylor polynomials T1(x), T2(x) and T3(x) for the
function f(x)¼ sin x at each of the following points:
(a) a¼ 0; (b) a ¼ p2�
Solution Here
f xð Þ ¼ sin x; f 0ð Þ ¼ 0; fp2
� �
¼ 1;
f 0 xð Þ ¼ cos x; f 0 0ð Þ ¼ 1; f 0p2
� �
¼ 0;
f 00 xð Þ ¼ �sin x; f 00 0ð Þ ¼ 0; f 00p2
� �
¼ �1;
f 000 xð Þ ¼ �cos x; f 000 0ð Þ ¼ �1; f 000p2
� �
¼ 0:
Hence:
(a) T1 xð Þ ¼ x; T2 xð Þ ¼ x and T3 xð Þ ¼ x� x3
3! ;
(b) T1 xð Þ ¼ 1; T2 xð Þ ¼ 1� 12
x� p2
� �2and T3 xð Þ ¼ 1� 1
2x� p
2
� �2: &
Problem 2 Determine the Taylor polynomials T1(x), T2(x) and T3(x)
for each of the following functions f at the given point a:
(a) f xð Þ ¼ ex; a ¼ 2; (b) f xð Þ ¼ cos x; a ¼ 0:
Problem 3 Determine the Taylor polynomial of degree 4 for each of
the following functions f at the given point a:
(a) f xð Þ ¼ 7� 6xþ 5x2 þ x3; a ¼ 1; (b) f xð Þ ¼ 11�x
; a ¼ 0;
(c) f xð Þ ¼ loge 1þ xð Þ; a ¼ 0; (d) f xð Þ ¼ sin x; a ¼ p4
;
(e) f xð Þ ¼ 1þ 12
x� 12
x2 � 16
x3 þ 14
x4; a ¼ 0:
Problem 4 Determine the percentage error involved in using the
Taylor polynomial of degree 3 for the function f (x)¼ tan x at 0 to eva-
luate tan 0.1. (Use your calculator to evaluate tan 0.1.)
Strictly speaking, we shoulduse more complicatednotation to indicate that theTaylor polynomial Tn(x)depends on n, a and f.
We do not usually multiplyout such brackets, since thatwould make the results lessclear.
316 8: Power series
8.1.2 Approximation by Taylor polynomials
We now look at some specific examples of functions to investigate the asser-
tion that Taylor polynomials provide good approximations for a large class of
functions.
The function f ðxÞ ¼ 1þ 12 x � 1
2 x2 � 16 x3 þ 1
4 x4
It follows from the result of Problem 3(e) above that the Taylor polynomials of
degrees 1, 2, 3 and 4 for f at 0 are:
T1 xð Þ ¼ 1þ 1
2x; T2 xð Þ ¼ 1þ 1
2x� 1
2x2;
T3 xð Þ ¼ 1þ 1
2x� 1
2x2� 1
6x3; and T4 xð Þ ¼ 1þ 1
2x� 1
2x2� 1
6x3þ 1
4x4:
Since f nð Þ 0ð Þ ¼ 0 for n� 5, it follows that for n� 5 the Taylor polynomial of
degree n for f at 0 is just the same as the Taylor polynomial of degree 4 for f at 0.
The graphs of these Taylor polynomials are as follows:
In this case, the polynomials T2(x) and T3(x) provide good approximations to
f(x) near 0, and Tn(x)¼ f(x) for all n� 4.
The function f(x)¼ sin x
By calculating higher derivatives of the function f(x)¼ sin x at 0, we can show
that the following are Taylor polynomials for f at 0
T1 xð Þ ¼ T2 xð Þ ¼ x; T3 xð Þ ¼ T4 xð Þ ¼ x� x3
3!;
T5 xð Þ ¼ T6 xð Þ ¼ x� x3
3!þ x5
5!; T7 xð Þ ¼ T8 xð Þ ¼ x� x3
3!þ x5
5!� x7
7!:
In general, if f is a polynomialof degree N, then
Tn xð Þ ¼ f xð Þ for all n � N:
8.1 Taylor polynomials 317
The following graphs illustrate how the approximation to f(x) given by Tn(x)
gets better as n increases.
For example, the graph of T5 appears to be very close to the graph of sine
over the interval �p2
, p2
� �
, so that sin x and T5(x) do not differ by very much for
values of x in this interval. Thus T5(x) seems to be a good approximation to
sin x in this interval.
It appears that, as the degree of the Taylor polynomial increases, so its
graph becomes a good approximation to that of sine over more and more of
R . For instance, in the above diagrams the shaded area covers the interval of the
x-axis on which the Taylor polynomial Tn(x) agrees with sin x to three decimal
places.
The function f ðxÞ ¼ 11�x
By repeated differentiation it is easy to verify that, for k¼ 1, 2, . . .
f kð Þ xð Þ ¼ k!
1� xð Þkþ1;
thus, in particular, f kð Þ 0ð Þ ¼ k!.Hence the Taylor polynomial of degree n for f at 0 is
Tn xð Þ ¼X
n
k¼0
f kð Þ 0ð Þk!
xk
¼X
n
k¼0
xk ¼ 1þ xþ x2 þ � � � þ xn:
The following diagram shows the graphs of the Taylor polynomials for f at 0 of
degrees 1, 2, 4 and 7.
318 8: Power series
For jxj � 1, the sequence{Tn(x)} does not converge,and so cannot provide anapproximation to f(x).
The graphs show that the nature of the approximation is very different from
the previous examples. For sine, the interval over which the approximation is
good seems to expand indefinitely as the degree of the polynomials increases;
but for f ðxÞ ¼ 11�x
the interval of good approximation always seems to be
contained in the interval (�1, 1).
Notice, however, that for this function f, the Taylor polynomials Tn(x) are
just the nth partial sums of the geometric seriesP
1
k¼0
xk; this series converges
with sum 11�x
for jxj< 1, and diverges for jxj � 1. It follows that, if jxj< 1, then
Tn xð Þ ! f xð Þ as n!1;
so that, if jxj< 1, the polynomials Tn(x) do provide better and better approx-
imations to f(x) as n increases.
Which functions can be approximated by Taylorpolynomials?
In the next section we obtain a criterion for determining those functions f for
which the Taylor polynomials provide useful approximating polynomials, and
8.1 Taylor polynomials 319
the intervals on which the approximation occurs. We also introduce certain
basic power series which correspond to the functions in the following problem.
Problem 5 Determine the Taylor polynomial of degree n for each of
the following functions at 0:
(a) f xð Þ ¼ 11�x
; (b) f xð Þ ¼ loge 1þ xð Þ; (c) f xð Þ ¼ ex;
(d) f xð Þ ¼ sin x; (e) f xð Þ ¼ cos x:
8.2 Taylor’s Theorem
8.2.1 Taylor’s Theorem and approximation
In Section 8.1 we demonstrated how to find the Taylor polynomial Tn(x) of
degree n for a function f at a point a. This polynomial and its first n derivatives
agree with f and its first n derivatives at a, and the polynomial appears to
approximate f at points near a. But how good an approximation is it? What
error is involved if we replace f by a Taylor polynomial at a? The answer to
these questions is given by the following important result.
Theorem 1 Taylor’s Theorem
Let f be (nþ 1)-times differentiable on an open interval containing the
points a and x. Then
f xð Þ ¼ f að Þ þ f 0 að Þ x� að Þ
þ f 00 að Þ2!
x� að Þ2þ � � � þ f nð Þ að Þn!
x� að ÞnþRn xð Þ;
where
Rn xð Þ ¼ f nþ1ð Þ cð Þnþ 1ð Þ! x� að Þnþ1;
and c is some point between a and x.
Remarks
1. When n¼ 0, Taylor’s Theorem reduces to the assertion
f xð Þ ¼ f að Þ þ f 0 cð Þ x� að Þ;which we can rewrite (for x 6¼ a) in the form
f xð Þ � f að Þx� a
¼ f 0 cð Þ; for some c between a and x:
But this is just the Mean Value Theorem! It follows that Taylor’s Theorem
can be considered as a generalisation of the Mean Value Theorem.
2. The result of Theorem 1 can be expressed in the form
f xð Þ ¼ Tn xð Þ þ Rn xð Þ;where Rn(x) is thought of as a ‘remainder term’ or ‘error term’ involved in
approximating f(x) by the estimate Tn(x).
Strictly speaking, we shoulduse more complicatednotation to indicate that theremainder term Rn(x) dependson n, a and f.
320 8: Power series
Proof For simplicity, we assume that x> a; the proof is similar if x< a.
We use the auxiliary function
h tð Þ ¼ f tð Þ � Tn tð Þ � A t � að Þnþ1; t 2 a; x½ �; (1)
where Tn is the Taylor polynomial of degree n for f at a, and A is a constant
chosen so that
h að Þ ¼ h xð Þ: (2)
Now
f að Þ ¼ Tn að Þ; f 0 að Þ ¼ T 0n að Þ; . . .; and f nð Þ að Þ ¼ T nð Þn að Þ;
so that
h að Þ ¼ 0; h0 að Þ ¼ 0; . . .; and h nð Þ að Þ ¼ 0:
The function h is continuous on the closed interval [a, x] and differentiable on
the open interval (a, x); also, h(a)¼ h(x). Hence, by Rolle’s Theorem, there
exists some number c1 between a and x for which
h0 c1ð Þ ¼ 0:
Next, we apply Rolle’s Theorem to the function h0 on the interval [a, c1]. The
function h0 is continuous on [a, c1] and differentiable on (a, c1); also,
h0(a)¼ h0(c1)¼ 0. Hence, by Rolle’s Theorem, there exists some number c2
between a and c1 for which
h00 c2ð Þ ¼ 0:
In turn, we apply Rolle’s Theorem to the functions
h00; h000; . . .; h nð Þ
on the intervals
a; c2½ �; a; c3½ �; . . . ; a; cn½ �;where c2 > c3 > c4 > � � � > cn > a:
At the last stage, we find that there exists some point c between a and cn for
which
h nþ1ð Þ cð Þ ¼ 0: (3)
By differentiating equation (1) (nþ 1) times, we obtain
h nþ1ð Þ tð Þ ¼ f nþ1ð Þ tð Þ � A nþ 1ð Þ!: (4)
From (3) and (4), we deduce that
0 ¼ f nþ1ð Þ cð Þ � A nþ 1ð Þ!;so that
A ¼ f nþ1ð Þ cð Þnþ 1ð Þ! : (5)
Finally, it follows from equations (1) and (2), with h(a)¼ 0, that
f xð Þ ¼ Tn xð Þ þ A x� að Þnþ1:
Hence, by equation (5)
f xð Þ ¼ Tn xð Þ þ f nþ1ð Þ cð Þnþ 1ð Þ! x� að Þnþ1;
as required. &
You may omit this proof at afirst reading.
This choice of A is made sothat we can apply Rolle’sTheorem to h on [a, x].
8.2 Taylor’s Theorem 321
Problem 1 Obtain an expression for R1(x) when Taylor’s Theorem is
applied to the function f xð Þ ¼ 11�x
at a¼ 0. Calculate the value of c
when x ¼ 34.
Problem 2 What can you say about Rn(x) when f is a polynomial of
degree at most n?
Problem 3 By applying Taylor’s Theorem to the function f (x)¼ cos x
at a¼ 0, prove that cos x ¼ 1� 12
x2 þ R3 xð Þ, x 2 R , where
R3 xð Þj j � 124
x4:
In most applications of Taylor’s Theorem, we do not know the value of c
explicitly. However, for many purposes this does not matter, since we can
show that jRn(x)j is small by finding an estimate for f nþ1ð Þ cð Þ�
�
�
� which is valid
for all c between a and x, and then applying the following result.
Corollary 1 Remainder Estimate
Let f be (n+1)-times differentiable on an open interval containing the points
a and x. If
f nþ1ð Þ cð Þ�
�
�
� � M;
for all c between a and x, then
f xð Þ ¼ Tn xð Þ þ Rn xð Þ;
where
Rn xð Þj j � M
nþ 1ð Þ! x� aj jnþ1:
Proof From Taylor’s Theorem, we have
Rn xð Þ ¼ f nþ1ð Þ cð Þnþ 1ð Þ! x� að Þnþ1;
where c is some point between a and x. It follows that
Rn xð Þj j ¼f nþ1ð Þ cð Þ�
�
�
�
nþ 1ð Þ! x� aj jnþ1
� M
nþ 1ð Þ! x� aj jnþ1:&
Example 1 By applying the Remainder Estimate to the function f (x)¼ sin x,
with a¼ 0 and n¼ 3, calculate sin 0.1 to four decimal places.
Solution Here
f xð Þ ¼ sin x; f 0ð Þ ¼ 0;
f 0 xð Þ ¼ cos x; f 0 0ð Þ ¼ 1;
f 00 xð Þ ¼ �sin x; f 00 0ð Þ ¼ 0;
f 000 xð Þ ¼ �cos x; f 000 0ð Þ ¼ �1:
Strictly speaking, we shoulduse more complicatednotation to indicate that theupper bound M depends on n,a and f.
322 8: Power series
Hence the Taylor polynomial of degree 3 for f at 0 is
T3 xð Þ ¼ f 0ð Þ þ f 0 0ð Þxþ f 00 0ð Þ2!
x2 þ f 000 0ð Þ3!
x3
¼ x� 1
6x3:
Also, f 4ð Þ xð Þ ¼ sin x, so that
f 4ð Þ cð Þ�
�
�
� ¼ sin cj j � 1; for c 2 R :
Taking M¼ 1 in the Remainder Estimate, we deduce that
R3 0:1ð Þj j � 1
4!� 0:1ð Þ4
¼ 1
24� 10�4
<1
2� 10�5:
Now
sin 0:1 ’ T3 0:1ð Þ
¼ 0:1� 1
6� 10�3
¼ 0:1� 0:0001666 . . .
¼ 0:0998333 . . .:
It follows from the Remainder Estimate that T3(0.1) gives an estimate for sin 0.1
to four decimal places. Hence sin 0.1¼ 0.0998 (to four decimal places). &
Problem 4 By applying the Remainder Estimate to the function
f xð Þ ¼ loge 1þ xð Þ, with a¼ 0 and n¼ 2, calculate loge 1:02 to four
decimal places.
In many practical situations we do not know how many terms of the power
series are needed in order to calculate the value of a given function to a
prescribed number of decimal places. In such cases, we can use the
Remainder Estimate to determine how many terms are needed.
Example 2 By applying the Remainder Estimate to the function f (x)¼ ex,
with a¼ 0, calculate the value of e to three decimal places.
Solution Since f kð Þ xð Þ ¼ ex; for k ¼ 0; 1; . . .; we have
f kð Þ 0ð Þ ¼ 1; for k ¼ 0; 1; . . .:
It follows that, for each n
Tn xð Þ ¼ 1þ xþ x2
2!þ � � � þ xn
n!:
Also, f nþ1ð Þ xð Þ ¼ ex, for all x, so that, for all c2 (0,1), we have
f nþ1ð Þ cð Þ�
�
�
� � e < 3:
It follows from the Remainder Estimate, with x¼ 1 and M¼ 3, that
Rn 1ð Þj j � 3
nþ 1ð Þ!� 1nþ1:
We proved that e< 3 in Sub-section 2.5.3.
8.2 Taylor’s Theorem 323
To calculate e to three decimal places, we must choose n so that
3
nþ 1ð Þ! < 2� 10�4; or 15;000 < nþ 1ð Þ!:
Since 7!¼ 5,040 and 8!¼ 40,320, we may safely choose n¼ 7.
It follows that
e ’ T7 1ð Þ
¼ 1þ 1
1!þ 1
2!þ 1
3!þ 1
4!þ 1
5!þ 1
6!þ 1
7!
¼ 1þ 1þ 0:5þ 0:16þ 0:1416þ 0:0083þ 0:00138
þ 0:000198412 . . .
¼ 2:7182 . . .:
Hence, e¼ 2.718 (to three decimal places). &
Problem 5 By applying the Remainder Estimate to the function f(x)¼cos x, with a¼ 0, calculate cos 0.2 rounded to four decimal places.
Our next example illustrates how to obtain an approximation valid over an
interval.
Example 3 Calculate the Taylor polynomial T3(x) for f xð Þ ¼ 1xþ2
at 1.
Show that T3(x) approximates f (x) with an error less than 5� 10�3 on the
interval [1, 2].
Solution Here
f xð Þ ¼ 1
xþ 2; f 1ð Þ ¼ 1
3;
f 0 xð Þ ¼ �1
xþ 2ð Þ2; f 0 1ð Þ ¼ � 1
9;
f 00 xð Þ ¼ 2
xþ 2ð Þ3; f 00 1ð Þ ¼ 2
27;
f 000 xð Þ ¼ �6
xþ 2ð Þ4; f 000 1ð Þ ¼ � 2
27:
Hence the Taylor polynomial of degree 3 for f at 1 is
T3 xð Þ ¼ 1
3� 1
9x� 1ð Þ þ 1
27x� 1ð Þ2� 1
81x� 1ð Þ3:
Also, f 4ð Þ xð Þ ¼ 24
xþ2ð Þ5, so that
f 4ð Þ cð Þ�
�
�
� � 24
35; for c 2 1; 2ð Þ:
Taking M ¼ 2435 in the Remainder Estimate, we deduce that
R3 xð Þj j � 24
35� 2� 1ð Þ4
4!
¼ 1
35¼ 0:0041 . . .; for x 2 1; 2½ �:
Since the remainder term is less than 0.005, it follows that T3(x) approximates
f (x) with an error less than 5� 10�3 on [1, 2]. &
324 8: Power series
Problem 6 Calculate the Taylor polynomial T4(x) for f (x)¼ cos x at p.
Show that T4(x) approximates f (x) with an error less than 3� 10�3 on the
interval 34p; 5
4p
� �
.
8.2.2 Taylor’s Theorem and power series
From Taylor’s Theorem, we know that, if a function f can be differentiated
as often as we please on an open interval containing the points a and x, then,
for any n
f xð Þ ¼ Tn xð Þ þ Rn xð Þ
¼X
n
k¼0
f kð Þ að Þk!
x� að Þk þ Rn xð Þ;
where
Rn xð Þ ¼ f nþ1ð Þ cð Þnþ 1ð Þ! x� að Þnþ1;
for some c between a and x. It follows that, if Rn(x)! 0 as n!1, then we can
express f (x) as a power series in (x� a).
Theorem 2 Let f have derivatives of all orders on an open interval con-
taining the points a and x. If Rn(x)! 0 as n!1, then
f xð Þ ¼X
1
n¼0
f nð Þ að Þn!
x� að Þn:
We call f the sum function of the power seriesP
1
n¼0
f nð Þ að Þn! x� að Þn, and we call
this power series the Taylor series for f at a.
Warning A dramatic example of what happens when the remainder term
does not tend to zero is given by the function
f xð Þ ¼ e�1
x2 ; x 6¼ 0;0; x ¼ 0:
For this function, f (0)¼ 0, f 0(0)¼ 0, f 00(0)¼ 0, . . .. Thus the Taylor polyno-
mial Tn(x) is identically zero for each n, although the function f is not identi-
cally zero. In this case, Rn(x)¼ f (x) for all n, and so the Taylor series for f at 0
X
1
n¼0
f nð Þ 0ð Þn!
xn ¼ 0þ 0xþ 0x2 þ � � �;
converges to f (x) only at 0!
We can use Theorem 2 to obtain the following basic power series.
Theorem 3 Basic power series
(a) 11�x¼1þ xþ x2 þ x3 þ � � � ¼
P
1
n¼0
xn; for xj j < 1;
(b) loge 1þ xð Þ ¼ x� x2
2þ x3
3� � � � ¼
P
1
n¼1
�1ð Þnþ1xn
n; for xj j < 1;
–3 –2 –1 1
1
2 3 x
y
y = e –1/x2
0
We indicate a proof of thisassertion in Section 8.6,Exercise 6 on Section 8.2.
8.2 Taylor’s Theorem 325
(c) ex ¼ 1þ xþ x2
2! þ x3
3!þ � � � ¼P
1
n¼0
xn
n!; for x 2 R ;
(d) sin x ¼ x� x3
3! þ x5
5! � � � � ¼P
1
n¼0
�1ð Þn x2nþ1
2nþ1ð Þ!; for x 2 R ;
(e) cos x ¼ 1� x2
2! þ x4
4! � � � � ¼P
1
n¼0
�1ð Þn x2n
2nð Þ!; for x 2 R :
Proof
(a) Here we use the fact that, for x 6¼ 1
1
1� x¼ 1þ xþ x2 þ x3 þ � � � þ xn þ xnþ1
1� x:
But the sequence xnþ1
1�x
n o
is null, if jxj< 1. Thus, by letting n!1, we
obtain
1
1� x¼X
1
n¼0
xn; for xj j < 1:
(b) Similarly
1
1þ t¼1� t þ t2 � � � � þ �1ð Þntnþ �1ð Þnþ1
tnþ1
1þ t:
Integrating both sides from 0 to x, where jxj< 1, we obtain
Z x
0
dt
1þ t¼Z x
0
1� t þ t2 � � � � þ �1ð Þntn þ �1ð Þnþ1t nþ1
1þ t
!
dt;
so that
loge 1þ tð Þ½ �x0¼ t � t2
2þ t3
3� � � � þ �1ð Þn t nþ1
nþ 1
�x
0
þ �1ð Þnþ1
Z x
0
t nþ1
1þ tdt:
Hence
loge 1þ xð Þ¼ x� x2
2þ x3
3� � � � þ �1ð Þn xnþ1
nþ 1
þ �1ð Þnþ1
Z x
0
tnþ1
1þ tdt:
When 0� x< 1, we have, by the Inequality Rule for integrals
�1ð Þnþ1
Z x
0
tnþ1
1þ tdt
�
�
�
�
�
�
�
�
¼Z x
0
tnþ1
1þ tdt �
Z x
0
tnþ1dt
¼ tnþ2
nþ 2
�x
0
¼ xnþ2
nþ 2� 1
nþ 2! 0 as n!1:
When�1< x< 0, we put T¼�t and X¼�x, so that 0< T< 1. Then
This identity is easily provedby multiplying both sides by1� x:
Theorem 2, Sub-section 7.4.1.
326 8: Power series
�1ð Þnþ1
Z x
0
tnþ1
1þ tdt
�
�
�
�
�
�
�
�
¼Z X
0
Tnþ1
1� TdT � 1
1� X
Z X
0
Tnþ1dT
¼ Xnþ2
1� Xð Þ nþ 2ð Þ �1
1� Xð Þ nþ 2ð Þ ! 0 as n!1:
Combining these two results, we deduce that
loge 1þ xð Þ ¼X
1
n¼1
�1ð Þnþ1 xn
n; for�1 < x < 1:
(c) This was one of our definitions of the exponential function.
(d) Let f (x)¼ sin x. Then, by Taylor’s Theorem, we have sin x ¼ Tn xð ÞþRn xð Þ, where
Rn xð Þ ¼ f nþ1ð Þ cð Þnþ 1ð Þ! x nþ1;
for some number c between 0 and x. We saw earlier that
f nþ1ð Þ cð Þ ¼ � sin c or � cos c;
so that, in particular, we can be sure that f nþ1ð Þ cð Þ�
�
�
� � 1. It follows from
the Remainder Estimate, with M¼ 1, that
Rn xð Þj j � 1
nþ 1ð Þ! xj jnþ1! 0 as n!1;
so that, in particular, Rn(x)! 0 as n!1.
Hence, by letting n!1 in the equation sin x ¼ Tn xð Þ þ Rn xð Þ, we obtain
sin x ¼ x� x3
3!þ x5
5!� x7
7!þ � � � ¼
X
1
n¼0
�1ð Þn x2nþ1
2nþ 1ð Þ!; for x 2 R :
(e) The proof is similar to that of part (d), so we omit it. &
Remark
Probably the first definitions of sin x and cos x that you met were expressed in
terms of a right-angled triangle, but those definitions only make sense for
x 2 0; p2
� �
. We have now shown that sin x and cos x can be represented by the
power series in parts (d) and (e) of Theorem 3, and we can use these power
series to define sin x and cos x for all x2R .
Finally, we use Taylor’s Theorem to prove an interesting limit which is a
generalisation of two limits that you met earlier
limn!1
1þ 1
n
� n
¼ e and limn!1
1þ x
n
� �n
¼ ex; for any x 2 R :
Example 4
(a) Calculate T1(x) and R1(x) for the function f xð Þ ¼ loge 1þ xð Þ; x 2 �1; 1ð Þ,at 0.
(b) By replacing x in part (a) by �x, where jxj> j�j, prove that, for any real
numbers � and �
For, 0<X< 1 and
11�T� 1
1�Xfor T2 [0, X].
See Sub-section 3.4.3.
Problem 5(d),Sub-section 8.1.2.
Sub-section 2.5.3.
8.2 Taylor’s Theorem 327
limx!1
1þ �x
� ��x
¼ e��:
(c) Deduce from part (a) that
n loge 1þ 1
n
�
! 1 as n!1:
Solution
(a) For the function f xð Þ ¼ loge 1þ xð Þ, x 2 �1; 1ð Þ, we have
f xð Þ ¼ loge 1þ xð Þ; f 0ð Þ ¼ 0;
f 0 xð Þ ¼ 1
1þ x; f 0 0ð Þ ¼ 1;
f 00 xð Þ ¼ �1
1þ xð Þ2:
Hence
T1 xð Þ ¼ f 0ð Þ þ f 0 0ð Þx ¼ x and R1 xð Þ ¼ f 00 cð Þ2!
x2 ¼ �x2
2 1þ cð Þ2;
for some number c between 0 and x.
(b) Replacing x by �x
in the equation
loge 1þ xð Þ ¼ x� x2
2 1þ cð Þ2; for xj j < 1;
we obtain
loge 1þ �x
� �
¼ �x� �2
2 1þ cð Þ2x2; for xj j > �j j:
Hence
�x loge 1þ �x
� �
¼ �� � �2�
2 1þ cð Þ2x; for xj j > �j j:
Since 1x! 0 as x!1, it follows that
limx!1
�x loge 1þ �x
� �
¼ ��;
so that
limx!1
loge 1þ �x
� ��x
¼ ��:
Since the exponential function is continuous on R , and
exp loge 1þ �x
� ��x�
¼ 1þ �x
� ��x
;
it follows, from the Composition Rule for limits, that
limx!1
1þ �x
� ��x
¼ e��:
(c) For x2 (�1, 1), we have, from part (a), that loge 1þ xð Þ ¼ x� x2
2 1þcð Þ2,for some number c between 0 and x. Putting x ¼ 1
n, we obtain
loge 1þ 1
n
�
¼ 1
n� 1
2n2 1þ cnð Þ2;
for some number cn between 0 and 1n.
328 8: Power series
Hence
n loge 1þ 1
n
�
¼ 1� 1
2n 1þ cnð Þ2
! 1 as n!1;
since
0 � 1
2n 1þ cnð Þ2� 1
2n
and 12n
� �
is a null sequence. &
8.3 Convergence of power series
8.3.1 The radius of convergence
In the previous section you saw that certain standard functions can be
expressed as the sum functions of power series; for example
1
1� x¼X
1
n¼0
xn and loge 1þ xð Þ ¼X
1
n¼1
�1ð Þnþ1xn
n; for xj j < 1:
These power series are the Taylor series for the given functions at 0.
Conversely, power series can be used to define functions. Thus, for instance,we defined the exponential function x 7! ex, x2R , to be the sum function for
the power seriesP
1
n¼0
xn
n!.
But there are many other functions which are defined as the sum functions of
power series (as distinct from a power series obtained as the Taylor series for a
given function). For example, the Bessel function
J0 xð Þ ¼X
1
n¼0
�1ð Þn x2
� �2n
n!ð Þ2; for x 2 R ;
arises in analyses of the vibrations of a circular drum and of the radiation from
certain types of radio antenna.
However, all such uses of power series depend on knowledge of those
numbers x for which a power seriesP
1
n¼0
an x� að Þn converges. In each of the
above examples, the series converges on an interval; indeed, all power series
have this property.
Theorem 1 Radius of Convergence Theorem
For a given power seriesP
1
n¼0
an x� að Þn, precisely one of the following
possibilities occurs:
(a) The series converges only when x¼ a;
(b) The series converges for all x;
Section 3.4.
See Exercise 6 on Section 8.4,in Section 8.6.
We give the proof ofTheorem 1 in Sub-section 8.3.3.
8.3 Convergence of power series 329
(c) There is a number R> 0, called the radius of convergence of the power
series, such that the series converges if jx� aj<R and diverges if
jx� aj>R.
For example:
(a)P
1
n¼0
n!xn converges only when x¼ 0;
(b)P
1
n¼0
xn
n! converges for all x;
(c)P
1
n¼0
xn converges if jxj< 1 and diverges if jxj> 1 – its radius of conver-
gence is 1.
Remarks
1. Sometimes we abuse our notation by writing R¼ 0, if a series converges
only for x¼ a, or R¼1, if a series converges for all x.
2. It is important to remember that Theorem 1, part (c), makes no assertion
about the convergence or divergence of the power series at the end-points
a�R, aþR of the interval (a�R, aþR).
For example, we shall see shortly that the three power series
X
1
n¼1
xn;X
1
n¼1
xn
nand
X
1
n¼1
xn
n2
all have radius of convergence 1. However, the intervals on which these
series converge are, respectively, (�1, 1), [�1, 1) and [�1, 1].
The interval of convergence of the power series is the interval (a�R, aþR),
together with any end-points of the interval at which the series converges.
The following diagram illustrates the various types of interval of conver-
gence ofP
1
n¼0
an x� að Þn:
Theorem 1 is not concerned with the problem of evaluating the radius of con-
vergence of a power series. However, a power series is simply a particular type of
series, and so all the convergence tests for series can be applied to power series.
In practice, we can tackle most commonly arising power series by using the
following version of the Ratio Test.
Theorem 2 Ratio Test for Radius of Convergence
Suppose thatP
1
n¼0
an x� að Þn is a given power series, and that
�
�
�
anþ1
an
�
�
�! L; as n!1:
(a) If L is1, the series converges only for x¼ a.
For all non-zero x, the seriesP
1
n¼0
n!xn diverges, by the Non-
null Test, as�
1n!xn
�
¼n 1
xð Þn
n!
o
is
a basic null sequence.
Example 2.
Sections 3.1–3.3.
We give the proof ofTheorem 2 in Sub-section 8.3.3.
R¼ 0.
330 8: Power series
(b) If L¼ 0, the series converges for all x.
(c) If L> 0, the series has radius of convergence 1L:
Example 1 Determine the radius of convergence of each of the following
power series:
(a)P
1
n¼1
nn xþ1ð Þnn! ; (b)
P
1
n¼1
x�2ð Þnn! :
Solution
(a) Here an ¼ nn
n! for all n, so
�
�
�
anþ1
an
�
�
�¼ nþ 1ð Þnþ1
nn� n!
nþ 1ð Þ! ¼ 1þ 1
n
� n
; for all n:
Thus�
�
�
anþ1
an
�
�
�! e as n!1:
Hence, by the Ratio Test, the radius of convergence is 1e�
(b) Here an ¼ 1n!, for all n, so
�
�
�
anþ1
an
�
�
�¼ n!
nþ 1ð Þ! ¼1
nþ 1; for all n:
Thus�
�
�
anþ1
an
�
�
�! 0 as n!1:
Hence, by the Ratio Test, the power seriesP
1
n¼1
x�2ð Þnn! converges for all x. &
Problem 1 Determine the radius of convergence of each of the follow-
ing power series:
(a)P
1
n¼0
2n þ 4nð Þxn; (b)P
1
n¼1
n!ð Þ22nð Þ! xn;
(c)P
1
n¼1
nþ 2�nð Þ x� 1ð Þn; (d)P
1
n¼1
xn
n!ð Þ1n:
Hint for part (d): Use Stirling’s Formula and the result of Example 1,
Sub-section 7.5.1.
If we wish to find the interval of convergence (as opposed to the radius of
convergence), then we may need to use some of the other tests for series in
order to determine the behaviour at the end-points of the interval.
Example 2 Determine the interval of convergence of each of the following
power series:
(a)P
1
n¼1
xn; (b)P
1
n¼1
xn
n; (c)
P
1
n¼1
xn
n2.
Solution
(a) Here an¼ 1, for all n, so�
�
�
anþ1
an
�
�
�¼ 1; for all n:
R¼1.
R¼ 1L�
8.3 Convergence of power series 331
Hence, by the Ratio Test, the radius of convergence is 1; in other wordsX
1
n¼1
xn converges for�1 < x < 1:
Next, we consider the behaviour of the power series at the end-points of
this interval, namely �1 and 1. Since the sequences 1nf gand �1ð Þnf g are
both non-null, it follows thatP
1
n¼1
xn diverges when x ¼ �1, by the Non-
null Test.
Hence the interval of convergence ofP
1
n¼1
xn is (�1, 1).
(b) Here an ¼ 1n, for all n, so�
�
�
anþ1
an
�
�
�¼ n
nþ 1¼ 1
1þ 1n
; for all n:
Thus�
�
�
anþ1
an
�
�
�! 1 as n!1:
Hence, by the Ratio Test, the radius of convergence is 1; in other wordsX
1
n¼1
xn
nconverges for�1 < x < 1:
But we know thatP
1
n¼1
1n
diverges andP
1
n¼1
�1ð Þnn
converges. It follows that the
interval of convergence of the power seriesP
1
n¼1
xn
nis �1; 1½ Þ.
(c) Here an ¼ 1n2, for all n, so�
�
�
anþ1
an
�
�
�¼ n2
ðnþ 1Þ2¼ 1
ð1þ 1nÞ2; for all n:
Thus�
�
�
anþ1
an
�
�
�! 1 as n!1:
Hence, by the Ratio Test, the radius of convergence is 1; in other words
X
1
n¼1
xn
n2converges for �1 < x < 1:
But we know thatP
1
n¼1
1n2 converges; so, by the Absolute Convergence Test,
the seriesP
1
n¼1
�1ð Þnn2 also converges. It follows that the interval of conver-
gence of the power seriesP
1
n¼1
xn
n2 is �1; 1½ �. &
Now, the Radius of Convergence Theorem tells us that, if a power seriesP
1
n¼0
an x� að Þn has radius of convergence R, then it converges for jx� aj<R. In
fact, we can say more than this – namely, that, for all points x with jx� aj<R,
the power series is actually absolutely convergent!
Theorem 3 Absolute Convergence Theorem
Let the power seriesP
1
n¼0
an x� að Þn have radius of convergence R. Then it is
absolutely convergent for all x with jx� aj<R.
By Example 2 ofSub-section 3.2.1, andat the start of Sub-section 3.3.2, respectively.
Theorem 1, Sub-section 3.3.1.
332 8: Power series
Such a choice is alwayspossible – for example,choose X to be the midpoint ofx and the nearest end-point ofthe interval (a�R, aþR).
ForP
1
n¼0
x� aX� a
�
�
�
�
nis a geometric
series.
A similar result holds forpower series of the formP
1
n¼0
an x� að Þn; a 6¼ 0.
Equivalently
limx!1�
P
1
n¼0
anxn
�
¼P
1
n¼0
an:
You may omit this proof at afirst reading.
Proof Let x be a number such that jx� aj<R, and chose a number X such
that jx� aj< jX� aj<R.
Now, the seriesP
1
n¼0
an X � að Þn is convergent, so that an X � að Þn! 0 as
n!1. In particular, there exists some number N such that an X � að Þnj j < 1,
for all n>N. It follows that
an x� að Þnj j ¼ x� a
X � a
�
�
�
�
�
�
n
� an X � að Þnj j
� x� a
X � a
�
�
�
�
�
�
n
; for n > N:
But the seriesP
1
n¼0
x� aX� a
�
�
�
�
nis convergent, so that the series
P
1
n¼0
an x� að Þnj j is also
convergent, by the Comparison Test for series.
It follows that the power seriesP
1
n¼0
an x� að Þn is absolutely convergent, as
required. &
Problem 2 Determine the interval of convergence of each of the
following power series:
(a)P
1
n¼0
nxn; (b)P
1
n¼1
1n3n xn:
Problem 3 Determine the radius of convergence of the power series
1þ �xþ � �� 1ð Þ2!
x2 þ � � � ¼X
1
n¼0
� �� 1ð Þ . . . �� nþ 1ð Þn!
xn;
where� 6¼ 0; 1; 2; . . .:
8.3.2 Abel’s Limit Theorem
We can sometimes use the ideas of power series to sum interesting and
commonly arising series such as
X
1
n¼1
�1ð Þnþ1
n¼1� 1
2þ 1
3� 1
4þ � � � and
X
1
n¼0
�1ð Þn
2nþ 1¼1� 1
3þ 1
5� 1
7þ � � �:
A major tool in this connection in the following result.
Theorem 4 Abel’s Limit Theorem
Let f (x) be the sum function of the power seriesP
1
n¼0
anxn, which has radius of
convergence 1; and letP
1
n¼0
an be convergent. Then
limx!1�
f xð Þ ¼X
1
n¼0
an:
Proof This proof is a wonderful illustration of the sheer power and magic of
the "! � method for proving results in Analysis! We use the standard
approach for proving that a limit exists as x! 1�.
Let sn ¼ a0 þ a1 þ � � � þ an�1 be the nth partial sum of the seriesP
1
n¼0
an; and
let s denote the sumP
1
n¼0
an.
8.3 Convergence of power series 333
Now,
a0 ¼ s1 and an ¼ snþ1 � sn; for all n � 1: (1)
It follows that, for xj j51
1� xð ÞX
1
n¼0
snþ1xn ¼ 1� xð Þ s1þ s2xþ s3x2þ �� �þ snþ1xnþ �� �� �
¼ s1þ s2xþ s3x2þ s4x3 � � � þ snþ1xnþ� s1x� s2x2� s3x3� � � �� snxn� snþ1xnþ1� �� �¼ s1þ s2� s1ð Þxþ s3� s2ð Þx2þ �� �þ snþ1� snð Þxnþ �� �
¼X
1
n¼0
anxn ¼ f ðxÞ:
We now study closely the identity
1� xð ÞX
1
n¼0
snþ1xn ¼ f ðxÞ: (2)
Next, we want to prove that:
for each positive number ", there is a positive number � such that
f xð Þ � sj j < ", for all x satisfying 1� � < x< 1.
Now, we use equation (2) in the following way to get a convenient expres-
sion for f (x)� s
f xð Þ � s ¼X
1
n¼0
anx n �X
1
n¼0
an
¼ 1� xð ÞX
1
n¼0
snþ1x n � s
¼ 1� xð ÞX
1
n¼0
snþ1x n � 1� xð ÞX
1
n¼0
sx n
¼ 1� xð ÞX
1
n¼0
snþ1 � sð Þx n: (3)
Next, choose a number N such that snþ1 � sj j < 12" for all n>N. We can
then apply the Triangle Inequality to equation (3), for x 2 0; 1ð Þ, to see that
f xð Þ� sj j ¼ 1� xð ÞX
N
n¼0
snþ1� sð ÞxnþX
1
n¼Nþ1
snþ1� sð Þxn
�
�
�
�
�
�
�
�
�
�
� 1� xð ÞX
N
n¼0
snþ1� sð Þxn
�
�
�
�
�
�
�
�
�
�
þ 1� xð ÞX
1
n¼Nþ1
snþ1� sð Þxn
�
�
�
�
�
�
�
�
�
�
� 1� xð ÞX
N
n¼0
snþ1� sj jxnþ 1� xð Þ� 1
2"�
X
1
n¼Nþ1
xn
Using (1).
This is simply the definitionof lim
x!1�f xð Þ ¼ s:
Here we use the definitionof s, equation (2), and the factthat the sum of the seriesP
1
n¼0
x n is 11�x
:
Such a choice of N is possible,since snþ1 ! s as n!1.
We split the infinite sum intotwo parts.
Using the Triangle Inequality.
For snþ1 � sj j < 12", for all
n<N; also, we use that
0< x< 1 so thatP
1
n¼0
xn ¼ 11�x
.
334 8: Power series
� 1� xð ÞX
N
n¼0
snþ1 � sj jxn þ 1
2"
� 1� xð ÞX
N
n¼0
snþ1 � sj j þ 1
2" (4)
Finally, since the linear function x 7! 1� xð ÞP
N
n¼0
snþ1 � sj j is continuous on
R , and takes the value 0 at 1, it follows that we can choose a positive number �(with � 2 (0, 1)) such that
1� xð ÞX
N
n¼0
snþ1 � sj j< 1
2"; for all x satisfying 1� � < x < 1:
If we substitute this upper bound 12" for 1� xð Þ
P
N
n¼0
snþ1 � sj j into the
inequality (4), we find that we have proved that
for each positive number ", there is a positive number � such that
f xð Þ � sj j < "; for all x satisfying 1� � < x < 1:
This is what we set out to prove. &
As an application of Abel’s Limit Theorem, we use it to evaluate
X
1
n¼1
�1ð Þnþ1
n¼ 1� 1
2þ 1
3� 1
4þ � � �:
Let f xð Þ ¼P
1
n¼1
�1ð Þnþ1
nxn. We have seen already that this power series converges
in (�1, 1) to the sum loge 1þ xð Þ. Further, we saw earlier that the seriesP
1
n¼1
�1ð Þnþ1
nis convergent, by the Alternating Test. It follows, from Abel’s
Limit Theorem, that
X
1
n¼1
�1ð Þnþ1
n¼ lim
x!1�
X
1
n¼1
�1ð Þnþ1
nxn ¼ lim
x!1�loge 1þ xð Þ ¼ loge 2:
Remark
It is always necessary to check that the conditions of Abel’s Theorem apply
before using it. For example, a thoughtless application to the identity
1� xþ x2 � x3 þ � � � ¼X
1
n¼0
�1ð Þnxn ¼ 1
1þ x; where xj j < 1;
of taking the limit as x! 1�, would give the following absurd conclusion
1� 1þ 1� 1þ � � � ¼X
1
n¼0
�1ð Þn ¼ 1
2!
Problem 4 Use Abel’s Limit Theorem to evaluateP
1
n¼0
�1ð Þn2nþ1¼ 1� 1
3þ 1
5� 1
7þ � � �.
Hint: Use the facts that tan�1 x ¼ x� x3
3þ x5
5� x7
7þ � � � for jxj< 1, and
that the latter series has radius of convergence 1.
Again we use that 0< x< 1.
We add in the extrarequirement that � < 1 in orderto ensure that we are onlyconsidering values of xin (0, 1).
Part (b) of Theorem 3, Sub-section 8.2.2.
We did not check that the
seriesP
1
n¼0
�1ð Þnxn was
convergent at 1!
We prove these facts in Sub-section 8.4.1.
8.3 Convergence of power series 335
8.3.3 Proofs
We now supply the proofs that we omitted from Sub-section 8.3.1.
Theorem 1 Radius of Convergence Theorem
For a given power seriesP
1
n¼0
an x� að Þn, precisely one of the following
possibilities occurs:
(a) The series converges only when x¼ a;
(b) The series converges for all x;
(c) There is a number R> 0 such that the series converges if jx� aj<R and
diverges if jx� aj>R.
Proof For simplicity, we shall assume that a¼ 0.
Clearly the possibilities (a) and (b) are mutually exclusive; we shall there-
fore assume that for a given power series neither possibility occurs, and then
prove that possibility (c) must occur.
First, let
S ¼ x :X
1
n¼0
anxn converges
( )
:
Since the possibility (b) has been excluded, we deduce that S must be bounded.
For, ifP
1
n¼0
anXn diverges, then, in view of the Absolute Convergence Theorem
for power series,P
1
n¼0
anxn cannot converge for any x with jxj> jXj, since
otherwiseP
1
n¼0
anXn would then have to be convergent – which is not the case.
Also, S is non-empty, since the power seriesP
1
n¼0
anxn converges at 0.
It then follows, from the Least Upper Bound Property of R , that the set S
must have a least upper bound, R say. We now prove that the seriesP
1
n¼0
anxn is
convergent if jxj<R and divergent if jxj>R.
First, notice that R> 0. For, since possibility (a) has been excluded, there must
be at least one non-zero value of x, x1 say, such thatP
1
n¼0
anxn1 is convergent; hence
we have x12 S. It follows, from the Absolute Convergence Theorem, thatP
1
n¼0
anxn must converge for those x with jxj< jx1j, so that � x1j j; x1j jð Þ lies in S.
Now, choose any x for which jxj<R. Then by the definition of R as sup S,
there exists some number x2 with jxj< x2<R such thatP
1
n¼0
anxn2 converges. It
follows, from the Absolute Convergence Theorem, thatP
1
n¼0
anxn converges.
Finally, choose any x for which xj j > R ¼ sup S. It follows that x =2 S, so thatP
1
n¼0
anxn must be divergent. &
You may omit this sub-section at a first reading.
For the general proof, replacex throughout by (x� a).
Theorem 3, Sub-section 8.3.1.
Here we identify a number Rthat will be the desired radiusof convergence.
336 8: Power series
In view of the Absolute Convergence Theorem, we can slightly strengthen
the conclusion of the Radius of Convergence Theorem, as follows.
Corollary If the power seriesP
1
n¼0
an x� að Þn has radius of convergence
R> 0, then it is absolutely convergent if jx� aj<R. If the series converges
for all x, then it is absolutely convergent for all x.
Finally we prove the Ratio Test for Radius of Convergence of power series.
Theorem 2 Ratio Test for Radius of Convergence
Suppose thatP
1
n¼0
an x� að Þn is a given power series, and that
�
�
�
anþ1
an
�
�
�! L as n!1:
(a) If L is1, the series converges only for x = a.
(b) If L¼ 0, the series converges for all x.
(c) If L> 0, the series has radius of convergence 1L:
Proof For simplicity, we shall assume that a¼ 0.
(a) Suppose that janþ1
anj ! 1 as n!1. Then, for any non-zero value of x, the
sequence janþ1xnþ1
anxn j ! 1, so that the sequence {anxn} is unbounded. It
follows, from the Non-null Test, that the seriesP
1
n¼0
anxn must be divergent.
(b) Suppose that janþ1
anj ! 0 as n!1. Then, for any non-zero value of x
anþ1xnþ1
anxn
�
�
�
�
�
�
�
�
¼�
�
�
anþ1
an
�
�
�� xj j
! 0� xj j ¼ 0;
so thatP
1
n¼0
anxn is absolutely convergent, and so is convergent.
(c) Suppose that�
�
�
anþ1
an
�
�
�! L as n!1, where L> 0.
First, suppose that xj j > 1L. Then
anþ1xnþ1
anxn
�
�
�
�
�
�
�
�
¼�
�
�
anþ1
an
�
�
�� xj j
! L� xj j > 1;
so thatP
1
n¼0
anxn is not absolutely convergent. It follows, from the above
Corollary to the Radius of Convergence Theorem, that the radius of
convergence ofP
1
n¼0
anxn must be less than or equal to 1L.
Next, suppose that xj j < 1L. Then
anþ1xnþ1
anxn
�
�
�
�
�
�
�
�
¼�
�
�
anþ1
an
�
�
�� xj j
! L� xj j< 1;
Theorem 3, Sub-section 8.3.1.
This result is sometimes alsocalled the AbsoluteConvergence Theorem.
For the general proof, replacex throughout by (x�a).
8.3 Convergence of power series 337
so thatP
1
n¼0
anxn is absolutely convergent, and so is convergent.
Hence the radius of convergence ofP
1
n¼0
anxn must be at least equal to 1L.
Combining these two facts, it follows that the desired radius of conver-
gence is exactly 1L. &
8.4 Manipulating power series
It would be tedious to have to apply Taylor’s Theorem every time that we wished
to determine the Taylor series of a given function. While sometimes this really
has to be done, in many commonly arising situations we can use standard rules
for power series and the list of basic power series to avoid most of the effort.
We now set out to establish the rules for manipulating power series.
8.4.1 Rules for power series
Many of the rules for manipulating power series are similar to the correspond-
ing rules for manipulating ‘ordinary’ series.
Theorem 1 Combination Rules
Let
f xð Þ ¼X
1
n¼0
an x� að Þn; for x� aj j < R; and
g xð Þ ¼X
1
n¼0
bn x� að Þn; for x� aj j < R0:
Then:
Sum Rule f þ gð Þ xð Þ ¼P
1
n¼0
an þ bnð Þ x� að Þn; for x� aj j < r;
where r ¼ min R;R0f g;
Multiple Rule lf xð Þ ¼P
1
n¼0
lan x� að Þn; for x� aj j < R; where l 2 R :
We do not supply a proof, as Theorem 1 is a simple restatement of the Sum and
Multiple Rules for ‘ordinary’ series.
We can use the Combination Rules to find the Taylor series at 0 for the
function cosh x. We start with the power series for the exponential function
ex ¼ 1þ xþ x2
2!þ x3
3!þ � � � ¼
X
1
n¼0
xn
n!; for x 2 R :
It follows that e�x ¼ 1� xþ x2
2! � x3
3! þ � � � ¼P
1
n¼0
�1ð Þn xn
n!; for x 2 R . We
may then use the Sum Rule to obtain
ex þ e�x ¼ 2� 1þ x2
2!þ x4
4!þ � � �
�
¼ 2�X
1
n¼0
x2n
2nð Þ!; for x 2 R ;
Theorem 3, Sub-section 8.2.2.
Thus, if the power series forboth f and g converge at aparticular point x, then so doesthe series for fþ g.
Theorem 2, Sub-section 3.1.4.
Theorem 3, Sub-section 8.2.2.
For the odd-powered termscancel.
338 8: Power series
so that, by using the Multiple Rule with l ¼ 12, we obtain
cosh x¼ 1
2exþ e�xð Þ ¼ 1þ x2
2!þ x4
4!þ � � � ¼
X
1
n¼0
x2n
2nð Þ!; for x 2 R :
Problem 1 Find the Taylor series at 0 for each of the following func-
tions, and state its radius of convergence:
(a) f xð Þ ¼ sinh x; x 2 R ;
(b) f xð Þ ¼ loge 1� xð Þ þ 2 1� xð Þ�1; xj j < 1:
Problem 2 Find the Taylor series at 0 for each of the following func-
tions, and state its radius of convergence:
(a) f xð Þ ¼ sinh xþ sin x; x 2 R ; (b) f xð Þ ¼ loge1þx1�x
� �
; xj j < 1;(c) f xð Þ ¼ 1
1þ2x2 ; x 2 R :
Remark
In Theorem 1, the radius of convergence of the power series for fþ g may be
larger than r ¼ min R;R0f g. For example, we can use the basic power series
and the Combination Rules to verify that the Taylor series at 0 for the functions
f xð Þ ¼ 11�x
and gðxÞ ¼ �11�xþ 1
1�x2
are
f xð Þ ¼ 1þ xþ x2 þ x3 þ x4þ � � � ¼X
1
n¼0
xn and
g xð Þ ¼ � 1
2x� 3
4x2 � 7
8x3 � 15
16x4� � � � ¼
X
1
n¼0
�1þ 1
2n
�
xn;
each with radius of convergence 1. It follows, from Theorem 1, that the power
series for the function f þ gð Þ xð Þ ¼ 11�x
2is
f þ gð Þ xð Þ ¼ 1þ 1
2xþ 1
4x2 þ 1
8x3 þ 1
16x4 þ � � � ¼
X
1
n¼0
1
2nxn;
and that this power series has radius of convergence at least 1. In fact, it has
radius of convergence 2.
We can find the Taylor series at 0 for the function f xð Þ¼ 1þ x1� x
, for jxj< 1, by
writing 1þx1�x
in the form2� 1�xð Þ
1�x¼ 2
1�x� 1. Since the Taylor series at 0 for the
function 21�x
is 2þ 2xþ 2x2 þ 2x3þ � � � ¼P
1
n¼0
2xn, with radius of conver-
gence 1, it follows that the Taylor series for f at 0 is 1þ 2xþ 2x2þ2x3þ � � � ¼ 1þ
P
1
n¼1
2xn, with radius of convergence 1.
However we could obtain the same result by multiplying together the Taylor
series for the functions x 7! 1þ x and x 7! 11�x
, since
1þ xð Þ 1þ xþ x2þ x3þ� � �� �
¼ 1þ xþ x2þ x3þ� � �þ x 1þ xþ x2þ x3þ� � �� �
¼ 1þ 2xþ 2x2þ 2x3þ� � �:In fact, we can justify such multiplication together of power series to obtain
further power series.
We simply add the powerseries term-by-term.
We now ‘collect together’ themultiples of successivevarious powers of x.
8.4 Manipulating power series 339
Theorem 2 Product Rule
Let
f xð Þ ¼X
1
n¼0
an x� að Þn; for x� aj j < R; and
g xð Þ ¼X
1
n¼0
bn x� að Þn; for x� aj j < R0:
Then
fgð Þ xð Þ ¼X
1
n¼0
cn x� að Þn; for x� aj j < r;where r ¼ min R;R0f g;
and
c0 ¼ a0b0; c1 ¼ a0b1 þ a1b0 and
cn ¼ a0bn þ a1bn�1 þ � � � þ an�1b1 þ anb0:
Theorem 2 is an immediate consequence of the Product Rule for series, applied
to the two seriesP
1
n¼0
an x� að Þn andP
1
n¼0
bn x� að Þn, both of which are absolutely
convergent for jx� aj< r – by the Absolute Convergence Theorem.
Example 1 Determine the Taylor series at 0 for the function f xð Þ ¼ 1þ x
1�xð Þ2.
Solution We use our knowledge of the power series at 0 for the functions
x 7! 11� x
and x 7! 1þ x1� x
, both of which have radius of convergence 1. Thus, by
the Product Rule, we have
1þ x
1� xð Þ2¼ 1
1� x� 1þ x
1� x
¼ 1þ xþ x2 þ x3 þ x4 þ � � �� �
� 1þ 2xþ 2x2 þ 2x3 þ 2x4 þ � � �� �
¼X
1
n¼0
cnxn;
where c0¼ 1� 1¼ 1, c1¼ 1� 2þ 1� 1¼ 3 and cn¼ 1� 2þ 1� 2þ . . .þ1� 2þ 1� 1¼ 2nþ 1.
Thus the required power series for f at 0 is 1þ x
1� xð Þ2 ¼P
1
n¼0
2nþ 1ð Þxn: &
Problem 3 Determine the Taylor series at 0 for:
(a) f xð Þ ¼ 1þ xð Þ loge 1þ xð Þ; for xj j < 1;
(b) f xð Þ ¼ 1þx
1�xð Þ3 ; for xj j < 1:
Next, we have seen already that the hyperbolic functions have the following
Taylor series at 0
sinh x ¼ xþ x3
3!þ x5
5!þ � � � and cosh x ¼ 1þ x2
2!þ x4
4!þ � � � ;
each with radius of convergence1. Notice that the derivative of the function
sinh x is cosh x, and that term-by-term differentiation of the series
xþ x3
3! þ x5
5! þ � � � gives the series 1þ x2
2! þ x4
4! þ � � �. It looks as though we can
As in Theorem 1, the radius ofconvergence of the product
seriesP
1
n¼0
cn x� að Þn may be
greater than r.
Theorem 5, Sub-section 3.3.4.
Theorem 3, Sub-section 8.3.1.
Also, its radius ofconvergence is at least 1.(In fact it is easy to check thatits radius of convergence isexactly 1.)
At the start of this sub-section.
340 8: Power series
obtain the Taylor series for the derivative of a function f simply by differentiat-
ing the Taylor series for f itself.
Our next result states that we can differentiate or integrate the Taylor series
of a function f term-by-term to obtain the Taylor series of the corresponding
function f 0 orR
f , respectively.
Theorem 3 Differentiation and Integration Rules
Let f xð Þ ¼P
1
n¼0
an x� að Þn; for x� aj j < R. Then:
Differentiation Rule f 0 xð Þ ¼P
1
n¼1
nan x� að Þn�1; for x� aj j < R;
Integration RuleR
f xð Þdx ¼P
1
n¼0
anx� að Þnþ1
nþ 1þ constant;
for x� aj j < R:
All three series have the same radius of convergence.
For example, consider the Taylor series at 0 for the function tan�1. We know that
1
1þ x2¼ 1� x2 þ x4 � x6 þ � � �
¼X
1
n¼0
�1ð Þnx2n; with radius of convergence 1;
and thatd
dxtan�1 x ¼ 1
1þ x2; for x 2 R :
It follows, from the Integration Rule, that
tan�1 x¼ x� x3
3þ x5
5� x7
7þ�� �þc; for xj j< 1;and some constant c:
Substituting x¼ 0 into this equation, we find that c¼ tan�1 0¼ 0. It follows that
tan�1 x ¼ x� x3
3þ x5
5� x7
7þ � � �; for xj j < 1:
Problem 4 Find the Taylor series at 0 for the following functions:
(a) f xð Þ¼ 1� xð Þ�2; for xj j< 1; (b) f xð Þ¼ 1� xð Þ�3; for xj j< 1;(c) f xð Þ¼ tanh�1 x; for xj j< 1:
Problem 5 Find the first three non-zero terms in the Taylor series at 0
for the function f (x)¼ ex(1� x)�2, jxj< 1. State its radius of convergence.
Problem 6 Let f be the function f xð Þ ¼ xþ x3
1:3þ x5
1:3:5þ � � � þx2nþ1
1:3:���: 2nþ1ð Þ þ � � �; x 2 R .
Determine the Taylor series at 0 for each of the following functions:
(a) f 0 xð Þ; (b) f 0 xð Þ � xf xð Þ:
Problem 7 Determine the Taylor series at 0 for the function
f xð Þ ¼ e�x2
, x 2 R .
Deduce thatR 1
0e�x2
dx ¼ 1� 13þ 1
10� 1
42þ � � � þ �1ð Þn
2nþ 1ð Þ�n!þ � � �:
To find the constant, put x¼ a.
They may, however, behavedifferently at the end-points oftheir intervals of convergence.
This is a geometric series.
By the Integration Rule, thefinal Taylor series musthave the same radius ofconvergence as the originalTaylor series, namely 1.
8.4 Manipulating power series 341
We have now found a whole variety of techniques for identifying Taylor series:
Taylor’s Theorem;
Combination, Product, Differentiation and Integration Rules.
But how do we know that different techniques will always give us the same
expression as the Taylor series? The following result states that there is only
one Taylor series for a function f at a given point a – any valid method will give
the same coefficients.
Theorem 4 Uniqueness Theorem
IfX
1
n¼0
an x�að Þn¼X
1
n¼0
bn x�að Þn; for x�aj j<R; then an¼ bn:
Proof Let f xð Þ ¼P
1
n¼0
an x� að Þn and g xð Þ ¼P
1
n¼0
bn x� að Þn, for jx� aj<R.
If we differentiate both equations n times, using the Differentiation Rule,
and put x¼ a, we obtain
f nð Þ að Þ ¼ n!� an and g nð Þ að Þ ¼ n!� bn;
Since f(x)¼ g(x), for jx� aj<R, we must have f (n)(a)¼ g(n)(a). It follows that
an¼ bn, for all n� 0. &
8.4.2 General Binomial Theorem
You will have already met the Binomial Theorem, which states that, for each
positive integer n
1þ xð Þn¼X
n
k¼0
n
k
�
xk; wheren
k
�
¼ n n� 1ð Þ . . . n� kþ 1ð Þk!
¼ n!
k! n� kð Þ! :
In the Binomial Theorem, the power n is a positive integer and the result is true
for all x2R . In fact, a similar result holds for more general values of the power
but with a restriction on the values of x for which it is valid.
Theorem 5 General Binomial Theorem
For any �2R
1þ xð Þ�¼X
1
n¼0
�n
�
xn; where�n
�
¼� ��1ð Þ . . . ��nþ1ð Þn!
and xj j< 1:
For example
1þ 2xð Þ�6 ¼X
1
n¼0
�6
n
�
2xð Þn; where
�6
n
�
¼ �6ð Þ �7ð Þ . . . �6� nþ 1ð Þn!
and xj j < 1
2;
See, for example, Appendix 1.
By convention,�0
�
¼ 1.
342 8: Power series
so that
1þ 2xð Þ�6 ¼ 1� 12xþ 84x2 � � � �; for xj j < 1
2:
Another important type of application occurs when the power � is not an
integer. For example
1þ xð Þ�13¼X
1
n¼0
� 13
n
�
xn; where� 1
3
n
�
¼� 1
3
� �
� 43
� �
� 73
� �
. . . � 13� nþ 1
� �
n!;
so that
1þ xð Þ�13¼ 1� 1
3xþ 2
9x2 � 14
81x3þ � � �; for xj j < 1:
Problem 8 Use the General Binomial Theorem to determine the first
four non-zero terms in the Taylor series at 0 for the function f xð Þ ¼1þ 6xð Þ
14, xj j< 1
6. State the radius of convergence of the Taylor series.
Problem 9
(a) Determine the Taylor series at 0 for the function f xð Þ ¼ 1� xð Þ�12,
jxj< 1.
(b) Hence determine the Taylor series at 0 for the function
f xð Þ ¼ sin�1 x, jxj< 1.
In our proof of the General Binomial Theorem, we use the following lemma.
Lemma Forany�2R andanyn2N , n�n
�
þ nþ1ð Þ �nþ1
�
¼� �n
�
.
Proof Since�k
�
¼ � ��1ð Þ: : : : : ��kþ1ð Þk! , for any k2N , we may rewrite the
left-hand side of the desired identity in the form
n�
n
�
þ nþ1ð Þ�
nþ1
�
¼ n�
n
�
þ nþ1ð Þ� ��1ð Þ : : : ��nð Þnþ1ð Þ!
¼ n�
n
�
þ� ��1ð Þ . . . ��nð Þn!
¼ n�
n
�
þ ��nð Þ� ��1ð Þ . . . ��nþ1ð Þn!
¼ n�
n
�
þ ��nð Þ�
n
�
¼��
n
�
: &
Theorem 5 General Binomial Theorem
For any �2R
1þxð Þ�¼X
1
n¼0
�n
�
xn; where�n
�
¼� ��1ð Þ . . . ��nþ1ð Þn!
and xj j< 1:
You may omit the remainderof this sub-section at a firstreading.
Here we cancel the term(nþ 1).
We now bring the term(�� n) in front of the fraction.
The last fraction is just�n
�
.
8.4 Manipulating power series 343
Proof Let
f xð Þ ¼X
1
n¼0
�n
�
xn and g xð Þ ¼ f xð Þ 1þ xð Þ��; for xj j < 1:
We want to prove that g(x)¼ 1, for all x with jxj< 1.
Using the rules for differentiation, we may differentiate the expression for
g to obtain
g0 xð Þ ¼ f 0 xð Þ 1þ xð Þ����f xð Þ 1þ xð Þ���1
¼ 1þ xð Þf 0 xð Þ � �f xð Þ½ � 1þ xð Þ���1:
Now
1þ xð Þf 0 xð Þ ¼ 1þ xð ÞX
1
n¼1
n�
n
�
xn�1
¼X
1
n¼1
n�
n
�
xn�1 þX
1
n¼1
n�
n
�
xn
¼X
1
n¼0
nþ 1ð Þ�
nþ 1
�
xn þX
1
n¼0
n�
n
�
xn
¼X
1
n¼0
nþ 1ð Þ�
nþ 1
�
þ n�
n
� �
xn
¼ �X
1
n¼0
�
n
�
xn
¼ �f xð Þ:It follows from the earlier expression for g0(x) that
g0 xð Þ ¼ 1þ xð Þf 0 xð Þ � �f xð Þ½ � 1þ xð Þ���1
¼ 0� 1þ xð Þ���1¼ 0:
So, g( x) is a constant. Hence
g xð Þ ¼ g 0ð Þ¼ f 0ð Þ ¼ 1;
as required. &
8.4.3 Proofs
Differentiation Rule The series
f xð Þ ¼X
1
n¼0
an x� að Þn andX
1
n¼1
nan x� að Þn�1
have the same radius of convergence, R say. Also, f is differentiable on
(a�R, aþR), and
f 0 xð Þ ¼X
1
n¼1
nan x� að Þn�1:
Here we differentiate thepower series term-by-term.Then we split the initialbracket.
We now replace n by nþ 1 inthe first sum, and check whatthe limits of summation are inboth sums.
Now we combine the twosums into one.
We can apply the result of theLemma to this square bracket.
By the definition of f(x).
You may omit this Sub-sectionat a first reading.
This is one part of Theorem 3,Sub-section 8.4.1.
344 8: Power series
Proof For simplicity, we assume that a¼ 0.
Let the seriesP
1
n¼0
anxn andP
1
n¼1
nanxn�1 have radii of convergence R and R0,
respectively.
We start by proving that R¼R0.
We first show that, if jxj<R, then the power seriesP
1
n¼1
nanxn�1 is conver-
gent. (By the Corollary to the Radius of Convergence Theorem, this shows
that R0 �R.)
To prove this, choose a real number c with jxj< c<R. Then the seriesP
1
n¼0
ancn is convergent, and so the sequence {ancn} is a null sequence. Thus
there is a number K such that
ancnj j � K; for n ¼ 0; 1; 2; . . .: (1)
Then
nanxn�1�
�
�
�¼ ancnj j � n
c� x
c
�
�
�
�
�
�
n�1
� K
c� n� x
c
�
�
�
�
�
�
n�1
; (2)
by (1). Since xc
�
�
�
� < 1, the seriesP
1
n¼1
n xc
�
�
�
�
n�1converges. Hence, by the Comparison
Test for series, the seriesP
1
n¼1
nanxn�1 is absolutely convergent, and so is
convergent. This proves that R0 �R.
Next, suppose that R0>R. Let c be any number such that R< c<R0. ThenP
1
n¼1
nancn�1 is absolutely convergent, by the Absolute Convergence Theorem.
But
ancnj j ¼ nancn�1�
�
�
�� c
n
� c� nancn�1�
�
�
�:
Hence, by the Comparison Test for series, the seriesP
1
n¼1
ancn is absolutely
convergent, and so is convergent. This contradicts the definition of R, and thus
shows that R0 6> R.
It follows that R¼R0.Next, we show that f is differentiable on (�R, R), and that f 0 is of the stated
form.
So, choose any point c2 (�R, R). Then choose a positive number r<R such
that c 2 (�r, r). Then, for all non-zero h such that jhj< r� jcj
f cþ hð Þ � f cð Þh
�X
1
n¼1
nancn�1 ¼ 1
h
X
1
n¼1
an cþ hð Þn� cn�hncn�1� �
: (3)
Now, we apply Taylor’s Theorem to the function x 7! xn on the interval with
end-points c and cþ h. We obtain
cþ hð Þn¼ cn þ nhcn�1 þ 1
2n n� 1ð Þh2cn�2
n ;
Sub-section 8.3.3.
For example, c ¼ 12
xj j þ Rð Þ.
Theorem 1, Sub-section 3.2.1.
For example, c ¼ 12
Rþ R0ð Þ.
For example, r ¼ 12
cj j þ Rð Þ.
8.4 Manipulating power series 345
where cn is some number between c and cþ h (and, in particular, with
cnj j � r).
Now
cþ hð Þn�cn � nhcn�1�
�
�
� � 1
2n n� 1ð Þh2r n�2: (4)
It follows from (3) and (4) and the Triangle Inequality that
f cþ hð Þ � f cð Þh
�X
1
n¼1
nancn�1
�
�
�
�
�
�
�
�
�
�
� 1
2hj jX
1
n¼2
n n� 1ð Þ anj jrn�2: (5)
Since the seriesP
1
n¼1
nanxn�1 has radius of convergence R, it follows that so
does the seriesP
1
n¼2
n n� 1ð Þanxn�2; we conclude that the seriesP
1
n¼2
n n� 1ð Þ
anrn�2 is (absolutely) convergent.
Hence, it follows from (5) and the Limit Inequality Rule that
limh!0
f cþ hð Þ � f cð Þh
¼X
1
n¼1
nancn�1: &
Integration Rule The series
f xð Þ ¼X
1
n¼0
an x� að Þn and F xð Þ ¼X
1
n¼0
an
nþ 1x� að Þnþ1
have the same radius of convergence, R say. Also, f is integrable on
(a�R, aþ R), andZ
f xð Þdx ¼ F xð Þ:
Proof The two power series have the same radius of convergence, by the
Differentiation Rule for power series, applied to F.
It also follows, from the Differentiation Rule for power series, that F is
differentiable and F0 ¼ f, so that F is a primitive of f. &
8.5 Numerical estimates for p
One of the most interesting problems in the history of mathematics throughout
the last two thousand years has been the determination to any pre-assigned
degree of accuracy of naturally arising irrational numbers such asffiffiffi
2p
, e and p.
Here we look at various way of estimating p, and prove that p is irrational.
8.5.1 Calculating p
It is easy to check that p is slightly larger than 3 by wrapping a string round a
circle of radius r and measuring the length of the string corresponding to one
circumference; for this length 2pr is just slightly larger than 6r.
For we have shown that term-by-term differentiation doesnot alter the radius ofconvergence of a powerseries.
This is the other part ofTheorem 3, Sub-section 8.4.1.
That is, F is a primitive of f on(a�R, aþR)
346 8: Power series
Over hundreds of years mathematicians devised more sophisticated methods
for estimating the value of p. Some of their results are as follows:
the Babylonians c. 2000 BC p ¼ 3 18¼ 3:125
the Egyptians c. 2000 BC p ¼ 3 1381’ 3:160
the Old Testament c. 550 BC p ¼ 3
Archimedes c. 300–200 BC between 3 17
and 3 1071
, p ’ 3:141
the Chinese c. 400–500 AD p ’ 3:1415926
the Hindus c. 500–600 AD p ’ffiffiffiffiffi
10p
’ 3:16
With the development of the Calculus in the seventeenth century, new for-
mulas for estimating p were discovered, including:
Wallis’s Formula p2¼ 2
1� 2
3� 4
3� 4
5� 6
5� 6
7� � � � � 2n
2n�1� 2n
2nþ1� � � �
Leibniz’s Series tan�1 x ¼ x� x3
3þ x5
5� x7
7þ � � �
Gregory’s Series p4¼ 1� 1
3þ 1
5� 1
7þ � � �
However, neither Wallis’s Formula nor Gregory’s Series is very useful for
calculating p beyond a few decimal places, as they converge far too slowly.
However, we can use the Taylor series for tan�1 effectively for calculating pby choosing values of x closer to 0; in fact, the smaller the value of x, the faster
the series converges, and so the fewer the number of terms we need to calculate
p to a given degree of accuracy.
For example, if we choose x ¼ 1ffiffi
3p in Leibniz’s Series, we obtain
p6¼ tan�1 1
ffiffiffi
3p�
¼ 1ffiffiffi
3p � 1
3
1ffiffiffi
3p� 3
þ 1
5
1ffiffiffi
3p� 5
� 1
7
1ffiffiffi
3p� 7
þ � � �
¼ 1ffiffiffi
3p 1� 1
9þ 1
45� 1
189þ � � �
�
:
This formula can be used to calculate p to several decimal places without much
effort; we gain about one extra place for every two extra terms.
To obtain series that are more effective for calculating p, we can use the
Addition Formula for tan�1
tan�1 xþ tan�1 y ¼ tan�1 xþ y
1� xy
�
;
provided that tan�1 xþ tan�1 y lies in the interval � p2; p
2
� �
.
Problem 1 Use the Addition Formula for tan�1 to prove that
tan�1 1
3
�
þ tan�1 1
4
�
þ tan�1 2
9
�
¼ p4:
Similar applications of the Addition Formula give further expressions for p4,
such as
6 tan�1 1
8
�
þ 2 tan�1 1
57
�
þ tan�1 1
239
�
¼ p4
;
4 tan�1 1
5
�
� tan�1 1
239
�
¼ p4:
1 Kings vii, 23;2 Chronicles iv, 2.
Theorem 5, Sub-section 7.4.2.
Just before Problem 4, Sub-section 8.4.1.
Problem 4, Sub-section 8.3.2.
This is known as Sharp’sFormula.
Problem 3(a) on Section 4.3,in Section 4.5.
This is known as Machin’sFormula.
8.5 Numerical estimates for p 347
Such formulas have been used to calculate p to great accuracy: 1 million decimal
places were achieved in 1974. Recently, improved methods of numerical ana-
lysis have been used to calculate p correct to several million decimal places.
In 1770, Lambert showed that p is irrational; that is, p is not the solution of a
linear equation a0þ a1x¼ 0, where the coefficients a0 and a1 are integers.
Then, in 1882, Lindemann showed that p is transcendental; that is, p is not a
solution of any polynomial equation a0þ a1xþ a2x2þ � � � þ anxn¼ 0, where
all the coefficients are integers.
We end with a couple of mnemonics that can be used to recall the first few
digits for p; the word lengths give the successive digits:
May I have a large container of coffee?3: 1 4 1 5 9 2 6
and
How I need a drink; alcoholic of course;3: 1 4 1 5 9 2 6
after all those formulas involving tangent functions!
5 3 5 8 9 7 9
8.5.2 Proof that p is irrational
We now prove that p is irrational. Our proof is rather intricate; and, surprisingly,
it uses many of the properties of integrals that you met in Chapter 7! It depends
on the properties of a suitably chosen integral that we examine in Lemmas 1–3.
Lemma 1 Let In ¼R 1
�11� x2ð Þncos 1
2px
� �
dx, n¼ 0, 1, 2, . . .. Then:
(a) In > 0; (b) In � 2.
Proof
(a) The function f xð Þ ¼ 1� x2ð Þncos 12px
� �
, for x2 [�1, 1], is non-negative,
for n¼ 0, 1, 2, . . .. It follows, from the Inequality Rule for integrals, that
In� 0, for each n.
To prove that In> 0 we need to examine the integral in more detail.
By the Inequality Rule for integrals, we haveZ 1
�1
f xð Þdx ¼Z �1
2
�1
þZ 1
2
�12
þZ 1
12
!
f xð Þdx
�Z 1
2
�12
f xð Þdx:
Now, if x 2 �12; 1
2
� �
, then 1� x2ð Þncos 12px
� �
� 34
� �ncos 1
4p
� �
, so that
Z 1
�1
f xð Þdx �Z 1
2
�12
3
4
� n
cos1
4p
�
dx
¼ 3
4
� n
cos1
4p
�
> 0;
as required.
For interest only, we list thefirst 1000 decimal places ofp in Appendix 3.
This solved the ancient Greekproblem of finding a ruler andcompass method to constructa square with an area equal tothat of a given circle.
You may omit this at a firstreading.
ForR�1
2
�1f xð Þdx � 0 and
R 112
f xð Þdx � 0:
By the Inequality Rule forintegrals.
For the length of the intervalof integration is 1.
348 8: Power series
(b) For x 2 �1; 1½ �, 1� x2ð Þncos 12px
� �
� 1� 1 ¼ 1, so thatZ 1
�1
f xð Þdx �Z 1
�1
1dx ¼ 2: &
Problem 2 Prove that pI0 ¼ 4 and p3I1 ¼ 32.
Next, we obtain a reduction formula for In, using integration by parts.
Lemma 2 The integral In satisfies the following reduction formula
p2In ¼ 8n 2n� 1ð ÞIn�1 � 16n n� 1ð ÞIn�2:
Proof Using integration by parts twice, we obtain
In ¼ 1� x2� �n2
psin
1
2px
� �1
�1
� 2
p
Z 1
�1
n �2xð Þ 1� x2� �n�1
sin1
2px
�
dx
¼ 4n
p
Z 1
�1
x 1� x2� �n�1
sin1
2px
�
dx
¼ 4n
px 1� x2� �n�1 � 2
p
�
cos1
2px
� �1
�1
þ 8n
p2
Z 1
�1
1� x2� �n�1� 2x2 n� 1ð Þ 1� x2
� �n�2n o
cos1
2px
�
dx
¼ 8n
p2In�1 �
16
p2n n� 1ð Þ
Z 1
�1
x2 1� x2� �n�2
cos1
2px
�
dx
¼ 8n
p2In�1 �
16
p2n n� 1ð Þ
Z 1
�1
1� x2� �n�2� 1� x2
� �n�1n o
cos1
2px
�
dx
¼ 8n
p2In�1 �
16
p2n n� 1ð Þ In�2 � In�1f g:
Multiplying both sides by p2, we obtain
p2In ¼ 8nIn�1 � 16n n� 1ð ÞIn�2 þ 16n n� 1ð ÞIn�1
¼ 8n 2n� 1ð ÞIn�1 � 16n n� 1ð ÞIn�2: &
We now use the result of Lemma 2 to prove the crucial tool in our proof that
p is irrational.
Lemma 3 For n¼ 0, 1, 2, . . ., there exist integers a0, a1, a2, . . ., an such
that
p2nþ1
n!In ¼ a0 þ a1pþ a2p2þ � � � þanpn ¼
X
n
k¼0
akpk
!
:
Proof For simplicity, let Jn ¼ p2nþ1
n! In. It follows from Problem 2 that
J0 ¼ pI0 ¼ 4 and J1 ¼ p3I1 ¼ 32: (1)
By the Inequality Rule forintegrals.
The value of the first term iszero.
The value of the first term iszero.
For
x2 1� x2� �n�2
¼ 1� 1� x2� �� �
1� x2� �n�2
¼ 1� x2� �n�2� 1� x2
� �n�1:
8.5 Numerical estimates for p 349
Next, we rewrite the reduction formula in Lemma 2 in terms of Jn as
Jn ¼p2nþ1
n!In
¼ 8n 2n� 1ð Þn!
p2n�1In�1 �16n n� 1ð Þ
n!p2n�1In�2
¼ 8 2n� 1ð ÞJn�1 � 16p2Jn�2: (2)
The desired result now follows by Mathematical Induction. We know that
it holds for n¼ 0 and for n¼ 1, by the statements (1) above. Using the
reduction formula (2) we can prove that the statement of the Lemma holds
for all n� 2. &
Finally, we can use the fact that, for any integer p, the sequence p2nþ1
n!
n o
is
null, to prove that p is irrational.
Theorem 1 p is irrational.
Proof Suppose, on the contrary, that p is rational; so that p ¼ pq, for p, q2N .
Then, the conclusion of Lemma 3 can be written in the form
p2nþ1
q2nþ1n!In ¼
X
n
k¼0
ak
pk
qk;
where the coefficients ak are integers.
If we now multiply both sides by the expression q2nþ1, we find that
p2nþ1
n!In ¼
X
n
k¼0
akpkq2nþ1�k;
so that
p2nþ1
n!In is an integer: (3)
However, we know that the sequence p2nþ1
n!
n o
is a null sequence; and we know,
from Lemma 1, that jInj� 2. It follows, from the Squeeze Rule for sequences, that
p2nþ1
n!In ! 0 as n!1: (4)
It follows from (4) that eventually p2nþ1
n! In < 1; and so, from (3), that in fact In
must equal 0. This contradicts the result of Lemma 1, part (a). This contra-
diction proves that, in fact, p must be rational. &
8.6 Exercises
Section 8.1
1. Determine the tangent approximation to the function f(x)¼ 2� 3xþ x2þ ex
at the given point a:
(a) a¼ 0; (b) a¼ 1.
We omit ‘the gory details’!
The sequencep2ð Þnn!
�
is a
basic null sequence.
This is a proof bycontradiction.
350 8: Power series
2. Determine the Taylor polynomial of degree 3 for each of the following
functions f at the given point a:
(a) f(x)¼ loge(1þ x), a¼ 2; (b) f xð Þ ¼ sin x, a ¼ p6;
(c) f xð Þ ¼ 1þ xð Þ�2, a ¼ 1
2; (d) f xð Þ ¼ tan x, a ¼ p
4:
3. Determine the Taylor polynomial of degree 4 for each of the following
functions f at the given point a:
(a) f(x)¼ cosh x, a¼ 0; (b) f(x)¼ x5, a¼ 1.
4. Determine the percentage error involved in using the Taylor polynomial of
degree 3 for the function f(x)¼ ex at 0 to evaluate e0.1.
Section 8.2
1. Obtain an expression for the remainder term R1(x) when Taylor’s Theorem is
applied to the function f(x)¼ ex at 0. Show that, when x¼ 1 and n¼ 1, then
the value of c in the statement of Taylor’s Theorem is approximately 0.36.
2. By applying Taylor’s Theorem to the function f(x)¼ sin x with a ¼ p4, prove
that
sin x ¼ 1ffiffiffi
2p 1þ x� p
4
� �
� 1
2x� p
4
� �2�
þ R2 xð Þ; x 2 R ;
where R2 xð Þj j � 16
x� p4
�
�
�
�
3:
3. By applying the Remainder Estimate to the function f(x)¼ sinh x with
a¼ 0, calculate sinh 0.2 to four decimal places.
4. Calculate the Taylor polynomial T3(x) for the function f xð Þ ¼ xxþ3
at 2.
Show that T3(x) approximates f(x) to within 10�4 on the interval 2; 52
� �
:
5. (a) Determine the Taylor polynomial of degree n for the function
f(x)¼ loge x at 1.
(b) Write down the remainder term Rn(x) in Taylor’s Theorem for this
function, and show that Rn(x)! 0 as n!1 if x2 (1, 2).
(c) By using Theorem 2 in Section 8.2, determine a Taylor series for f at 1
which is valid when x2 (1, 2).
6. Let f xð Þ ¼ e�1
x2 ; x 6¼ 0;0; x ¼ 0:
(a) Prove that f 0 0ð Þ ¼ 0:(b) Prove that, for x 6¼ 0, f 0(x) is of the form
(a polynomial of degree at most 3 in 1x)�e�
1
x2 .
(c) Prove that, for x 6¼ 0, f (n)(x) is of the form
(a polynomial of degree at most 3n in 1x)�e�
1
x2 .
(d) Prove that, for any positive integer n, f (n)(0)¼ 0.
Section 8.3
1. Determine the radius of convergence of each of the following power series:
(a)P
1
n¼1
3nð Þ!n!ð Þ2 xþ 1ð Þn; (b)
P
1
n¼1
nn
n! x� 1e
� �n;
For this function, the Taylorpolynomial Tn(x) isidentically zero for each n,although the function f is notidentically zero. In this case,Rn(x)¼ f(x), for all n; andso the Taylor series for f
at 0,P
1
n¼0
f nð Þ 0ð Þn! xn, converges to
f(x) only at 0.
8.6 Exercises 351
(c)P
1
n¼1
nþ 1ð Þ�nxn; (d) 1
2xþ 1:3
2:5 x2 þ 1:3:52:5:8 x3 þ � � �;
(e) 1þ a:b1:c xþ a aþ1ð Þ:b bþ1ð Þ
1:2:c cþ1ð Þ x2 þ � � �, where a, b, c > 0:
2. Determine the Taylor series for the function f(x)¼ loge(2þ x) at the given
point a; in each case, indicate the general term and state the radius of
convergence of the series:
(a) a¼ 1; (b) a¼�1.
Hint: Use the Taylor series at 0 for the function x 7! loge 1þ xð Þ, with
t¼ x� 1 and t¼ xþ 1, respectively.
3. Determine the interval of convergence of each of the following power
series:
(a)P
1
n¼1
2n
n! xn; (b)P
1
n¼1
�1ð Þn 2n
nx� 1ð Þn:
4. Give an example (if one exists) of each of the following:
(a) a power seriesP
1
n¼0
anxn which diverges at both x¼ 1 and x¼�2;
(b) a power seriesP
1
n¼0
anxn which diverges at x¼ 1 but converges at x¼�2;
(c) a power seriesP
1
n¼0
anxn which converges at x¼ 1 but diverges at x¼�2.
5. Give an example (if one exists) of each of the following:
(a) a power seriesP
1
n¼0
anxn which converges only if �3 < x < 3;
(b) a power seriesP
1
n¼0
anxn which converges only if �3 � x < 3;
(c) a power seriesP
1
n¼0
anxn which converges only if �3 < x � 3;
(d) a power seriesP
1
n¼0
anxn which converges only if �3 � x � 3.
Section 8.4
1. Determine the Taylor series for each of the following functions at 0; in each
case, indicate the general term and state the radius of convergence of the
series:
(a) f xð Þ ¼ loge 1þ xþ x2ð Þ ¼ loge1�x3
1�x
� �
; (b) f xð Þ ¼ cosh x1�x
:
2. Determine the first three non-zero terms in the Taylor series for each of the
following functions at 0; and state the radius of convergence:
(a) f(x)¼ cos(ex� 1); (b) f(x)¼ loge(1þ sin x); (c) f(x)¼ ex sin x.
3. For the function f xð Þ ¼ xþ 23
x3 þ 2:43:5 x5 þ 2:4:6
3:5:7 x7 þ � � �, for jxj< 1, prove
that
1� x2� �
f 0 xð Þ � xf xð Þ ¼ 1:
4. By using the Integration Rule for power series, prove that
sinh�1 x ¼ x� 1
2
x3
3þ 1:3
2:4
x5
5� 1:3:5
2:4:6
x7
7þ � � �; for xj j < 1:
This series is often called thehypergeometric series.
In fact, f xð Þ ¼ sin�1 xffiffiffiffiffiffiffiffi
1�x2p .
352 8: Power series
5. By using the Integration Rule for power series, find an infinite series with
sumR 1
0sin t2ð Þdt.
6. The Bessel function J0 is defined by the power series
J0 xð Þ¼ 1� x2
22: 1!ð Þ2þ x4
24: 2!ð Þ2� x6
26: 3!ð Þ2þ�� �þ �1ð Þnx2n
22n: n!ð Þ2þ�� � ; for x2R :
(a) Determine power series for J00 xð Þ and J000 xð Þ, indicating the general term
in each.
(b) Prove that xJ000 xð Þ þ J00 xð Þ þ xJ0 xð Þ ¼ 0.
J0 is not the Taylor series at 0of any familiar function.
8.6 Exercises 353
Appendix 1: Sets, functions and proofs
Sets
A set is a collection of objects, called elements. We use the symbol 2 to mean ‘is a
member of’ or ‘belongs to’; thus ‘x2A’ means ‘the element x is a member of the set A’.
Similarly, we use the symbol 62 to mean ‘is not a member of’ or ‘does not belong to’;
thus ‘x 62A’ means ‘the element x is not a member of the set A’.
We often use curly brackets (or braces) to list the elements of a set in some way. Thus
{a, b, c} denotes the set whose three elements are a, b and c. Similarly, {x : x2¼ 2}
denotes the set whose elements x are such that x2¼ 2; we would read this symbol in
words as ‘the set of x such that x2¼ 2’. When we define a set in this latter way, the
symbol x is a dummy variable; that is, if we replace that symbol x by any other symbol,
such as y, the set is exactly the same set; thus, for instance, the sets {x : x2¼ 2} and
{y : y2¼ 2} are identical.
Two sets are equal if they contain the same elements. We say that a set A is a subset
of a set B, if all the elements of A are elements of B, and we denote this by writing
‘A�B’. We may wish to indicate specifically the possibility that the subset A is equal to
B by writing ‘A�B’, or that A is a proper subset of B (that is, A is a subset of B but
A 6¼B), by writing ‘A�6¼
B’.
Notice that, to show that two sets A and B are equal, it is necessary to prove that A is
a subset of B and that B is a subset of A. In symbols
A ¼ B is equivalent to the two properties both holding: A � B and B � A:
The empty set is the set that contains no elements; it is denoted by the symbol ‘Ø’.
It may seem strange to define such a thing; but, in practice, it is often a convenient set to use.
The union of two sets A and B consists of the set of elements that belong to at least
one of A and B; in symbols
A [ B ¼ x : x 2 A or x 2 Bf g:
The intersection of two sets A and B consists of the set of elements that belong to
both A and B; in symbols
A \ B ¼ x : x 2 A and x 2 Bf g:
We denote the set of elements that belong to A but not to B by A�B; that is
A� B ¼ x : x 2 A; x =2Bf g:There are some special symbols used that are used to denote commonly arising sets of
real numbers:
N , the set of all natural numbers; thus N ¼ {1, 2, 3, . . .};
Z, the set of all integers; thus Z ¼ {0, �1, �2, �3, . . .};
Q , the set of all rational numbers; that is, numbers of the form pq, where p, q2Z
but q 6¼ 0;
R , the set of all real numbers;
Rþ, the set of all positive real numbers.
On the real line R , an interval I is a set of real numbers such that, if a and b both lie in I,
then all numbers between a and b also lie in I.
Thus the colon ‘:’ means‘such that’.
Here we also allow x to belongto both A and B.
354
There are nine types of intervals in R :
a; b½ � ¼ x : x 2 R ; a � x � bf g;a; bð Þ ¼ x : x 2 R ; a < x < bf g;a; b½ Þ ¼ x : x 2 R ; a � x < bf g;a; bð � ¼ x : x 2 R ; a < x � bf g;a;1½ Þ ¼ x : x 2 R ; x � af g;a;1ð Þ ¼ x : x 2 R ; x > af g;�1; bð � ¼ x : x 2 R ; x � bf g;�1; bð Þ ¼ x : x 2 R ; x < bf g;
R .
An interval is said to be closed if it contains both of its end-points, open if it contains
neither of its end-points, and half-closed or half-open if it is neither closed nor open.
Functions
A function is a mapping of some element of R onto another element of R . Thus a
function f is defined by specifying:
a set A, called the domain of f;
a set B, called the codomain of f;
a rule x 7! f xð Þ that associates with each element x of A a unique element f(x) of B.
Symbolically, we write this in the following way:
f : A! B
x 7! f xð Þ:
The element f(x) is called the image of x under f, and the set f (A)¼ {y : y¼ f (x) for
some x2A} the image of A under f.
A particularly simple function is the identity function that maps each element of a
set to itself. Thus, the identity function on a set A, iA, is the function
iA : A! B
x 7! x:
Sometimes every point of the codomain is in the image. We say that a function
f : A!B is an onto function if f (A)¼B.
Sometimes every point of the image is the image of precisely one point of the
domain. We say that a function f : A!B is a one–one function if:
whenever f(x)¼ f(y) for elements x, y of A, then necessarily x¼ y.
Notice that this does NOT mean that f is a one–one mapping of A onto B.
A given function f : A!B may be onto, one–one, both onto and one–one, or neither
onto nor one–one. A function f : A!B is called a bijection if it is both onto and
one–one.
If a function f : A!B is one–one, then it has an inverse function f�1 : f (B)!A with
the defining rule
f�1 yð Þ ¼ x; where y ¼ f xð Þ:
Notice that a function f only has an inverse function f�1 if it is bijective.
Sometimes we want to ‘change the domain’ of a function to a larger set or a
smaller set.
The first four types ofintervals are boundedintervals.
The final five types ofintervals are unboundedintervals.
Sets, functions and proofs 355
If f : A!B, and C is a subset of A, we say that the function g : C!B is the restriction
of f to C if
g xð Þ ¼ f xð Þ; for all x 2 C:
Similarly, if f : A!B, and A is a subset of C, we say that the function g : C!B is the
extension of f to C if
g xð Þ ¼ f xð Þ; for all x 2 A:
Finally, the composite function or composite g f of two functions
f : A! B and g : C ! D;
where f (A)�C, is the function
g f : A! D
x 7! g f xð Þð Þ:
Proofs
Logical implications are so important in Mathematics that we use some special sym-
bols, namely),( and,, to denote implications.
Thus, if we are discussing two statements P and Q, we write
P)Q to mean: ‘if P is true, then Q is true’ or ‘P implies Q’;
P(Q to mean: ‘P is implied by Q’, or ‘if Q is true, then P is true’ or ‘Q implies P’;
P,Q to mean: ‘P is true if and only if Q is true’;
this is equivalent to the two separate statements
P) Q and Q) P:
In this case, P and Q are equivalent statement.
The implications P)Q and Q)P are called converse implications.
In order to prove that an assertion ‘P)Q’ is true, we need to verify that, in all
situations where the statement P holds, then the statement Q also holds. To prove that an
assertion ‘P)Q’ is not true, all that we need do is to find one example of a situation
where the statement P holds but the statement Q does not. Such a situation is called a
counter-example to the assertion.
Sometimes we prove assertions P by simply checking all possible cases; generally,
though, we devise logical arguments that deal with all cases at the same time.
Sometimes our logical arguments appear slightly convoluted to non-mathematicians;
two examples of these are:
Proof by contradiction: In order to show that a statement P holds, we start by
assuming that P is false; we then look at the consequences of that assumption, and
identify some specific consequence that is untrue or contradictory. It then follows that
P must hold, after all.
Proof by contraposition: In order to show that an implication P)Q is true, it is
sufficient to prove that
if Q does not hold, then P does not hold.
Sometimes mathematical proofs are long and complicated, and involve the use of a
variety of approaches to prove the desired result.
Notice that not all approaches to proving a result are necessarily valid! Some
examples of such approaches are:
Proof by picture: Here you make a false claim in your argument because it appears
to be true in the particular diagram that you have drawn.
Proof by example: For instance, you prove that some property holds for n¼ 1 and
n¼ 2, and then assert that it therefore holds for all positive integers.
Sometimes we use the termextension to denote a functiong : C!D where A�C,B�D, and g(x)¼ f(x), for allx2A.
Thus, to prove that P,Q, weneed to prove that P)Q andQ)P.
This is sometimes calleddisproof by counter-example.
This approach is sometimescalled proof by exhaustion.
This fact is sometimes(humorously) called proof byperspiration.
356 Appendix 1
Proof by omission: For instance, you prove one or two special cases of a result, and
claim that the general proof is ‘similar’.
Proof by superiority: Here you claim that the result is ‘obvious’, rather than sit down
to construct a logical proof of it.
Proof by mumbo-jumbo: Here you write down a jumble of formulas and relevant
words, ending up with the claim that you have proved the result.
Principle of Mathematical Induction
This is a standard method of proving statements involving an integer n, generally a
positive integer.
Principle of Mathematical Induction
Let P(n) denote a statement involving a positive integer n. If the following two
conditions are satisfied:
1. the statement P(1) is true,
2. whenever the statement P(k) is true for a positive integer k, then the statement
P(kþ 1) is also true,
then the statement P(n) is true for all positive integers n.
Sometimes we need to use an equivalent version of the Principle.
(Second) Principle of Mathematical Induction
Let P(n) denote a statement involving a positive integer n. If the following two
conditions are satisfied:
1. the statement P(1) is true,
2. whenever the statements P(1), P(2), . . ., P(k) are true for a positive integer k, then
the statement P(kþ 1) is also true,
then the statement P(n) is true for all positive integers n.
We have stated the Principle when P(n) applies to all integers n� 1; analogous
results hold when P(n) applies to all integers n�N, whatever integer N may be.
We end with some useful results that can be proved using the Principle of
Mathematical Induction.
Sums
X
n
k¼1
1 ¼ n;X
n
k¼1
k ¼ n nþ 1ð Þ2
;
X
n
k¼1
k2 ¼ n nþ 1ð Þ 2nþ 1ð Þ6
;X
n
k¼1
k3 ¼ n nþ 1ð Þ2
� �2
;
X
n
k¼0
2k þ 1ð Þ ¼ nþ 1ð Þ2;
sin Aþ sin Aþ Bð Þ þ � � � þ sin Aþ n� 1ð ÞBð Þ
¼sin 1
2nB
� �
sin 12
B� � sin Aþ n� 1
2B
� �
; B 6¼ 0:
Arithmetic progression
aþ aþ dð Þ þ aþ 2dð Þ þ � � � þ aþ n� 1ð Þdð Þ ¼ n aþ n�12
d� �
:
Thus, in order to use thePrinciple, we have a two stagestrategy:
1. check P(1),
2. check that P(k))P(kþ 1).
Sets, functions and proofs 357
Geometric progression
aþ ar þ ar2 þ � � � þ arn�1 ¼ a 1�rn
1�r; r 6¼ 1:
Binomial Theorem
1þ xð Þn ¼X
n
k¼0
n
k
� �
xk ¼ 1þ nxþ n n� 1ð Þ2!
x2 þ � � � þ xn;
aþ bð Þn ¼X
n
k¼0
n
k
� �
an�kbk
¼ an þ nan�1bþ n n� 1ð Þ2!
an�2b2 þ � � � þ bn:
It follows that the sum of the
geometric seriesP
1
k¼0
ark, for
jrj< 1, is a1�r
.
Here,n
k
� �
¼ n!k! n�kð Þ! and
0!¼ 1.
358 Appendix 1
Appendix 2: Standard derivativesand primitives
f(x) f 0(x) Domain
k 0 R
x 1 R
xn, n2Z� {0} nxn�1R
x�, �2R �x��1 (0,1)
ax, a> 0 ax loge a R
xx xx (1þ loge x) (0,1)
sin x cos x R
cos x �sin x R
tan x sec2 x R � nþ 12
� �
p :n 2 Z� �
cosec x �cosec x cot x R � {np : n2Z}
sec x sec x tan x R � nþ 12
� �
p :n 2 Z� �
cot x �cosec2 x R � {np : n2Z}
sin�1 x 1ffiffiffiffiffiffiffiffi
1�x2p (�1, 1)
cos�1 x �1ffiffiffiffiffiffiffiffi
1�x2p (�1, 1)
tan�1 x 11þx2 R
ex exR
loge x 1x
(0,1)
sinh x cosh x R
cosh x sinh x R
tanh x sech2 x R
sinh�1 x 1ffiffiffiffiffiffiffiffi
1þx2p R
cosh�1 x 1ffiffiffiffiffiffiffiffi
x2�1p (1,1)
tanh�1 x 11�x2 (�1, 1)
359
f (x) Primitive F(x) Domain
xn, n2Z� {�1} xnþ1
nþ1R
x�, �2R � {�1} x�þ1
�þ1(0,1)
ax, a> 0 ax
logeaR
sin x �cos x R
cos x sin x R
tan x loge (sec x) � 12p; 1
2p
� �
ex exR
1x
loge x (0,1)1x
loge jxj (�1, 0)
loge x x loge x� x (0,1)
sinh x cosh x R
cosh x sinh x R
tanh x loge (cosh x) R
1a2�x2, a> 0 1
2aloge
aþxa�x
� �
(�a, a)1
a2þx2, a> 0 1a
tan�1 xa
� �
R
1ffiffiffiffiffiffiffiffiffi
a2�x2p , a> 0
sin�1 xa
� �
�cos�1 xa
� �
(
(�a, a)
(�a, a)
1ffiffiffiffiffiffiffiffiffi
x2�a2p , a> 0 loge xþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � a2p� �
cosh�1 xa
� �
(
(a,1)
(a,1)
1ffiffiffiffiffiffiffiffiffi
a2þx2p , a> 0 loge xþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ x2p
� �
sinh�1 xa
� �
R
Rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 � x2p
, a> 0 12
xffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 � x2p
þ 12
a2 sin�1 xa
� �
(�a, a)ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � a2p
, a> 0 12
xffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � a2p
� 12
a2 loge xþffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � a2p� �
(a,1)ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ x2p
, a> 0 12
xffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ x2p
þ 12
a2 loge xþffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ x2p
� �
R
eax sin bx, a,b 6¼ 0 eax
a2þb2 a sin bx� b cos bxð Þ R
eax cos bx, a,b 6¼ 0 eax
a2þb2 a cos bxþ b sin bxð Þ R
360 Appendix 2
Appendix 3: The first 1000 decimal places offfiffiffi
2p
, e and p
More than a million decimal places of these numbers, and various similar commonly
arising numbers, are easily available on the internet.
ffiffiffi
2p¼ 1.
4142135623 7309504880 1688724209 6980785696 7187537694 8073176679
7379907324 7846210703 8850387534 3276415727 3501384623 0912297024
9248360558 5073721264 4121497099 9358314132 2266592750 5592755799
9505011527 8206057147 0109559971 6059702745 3459686201 4728517418
6408891986 0955232923 0484308714 3214508397 6260362799 5251407989
6872533965 4633180882 9640620615 2583523950 5474575028 7759961729
8355752203 3753185701 1354374603 4084988471 6038689997 0699004815
0305440277 9031645424 7823068492 9369186215 8057846311 1596668713
0130156185 6898723723 5288509264 8612494977 1542183342 0428568606
0146824720 7714358548 7415565706 9677653720 2264854470 1585880162
0758474922 6572260020 8558446652 1458398893 9443709265 9180031138
8246468157 0826301005 9485870400 3186480342 1948972782 9064104507
2636881313 7398552561 1732204024 5091227700 2269411275 7362728049
5738108967 5040183698 6836845072 5799364729 0607629969 4138047565
4823728997 1803268024 7442062926 9124859052 1810044598 4215059112
0249441341 7285314781 0580360337 1077309182 8693147101 7111168391
6581726889 4197587165 8215212822 9518488472 . . .
e¼ 2.
7182818284 5904523536 0287471352 6624977572 4709369995 9574966967
6277240766 3035354759 4571382178 5251664274 2746639193 2003059921
8174135966 2904357290 0334295260 5956307381 3232862794 3490763233
8298807531 9525101901 1573834187 9307021540 8914993488 4167509244
7614606680 8226480016 8477411853 7423454424 3710753907 7744992069
5517027618 3860626133 1384583000 7520449338 2656029760 6737113200
7093287091 2744374704 7230696977 2093101416 9283681902 5515108657
4637721112 5238978442 5056953696 7707854499 6996794686 4454905987
9316368892 3009879312 7736178215 4249992295 7635148220 8269895193
6680331825 2886939849 6465105820 9392398294 8879332036 2509443117
3012381970 6841614039 7019837679 3206832823 7646480429 5311802328
7825098194 5581530175 6717361332 0698112509 9618188159 3041690351
5988885193 4580727386 6738589422 8792284998 9208680582 5749279610
4841984443 6346324496 8487560233 6248270419 7862320900 2160990235
3043699418 4914631409 3431738143 6405462531 5209618369 0888707016
7683964243 7814059271 4563549061 3031072085 1038375051 0115747704
1718986106 8739696552 1267154688 9570350354 . . .
Numerical Analysts enjoyadding more digits to suchdecimals, using ever-moresophisticated computingtechniques!
For most practical purposesffiffiffi
2p’ 1:414:
For most practical purposese ’ 2:718:
A useful fact sometimes isthat e3’ 20.
361
p¼ 3.
1415926535 8979323846 2643383279 5028841971 6939937510 5820974944
5923078164 0628620899 8628034825 3421170679 8214808651 3282306647
0938446095 5058223172 5359408128 4811174502 8410270193 8521105559
6446229489 5493038196 4428810975 6659334461 2847564823 3786783165
2712019091 4564856692 3460348610 4543266482 1339360726 0249141273
7245870066 0631558817 4881520920 9628292540 9171536436 7892590360
0113305305 4882046652 1384146951 9415116094 3305727036 5759591953
0921861173 8193261179 3105118548 0744623799 6274956735 1885752724
8912279381 8301194912 9833673362 4406566430 8602139494 6395224737
1907021798 6094370277 0539217176 2931767523 8467481846 7669405132
0005681271 4526356082 7785771342 7577896091 7363717872 1468440901
2249534301 4654958537 1050792279 6892589235 4201995611 2129021960
8640344181 5981362977 4771309960 5187072113 4999999837 2978049951
0597317328 1609631859 5024459455 3469083026 4252230825 3344685035
2619311881 7101000313 7838752886 5875332083 8142061717 7669147303
5982534904 2875546873 1159562863 8823537875 9375195778 1857780532
1712268066 1300192787 6611195909 2164201989 . . .
For most practical purposesp ’ 3:1416:
A useful fact sometimes isthat p2’ 10.
362 Appendix 3
Appendix 4: Solutions to the problems
Chapter 1
Section 1.1
1. �1 < � 1720< � 45
53< 0 < 45
53< 17
20< 1:
2. Let a and b be two distinct rational numbers, where a< b. Let c ¼ 12
aþ bð Þ; then
c is rational, and
c� a ¼ 1
2b� að Þ > 0; and
b� c ¼ 1
2b� að Þ > 0;
so that a< c< b.
3. 17¼ 0:142857142857 . . .; 2
13¼ 0:153846153846 . . ..
4. (a) Let x ¼ 0:231.
Multiplying both sides by 103, we obtain
1000x ¼ 231:231 ¼ 231þ x:
Hence
999x ¼ 231) x ¼ 231
999¼ 77
333:
(b) Let x ¼ 0:81.
Multiplying both sides by 102, we obtain
100x ¼ 81:81 ¼ 81þ x:
Hence
99x ¼ 81 ) x ¼ 81
99¼ 9
11:
Thus
2:281 ¼ 2þ 2
10þ 9
110¼ 251
110:
5. 1720¼ 0:85 and 45
53¼ 0:84 . . ., so that 45
53< 17
20.
6. Suppose that there exists a rational number x such that x3¼ 2. Then we can write
x ¼ pq. By cancelling, if necessary, we may assume that p and q have no common
factor. The equation x3¼ 2 now becomes
p3 ¼ 2q3:
Now, the cube of an odd number is odd, because
2k þ 1ð Þ3 ¼ 8k3 þ 12k2 þ 6k þ 1
¼ 2 4k3 þ 6k2 þ 3k� �
þ 1;
and so p must be even. Hence we can write p¼ 2r, say. Our equation now becomes
363
2rð Þ3¼ 2q3;
so that
q3 ¼ 4r3:
Hence q is also even, so that p and q do have a common factor 2, which contradicts
our earlier statement that p and q have no common factors.
Thus, no such number x exists.
7. We may take, for example, x¼ 0.34 and y¼ 0.34001000100001. . ..
8. Let a and b have decimal representations
a ¼ a0 � a1a2a3 . . . and b ¼ b0 � b1b2b3 . . .;
where we arrange that a does not end in recurring 9s, whereas b does not terminate
(this latter can be arranged by replacing a terminating representation by an equiva-
lent representation that ends in recurring 9s).
Since a< b, there must be some integer n such that
a0 ¼ b0; a1 ¼ b1; . . . ; an�1 ¼ bn�1; but an < bn:
Then x¼ a0 � a1a2a3 . . . an� 1bn is rational, and a< x< b as required.
Next, let c ¼ 12
xþ bð Þ, so that x< c< b. By repeating the same procedure, this
time to the interval between the numbers c and b, we can find a rational number y
with c< y< b.
We have thus constructed two distinct rational numbers between a and b, as
required.
Section 1.2
1. Rule 1 For any a, b2R , a� b, b� a� 0.
Rule 2 For any a, b, c2R , a� b, aþ c� bþ c.
Rule 3
� For any a, b2R and any c> 0, a� b, ac� bc;
� For any a, b2R and any c< 0, a� b, ac� bc.
Remark
Note that the following results are NOT true in general:
� For any a, b2R and any c� 0, a� b, ac� bc;
� For any a, b2R and any c� 0, a� b, ac� bc.
For, if c¼ 0, then we can make no assertion as to whether a� b or a� b from
the information that ac� bc or ac� bc. To see this, take in turn a¼ 2, b¼ 3 and
a¼ 2, b¼ 1.
Rule 4 (Reciprocal Rule) For any positive a, b2R , a � b, 1a� 1
b.
Rule 5 (Power Rule) For any non-negative a, b2R , and any p> 0, a� b, ap� bp.
2. (a) xþ 3> 5.
(b) 2� x< 0.
(c) 5xþ 2> 12.
(d) �1ð5xþ 2Þ >
�112
.
3. (a) Rearranging the inequality, we obtain
4x� x2 � 7
x2 � 1� 3, 4x� x2 � 7
x2 � 1� 3 � 0 , 4x� 4x2 � 4
x2 � 1� 0
, x2 � xþ 1
x2 � 1� 0 ,
x� 12
� �2þ 34
x2 � 1� 0:
Here a0, b0 are non-negativeintegers, and a1, b1, a2, b2, . . .are digits.
Note that x does not end inrecurring 9s, from the way inwhich it was constructed.
364 Appendix 4
Since x� 12
� �2þ 34> 0, for all x, the inequality holds if and only if x2� 1< 0.
Hence the solution set is
x :4x� x2 � 7
x2 � 1� 3
� �
¼ �1; 1ð Þ:
(b) Rearranging the inequality, we obtain
2x2 � xþ 1ð Þ2 , 2x2 � x2 þ 2xþ 1
, x2 � 2x� 1 � 0
, x� 1ð Þ2� 2:
Hence the solution set is
x : 2x2 � xþ 1ð Þ2n o
¼ x : x� 1 � �ffiffiffi
2pn o
[ x : x� 1 �ffiffiffi
2pn o
¼ �1; 1�ffiffiffi
2p� i
[ 1þffiffiffi
2p
;1h �
:
4. We can obtain an equivalent inequality by squaring, provided that both sides are non-
negative.
Now,ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2x2 � 2p
is defined when 2x2� 2� 0; that is, for x2� 1. So,ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2x2 � 2p
is
defined and non-negative if x lies in (�1, �1][ [1,1).
Hence, for x� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2x2 � 2p
> x, 2x2 � 2 > x2
, x2 > 2:
So the part of the solution set in [1,1) isffiffiffi
2p
; 1� �
.
Suppose, next, that x��1. Thenffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2x2 � 2p
� 0 > x;
so that the whole of (�1, �1] lies in the solution set.
Hence the complete solution set is
x :ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2x2 � 2p
> xn o
¼ �1;�1ð � [ffiffiffi
2p
;1� �
:
5. (a) 2x2 � 13
< 5, �5 < 2x2 � 13 < 5 ðby Rule 6Þ, 8 < 2x2 < 18
, 4 < x2 < 9
, 2 < xj j < 3:
Hence the solution set is
x : 2x2 � 13
< 5 �
¼ �3;�2ð Þ [ 2; 3ð Þ:
(b) x� 1j j � 2 xþ 1j j , x� 1ð Þ2� 4 xþ 1ð Þ2
, x2 � 2xþ 1 � 4x2 þ 8xþ 4
, 0 � 3x2 þ 10xþ 3
, 0 � 3xþ 1ð Þ xþ 3ð Þ:Hence the solution set is
x : x� 1j j � 2 xþ 1j jf g ¼ �1;�3ð � [ � 1
3;1
�
:
Solutions to the problems 365
Section 1.3
1. (a) Suppose that aj j � 12.
The Triangle Inequality gives
aþ 1j j � aj j þ 1
� 1
2þ 1 ðsince aj j � 1
2Þ
¼ 3
2:
Hence
aþ 1j j � 3
2:
(b) Suppose that bj j < 12.
The ‘reverse form’ of the Triangle Inequality gives
b3 � 1
� jjb3j � 1j¼ bj j3�1
� 1� bj j3:Now
bj j < 1
2) bj j3< 1
8) 1� bj j3 > 7
8;
so, from the previous chain of inequalities
b3 � 1
>7
8:
2. Rearranging the inequality, we obtain
3n
n2 þ 2< 1, 3n < n2 þ 2
, 0 < n2 � 3nþ 2
, 0 < n� 1ð Þ n� 2ð Þ;and this final inequality certainly holds for n> 2.
3. The result is true for all those a and b for which aþ b� 0.
We now consider those a and b for which aþ b> 0. Rearranging the given
inequality, we obtain
aþ bffiffiffi
2p �
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ b2p
, aþ bð Þ2
2� a2 þ b2
, a2 þ 2abþ b2 � 2a2 þ 2b2
, 0 � a2 � 2abþ b2
, 0 � a� bð Þ2;
and this final inequality certainly holds.
This completes the proof.
4. (a) Using the rules for rearranging inequalities, we obtain
1
2aþ 2
a
�
< a, 1
2� 2
a< a� 1
2a
, 1
a<
1
2a
, 2 < a2 ðsince a > 0Þ:
366 Appendix 4
Since the final inequality is true, the first inequality must also be true. Hence
1
2a þ 2
a
�
< a:
(b) In Example 3 and the subsequent remark, we saw that, if a 6¼ b, then
ab <a þ b
2
� 2
: ( )
Now let b ¼ 2a . Then a 6¼ b, since a > 2
a (as a2 > 2); so, by ( ), it follows that
a � 2
a <
1
2a þ 2
a
� � 2
;
which gives the required inequality.
Alternatively , use a direct argument after squaring the expression 12
a þ 2a
� �
:
5. Since c , d � 0, we can choose non-negative numbers a and b with c ¼ a2 and d ¼ b2 .
Substituting into the resultffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ b2p
� a þ b, for a, b � 0, of Example 4, we obtainffiffiffiffiffiffiffiffiffiffiffi
cþ dp
�ffiffiffi
cpþ
ffiffiffi
dp
:
6. In Problem 5 we saw thatffiffiffiffiffiffiffiffiffiffiffi
cþ dp
�ffiffiffi
cpþ
ffiffiffi
dp
; for numbers c; d � 0: ( )
Following the good advice in the margin note before the Problem, we use this!
For a, b, c� 0, we deduce from () that
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
aþ bþ cp
¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
aþ bþ cð Þp
�ffiffiffi
apþ
ffiffiffiffiffiffiffiffiffiffiffi
bþ cp
ðby an application of ðÞÞ�
ffiffiffi
apþ
ffiffiffi
bpþ
ffiffiffi
cp
ðby a further application of ðÞÞ:
7. If we substitute x ¼ 1n
in the Binomial Theorem for (1þ x)n, when n� 2 we get
1þ 1
n
� n
¼ 1þ n1
n
�
þ n n� 1ð Þ2!
1
n
� 2
þ � � � þ 1
n
� n
� 1þ 1þ n� 1
2n
¼ 2þ 1
2� 1
2n¼ 5
2� 1
2n:
When n¼ 1, the inequality also holds. This completes the proof.
8. Let P(n) be the statement
PðnÞ : 4n > n4:
STEP 1 First we show that P(5) is true: 45� 54.
Since 45¼ 1024 and 54¼ 625, P(5) is certainly true.
STEP 2 We now assume that P(k) holds for some k� 5, and deduce that P(kþ 1) is
then true.
So, we are assuming that 4k> k4. Multiplying this inequality by 4 we get
4kþ1 > 4k4;
so it is therefore sufficient for our purposes to prove that 4k4� (kþ 1)4.
Now
4k4 � k þ 1ð Þ4, 4 � 1þ 1
k
� 4
: ( )
As k increases, the expression 1þ 1k
decreases. Since k� 5, 1þ 1k�
1þ 15¼ 6
5; it follows that
This assumption is just P(k).
Since P(kþ 1) is:4kþ 1> (kþ 1)4.
Solutions to the problems 367
1þ 1
k
� 4
� 6
5
� 4
¼ 1296
625¼ 2:0736 < 4:
Thus the inequality () certainly holds for k� 5, and so it follows that
4kþ 1� (kþ 1)4 also holds for k� 5.
In other words: P(k) true for some k� 5)P(kþ 1) true.
It follows, by the Principle of Mathematical Induction, that 4n> n4,
for n� 5.
9. Substituting x ¼ �1ð2nÞ into Bernouilli’s Inequality (1þ x)n� 1þ nx (which we may
do since �1ð2nÞ � �1 for all natural numbers n), we obtain
1� 1
2n
� n
� 1� n� 1
2n
¼ 1
2:
If we take the nth root of both sides of this final inequality (which is permissible, by
the Power Rule), we find that
1� 1
2n� 1
21n
;
so that
21n � 1
1� 12n
¼ 2n
2n� 1
¼ 1þ 1
2n� 1:
10. If we apply the Cauchy–Schwarz Inequality, withffiffiffiffiffi
akp
in place of ak and 1ffiffiffiffi
akp in
place of bk, we have
a1 þ a2 þ � � � þ anð Þ 1
a1
þ 1
a2
þ � � � þ 1
an
�
� ffiffiffiffiffi
a1
p � 1ffiffiffiffiffi
a1p þ ffiffiffiffiffi
a2
p � 1ffiffiffiffiffi
a2p þ � � � þ ffiffiffiffiffi
an
p � 1ffiffiffiffiffi
anp
� 2
¼ 1þ 1þ � � � þ 1ð Þ2 ðwith n terms in the bracketÞ
¼ n2:
11. Applying Theorem 3 to the nþ 1 positive numbers 1; 1þ 1n; 1þ 1
n; . . . ; 1þ 1
n,
we obtain
1þ 1
n
� n�
1nþ1
� 1
nþ 1� nþ 1þ n� 1
n
�
¼ nþ 2
nþ 1
¼ 1þ 1
nþ 1;
taking the (nþ 1)th power of this last inequality, by the Power Rule, we deduce
that
1þ 1
n
� n
� 1þ 1
nþ 1
� nþ1
:
368 Appendix 4
Section 1.4
1. (a)
E1 is bounded above. For example, M¼ 1 is an upper bound of E1, since
x � 1; for all x 2 E1:
Also, max E1¼ 1, since 12E1.
(b)
E2 is bounded above. For example, M¼ 1 is an upper bound of E2, since
1� 1
n� 1; for n ¼ 1; 2; . . .:
However, E2 has no maximum element. If x2E2, then x ¼ 1� 1n
for some
positive integer n; and so there is another element of E2, such as 1� 1nþ1
, with
1� 1
n< 1� 1
nþ 1since
1
n>
1
nþ 1
�
:
Hence x is not a maximum element of E2.
(c)
The set E3 is not bounded above, and so it cannot have a maximum element. For
each number M, there is a positive integer n such that n2>M (for instance, take
n>M, which implies that n2> n>M).
Hence M cannot be an upper bound of E3.
2. (a) The set E1¼ (�1, 1] is not bounded below, and so it cannot have a minimum
element. For each number m, there is a (negative) number x such that x<m.
Since x2E1, m cannot be a lower bound of E1.
(b) E2 is bounded below by 0, since
1� 1
n� 0; for n ¼ 1; 2 . . .:
Also, 02E2, and so min E2¼ 0.
(c) E3 is bounded below by 1, since
n2 � 1; for n ¼ 1; 2 . . .:
Also, 12E3, and so min E3¼ 1.
3. ƒ is increasing on the interval [�3, �2). Since �3� x<�2 so that x22 (4, 9], we
have f xð Þ ¼ 1x2 2 1
9, 1
4
� �
. Hence ƒ is bounded above and bounded below.
Next, since f ð�3Þ ¼ 19
and 19
is a lower bound for ƒ on the interval [�3, �2), it
follows that ƒ has a minimum value of 19
on this interval.
Finally, 14
is an upper bound for ƒ on the interval [�3, �2) but there is no point x
in [�3, �2) for which f ðxÞ ¼ 14. So 1
4cannot be a maximum of f on the interval.
However, if y is any number in 19; 1
4
� �
, there is a number x > � 1ffiffi
yp in
� 1ffiffi
yp ;�2
� �
�3;�2½ Þ such that f xð Þ ¼ 1x2 > y, so that no number in 1
9, 1
4
� �
will
serve as a maximum of f on the interval. It follows that f has no maximum value
on [�3, �2).
Solutions to the problems 369
4. (a) The set E1¼ (�1, 1] has a maximum element 1, and so
sup E1 ¼ max E1 ¼ 1:
(b) We know that E2 ¼ f1� 1n
: n ¼ 1, 2, . . .g is bounded above by 1, since
1� 1
n� 1; for n ¼ 1; 2 . . .:
To show that M¼ 1 is the least upper bound of E2, we prove that, if M0< 1, then
there is an element 1� 1n
of E2 such that
1� 1
n> M0:
However, since M0< 1, we have
1� 1
n> M0 , 1�M0 >
1
n
, n >1
1�M0ðsince 1�M0 > 0Þ:
Choosing a positive integer n so large that n > 11�M0, we obtain 1� 1
n> M0, as
required. Hence 1 is the least upper bound of E2.
(c) The set E3¼ {n2: n¼ 1, 2, . . .} is not bounded above, and so it cannot have a
least upper bound.
5. (a) The set E1¼ (1, 5] is bounded below by 1, since
1 � x; for all x 2 E1:
To show that m¼ 1 is the greatest lower bound of E1, we prove that, if m0> 1,
then there is an element x in E1 which is less than m0. Since m0> 1, there is a
number x of the form x¼ 1.00 . . . 01 such that
1 < x < m0;
and clearly x2E1.
Hence 1 is the greatest lower bound of E1.
(b) The set E2 ¼ f 1n2 : n ¼ 1, 2, . . .g is bounded below by 0, since
0 <1
n2; for n ¼ 1; 2; . . .:
To show that m¼ 0 is the greatest lower bound of E2, we prove that, if m0> 0,
then there is an element 1n2 in E2 such that 1
n2 < m0. Since m0> 0, we have
1
n2< m0 , n2 >
1
m0
, n >1ffiffiffiffiffi
m0p :
Choosing a positive integer n so large that n > 1ffiffiffiffi
m0p , we obtain 1
n2 < m0, as
required. Hence 0 is the greatest lower bound of E2.
6. ƒ is decreasing on the interval [1, 4). Since 1� x< 4 so that x22 [1, 16), we have
f xð Þ ¼ 1x2 2 1
16, 1
� �
. Hence ƒ is bounded above by 1 and bounded below by 116
.
Next, since f(1)¼ 1 and 1 is an upper bound for ƒ on the interval [1, 4), it follows
that ƒ has least upper bound 1 on [1, 4).
Finally, 116
is a lower bound for ƒ on the interval [1, 4) but there is no point x in
[1, 4) for which f ðxÞ ¼ 116
.
However, if y is any number in 116
, 1� �
, there is a number x < 1ffiffi
yp in [1, 4) such that
f xð Þ ¼ 1x2 < y, so that no number in 1
16, 1
� �
will serve as a lower bound of ƒ on [1, 4).
It follows that ƒ has 116
as its greatest lower bound on [1, 4).
370 Appendix 4
Chapter 2
Section 2.1
1. (a) (i) 4, 7, 10, 13, 16;
(ii) 13
, 19
, 127
, 181
, 1243
;
(iii) �1, 2, �3, 4, �5.
(b) (i) a1¼ 1, a2¼ 2, a3¼ 6, a4¼ 24, a5¼ 120;
(ii) a1¼ 2, a2¼ 2.25, a3¼ 2.37, a4¼ 2.44, a5¼ 2.49.
2. (a)
(b)
(c)
(d)
3. (a) n! is monotonic, because
an ¼ n! and anþ1 ¼ nþ 1ð Þ!;
Solutions to the problems 371
so that
anþ1� an ¼ nþ 1ð Þ!� n!
¼ n� n! > 0; for n ¼ 1; 2; . . .:
Thus {n!} is increasing.
Alternatively, an> 0, for all n, and
anþ1
an
¼ nþ 1ð Þ!n!
¼ nþ 1 � 1; for n ¼ 1; 2; . . .;
so that {n!} is increasing.
(b) {2�n} is monotonic, because
an ¼ 2�n and anþ1 ¼ 2� nþ1ð Þ;
so that
anþ1 � an ¼1
2nþ1� 1
2n
¼ 1
2n
1
2� 1
�
< 0; for n ¼ 1; 2; . . .:
Thus {2�n} is decreasing.
Alternatively, an> 0, for all n, and
anþ1
an
¼ 2n
2nþ1¼ 1
2< 1; for n ¼ 1; 2; . . .;
so that {2�n} is decreasing.
(c) nþ 1n
�
is monotonic, because
an ¼ nþ 1
nand anþ1 ¼ nþ 1þ 1
nþ 1;
so that
anþ1 � an ¼ nþ 1þ 1
nþ 1
�
� nþ 1
n
�
¼ n nþ 1ð Þ � 1
n nþ 1ð Þ > 0; for n ¼ 1; 2; . . .:
Thus nþ 1n
�
is increasing.
4. (a) TRUE: 2n> 1000, for n> 9, since {2n} is increasing and 210¼ 1024.
(b) FALSE: All the terms a1, a3, a5, . . . are negative.
(c) TRUE: 1n< 0:025, for all n > 1
0:025¼ 40:
(d) TRUE: an> 0 for all n, and anþ1
an¼ 1
4nþ1
n
� �4:
Now
1
4
nþ 1
n
� 4
� 1, nþ 1
n
� 4
� 4 , 1þ 1
n� 4
14
, 1
n�
ffiffiffi
2p� 1 , n � 1
ffiffiffi
2p� 1’ 2:414:
Soanþ1
an
� 1; for n > 2:
Hence
anþ1 � an; for n > 2;
so that
n4
4n
�
is eventually decreasing.
372 Appendix 4
Section 2.2
1. (a) 1n< 1
100, n > 100, by the Reciprocal Rule (for positive numbers). Hence we
may take X¼ 100.
(b) 1n< 3
1000, n > 1000
3¼ 333:3333 . . ., by the Reciprocal Rule (for positive num-
bers). Hence we may take X¼ 333.
2. (a) �1ð Þn
n2
<1
100, 1
n2<
1
100
, n2 > 100
, n > 10:
Hence we may take X¼ 10.
(b) �1ð Þn
n2
<3
1000, 1
n2<
3
1000
, n2 >1000
3
, n >
ffiffiffiffiffiffiffiffiffiffi
1000
3
r
’ 18:25:
Hence we may take X¼ 18.
3. (a) The sequencen
12n� 1
o
is a null sequence.
To prove this, we want to show that:
for each positive number ", there is a number X such that
1
2n� 1
< "; for all n > X: ( )
We know that
1
2n� 1
< ", 1
2n� 1< " ðsince 2n� 1 > 0Þ
, 2n� 1 >1
"
, n > 12
1þ 1
"
�
;
it follows that () holds if we take X ¼ 12
�
1þ 1"
�
. Hencen
12n�1
o
is null.
(b) The sequence�1ð Þn10
n o
is not a null sequence.
To prove this, we must find a positive value of " such that the sequence does
not eventually lie in the horizontal strip on the sequence diagram from �" up
to ". We can take " ¼ 120
, as the following sequence diagram shows:
Any value for X greater than100 will also be valid.
Any value for X greater than333 will also be valid.
Any value for X greater than 10will also be valid.
Any value for X greater than 18will also be valid.
Solutions to the problems 373
There is NO value of X such that the following statement holds
�1ð Þn
10
<1
20; for all n > X:
(c) The sequence�1ð Þnn4þ1
n o
is a null sequence.
To prove this, we want to show that:
for each positive number ", there is a number X such that
�1ð Þn
n4 þ 1
< "; for all n > X: ( )
We know that
�1ð Þn
n4 þ 1
< ", 1
n4 þ 1< "
, n4 þ 1 >1
"
, n4 >1
"� 1:
Now, if "� 1, then 1" � 1 � 0, so that
n4 >1
"� 1; for n ¼ 1; 2; . . .;
hence () holds with X¼ 1.
On the other hand, if 0<"< 1, then 1" � 1 > 0, so that
n4 >1
"� 1, n >
1
"� 1
� 14
;
hence () holds with X ¼ 1" � 1� �1
4.
Thus () holds in either case. Hence�1ð Þn
n4 þ 1
n o
is null.
4. (a) We know thatn
12n�1
o
is null, and so 1
2n�1ð Þ3n o
is also null, by the Power Rule.
(b) The sequences 1n
�
andn
12n�1
o
are null, so 6ffiffi
n5p
n o
and 5
2n�1ð Þ7n o
are also null, by
the Power Rule and Multiple Rule.
Hence the sequence 6ffiffi
n5p þ 5
2n�1ð Þ7n o
is null, by the Sum Rule.
(c) The sequences 1n
�
andn
12n�1
o
are null, so 1n4
�
and 1
2n�1ð Þ13
� �
are also null, by
the Power Rule.
Hence 1
3n4 2n�1ð Þ13
� �
is also null, by the Product Rule and Multiple Rule.
5. Here, using the Hint, we have
n1
2
� n
¼ n� 1
2n
� n� 1
n2¼ 1
n:
Thus, the sequence 1n
�
dominates the sequence n 12
� �n �
, and is itself null. It
follows, from the Squeeze Rule, that the sequence n 12
� �n �
is null.
6. (a) We guess that 1n2þn
n o
is dominated by 1n
�
.
To check this, we have to show that
1
n2 þ n� 1
n; for n ¼ 1; 2; . . .;
this certainly holds, because
n2 þ n � n; for n ¼ 1; 2; . . .:
Since 1n
�
is null, we deduce that 1n2þn
n o
is null, by the Squeeze Rule.
374 Appendix 4
(b) We guess that�1ð Þnn!
n o
is dominated by 1n
�
.
To check this, we have to show that
�1ð Þn
n!
� 1
n; for n ¼ 1; 2; . . . ;
this certainly holds, because
�1ð Þn
n!
¼ 1
n!; for n ¼ 1; 2; . . .;
and
n! � n; for n ¼ 1; 2; . . .:
Since 1n
�
is null, we deduce that �1ð Þn
n!
�
is null, by the Squeeze Rule.
(c) We guess that sin n2
n2 þ 2n
n o
is dominated by 1n2
�
.
To check this, we have to show that
sin n2
n2 þ 2n
� 1
n2; for n ¼ 1; 2; . . .;
this certainly holds, because
sin n2
� 1; for n ¼ 1; 2; . . .;
and
n2 þ 2n � n2; for n ¼ 1; 2; . . .:
Since 1n2
�
is null, we deduce that sin n2
n2 þ 2n
n o
is null, by the Squeeze Rule.
Section 2.3
1. (a)
The sequence appears to converge to 1.
(b) Since bn ¼ an � 1 ¼ nþ 1n� 1 ¼ 1
n, it follows that bnf g ¼ 1
n
�
is a null sequence.
2. (a)
Solutions to the problems 375
Here an � 1 ¼ n2 � 1n2 þ 1� 1 ¼ �2
n2 þ 1. We know that �2
n2 þ 1
n o
is a null sequence, so it
follows that {an� 1} is also a null sequence.
Hence {an} converges to 1.
(b)
Here an � 12¼ n3þ �1ð Þn
2n3 � 12¼ �1ð Þn
2n3 . We know that�1ð Þn2n3
n o
is a null sequence, so it
follows that an � 12
�
is also a null sequence.
Hence {an} converges to 12.
3. (a) The dominant term is n3, so we write an as
an ¼n3 þ 2n2 þ 3
2n3 þ 1
¼1þ 2
nþ 3
n3
2þ 1n3
:
Since 1n
�
and 1n3
�
are basic null sequences
limn!1
an ¼1þ 0þ 0
2þ 0¼ 1
2;
by the Combination Rules.
(b) The dominant term is 3n, so we write an as
an ¼n2 þ 2n
3n þ n3
¼n2
3n þ 23
� �n
1þ n3
3n
:
Since n2
3n
n o
, 23
� �n �
and n3
3n
n o
are basic null sequences
limn!1
an ¼0þ 0
1þ 0¼ 0;
by the Combination Rules.
(c) The dominant term is n!, so we write an as
an ¼n!þ �1ð Þn
2n þ 3n!
¼1þ �1ð Þn
n!2n
n! þ 3:
Since�1ð Þnn!
n o
and 2n
n!
�
are basic null sequences
limn!1
an ¼1þ 0
0þ 3¼ 1
3;
by the Combination Rules.
376 Appendix 4
4. (a) We know that
n1n � 1þ
ffiffiffiffiffiffiffiffiffiffiffi
2
n� 1
r
, n � 1þffiffiffiffiffiffiffiffiffiffiffi
2
n� 1
r
!n
:
Using the hint, with x ¼ffiffiffiffiffiffi
2n�1
q
, we obtain
1þffiffiffiffiffiffiffiffiffiffiffi
2
n� 1
r
!n
� n n� 1ð Þ2!
ffiffiffiffiffiffiffiffiffiffiffi
2
n� 1
r
!2
¼ n n� 1ð Þ2
� 2
n� 1¼ n;
as required.
(b) Since n� 1, we have n1n � 1. Combining this inequality with that in part (a), we
obtain
1 � n1n � 1þ
ffiffiffiffiffiffiffiffiffiffiffi
2
n� 1
r
; for n � 2:
Now,ffiffiffiffiffiffiffiffi
2n� 1
qn o1
n¼2is a null sequence, by the Power Rule, so that
limn!1
1þffiffiffiffiffiffiffiffiffiffiffi
2
n� 1
r
!
¼ 1:
Hence, by the Squeeze Rule, limn!1
n1n
� �
¼ 1:
Section 2.4
1. (a) {1þ (� 1)n} is bounded, since the terms 1þ (�1)n take only the values 0 and 2.
Hence
1þ �1ð Þnj j � 2; for n ¼ 1; 2; . . .:
(b) {(� 1)n n} is unbounded. Given any number K, there is a positive integer n such
that
�1ð Þnnj j > K;
for instance, n ¼ Kj j þ 1.
(c) 2nþ 1n
�
is bounded, since 2nþ 1n¼ 2þ 1
nso that
2nþ 1
n
¼ 2þ 1
n� 3; for n ¼ 1; 2; . . .:
(d) 1� 1n
� �n �
is bounded, since
0 � 1� 1
n� 1; for n ¼ 1; 2; . . .;
so that
1� 1
n
� n
¼ 1� 1
n
� n
� 1; for n ¼ 1; 2; . . .:
2. (a)ffiffiffi
npf g is unbounded, and hence divergent, by Corollary 1.
(b) n2 þ nn2 þ 1
n o
is convergent (with limit 1), and hence bounded, by Theorem 1. In fact
n2 þ n
n2 þ 1¼
1þ 1n
1þ 1n2
� 1þ 1
n� 2; for n ¼ 1; 2; . . .:
(c) {(� 1)nn2} is unbounded, and hence divergent, by Corollary 1.
(d) The terms of the sequence n �1ð Þn �
are
1; 2;1
3; 4;
1
5; 6; . . .;
so the sequence is unbounded: given any number K, there is an even positive
integer 2n such that 2n>K. Hence the sequence is divergent, by Corollary 1.
Solutions to the problems 377
3. (a) Each term of 2n
n
�
is positive, and n2n
�
is a basic null sequence. Hence 2n
n!1,
by the Reciprocal Rule.
(b) First, note that
2n � n100 ¼ 2n 1� n100
2n
�
; for n ¼ 1; 2; . . .;
and that n100
2n
n o
is a basic null sequence. It follows that n100
2n is eventually less
than 1, so that 2n� n100 is eventually positive.
Also
limn!1
1
2n � n100¼ lim
n!1
12n
1� n100
2n
¼ 0
1� 0¼ 0;
by the Combination Rules.
Hence 2n� n100!1, by the Reciprocal Rule.
(c) We know that 2n
n!1, by part (a), and that
2n
nþ 5n100 � 2n
n; for n ¼ 1; 2; . . .:
Hence 2n
nþ 5n100 !1, by the Squeeze Rule.
Remark
You could have used the Reciprocal Rule or the Sum and Multiple Rules.
(d) Each term of the sequence 2n þ n2
n10 þ n
n o
is positive, and
limn!1
2n þ n2
n10 þ n
� �1
¼ limn!1
n10 þ n
2n þ n2
¼ limn!1
n10
2n þ n2n
1þ n2
2n
¼ 0þ 0
1þ 0¼ 0;
by the Combination Rules.
Hence 2n þ n2
n10 þ n!1, by the Reciprocal Rule.
4. (a) (i) a2¼ 4, a4¼ 16, a6¼ 36, a8¼ 64, a10¼ 100;
(ii) a3¼ 9, a7¼ 49, a11¼ 121, a15¼ 225, a19¼ 361;
(iii) a1¼ 1, a4¼ 16, a9¼ 81, a16¼ 256, a25¼ 625.
(b) a1 ¼ 1; a3 ¼ 13; a5 ¼ 1
5;
a2¼ 2, a4¼ 4, a6¼ 6.
5. (a) If an ¼ �1ð Þnþ 1n, for n¼ 1, 2, . . ., then
a2k ¼ 1þ 1
2kand a2k�1 ¼ �1þ 1
2k � 1;
so that
limk!1
a2k ¼ 1; whereas limk!1
a2k�1 ¼ �1:
Hence {an} is divergent, by the First Subsequence Rule.
(b) If an ¼ 13n� 1
3n� �
, for n¼ 1, 2, . . ., then
a3k ¼ 0 and a3kþ1 ¼1
3; for k ¼ 1; 2; . . .;
so that
limk!1
a3k ¼ 0; whereas limk!1
a3kþ1 ¼ 13:
Hence {an} is divergent, by the First Subsequence Rule.
378 Appendix 4
(c) If an ¼ n sin 12np
� �
, for n¼ 1, 2, . . ., then
a1 ¼ 1; a2 ¼ 0; a3 ¼ �3; a4 ¼ 0; a5 ¼ 5; a6 ¼ 0; . . .:
Now
a4kþ1 ¼ 4k þ 1ð Þ sin 2kpþ 1
2p
�
¼ 4k þ 1; for k ¼ 1; 2; . . .;
so that a4kþ1!1.
Hence {an} is divergent, by the Second Subsequence Rule.
Section 2.5
1. Suppose that the sequence {an} is decreasing and bounded below, so that it is
necessarily convergent. We show that the limit of {an} is the number
m¼ inf {an : n¼ 1, 2, . . .}.
To see this, let ‘ denote limn!1
an. By the Limit Inequality Rule, since an�m we know
that ‘�m. Now assume that in fact ‘>m.
Since ‘>m, it follows that ‘ > 12ð‘þ mÞ > m. It follows, from the definition of
greatest lower bound, that there then exists some integer X such that aX <12‘þ mð Þ;
and so, since {an} is decreasing, that
an <1
2‘þ mð Þ; for all n > X:
We deduce from the Limit Inequality Rule that ‘ � 12‘þ mð Þ. We may rearrange this
inequality as 2‘� ‘þm, or ‘�m.
This contradicts our assumption that ‘>m. It follows that, in fact, ‘¼m.
2. We use the ideas in the discussion prior to the problem.
(a) Let a1 > 3. Then, from equation (5) in the discussion before the statement of the
problem
a2 � a1 ¼1
4a1 � 1ð Þ a1 � 3ð Þ
> 0;
so that a2> a1. This suggests that, in general, it might be that anþ1> an in the
case that a1> 3; we check this, using Mathematical Induction.
Let P(n) be the statement: If a1> 3, then anþ 1> an.
We have just seen that the statement P(1) is true.
Now assume that the statement P(k) is true; that is, that akþ1> ak for some
k � 1. It follows, from equation ( 5) in the discussion before the statement of the
problem, that
akþ1 � ak ¼1
4ak � 1ð Þ ak � 3ð Þ
> 0;
so that akþ 1> ak; in other words, the statement P(kþ 1) is true.
This proves that P(n) holds for all n� 1.
Thus, if a1> 3, the sequence {an} is increasing.
It follows, from the Monotone Convergence Theorem, that either {an} con-
verges to some (finite) limit ‘ or tends to infinity as n!1.
But the only possible limits of the sequence are 1 and 3, so that since a1> 3
and the sequence is increasing, clearly {an} cannot tend to a limit. It follows thatan!1 as n!1.
(b) Let 0� a1 < 1. Then, from equation ( 5) in the discussion before the statement of
the problem,
We shall show that thisassumption leads to acontradiction.
By letting n!1.
For we cannot have ‘<m and‘�m!
We need a discussion of thistype in order to identify theresult that we wish to verifyusing Mathematical Induction.
Solutions to the problems 379
a2 � a1 ¼1
4a1 � 1ð Þ a1 � 3ð Þ
> 0;
so that a2> a1. This suggests that, in general, it might be that anþ1> an in the
case that 0� a1< 1; we check this, using Mathematical Induction.
Let P(n) be the statement: If 0� a1< 1, then anþ1> an.
We have just seen that the statement P(1) is true.
Now assume that the statement P(k) is true; that is, that akþ1> ak for some
k � 1. It follows, from equation (5) in the discussion before the statement of the
problem, that
akþ1 � ak ¼1
4ak � 3ð Þ ak � 1ð Þ
> 0;
so that akþ 1> ak; in other words, the statement P(kþ 1) is true.
This proves that P(n) holds for all n� 1.
Thus, if 0� a1< 1, the sequence {an} is increasing.
It follows, from the Monotone Convergence Theorem, that either {an} con-
verges to some (finite) limit ‘ or tends to infinity as n!1.
Now, we can use equation (3) before the problem, with n¼ 1, to see that the
assumption a1< 1 implies that
a2 � 1 ¼ 1
4a2
1 � 1� �
< 0;
so that a2< 1. We can then prove, by Mathematical Induction, using an argu-
ment similar to that in part (a), that, if a1< 1, then an< 1 for all n� 1.
But the only possible limits of the sequence are 1 and 3, so that since an< 1 for
all n� 1, clearly {an} cannot tend to 3 or to infinity. It follows that an! 1 as
n!1.
(c) Let a1< 0. Since anþ1 ¼ 14
a2n þ 3
� �
, it is clear that the behaviour of the sequence
as n!1 depends only on the magnitude of a1 and not on its sign.
It follows from this observation that:
if �1< a1< 0, then an! 1 as n!1;
if a1¼�1, then an! 1 as n!1;
if �3< a1<�1, then an! 1 as n!1;
if a1¼�3, then an! 3 as n!1;
if a1<�3, then an!1 as n!1.
3. (a) The Binomial Theorem gives
1þ x
n
� �n
¼ 1þ nx
n
� �
þ n n� 1ð Þ2!
x
n
� �2
þ � � � þ x
n
� �n
¼ 1þ xþ 1
2!1� 1
n
�
x2 þ � � � þ xn
nn:
The general term in this expansion is
n n� 1ð Þ . . . n� k þ 1ð Þk!
x
n
� �k
¼ 1
k!1� 1
n
�
1� 2
n
�
. . . 1� k � 1
n
�
xk:
If k and x are fixed, this general term increases as n increases. Hence the
sequence 1þ xn
� �n �
is increasing if x> 0.
(b) By the Binomial Theorem
1þ 1
n
� k
� 1þ k1
n
�
¼ 1þ k
n; for k ¼ 1; 2; . . .: ( )
380 Appendix 4
To prove that 1þ xn
� �n �
is bounded above, we choose an integer k� x and use
the inequality (), as follows
1þ x
n
� �n
� 1þ k
n
� n
� 1þ 1
n
� k !n
¼ 1þ 1
n
� n� k
� ek;
since the sequence 1þ 1n
� �n �
is increasing with limit e. Hence 1þ xn
� �n �
is
bounded above by ek.
(c) Since 1þ xn
� �n �
is increasing and bounded above, it must be convergent, by the
Monotone Convergence Theorem.
4. The first n terms of the sequence 1þ 1n
� �n �
are
2
1
� 1
;3
2
� 2
;4
3
� 3
; . . . ;nþ 1
n
� n
:
Taking the product of these terms, each of which is less than e, we obtain
213243 . . . nþ 1ð Þn
112233 . . . nn< en;
it follows, by cancellation, that
nþ 1ð Þn
n!< en:
Hence
n! >nþ 1ð Þn
en¼ nþ 1
e
� n
; for n ¼ 1; 2; . . .:
5. We use the formulas:
(a) Area ¼ 12� 1� 1� sin 1
3p
� �
¼ 12�
ffiffi
3p
2¼
ffiffi
3p
4.
(b) Area ¼ 12� 1� 1� sin 1
6p
� �
¼ 12� 1
2¼ 1
4.
(c) Area ¼ 12� 2 tan 1
6p
� �
� 1 ¼ 12� 2� 1
ffiffi
3p ¼ 1
ffiffi
3p .
(d) Area ¼ 12� 2 tan 1
12p
� �
� 1 ¼ tan 112p
� �
.
To determine tan 112p
� �
, we use the formula
tan1
6p
�
¼2 tan 1
12p
� �
1� tan2 112p
� � :
Since tan 16p
� �
¼ 1ffiffi
3p , we obtain
tan2 1
12p
�
þ 2ffiffiffi
3p
tan 112p
� �
� 1 ¼ 0;
so that
tan1
12p
�
¼�2
ffiffiffi
3p�
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2ffiffiffi
3p� �2þ4
q
2
¼ �ffiffiffi
3p� 2:
Solutions to the problems 381
Since tan 112p
� �
> 0, we must have tan 112p
� �
¼ 2�ffiffiffi
3p
; it follows that
Area ¼ 2�ffiffiffi
3p
:
Chapter 3
Section 3.1
1. (a) Using the formula for summing a geometric series, with a ¼ r ¼ � 13, we obtain
sn ¼ � 1
3
�
þ � 1
3
� 2
þ � 1
3
� 3
þ � � � þ � 1
3
� n
¼ � 1
3�
1� � 13
� �n
1� � 13
� �
¼ � 1
41� � 1
3
� n�
:
Since � 13
� �n �
is a basic null sequence
limn!1
sn ¼ �1
4;
and so
X
1
n¼1
� 1
3
� n
is convergent;with sum� 1
4:
(b) Here
sn ¼ 1þ �1ð Þ1þ �1ð Þ2þ � � � þ �1ð Þn�1
¼ 1� 1þ 1� 1þ � � � þ �1ð Þn�1;
so that
sn ¼1; n odd;0; n even:
�
Hence s2kþ1! 1 and s2k! 0 as k!1, so that {sn} is divergent, by the First
Subsequence Rule. Thus the seriesP
1
n¼0
�1ð Þn is divergent.
(c) Using the formula for summing a geometric series with a¼ 2 and r ¼ 12,
we obtain
sn ¼ 2þ 1þ 1
2
�
þ 1
2
� 2
þ � � � þ 1
2
� n�3
¼ 2�1� 1
2
� �n
1� 12
� �
¼ 4 1� 1
2
� n�
:
Since 12
� �n �
is a basic null sequence, limn!1
sn ¼ 4, and so
X
1
n¼�1
1
2
� n
is convergent;with sum 4:
2. (a) We interpret 0.111. . . as
1
101þ 1
102þ 1
103þ � � �:
382 Appendix 4
This is a geometric series with a ¼ 110
and r ¼ 110
. Since 110< 1, the series is
convergent with sum
a
1� r¼
110
1� 110
¼ 1
9;
hence
0:111 . . . ¼ 1
9:
(b) We interpret 0.86363. . . as
8
10þ 1
10
63
1001þ 63
1002þ � � �
�
:
The series in the bracket is a geometric series with a ¼ 63100
and r ¼ 1100
. Since1
100< 1, this series is convergent with sum
a
1� r¼
63100
1� 1100
¼ 63
99¼ 7
11;
hence
0:86363 . . . ¼ 8
10þ 7
110¼ 19
22:
(c) We interpret 0.999. . . as
9
101þ 9
102þ 9
103þ � � �:
This is a geometric series with a ¼ 910
and r ¼ 110
. Since 110< 1, this series is
convergent with sum
a
1� r¼
910
1� 110
¼ 1;
hence
0:999 . . . ¼ 1:
3. s1 ¼1
1� 2¼ 1
2;
s2 ¼1
1� 2þ 1
2� 3¼ 1
2þ 1
6¼ 2
3;
s3 ¼1
1� 2þ 1
2� 3þ 1
3� 4¼ 2
3þ 1
12¼ 3
4;
s4 ¼1
1� 2þ 1
2� 3þ 1
3� 4þ 1
4� 5¼ 3
4þ 1
20¼ 4
5:
4. Since
1
n nþ 2ð Þ ¼1
2
1
n� 1
nþ 2
�
; for n ¼ 1; 2; . . .;
we have
X
1
n¼1
1
n nþ 2ð Þ ¼X
1
n¼1
1
2
1
n� 1
nþ 2
�
;
and so
sn¼1
1�3þ 1
2�4þ 1
3�5þ 1
4�6þ ��� þ 1
n nþ2ð Þ
¼ 1
21�1
3
�
þ 1
2�1
4
�
þ 1
3�1
5
�
þ 1
4�1
6
�
þ�� �þ 1
n� 1
nþ2
� � �
:
Solutions to the problems 383
Most of the terms in alternate brackets cancel out, leaving
sn ¼1
21þ 1
2� 1
nþ 1� 1
nþ 2
�
¼ 3
4� 1
2 nþ 1ð Þ �1
2 nþ 2ð Þ :
Since 1nþ 1
n o
and 1nþ 2
n o
are null sequences, limn!1
sn ¼ 34, and so
X
1
n¼1
1
n nþ 2ð Þ is convergent, with sum3
4:
5. The seriesP
1
n¼1
34
� �nis a geometric series, with a ¼ r ¼ 3
4. Hence, it is convergent, with
sum34
1�34
¼ 3.
The seriesP
1
n¼1
1n nþ 1ð Þ is convergent, with sum 1 (cf. Sub-section 3.1.3). Hence, by
the Combination Rules
X
1
n¼1
3
4
� n
� 2
n nþ 1ð Þ
�
is convergent, with sum 3� 2� 1ð Þ ¼ 1:
6. By the Combination Rules for sequences
limn!1
n2
2n2 þ 1¼ lim
n!1
1
2þ 1n2
¼ 1
2;
so that the sequence n2
2n2 þ 1
n o
is not null.
Hence, by the Non-null Test,P
1
n¼1
n2
2n2 þ 1is divergent.
Section 3.2
1. Let sn ¼ 1þ 122 þ 1
32 þ � � � þ 1n2 and tn ¼ 1þ 1
2þ 1
3þ � � � þ 1
n. The values of these
partial sums are as listed below:
n 1 2 3 4 5 6 7 8
sn 1 1.25 1.36 1.42 1.46 1.49 1.51 1.53
tn 1 1.50 1.83 2.08 2.28 2.45 2.59 2.72
2. (a) We use the Comparison Test. Since
n3 þ n � n3; for n ¼ 1; 2; . . .;
384 Appendix 4
we have1
n3 þ n� 1
n3; for n ¼ 1; 2; . . .:
SinceP
1
n¼1
1n3 is convergent, we deduce, from the Comparison Test, that
X
1
n¼1
1
n3 þ nis convergent:
(b) Let
an ¼1
nþffiffiffi
np and bn ¼
1
n; for n ¼ 1; 2; . . .;
so that
limn!1
an
bn
¼ limn!1
n
nþffiffiffi
np
¼ limn!1
1
1þ 1ffiffi
np
¼ 1 6¼ 0:
SinceP
1
n¼1
1n
is divergent, we deduce, from the Limit Comparison Test, that
X
1
n¼1
1
nþffiffiffi
np is divergent:
(c) Let
an ¼nþ 4
2n3 � nþ 1and bn ¼
1
n2; for n ¼ 1; 2; . . .;
so that
limn!1
an
bn
¼ limn!1
n3 þ 4n2
2n3 � nþ 1
¼ limn!1
1þ 4n
2� 1n2 þ 1
n3
¼ 1
26¼ 0:
SinceP
1
n¼1
1n2 is convergent, we deduce, from the Limit Comparison Test, that
X
1
n¼1
nþ 4
2n3 � nþ 1is convergent:
(d) We use the Comparison Test. Since
0 � cos2 2nð Þ � 1; for n ¼ 1; 2; . . .;
we have
0 � cos2 2nð Þn3
� 1
n3; for n ¼ 1; 2; . . .:
SinceP
1
n¼1
1n3 is convergent, we deduce, from the Comparison Test, that
X
1
n¼1
cos2 2nð Þn3
is convergent:
3. (a) Let an ¼ n3
n!, for n¼ 1, 2, . . ., so that
anþ1
an
¼ nþ 1ð Þ3
nþ 1ð Þ!
!
� n!
n3
�
¼ nþ 1ð Þ2
n3
¼ n2 þ 2nþ 1
n3¼ 1
nþ 2
n2þ 1
n3:
Solutions to the problems 385
Hence, by the Combination Rules for sequences
limn!1
anþ1
an
¼ 0;
it follows, from the Ratio Test, thatP
1
n¼1
n3
n! is convergent.
(b) Let an ¼ n22n
n! , for n¼ 1, 2, . . ., so that
anþ1
an
¼ nþ 1ð Þ22nþ1
nþ 1ð Þ!
!
� n!
n22n
�
¼ 2 nþ 1ð Þn2
¼ 21
nþ 1
n2
�
:
Hence, by the Combination Rules for sequences
limn!1
anþ1
an
¼ 0;
it follows, from the Ratio Test, thatP
1
n¼1
n22n
n! is convergent.
(c) Let an ¼ 2nð Þ!nn , for n¼ 1, 2, . . ., so that
anþ1
an
¼ 2nþ 2ð Þ!nþ 1ð Þnþ1
!
� nn
2nð Þ!
�
¼ 2nþ 2ð Þ 2nþ 1ð Þnn
nþ 1ð Þnþ1
¼ 2 2nþ 1ð Þnn
nþ 1ð Þn
¼ 4nþ 2
1þ 1n
� �n :
We know that limn!1
1þ 1n
� �n¼ e and that 14nþ 2
n o
is null, so that1þ1
nð Þn4nþ 2
� �
is null,
by the Product Rule for sequences. It follows, from the Reciprocal Rule for
sequences, that
anþ1
an
!1 as n!1:
It follows, from the Ratio Test, thatP
1
n¼1
2nð Þ!nn is divergent.
Remark
Notice that
2nð Þ!nn� 2n
n
�
� 2n� 1
n
�
� � � � � nþ 1
n
�
� 1;
it follows, from the Non-null Test, thatP
1
n¼1
2nð Þ!nn is divergent.
4. (a) Let an ¼ 1n loge n
, n¼ 2, . . .. Then an is positive; and, since {nlogen} is an increas-
ing sequence, anf g ¼ 1n loge n
n o
is a decreasing sequence.
Next, let bn ¼ 2na2n ; thus
bn ¼ 2n � 1
2n loge 2nð Þ
¼ 1
n loge 2¼ 1
loge 2� 1
n:
386 Appendix 4
SinceP
1
n¼2
1n
is a basic divergent series, it follows by the Multiple Rule thatP
1
n¼2
bn must be divergent. Hence, by the Condensation Test, the original series
P
1
n¼2
1n loge n
must also be divergent.
(b) Let an ¼ 1
n loge nð Þ2, n¼ 2, . . .. Then an is positive; and, since {n(loge n)2} is an
increasing sequence, anf g ¼ 1
n loge nð Þ2n o
is a decreasing sequence.
Next, let bn ¼ 2na2n ; thus
bn ¼ 2n � 1
2n loge 2nð Þð Þ2
¼ 1
n loge 2ð Þ2¼ 1
loge 2ð Þ2� 1
n2:
SinceP
1
n¼2
1n2 is a basic convergent series, it follows by the Multiple Rule that
P
1
n¼2
bn must be convergent. Hence by the Condensation Test, the original series
P
1
n¼2
1
n loge nð Þ2 must also be convergent.
Section 3.3
1. (a) Let an ¼ �1ð Þnþ1n
n3 þ 1, for n¼ 1, 2, . . . , so that
anj j ¼n
n3 þ 1; for n ¼ 1; 2; . . .:
Now
n
n3 þ 1� n
n3¼ 1
n2; for n ¼ 1; 2; . . .;
andP
1
n¼1
1n2 is a basic convergent series. Hence, by the Comparison Test
X
1
n¼1
n
n3 þ 1is convergent:
It follows, from the Absolute Convergence Test, that
X
1
n¼1
�1ð Þnþ1n
n3 þ 1is convergent:
(b) Let an ¼ cos n2n , for n¼ 1, 2, . . .; then
anj j �1
2n; for n ¼ 1; 2; . . .;
since cos nj j � 1.
NowP
1
n¼1
12n is a basic convergent series (in fact, a convergent geometric
series). Hence, by the Comparison Test
X
1
n¼1
cos n
2nis absolutely convergent:
It follows, from the Absolute Convergence Test, that
X
1
n¼1
cos n
2nis convergent:
Solutions to the problems 387
2. The series1
2þ 1
4� 1
8þ 1
16þ 1
32� 1
64þ � � �
is absolutely convergent, and hence convergent, by the Comparison Test, since the
seriesP
1
n¼1
12n is a convergent geometric series.
By the infinite form of the Triangle Inequality, we have
1
2þ 1
4� 1
8þ 1
16þ 1
32� 1
64þ � � �
� 1
2þ 1
4þ 1
8þ 1
16þ 1
32þ 1
64þ � � �
¼ 1:
It follows that the sum of the original series lies in the interval [�1, 1].
3. (a) The sequence�1ð Þnþ1
n3
n o
is of the form
�1ð Þnþ1bn
�
, where
bn ¼1
n3; for n ¼ 1; 2; . . .:
Now:
1. 1n3 � 0, for n¼ 1, 2, . . .;
2. 1n3
�
is a basic null sequence;
3. 1n3
�
is decreasing, because {n3} is increasing.
Hence, by the Alternating Test,P
1
n¼1
�1ð Þnþ1
n3 is convergent.
(b) The sequence �1ð Þnþ1 nnþ2
n o
is is not a null sequence, since
limn!1
n
nþ 2¼ 1;
and so the odd subsequence tends to 1.
Hence, by the Non-null Test
X
1
n¼1
�1ð Þnþ1 n
nþ 2is divergent:
(c) The sequence�1ð Þnþ1
n13þn
12
� �
is of the form �1ð Þnþ1bn
n o
, where
bn ¼1
n13 þ n
12
; for n ¼ 1; 2; . . .:
Now:
1. 1
n13 þ n
12
� 0, for n¼ 1, 2, . . .;
2. 1
n13 þ n
12
� �
is a null sequence, by the Squeeze Rule, since
0 � 1
n13 þ n
12
� 1
n12
;
where 1
n12
n o
is a basic null sequence;
3. 1
n13 þ n
12
� �
is decreasing, because n13 þ n
12
n o
is increasing.
Hence, by the Alternating Test,P
1
n¼1
�1ð Þnþ1
n13 þ n
12
is convergent.
4. Let sn and tn denote the nth partial sums of the series
1� 1
2� 1
4þ 1
3� 1
6� 1
8þ 1
5� 1
10� 1
12þ � � � and 1� 1
2þ 1
3� 1
4þ 1
5� 1
6þ � � � ;
respectively. Denote by Hn the nth partial sumP
n
k¼1
1k¼ 1þ 1
2þ � � � þ 1
nof the harmonic
series.
388 Appendix 4
Th e t e r m s o f t he s e r i e sP
1
n ¼1
an come ‘in a natural way’ in t hr ees, so it seems
sensible to loo k at the p arti al sum s s3n
s3 n ¼ 1 � 1
2 � 1
4 þ 1
3 � 1
6 � 1
8 þ 1
5 � 1
10 � 1
12 þ �� �þ 1
2n � 1 � 1
4n � 2 � 1
4n
¼ 1 � 1
2 � 1
4
�
þ 1
3 � 1
6 � 1
8
�
þ 1
5 � 1
10 � 1
12
�
þ�� �þ 1
2 n � 1 � 1
4 n � 2 � 1
4 n
�
¼ 1 þ 1
3 þ 1
5 þ �� �þ 1
2n � 1
�
� 1
2 þ 1
4 þ 1
6 þ 1
8 þ �� �þ 1
4n � 2 þ 1
4n
�
¼ 1 þ 1
2 þ 1
3 þ 1
4 þ �� �þ 1
2n � 1 þ 1
2n
�
� 1
2 þ 1
4 þ �� �þ 1
2n
�
� 1
21 þ 1
2 þ 1
3 þ �� �þ 1
2 n
�
¼ H2n �1
2 Hn �
1
2 H 2n
¼ 1
2H2n � Hnð Þ:
But as we saw in Example 2
t2n ¼ H2 n � H n ;
so that
s3n ¼1
2 t2 n :
By assumption, tn ! log e 2 and so t2n ! log e 2 as n !1. It follows, by the Multiple
Rule for sequences, that
s3n !1
2loge 2 as n !1:
Finally
s3n �1 ¼ s3n þ1
4n ! 1
2loge 2 as n !1;
and
s3n �2 ¼ s3n þ1
4n � 2 þ 1
4n ! 1
2loge 2 as n !1:
It follows from these three results that sn ! 12
loge 2 as n !1.
5. We follow the pattern of the solution to Example 3, by finding a subsequence of partial
sums of the seriesP
1
n¼1
�1ð Þnþ1
n¼1� 1
2þ 1
3� 1
4þ 1
5� 1
6þ � � � that are greater than 2, 3, . . .
in turn and ensuring that the other partial sums are not much less than these values.
First, note that all terms of the seriesP
1
n¼1
�1ð Þnþ1
nare at most 1 in modulus.
Next, since the ‘positive’ part of the seriesP
1
n¼1
�1ð Þnþ1
nhas partial sums that tend to
1, we start to construct the desired rearranged series as follows. Take enough of the
‘positive’ terms 1; 13; 1
5; . . .; 1
2N1 � 1so that the sum 1þ 1
3þ 1
5þ � � � þ 1
2N1 � 1is greater
than 2, choosing N1 so that it is the first integer such that this sum is greater than 2.
Then these N1 terms will form the first N1 terms in our desired rearranged series.
Next, take one ‘negative’ term 12
and consider the expression
1þ 1
3þ 1
5þ � � � þ 1
2N1 � 1
�
� 1
2
�
:
It is certainly less than the partial sum tN1¼ 1þ 1
3þ 1
5þ � � � þ 1
2N1 � 1of the rearranged
series, for which tN1> 2. But, since all terms are at most 1 in magnitude, it follows
that tN1þ1 > 1.
Solutions to the problems 389
Now add in some more ‘positive’ terms 12N1þ1
; 12N1þ3
; . . .; 12N2�1
so that the sum
1þ 1
3þ 1
5þ � � � þ 1
2N1 � 1
�
� 1
2
�
þ 1
2N1 þ 1þ 1
2N1 þ 3þ � � � þ 1
2N2 � 1
�
is greater than 3, choosing N2 so that it is the first available integer such that this sum
is greater than 3. Then these N2þ 1 terms will form the first N2þ 1 terms in our
desired rearranged series.
We then add in just one ‘negative’ term; this makes the next partial sum of the
rearranged series satisfy tN2þ2 > 2. Then we add in sufficient positive terms to make
the partial sum tN3þ2 > 4. And so on indefinitely.
In each set of two steps in this process we must use at least one of the ‘positive’
terms and one of the ‘negative’ terms, so that eventually all the ‘positive’ terms and
all the ‘negative’ terms of the original series will be taken exactly once in the new
series, which we denote byP
1
n¼1
bn. So certainly the seriesP
1
n¼1
bn is a rearrangement of
the original seriesP
1
n¼1
�1ð Þnþ1
n.
From our construction, we have ensured that:
for all n>Nk, for any k, the nth partial sums ofP
1
n¼1
bn are >k.
It follows, then, that the nth partial sums ofP
1
n¼1
bn tend to infinity as n!1.
6. (a) The sequence 12
n �
is not null; hence, by the Non-null Test, the seriesP
1
n¼1
12
n is
divergent.
ThusP
1
n¼1
12
n is neither convergent nor absolutely convergent.
(b) We have
5nþ 2n
3n¼ 5n
1
3
� n
þ 2
3
� n
; for n ¼ 1; 2; . . .:
Now,P
1
n¼1
n 13
� �nand
P
1
n¼1
23
� �nare both basic convergent series; hence, by the
Combination Rules for series
X
1
n¼1
5nþ 2n
3nis convergent ðand absolutely convergentÞ:
(c) The sequence fang ¼n
32n3 � 1
o
is null (so the Non-null Test is inappropriate),
and contains only positive terms.
The Ratio Test fails, since
limn!1
anþ1
an
¼ limn!1
2n3 � 1
2 nþ 1ð Þ3�1¼ 1:
Instead, we use the Limit Comparison Test, with
bn ¼1
n3; for n ¼ 1; 2; . . .:
Then
limn!1
an
bn
¼ limn!1
32n3�1
1n3
¼ limn!1
3n3
2n3 � 1
¼ 3
26¼ 0:
Since all terms are non-negative.
390 Appendix 4
SinceP
1
n¼1
1n3 is a basic convergent series, we deduce, from the Limit Comparison
Test, thatX
1
n¼1
3
2n3 � 1is convergent:
Since the terms in the series are non-negative, it follows that it is also absolutely
convergent.
(d) The sequence�1ð Þnþ1
n13
n o
is null, butP
1
n¼1
1
n13
is a basic divergent series. HenceP
1
n¼1
�1ð Þnþ1
n13
is not absolutely convergent.
However,�1ð Þnþ1
n13
n o
is of the form �1ð Þnþ1bn
n o
, where
bn ¼1
n13
; for n ¼ 1; 2; . . . :
Now:
1. 1
n13
� 0, for n¼ 1, 2, . . .;
2. 1
n13
n o
is a null sequence;
3. 1
n13
n o
is decreasing, because
n13
�
is increasing.
Hence, by the Alternating Test,P
1
n¼1
�1ð Þnþ1
n13
is convergent.
(e) The sequence an ¼ �1ð Þnþ1n2
n2 þ 1; n ¼ 1; 2; . . ., is not null, since
limn!1
n2
n2 þ 1¼ lim
n!1
1
1þ 1n2
¼ 1;
so that the odd subsequence tends to 1.
Hence,P
1
n¼1
anj j andP
1
n¼1
an are divergent, by the Non-null Test. It follows that
X
1
n¼1
�1ð Þnþ1n2
n2 þ 1is neither convergent nor absolutely convergent:
(f) Let an ¼ �1ð Þnþ1n
n3þ5; n ¼ 1; 2; . . .; then
anj j ¼n
n3 þ 5; n ¼ 1; 2; . . .;
so that
anj j �n
n3¼ 1
n2; for n ¼ 1; 2; . . .:
It follows, by the Comparison Test, thatP
1
n¼1
anj j is convergent; that is, thatP
1
n¼1
an
is absolutely convergent.
Hence, by the Absolute Convergence Test,P
1
n¼1
an is convergent.
(g) Since
n6
2n
�
is a basic null sequence of positive terms, it follows, from the
Reciprocal Rule for sequences, that
2n
n6!1 as n!1:
Hence 2n
n6
�
is not a null sequence; it follows, from the Non-null Test, that
X
1
n¼1
2n
n6is divergent:
Thus,P
1
n¼1
2n
n6 is neither convergent nor absolutely convergent.
SinceP
1
n¼1
1n2 is a basic
convergent series.
Solutions to the problems 391
(h) Let an ¼ �1ð Þnþ1n
n2þ2, n¼ 1, 2, . . ., so that
anj j ¼n
n2 þ 2; for n ¼ 1; 2; . . .:
Now we try the Limit Comparison Test, with
bn ¼1
n; for n ¼ 1; 2; . . .:
We obtain
limn!1
anj jbn
¼ limn!1
nn2þ2
1n
¼ limn!1
n2
n2 þ 2¼ 1 6¼ 0:
Thus,P
1
n¼1
anj j is divergent, sinceP
1
n¼1
1n
is divergent; it follows thatP
1
n¼1
an is not
absolutely convergent.
However,�1ð Þnþ1
n
n2 þ 2
n o
is of the form �1ð Þnþ1bn
n o
, where
bn ¼n
n2 þ 2; for n ¼ 1; 2; . . .:
Now:
1. nn2 þ 2
� 0, for n¼ 1, 2, . . .;
2. nn2 þ 2
n o
is a null sequence, since
n
n2 þ 2¼
1n
� �
1þ 2n2
� �! 0 as n!1;
3. nn2 þ 2
n o
is decreasing, sincen
n2þ2n
o
is increasing, because
nþ 1ð Þ2þ2
nþ 1� n2 þ 2
n¼ nþ 1þ 2
nþ 1
�
� nþ 2
n
�
¼ 1� 2
n nþ 1ð Þ � 0:
Hence, by the Alternating Test,P
1
n¼1
�1ð Þnþ1n
n2 þ 2is convergent.
(i) Let an ¼ 1
n loge nð Þ34
, n¼ 2, . . .. Then an is positive; and, since n loge nð Þ34
n o
is an
increasing sequence, anf g ¼n
1
n loge nð Þ34
o
is a decreasing sequence.
Next, let bn ¼ 2na2n ; thus
bn ¼ 2n � 1
2n loge 2nð Þ34
¼ 1
n loge 2ð Þ34
¼ 1
loge 2ð Þ34
� 1
n34
:
SinceP
1
n¼2
1
n34
is a basic divergent series, it follows (for example, by the Ratio Test)
thatP
1
n¼2
bn must be divergent. Hence, by the Condensation Test, the original
seriesP
1
n¼2
1
n loge nð Þ34
must also be divergent. Hence the series is not convergent, and,
since its terms are non-negative, is not absolutely convergent.
392 Appendix 4
Section 3.4
1. Substituting x¼ 2 into the power series for ex, we find that the seventh partial sum of
the power series for e2 gives
e2 ’ 1þ 2
1!þ 22
2!þ 23
3!þ 24
4!þ 25
5!þ 26
6!
¼ 1þ 2þ 4
2þ 8
6þ 16
24þ 32
120þ 64
720
¼ 1þ 2þ 2þ 4
3þ 2
3þ 4
15þ 4
45
¼ 7þ 16
45¼ 7:355:
Hence this method gives as an estimate for e2 to 3 decimal places the number 7.355.
Remark
In fact, e2¼ 7.389, to 3 decimal places.
2. From equation ( 5) we know that
0 < e� sn <1
n� 1ð Þ!�1
n� 1;
so that we want n to satisfy the requirement that
1
n� 1ð Þ!�1
n� 1< 5� 10�11; or n� 1ð Þ!� n� 1ð Þ > 1
5� 1011 ¼ 2� 1010:
But 13!� 13 ’ 8.095� 1010, so n¼ 14 is sufficient.
In other words, 14 terms will suffice.
Chapter 4
Section 4.1
1. (a) Let {xn} be any sequence in R that converges to 2. Then, by the Combination
Rules for sequences, we have
f xnð Þ ¼ x3n � 2x2
n
! 23 � 2� 22 ¼ 8� 8 ¼ 0 as n!1:But f(0)¼ 0, so that f(xn)! f(0) as n!1. Thus f is continuous at 2.
(b) Consideration of the graph of f near 1 suggests that f is not continuous at 1.
So, let {xn} be any sequence in (0, 1) that tends to 1. Then f(xn)¼ 0, for all n,
so that
f xnð Þ ! 0 as n!1:But f(1)¼ 1, so that f(xn) 6! f(1) as n!1. Thus f is not continuous at 1.
Remark
If we had chosen {xn} to be any sequence in (1, 2) that tends to 1, we would
have found that f(xn)! f(1) as n!1. However this does NOT prove that f is
continuous at 1, since the definition of continuity requires that f(xn)! f(1) for
ALL sequences that tend to 1.
2. (a) Let c be any point in R .
Let {xn} be any sequence in R that converges to c. Then f(xn)¼ 1, for all n,
so that
f xnð Þ!1 as n!1:
For example, xn ¼ 1� 12n
.
Solutions to the problems 393
But f(c)¼ 1, so that f(xn)! f(c) as n!1. Thus f is continuous at c.
Since c is an arbitrary point of R , it follows that f is continuous on R .
(b) Let c be any point in R .
Let {xn} be any sequence in R that converges to c. Then f(xn)¼ xn, for all n,
so thatf xnð Þ ! c as n!1:
But f(c)¼ c, so that f(xn)! f(c) as n!1. Thus f is continuous at c.
Since c is an arbitrary point of R , it follows that f is continuous on R .
3. The domain of f is the interval I¼ {x : x� 0} if n is even and R if n is odd.
Thus, we have to show that for each c in I:
for each sequence {xk} in I such that xk! c, then x1n
k ! c1n.
First, let c¼ 0. We have already seen (in the Power Rule for null sequences and the
subsequent Remark) that, for any null sequence {xk} in I, x1n
k
n o
is also a null
sequence. In other words, f(xk)! f(0) as k!1. Hence f is continuous at 0.
Next, let c 6¼ 0. We have to prove that if {xk� c} is a null sequence, then so is
x1n
k � c1n
n o
.
Now, using the hint, we have
x1n
k � c1n ¼ xk � c
xn�1
n
k þ xn�2
n
k c1n þ x
n�3n
k c2n þ � � � þ c
n�1n
:
By the result of part (b) of Exercise 3 on Section 2.3, we obtain
xn�1
n
k þ xn�2
n
k c1n þ x
n�3n
k c2n þ � � � þ c
n�1n
! cn�1
n þ cn�2
n c1n þ c
n�3n c
2n þ � � � þ c
n�1n as k!1
¼ ncn�1
n ;
since c 6¼ 0, we deduce that
x1n
k � c1n ! 0
ncn�1
n
as k!1
¼ 0:
In other words, x1n
k � c1n
n o
is null, as required. It follows that f is continuous at c.
4. (a) Let d be any point in R .
Let {xn} be any sequence in R that converges to d. Then f(xn)¼ c, for all n, so that
f xnð Þ ! c as n!1:But f(d)¼ c, so that f(xn)! f(d) as n!1. Thus f is continuous at d.
Since d is an arbitrary point of R , it follows that f is continuous on R .
(b) Let c be any point in R .
Let {xk} be any sequence in R that converges to c. Then f(xk)¼ xkn, for all k,
so that, by the Product Rule for sequences
f xkð Þ ! cn as k!1:But f(c)¼ cn, so that f (xk)! f (c) as k!1. Thus f is continuous at c.
Since c is an arbitrary point of R , it follows that f is continuous on R .
(c) Let c be any point in R .
Let {xn} be any sequence in R that converges to c. Then, by Theorem 4 of
Sub-section 2.3.3, we have
f xnð Þ ¼ xnj j ! cj j as n!1:But f(c)¼ jcj, so that f(xn)! f(c) as n!1. Thus f is continuous at c.
Since c is an arbitrary point of R , it follows that f is continuous on R .
394 Appendix 4
5. First, let c¼ 0. Consideration of the graph of f near 0 suggests that f is not continuous at 0.
So, let {xn} be any sequence in (0, 1) that tends to 0. Then f(xn)¼ 1, for all n, so that
f xnð Þ ! 1 as n!1:But f(0)¼ 0, so that f(xn) 6! f(0) as n!1. Thus f is not continuous at 0.
Next, let c> 0. We will show that f is continuous at c.
So, let {xn} be any sequence that tends to c. Since c> 0, it follows that eventually
xn> 0; thus xn> 0 for all n>N, for a suitable choice of N. Then f(xn)¼ 1, for all
n>N, so that
f xnð Þ ! 1 as n!1:But f(c)¼ 1, so that f(xn)! f(c) as n!1. Thus f is continuous at c.
Finally, let c< 0. We will show that f is continuous at c.
So, let {xn} be any sequence that tends to c. Since c< 0, it follows that eventually
xn< 0; thus xn< 0 for all n>N, for a suitable choice of N. Then f(xn)¼�1, for all
n>N, so that
f xnð Þ ! �1 as n!1:But f(c)¼�1, so that f(xn)! f(c) as n!1. Thus f is continuous at c.
It follows that f is continuous on R � {0}, and discontinuous at 0.
6. We have already seen, in Example 3 of Sub-section 4.1.1 , that the function
g : x 7! x12; for x � 0;
is continuous on [0, 1). Also, it follows from Problem 4 that the function
h : x 7! x3; for x 2 R ;
is continuous on R .
We may apply the Composition Rule to the functions g and h, since
g([0,1))R ; it follows that
h � g : x 7! x12
� �3
¼ x32; for x � 0;
is continuous on [0,1).
7. By the Sum and Product Rules, and by using the fact that the constant function and
the identity function are continuous on R , we see that
g : x 7! x2 þ xþ 1; x 2 R ;
is continuous on R . Since g(R) (0,1) and the square root function h is continuous
on R , it follows, by the Composition Rule, that the function
h � g : x 7!ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ xþ 1p
; x 2 R ;
is continuous on R .
Next, by the Sum, Multiple and Product Rules, and by using the fact that the constant
function and the identity function are continuous on R , we see that the functions
k : x 7! �5x and l : x 7!1þ x2; x 2 R ;
are continuous on R . Since l(x) 6¼ 0, it follows from the Product and Quotient Rules,
that
m : x 7! �5x
1þ x2; x 2 R ;
is continuous on R .
It then follows, by applying the Sum Rule to h� g and m, that the function
f xð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ xþ 1p
� 5x
1þ x2; x 2 R ;
is continuous on R .
Equivalently, all polynomialsare continuous on R .
Equivalently, all polynomialsare continuous on R .
Equivalently, rational functionsare continuous on theirdomains.
Solutions to the problems 395
8. (a) The graph of f suggests that we should find functions g and h that squeeze f near 0.
y = x sin 1x
y
x
So, we define g(x)¼� jxj, x2R , and h(x)¼ jxj, x2R . With these two chosen,
we check the conditions of the Squeeze Rule.
Now we know that
� 1 � sin1
x
�
� 1; for any x 6¼ 0:
It follows that
� xj j � x sin1
x
�
� xj j; for any x 6¼ 0;
so that
g xð Þ ¼ � xj j½ � � f xð Þ � xj j ¼½ � h xð Þ; for any x 2 R :
So condition 1 of the Squeeze Rule is satisfied.
Next, the functions f, g and h all take the value 0 at the point 0. Thus condition
2 of the Squeeze Rule is satisfied.
Finally, the functions g and h are both continuous at 0.
It then follows, from the Squeeze Rule, that f is continuous at 0.
(b) Consideration of the graph of f near 0 suggests that f is not continuous at 0.
y = sin 1x
1
–1
So, let {xn} be the sequence
xnf g ¼1
2nþ 12
� �
p
( )
:
Then xn! 0 as n!1, and
f xnð Þ ¼ sin 2nþ 1
2
�
p�
¼ sin1
2p
�
¼ 1
for all n, so that
f xnð Þ ! 1 as n!1:But f(0)¼ 0, so that f(xn) 6! f(0) as n!1. Thus f is not continuous at 0.
396 Appendix 4
9. Since the constant function and the identity function are continuous on R , it follows
from the Combination Rules that the function
g xð Þ ¼ x2 þ 1; x 2 R ;
is continuous on R .
Next, since the sine function is continuous on R , it follows by the Composition
Rule, that
sin � g xð Þ ¼ sin x2 þ 1� �
; x 2 R ;
is continuous on R .
It then follows from the Multiple and Sum Rules, that the function
f xð Þ ¼ x2 þ 1þ 3 sin x2 þ 1� �
; x 2 R ;
is continuous on R .
10. The cosine function is continuous on R , so that, by the Multiple Rule, the function
x 7!p2
cos x; x 2 R ;
is continuous on R .
Then, since the sine function, also, is continuous on R , we deduce, from the
Composition Rule, that
f xð Þ ¼ sinp2
cos x� �
; x 2 R ;
is continuous on R .
11. Since the identity function is continuous on R , it follows, from the Combination
Rules, that the function
g xð Þ ¼ x5 � 5x2; x 2 R ;
is continuous on R .
Then, since the cosine function is continuous on R , we deduce, from the
Composition Rule, that
h xð Þ ¼ cos x5 � 5x2� �
; x 2 R ;
is continuous on R .
Next, since the identity function is continuous on R , it follows, from the
Combination Rules, that the function
k xð Þ ¼ �x2; x 2 R ;
is continuous on R .
Then, since the exponential function is continuous on R , we deduce, from the
Composition Rule, that
l xð Þ ¼ e�x2
; x 2 R ;
is continuous on R .
Finally, it follows, from the Combination Rules, that
f xð Þ ¼ cos x5 � 5x2� �
þ 7e�x2
; x 2 R ;
is continuous on R .
Section 4.2
1. Consider the function
f xð Þ ¼ cos x� x; x 2 0; 1½ �:We shall show that f has a zero c in (0, 1).
Equivalently, all polynomialsare continuous on R .
Equivalently, all polynomialsare continuous on R .
Equivalently, all polynomialsare continuous on R .
Solutions to the problems 397
Certainly, f is continuous on [0, 1], and
f 0ð Þ ¼ cos 0� 0 ¼ 1 > 0;
f 1ð Þ ¼ cos 1� 1< 0:
Thus, by the Intermediate Value Theorem, there is some number c in (0, 1) such that
f (c)¼ 0, and so such that
cos c ¼ c:
2. If f (0)¼ 0 or f (1)¼ 1, we can take c¼ 0 or c¼ 1, respectively.
Otherwise, we have f (0)> 0 and f (1)< 1, since the image of [0, 1] under f lies
in [0, 1].
We now define the auxiliary function
g xð Þ ¼ f xð Þ � x; x 2 0; 1½ �;and show that g has a zero c in (0, 1).
Certainly, g is continuous on [0, 1], and
g 0ð Þ ¼ f 0ð Þ � 0 > 0;
g 1ð Þ ¼ f 1ð Þ � 1< 0:
Thus, by the Intermediate Value Theorem, there is some number c in (0, 1) such that
g(c)¼ 0, and so such that
f cð Þ ¼ c:
3. For the function f (x)¼ x5þ x� 1, x2 [0, 1], we have
f 0ð Þ ¼ �1 < 0; f 1ð Þ ¼ 1þ 1� 1 ¼ 1 > 0; and
f1
2
�
¼ 1
32þ 1
2� 1 ¼ � 15
32:
It follows, by the Intermediate Value Theorem applied to 12; 1
� �
, that f has a zero
in 12; 1
� �
.
Next, f 34
� �
¼ 2431024þ 3
4� 1 ¼ � 13
1024< 0. Thus, since f 3
4
� �
< 0 and f(1)> 0 it fol-
lows, by the Intermediate Value Theorem applied to 34; 1
� �
, that f has a zero in 34; 1
� �
.
Finally, f 78
� �
¼ 16;80732;768
� 18¼ 12;711
32;768> 0. Thus, since f 3
4
� �
< 0 and f 78
� �
> 0 it fol-
lows, by the Intermediate Value Theorem applied to 34; 7
8
� �
, that f has a zero in 34; 7
8
� �
.
This interval 34; 7
8
� �
is of length 18, as required.
4. We compile the following table of values for the continuous function p:
x �1 0 1 2
p(x) �3 1 �1 3
Since p(�1)< 0 and p(0)> 0, it follows, from the Intermediate Value Theorem, that
p has a zero in (�1, 0).
398 Appendix 4
Since p(0)> 0 and p(1)< 0, it follows, from the Intermediate Value Theorem,
that p has a zero in (0, 1).
Since p(1)< 0 and p(2)> 0, it follows, from the Intermediate Value Theorem,
that p has a zero in (1, 2).
5. For the polynomial p(x)¼ x5þ 3x4� x� 1, x2R , that is continuous on R , we have
M ¼ 1þmax 3j j; �1j j; �1j jf g ¼ 4;
so that, by the Zeros Localisation Theorem, all the zeros of p lie in (�4, 4).
We now compile a table of values of p(x), for x¼�4, �3, . . . , 4:
x �4 �3 �2 �1 0 1 2 3 4
p(x) �253 2 17 2 �1 2 77 482 1,787
By applying the Intermediate Value Theorem to p on each of the three intervals
[�4, �3], [�1, 0] and [0, 1], we deduce that p has a zero in each of the intervals
ð�4;�3Þ; ð�1; 0Þ and ð0; 1Þ:6. (a) First, we look separately at the two closed subintervals [�1, 0] and [0, 2], on
each of which f is strictly monotonic. The function f is continuous on each of
these subintervals, and so has a maximum and a minimum on each.
Since f is strictly decreasing on [�1, 0], it follows that
max f xð Þ : x 2 �1; 0½ �f g ¼ f �1ð Þ ¼ 1;
and this is attained in [�1, 0] only at � 1; also
min f xð Þ : x 2 �1; 0½ �f g ¼ f 0ð Þ ¼ 0;
and this is attained in [�1, 0] only at 0.
Since f is strictly increasing on [0, 2], it follows that
max f xð Þ : x 2 0; 2½ �f g ¼ f 2ð Þ ¼ 4;
and this is attained in [0, 2] only at 2; also
min f xð Þ : x 2 0; 2½ �f g ¼ f 0ð Þ ¼ 0;
and this is attained in [0, 2] only at 0.
Combining these results, we see that
max f xð Þ : x 2 �1; 2½ �f g ¼ 4;
and this is attained in [� 1, 2] only at 2; also
min f xð Þ : x 2 �1; 2½ �f g ¼ 0;
and this is attained in [�1, 2] only at 0.
(b) First, we look separately at the three closed subintervals 0; 12p
� �
, 12p; 3
2p
� �
and32p; 2p� �
on each of which f is strictly monotonic. The function f is continuous on
each of these subintervals, and so has a maximum and a minimum on each.
Since f is strictly increasing on 0; 12p
� �
, it follows that
max f xð Þ : x 2 0; 12p
� � �
¼ f 12p
� �
¼ 1;
and this is attained in 0; 12p
� �
only at 12p; also
min f xð Þ : x 2 0; 12p
� � �
¼ f 0ð Þ ¼ 0;
and this is attained in 0; 12p
� �
only at 0.
Since f is strictly decreasing on 12p; 3
2p
� �
, it follows that
max f xð Þ : x 2 12p; 3
2p
� � �
¼ f 12p
� �
¼ 1;
and this is attained in 12p; 3
2p
� �
only at 12p; also
min f xð Þ : x 2 12p; 3
2p
� � �
¼ f 32p
� �
¼ �1;
Solutions to the problems 399
and this is attained in 12 p; 3
2 p
� �
only at 32p .
Next, since f is strictly increasing on 32 p; 2p� �
, it follows that
max f xð Þ : x 2 32 p; 2p� � �
¼ f 2 pð Þ ¼ 0;
and this is attained in 32 p; 2p� �
only at 0; also
min f xð Þ : x 2 32p ; 2p� � �
¼ f 32p
� �
¼ �1;
and this is attained in 32 p; 2p� �
only at 32p.
Combining these results, we see that
max f xð Þ : x 2 0; 2p½ �f g ¼ 1;
and this is attained in [0, 2p] only at 12p; also
min f xð Þ : x 2 0; 2p½ �f g ¼ �1;
and this is attained in [0, 2p] only at 32p.
Section 4.3
1. (a) If 0 � x1 < x 2, then 2x 1 < 2x 2 and x14< x2
4. Hence
x 41 þ 2x 1 þ 3 < x 42 þ 2x2 þ 3;
so that f is strictly increasing on R , and is thus one–one on R .
(b) If 0 < x1 < x 2 , then x 12< x2
2 and 1x1> 1
x2, so that � 1
x1< � 1
x2: Hence
x21 �
1
x1
< x22 �
1
x2
;
it follows that f is strictly increasing on (0,1), and is thus one–one on (0,1).
2. We perform the three steps of the strategy.
1. We showed, in Problem 1(b) above, that f is strictly increasing on (0, 1).
2. The function
x 7! x2 � 1
x¼ x3 � 1
x; x 2 R � 0f g;
is a rational function, and therefore is continuous on its domain R � {0}. In
particular, f is continuous on (0,1).
3. Choose the increasing sequence {n}, which tends to1, the right-hand end-point
of (0,1). Then
f nð Þ ¼ n2 � 1
n!1 as n!1;
by the Reciprocal Rule for sequences. Thus the right-hand end-point of
J ¼ f 0;1ð Þð Þ is1.
4. Choose the decreasing sequence 1n
�
, which tends to 0, the left-hand end-point of
(0,1). Then
f1
n
�
¼ 1
n2� n! �1 as n!1;
by the Reciprocal Rule for sequences. Thus the left-hand end-point of
J ¼ f 0;1ð Þð Þ is �1.Hence f has a continuous inverse f�1: R! (0, 1), by the Inverse Function
Rule.
3. (a) Since sin 14p
� �
¼ 1ffiffi
2p and 1
4p lies in �1
2p; 1
2p
� �
, we have sin�1 1ffiffi
2p� �
¼ 14p.
Since cos 13p
� �
¼ 12, we have cos 2
3p
� �
¼ �12, and 2
3p lies in [0, p], so that
cos�1 �12
� �
¼ 23p.
Since tan 13p
� �
¼ffiffiffi
3p
and 13p lies in �1
2p; 1
2p
� �
, we have tan�1ffiffiffi
3p� �
¼ 13p.
400 Appendix 4
(b) Following the Hint, we put y¼ sin�1x. Then
cos 2 sin�1 x� �
¼ cos 2yð Þ¼ 1� 2 sin2 y
¼ 1� 2x2;
since x¼ sin y.
4. Following the Hint, we put a¼ logex and b¼ logey. Then x¼ ea and y¼ eb, so that
loge xyð Þ ¼ loge eaeb� �
¼ loge eaþb� �
¼ aþ b
¼ loge xþ loge y:
5. Let y¼ cosh�1 x, where x� 1. Then
x ¼ cosh y ¼ 12
ey þ e�yð Þ;so that
e2y � 2xey þ 1 ¼ 0:
This is a quadratic equation in ey, and so
ey ¼ x�ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � 1p
:
Both choices of þ or � give a positive expression on the right, but recall that y� 0
and so ey� 1. Since
xþffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � 1p� �
� x�ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � 1p� �
¼ 1;
we choose the þ sign, because xþffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � 1p
� 1, whereas x�ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � 1p
� 1.
Hence
y ¼ cosh�1 x ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � 1p� �
:
(The value y ¼ loge x�ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � 1p� �
gives the negative solution of the equation
cosh y¼ x.)
Section 4.4
1. (a) For x> 0, f xð Þ ¼ x� ¼ e� loge x.
Now, the functions
x 7! loge x; x 2 0;1ð Þ; and x 7!ex; x 2 R ;
are continuous; hence, by the Multiple Rule and the Composition Rule, f is
continuous.
(b) For x> 0, f xð Þ ¼ xx ¼ ex loge x.
Now, the functions
x 7! loge x; x 2 0;1ð Þ; x 7! x; x 2 R ;
and x 7! ex; x 2 R ;
are continuous; hence, by the Product Rule and the Composition Rule, f is
continuous.
2. Since ax ¼ ex loge a and ay ¼ ey loge a, we have
axay ¼ ex loge aey loge a
¼ ex loge aþy loge a
¼ e xþyð Þ loge a
¼ axþy:
Solutions to the problems 401
3. We use the inequality ex> 1þ x, for x> 0. Applying this inequality with x replaced
by xe
� �
� 1, we obtain
exeð Þ�1 > 1þ x
e� 1
� �
; for x > e;
¼ x
e¼ xe�1:
It follows that
exe > x;
so that, by Rule 5 with p¼ e
ex > xe:
Chapter 5
Section 5.1
1. (a) The function
f xð Þ ¼ x2 þ x
x
has domain R � {0}, so that f is defined on any punctured neighbourhood of 0.
Next, notice that f (x) ¼ xþ 1, for x 6¼ 0. It follows that, if {xn} lies in R � {0}
and xn! 0 as n!1, then f(xn)! 1 as n!1.
Hence
limx!0
f xð Þ ¼ 1:
(b) First, notice that
f xð Þ ¼ xj jx¼ 1; x > 0,
�1; x < 0.
�
The domain of f is R � {0}, so that f is defined on any punctured neighbour-
hood of 0.
However, the two null sequences 1n
�
and � 1n
�
both have non-zero terms,
but
limn!1
f1
n
�
¼ 1; whereas limn!1
f � 1
n
�
¼ �1:
Hence, limx!0
f xð Þ does not exist.
2. (a) The function f xð Þ ¼ffiffiffi
xp
is defined on [0, 1), which contains the point 2; it is
also continuous at 2.
Hence, by Theorem 2
limx!2
ffiffiffi
xp¼
ffiffiffi
2p
:
(b) The function f xð Þ ¼ffiffiffiffiffiffiffiffiffi
sin xp
is defined on [0, p], which contains the point p2; it is
also continuous at p2, by the Composition Rule for continuous functions.
Hence, by Theorem 2
limx!p
2
ffiffiffiffiffiffiffiffiffi
sin xp
¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
sin� p
2
�
r
¼ 1:
(c) The function f xð Þ ¼ ex
1þ xis defined on (�1,1), which contains the point 1; it is
also continuous at 1, by the Composition Rule for continuous functions.
Hence, by Theorem 2
limx!1
ex
1þ x¼ 1
2e:
402 Appendix 4
3. (a) Since
limx!0
sin x
x¼ 1 and lim
x!0
1
2þ x¼ 1
2;
we deduce from the Product Rule for limits that
limx!0
sin x
2xþ x2¼ 1
2:
(b) Let f(x)¼ x2 and g xð Þ ¼ sin xx
; then
limx!0
x2 ¼ 0 and limx!0
sin x
x¼ 1:
Also, f(x)¼ x2 6¼ 0 in R � {0}; hence, by the Composition Rule
limx!0
g f xð Þð Þ ¼ limx!0
sin x2ð Þx2
¼ 1:
(c) Let f xð Þ ¼ xsin x
and g xð Þ ¼ffiffiffi
xp
; then f is defined on the punctured neighbourhood
(�p, 0)[ (0, p) of 0, and, by the Quotient Rule for limits
limx!0
f xð Þ ¼ limx!0
sin x
x
� �1
¼ 1:
Also, g is defined on (0,1), and is continuous at 1; hence, by Theorem 2 and the
Composition Rule
limx!0
g f xð Þð Þ ¼ limx!0
x
sin x
� �12¼ g 1ð Þ ¼ 1:
4. Here
ð f � gÞðxÞ ¼f ð0Þ; x ¼ 0,
f ð1þ xÞ; x 6¼ 0,
�
¼f ð0Þ; x ¼ 0,
f ð0Þ; x ¼ �1,
f ð1þ xÞ; x 6¼ 0, �1 (so that 1 + x 6¼ 1, 0)
8
>
<
>
:
¼0; x ¼ 0,
0; x ¼ �1,
2þ ð1þ xÞ; x 6¼ 0, �1,
8
<
:
¼0; x ¼ 0, �1,
3þ x; x 6¼ 0, �1.
�
In particular, ( f� g)(0)¼ 0.
Next, limx!0
g xð Þ ¼ 1, so that
f limx!0
g xð Þ�
¼ f 1ð Þ ¼ �2:
Finally, since f(g(x))¼ 3þ x on the punctured neighbourhood (�1, 0)[ (0, 1) of 0
limx!0
f g xð Þð Þ ¼ limx!0
3þ xð Þ ¼ 3:
5. (a) The inequalities
� x2 � x2 sin1
x
�
� x2; for x 6¼ 0;
show that condition 1 of the Squeeze Rule holds, with
g xð Þ ¼ �x2 and h xð Þ ¼ x2; x 2 R :
Solutions to the problems 403
Since
limx!0
x2� �
¼ 0 and limx!0�x2� �
¼ 0;
it follows, from the Squeeze Rule, that
limx!0
x2 sin1
x
�
¼ 0:
(b) The inequalities
� xj j � x cos1
x
�
� xj j; for x 6¼ 0;
show that condition 1 of the Squeeze Rule holds, with
g xð Þ ¼ � xj j and h xð Þ ¼ xj j; x 2 R :
Since
limx!0� xj jð Þ ¼ 0 and lim
x!0xj jð Þ ¼ 0;
it follows, from the Squeeze Rule, that
limx!0
x cos1
x
�
¼ 0:
6. Let
f xð Þ ¼ sin1
x; �1� x < 0,
1þ x; 0 � x � 0.
(
Then, for n¼ 1, 2, . . ., we have
f � 1
np
�
¼ sin �npð Þ ¼ 0! 0 as n!1;
whereas
f � 1
2nþ 12
ð Þp
�
¼ sin � 2nþ 12
� �
p� �
¼ �1! �1 as n!1:
It follows that limx!0�
f xð Þ does not exist.
However, for any sequence {xn} in ( 0, 3) for which xn! 0, it follows, from the
fact that the function x 7! 1þ x is continuous on [0, 3), that
limx!0þ
f xð Þ ¼ f 0ð Þ ¼ 1:
7. (a) Since limx!0
sin xx¼ 1, we have, by Theorem 7, that
limx!0þ
sin x
x¼ 1:
Hence, by the Sum Rule
limx!0þ
sin x
xþ
ffiffiffi
xp�
¼ limx!0þ
sin x
xþ lim
x!0þ
ffiffiffi
xp
¼ 1þ 0
¼ 1:
(b) Let f xð Þ ¼ffiffiffi
xp
, x� 0, and g xð Þ ¼ sin xx
, x 6¼ 0. Then
limx!0þ
ffiffiffi
xp¼ 0 and lim
x!0
sin x
x¼ 1:
Also, f xð Þ ¼ffiffiffi
xp6¼ 0 in (0,1). Hence, by the Composition Rule for limits
limx!0þ
g f xð Þð Þ ¼ limx!0þ
sinffiffiffi
xpð Þffiffiffi
xp ¼ 1:
404 Appendix 4
8. Since 1þ x � ex � 11�x
, for jxj< 1, we have
x � ex � 1 � x
1� x; for xj j < 1; ( )
so that
1 � ex � 1
x� 1
1� x; for 0 < x < 1:
Then, since limx!0þ
1 ¼ 1 and limx!0þ
11�x¼ 1, it follows, from the Squeeze Rule, that
limx!0þ
ex � 1
x¼ 1:
Next, it follows, from the inequalities () above, that
1 � ex � 1
x� 1
1� x; for�1 < x < 0;
so that, by an argument similar to the one above, we have
limx!0�
ex � 1
x¼ 1:
Hence, by Theorem 7, we may combine these two one-sided limit results to obtain
limx!0
ex � 1
x¼ 1:
Section 5.2
1. (a) Let f (x)¼ jxj. Then f(x)> 0 for x 6¼ 0, and, since f is continuous at 0
limx!0
xj j ¼ 0:
Hence, by the Reciprocal Rule
1
f xð Þ ¼1
xj j ! 1 as x! 0:
(b) Let f(x)¼ x3� 1. Then f(x)> 0 for x2 (1,1), and, since f is continuous at 1
limx!1þ
x3 � 1� �
¼ 0:
Hence, by the Reciprocal Rule
1
f xð Þ ¼1
x3 � 1!1 as x! 1þ;
so that
� 1
f xð Þ ¼1
1� x3! �1 as x! 1þ:
(c) Let f xð Þ ¼ x3
sin x. Then f(x)> 0 for x 2 � 1
2p; 1
2p
� �
, and
limx!0
x3
sin x¼ lim
x!0
x2
sin xx
� �
¼limx!0
x2
� �
limx!0
sin xx
� �
¼ 0
1¼ 0;
by the Quotient Rule for limits.
Solutions to the problems 405
Hence, by the Reciprocal Rule
1
f xð Þ ¼sin x
x3!1 as x! 0:
2. (a) Let
f xð Þ ¼ 1
x2; x 2 R � 0f g; and g xð Þ ¼ 1
x2; x 2 R � 0f g:
Then f(x)!1 as x! 0, and g(x)!1 as x! 0; but
f xð Þ � g xð Þ ¼ 0! 0 as x! 0:
(b) Let
f xð Þ ¼ 1
x2; x 2 R � 0f g; and g xð Þ ¼ x2; x 2 R � 0f g:
Then f(x)!1 as x!1, and g(x)! 0 as x!1; but
f xð Þg xð Þ ¼ 1! 1 as x! 0:
3. (a) By the Sum Rule for limits as x!1, we have
limx!1
2x3 þ x
x3¼ lim
x!12þ 1
x2
�
¼ limx!1
2ð Þ þ limx!1
1
x2
�
¼ 2þ 0 ¼ 2:
(b) Since � 1� sin x � 1 for x2R , we have
� 1
x� sin x
x� 1
x; for x 2 0;1ð Þ:
Also
g xð Þ ¼ � 1
x! 0 as x!1
and
h xð Þ ¼ 1
x! 0 as x!1:
Hence, by the Squeeze Rule for limits as x!1 ,
f xð Þ ¼ sin x
x! 0 as x!1:
4. (a) Let f(x)¼ log ex and g(x)¼ log ex, for x2 (0,1). Then
f xð Þ ! 1 as x!1 and g xð Þ ! 1 as x!1:Hence, by the Composition Rule for limits
g f xð Þð Þ ¼ log e log exð Þ ! 1 as x!1:
(b) Let f xð Þ ¼ 1x
and g xð Þ ¼ ex
x, for x2 (0,1). Then
g f xð Þð Þ ¼ xe1x; for x 2 0;1ð Þ:
Now
f xð Þ ! �1 as x! 0�:
In order to use the Composition rule for limits, we need to determine the
behaviour of g(x) as x!�1; that is, the behaviour of g(�x) as x!1.
406 Appendix 4
Now
g �xð Þ ¼ � e�x
x¼ � 1
xex
andxex !1 as x!1:
It follows, by the Reciprocal Rule, that
g �xð Þ ! 0 as x!1:Hence, by the Composition Rule for limits,
g f xð Þð Þ ¼ xe1x ! 0 as x! 0�:
5. Let
f xð Þ ¼ 1; for all x 2 R ;
and
gðxÞ ¼ 2; x ¼ 1,
3; x 6¼ 1.
�
Then
f xð Þ ! 1 as x!1 ðso that ‘ ¼ 1Þ;and
g xð Þ ! 3 as x! 1 ðso that m ¼ 3Þ;
however as x!1, we have
g f xð Þð Þ ¼ g 1ð Þ ¼ 2
! 2 6¼ 3ð¼ mÞ:
Section 5.3
1. (a) an � 0j j < 0:1
, �1ð Þnn2
< 0:1
, 1n2 < 0:1
, n2 > 10:
It follows that j an� 0j< 0.1 for all n>X, so long as X �ffiffiffiffiffi
10p
’ 3:16:
(b) an � 0j j < 0:01
, �1ð Þnn2
< 0:01
, 1n2 < 0:01
, n2 > 100
, n > 10:
It follows that jan� 0j< 0.01 for all n>X, so long as X� 10.
(c) an � 0j j < "
, �1ð Þnn2
< "
, 1n2 < "
, n2 > 1"
, n > 1ffiffi
"p :
It follows that jan� 0 j<" for all n>X, so long as X � 1ffiffi
"p .
Solutions to the problems 407
2. (a) f xð Þ � 5j j < 0:1
, 2x� 2j j < 0:1
, x� 1j j < 0:05:
It follows that j f (x)� 5j< 0.1 whenever 0< jx� 1j<�, so long as �� 0.05.
(b) f xð Þ � 5j j < 0:01
, 2x� 2j j < 0:01
, x� 1j j < 0:005:
It follows that j f (x)� 5j< 0.01 whenever 0< jx� 1j<�, so long as �� 0.005.
(c) f xð Þ � 5j j < "
, 2x� 2j j < "
, x� 1j j <1
2":
It follows that j f (x)� 5j<" whenever 0< jx� 1j<�, so long as � � 12".
3. (a) The function f (x)¼ 5x� 2 is defined on every punctured neighbourhood of 3.
We must prove that:
for each positive number ", there is a positive number � such that
5x� 2ð Þ � 13j j < "; for all x satisfying 0 < x� 3j j < �;
that is
x� 3j j < 1
5"; for all x satisfying 0 < x� 3j j < �: ( )
Choose � ¼ 15" ; then the statement () holds.
Hence
limx!3
5x� 2ð Þ ¼ 13:
(b) The function f (x )¼ 1� 7x3 is defined on every punctured neighbourhood of 0.
We must prove that:
for each positive number ", there is a positive number � such that
1� 7x3� �
� 1
< "; for all x satisfying 0 < x� 0j j < �;
that is
7x3
< "; for all x satisfying 0 < xj j < �: ( )
Choose � ¼ffiffiffiffiffi
17"3
q
(so that 7�3¼ "); then the statement () holds.
Hence
limx!0
1� 7x3� �
¼ 1:
(c) The function f xð Þ ¼ x2 cos 1x3
� �
is defined on every punctured neighbourhood of 0.
We must prove that:
for each positive number ", there is a positive number � such that
x2 cos1
x3
�
� 0
< "; for all x satisfying 0 < x� 0j j < �;
that is
x2 cos1
x3
�
< "; for all x satisfying 0 < xj j < �: ( )
408 Appendix 4
Choose � ¼ ffiffiffi
"p
. Then, since cos 1x3
� �
� 1 for all non-zero x, it follows that, for
0< jxj<�
x2 cos1
x3
�
� x2
< �2 ¼ ";so that the statement () holds.
Hence
limx!0
x2 cos1
x3
�
¼ 0:
(d) We consider the cases that x< 0 and x> 0 separately; that is, the two one-sided
limits.
Since jxj ¼� x when x< 0, we have
limx!0�
xþ xj jx¼ lim
x!0�
0
x¼ 0;
also, since jxj ¼ x when x> 0, we have
limx!0þ
xþ xj jx¼ lim
x!0þ
2x
x¼ lim
x!0þ2ð Þ ¼ 2:
Since the two one-sided limits are not equal, it follows that limx!0
xþ xj jx
does not
exist.
4. The function f(x)¼ x3, x2R , is defined on every punctured neighbourhood of 1.
We must prove that:
for each positive number ", there is a positive number � such that
jx3� 1j<" for all x satisfying 0 < x� 1j j < �;
that is
x2 þ xþ 1
� x� 1j j < "; for all x satisfying 0 < x� 1j j < �: ( )
Choose � ¼ min 1; 17"f g, so that:
(i) for 0< jx� 1j<�, we have x2 (0, 2), and so
x2 þ xþ 1
¼ x2 þ xþ 1
< 22 þ 2þ 1 ¼ 7;
(ii) x� 1j j < 17":
It follows that
x2 þ xþ 1
� x� 1j j < 7� 1
7" ¼ "; for all x satisfying 0 < x� 1j j < �;
that is, the statement () holds.
Hence
limx!1
x3� �
¼ 1:
5. Since the limits for f and g exist, the functions f and g must be defined on some
punctured neighbourhoods (c� r1, c)[ ( c, cþ r1) and (c� r2, c)[ (c, cþ r2) of c, it
follows that the function fg is certainly defined on the punctured neighbourhood
(c� r, c)[ ( c, cþ r), where r¼min{r1, r2}.
We want to prove that:
for each positive number ", there is a positive number � such that
f xð Þg xð Þð Þ � ‘mð Þj j < "; for all x satisfying 0 < x� cj j < �: ( )
Solutions to the problems 409
In order to examine the first inequality in (), we write
f xð Þg xð Þ � ‘m as f xð Þ � ‘ð Þ g xð Þ � mð Þ þ m f xð Þ � ‘ð Þ þ ‘ g xð Þ � mð Þ;
so that
f xð Þg xð Þ � ‘mj j ¼ f xð Þ � ‘ð Þ g xð Þ �mð Þ þm f xð Þ � ‘ð Þ þ ‘ g xð Þ �mð Þj j� f xð Þ � ‘j j � g xð Þ �mj j þ mj j f xð Þ � ‘j j þ ‘j j g xð Þ �mj j: (1)
We know that, since limx!c
f xð Þ ¼ ‘, there is a positive number �1 such that
f xð Þ � ‘j j < "; for all x satisfying 0 < x� cj j < �1; (2)
and a positive number �2 such that
f xð Þ � ‘j j < 1; for all x satisfying 0 < x� cj j < �2; (3)
similarly, since limx!c
g xð Þ ¼ m, there is a positive number �3 such that
g xð Þ � mj j < "; for all x satisfying 0 < x� cj j < �3: (4)
We now choose �¼min {�1, �2, �3}. Then the statements (2), (3) and (4) hold for all
x satisfying 0< jx� cj<�, so that, from (1), we deduce
f xð Þg xð Þ � ‘mj j � f xð Þ � ‘j j � g xð Þ � mj j þ mj j f xð Þ � ‘j j þ ‘j j g xð Þ � mj j� 1� "þ mj j"þ ‘j j"¼ 1þ mj j þ ‘j jð Þ";
for all x satisfying 0< jx� cj<�.This is equivalent to the statement (*), in which the number " has been replaced
by K", where K¼ 1þ jmj þ j‘j. Since K is a constant, this is sufficient, by the
K" Lemma.
It follows that limx!c
f xð Þg xð Þ ¼ ‘m, as required.
6. Since limx!c
f xð Þ ¼ ‘ and ‘ 6¼ 0, we may take " ¼ 12‘j j in the definition of limit; it
follows that there is a positive number � such that
f xð Þ � ‘j j < 1
2‘j j for all x satisfying 0 < x� cj j < �;
in other words, for all x satisfying 0< jx� cj<�, we have
� 1
2‘j j < f xð Þ � ‘ < 1
2‘j j;
which we may rewrite in the form
‘� 1
2‘j j < f xð Þ < ‘þ 1
2‘j j: ( )
It follows, from (*), that, if ‘> 0, then
f xð Þ > ‘� 1
2‘j j
¼ ‘� 1
2‘ ¼ 1
2‘ > 0;
and, if ‘< 0, then
f xð Þ < ‘þ 1
2‘j j
¼ ‘� 1
2‘ ¼ 1
2‘ < 0:
In either case, we obtain that f(x) has the same sign as ‘ on the punctured neighbour-
hood {x: 0< jx� cj<�} of c.
410 Appendix 4
Section 5.4
1. The function f (x)¼ x3, x2R , is defined on R .
We must prove that:
for each positive number ", there is a positive number � such that
x3 � 1
< "; for all x satisfying x� 1j j < �;
that is
x2 þ xþ 1
� x� 1j j < "; for all x satisfying x� 1j j < �: ( )
Choose � ¼ min 1; 17"
�
, so that
(i) for jx� 1j<�, we have x2 (0, 2), and so
x2 þ xþ 1
¼ x2 þ xþ 1
< 22 þ 2þ 1 ¼ 7;
(ii) x� 1j j < 17":
It follows that
x2 þ xþ 1
� x� 1j j < 7� 1
7" ¼ "; for all x satisfying x� 1j j < �;
that is, the statement (*) holds.
Hence f is continuous at 1.
2. The function f xð Þ ¼ffiffiffi
xp
is defined on [0,1), and f 4ð Þ ¼ffiffiffi
4p¼ 2.
We must prove that:
for each positive number ", there is a positive number � such thatffiffiffi
xp� 2
< "; for all x satisfying x� 4j j < �: ðÞNow
ffiffiffi
xp� 2
¼ x� 4ffiffiffi
xpþ 2
¼ x� 4j jffiffiffi
xpþ 2
� 1
2x� 4j j; since
ffiffiffi
xp� 0:
So, choose �¼ ". It follows that, for all x satisfying jx� 4j<�, we have
ffiffiffi
xp� 2
� 1
2x� 4j j
<1
2�
¼ 1
2"
< ":
That is, the statement () holds.
Hence f is continuous at 4.
3. The graph of f suggests that, while f is ‘well behaved’ to the left of 2, we should
examine the behaviour of f to the right of 2. If we choose " to be any positive number
less than 14, say, then there will always be points x (with x > 2) as close as we please to
2 where f (x) is not within a distance " of f (2)¼ 1.
So, take " ¼ 14. Then, if f were continuous at 2, there would be some positive
number � such that
Solutions to the problems 411
f xð Þ � 1j j < 1
4; for all x satisfying x� 2j j < �: ( )
Now let xn ¼ 2þ 1n, for n¼ 1, 2, . . .. Then
xn ¼ 2þ 1
n
�
2 2; 2þ �ð Þ; for all n > X; where X ¼ 1
�;
but
f xnð Þ � 1j j ¼ 2þ 1
n� 1
2
�
� 1
¼ 1
2þ 1
n
65 1
4¼ ":
Thus, with this choice of ", no value of � exists such that requirement (*) holds.
This proves that f is discontinuous at 2.
4. Since f is continuous at an interior point c of I and f (c) 6¼ 0, it follows, by Theorem 2
of Sub-section 5.4.1, that there is a neighbourhood N¼ ( c� r, cþ r) of c on which
f xð Þj j > 1
2f cð Þj j: (1)
Since f (c) 6¼ 0, it follows from the inequality (1) that
f xð Þj j > 1
2f cð Þj j
> 0; for all x with x� cj j < r:
In particular, f (x) 6¼ 0 on N, so that 1f
is defined on N.
To prove that 1f
is continuous at c, we must prove that,
for each positive number ", there is a positive number � such that
1
f xð Þ �1
f cð Þ
< "; for all x satisfying x� cj j < �: ðÞ
Now
1
f xð Þ �1
f cð Þ
¼ f cð Þ � f xð Þf xð Þf cð Þ
¼ f xð Þ � f cð Þj jf xð Þj j � f cð Þj j :
(2)
Since f is continuous at c, we know that, for the given value of " in (*), there is a
positive number �1 such that
f xð Þ � f cð Þj j < "; for all x satisfying x� cj j < �1: (3)
Now choose �¼min{r, �1}, so that both (1) and (3) hold, for all x satisfying
jx� cj<�. It then follows, from (1), (2) and (3), that
1
f xð Þ �1
f cð Þ
¼ f xð Þ � f cð Þj jf xð Þj j � f cð Þj j
<"
12
f cð Þj j2
¼ 112
f cð Þj j2� ":
This is equivalent to the statement (*), in which the number " has been replaced by
K", where K ¼ 112 f cð Þj j2. Since K is a constant, this is sufficient, by the K" Lemma.
It follows that 1f
is continuous at c, as required.
412 Appendix 4
Section 5.5
1. Let " be any given positive number. We want to find a positive number � such that
f xð Þ � f cð Þj j < "; for all x and c in ½2; 3� satisfying x� cj j < �;
that is, that
x2 � c2
< "; for all x and c in ½2; 3� satisfying x� cj j < �: ( )
Now, since x2� c2¼ (x� c)(xþ c), we have
x2 � c2
¼ x� cj j � xþ cj j:But, for x and c in [2, 3], we know that
4 � xþ c � 6:
Hence, in order to prove the statement (*), it is sufficient to prove that there exists a
positive number � such that
6 x� cj j < "; for all x and c in ½2; 3� satisfying x� cj j < �: ( )In view of (**), take � ¼ 1
6" (note that this choice of � depends only on ", and
not at all on x or c). With this choice of �, it follows that if x, c2 [2, 3] and satisfy
jx� cj<�, then
6 x� cj j < 6�
¼ ":In other words, the statement (**) holds, and hence the statement (*) also holds.
Hence the choice � ¼ 16" serves our purpose.
Chapter 6
Section 6.1
1. (a) Let c be any point of R . Then, for any non-zero h, the difference quotient Q(h)
at c is
Q hð Þ ¼ f cþ hð Þ � f cð Þh
¼ cþ hð Þn� cn
h
¼ 1
hncn�1hþ 1
2n n� 1ð Þcn�2h2 þ � � � þ nchn�1 þ hn
�
n terms in the bracketð Þ
¼ ncn�1 þ 1
2n n� 1ð Þcn�2hþ � � � þ nchn�2 þ hn�1
! ncn�1 as h! 0:
It follows that f is differentiable at c, and that f 0(c)¼ ncn�1.
(b) Let f (x)¼ k, and let c be any point of R . Then, for any non-zero h, the difference
quotient Q(h) at c is
Q hð Þ ¼ f cþ hð Þ � f cð Þh
¼ k � k
h
¼ 0
! 0 as h! 0:
It follows that f is differentiable at c, and that f 0(c)¼ 0.
Solutions to the problems 413
2. Let c be any point of R , with c 6¼ 0. Then, for any non-zero h, the difference quotient
Q(h) at c is
Q hð Þ ¼ f cþ hð Þ � f cð Þh
¼ 1
h
1
cþ h� 1
c
�
¼ c� cþ hð Þh cþ hð Þc
¼ �1
cþ hð Þc
! � 1
c2as h! 0:
It follows that f is differentiable at c, and that f 0 cð Þ ¼ � 1c2.
3. To prove that f (x)¼ x4, x2R , is differentiable at 1, with derivative f 0(1)¼ 4, we
have to show that:
for each positive number ", there is a positive number � such that
x4 � 1
x� 1� 4
< "; for all x satisfying 0 < x� 1j j < �;
that is
x3 þ x2 þ x� 3
< "; for all x satisfying 0 < x� 1j j < �:
Now, x3þ x2þ x� 3¼ (x� 1)(x2þ 2xþ 3); thus, for 0<jx� 1j< 1, we have
x2 (0, 2), so that
x2 þ 2xþ 3
< 22 þ 2� 2þ 3 ¼ 11:
Next, choose � ¼ min 1; 111"
�
. With this choice of �, it follows that, for all x
satisfying 0<jx� 1j<�, we have
x3 þ x2 þ x� 3
¼ x� 1j j � x2 þ 2xþ 3
<1
11"� 11 ¼ ":
It follows that f is differentiable at 1, with derivative f 0(1)¼ 4.
4. (a) Let f xð Þ ¼ jxj12. The graph of f near 0 suggests that f is not differentiable at 0.
Then, for any non-zero h, the difference quotient Q(h) at 0 is
Q hð Þ ¼ f hð Þ � f 0ð Þh
¼ hj j12� 0
h
¼ hj j12
h:
For h> 0, it follows that Q hð Þ ¼ 1
h12
. So, in particular, if we take h ¼ 1n2 for n2N ,
we find that
Q1
n2
�
¼ n!1 as n!1:
It follows that f is not differentiable at 0.
(b) Let f(x)¼ [x], the integer part of x. The graph of f near 1 suggests that f is not
differentiable at 1.
We use the fact that
x4�1¼ x2�1� �
x2þ1� �
¼ x�1ð Þ xþ1ð Þ x2þ1� �
¼ x�1ð Þ x3þx2þxþ1� �
:
414 Appendix 4
Then, for any h with �1< h< 0, the difference quotient Q(h) at 1 is
Q hð Þ ¼ f 1þ hð Þ � f 1ð Þh
¼ 0� 1
h
¼ � 1
h:
It follows that
limh!0�
Q hð Þ ¼ limh!0�
� 1
h
�
¼ 1:Hence f is not differentiable at 1.
5.
(a) For 2> h> 0, the difference quotient for f at �2 is
Q hð Þ ¼ f �2þ hð Þ � f �2ð Þh
¼ � �2þ hð Þ2þ 4
h
¼ 4� h
! 4 as h! 0þ:
Hence f is differentiable on the right at �2, and fR0 (�2)¼ 4.
f is not differentiable at �2, since it is not defined to the left of 2.
(b) For �1< h< 1, the difference quotient for f at 0 is
Q hð Þ ¼ f hð Þ � f 0ð Þh
¼�h2�0
h; h < 0;
h4�0h; h > 0;
(
¼�h; h < 0;
h3; h > 0:
�
Thus
limh!0�
Q hð Þ ¼ 0 and limh!0þ
Q hð Þ ¼ 0;
so that fL0 (0) and fR
0 (0) both exist and equal 0.
It follows that f is differentiable at 0, and f 0(0)¼ 0.
Solutions to the problems 415
(c) The difference quotient for f at 1 is
Q hð Þ ¼ f 1þ hð Þ � f 1ð Þh
¼1þhð Þ4�1
h; �1 < h < 0,
1þhð Þ3�1
h; 0 < h < 1,
8
<
:
¼ 4þ 6hþ 4h2 þ h3; �1 < h < 0,
3þ 3hþ h2; 0 < h < 1.
�
Thus
limh!0�
Q hð Þ ¼ 4 and limh!0þ
Q hð Þ ¼ 3;
so that fL0 (1)¼ 4 and fR
0 (1)¼ 3.
It follows that f is not differentiable at 1.
(d) The difference quotient for f at 2 is
Q hð Þ ¼ f 2þ hð Þ � f 2ð Þh
¼2þhð Þ3�8
h; �1 < h < 0,
0�8h; 0 < h < 1,
(
¼12þ 6hþ h2; �1 < h < 0,
� 8h; 0 < h < 1.
(
Since limh!0þ
Q hð Þ does not exist, it follows that f is not differentiable at 2.
However f is differentiable on the left at 2, and fL0 (2)¼ 12.
6. For n2N
f1
2nþ 12
� �
p
!
¼ sin 2nþ 1
2
�
p
¼ 1! 1 as n!1:Hence, although 1
2nþ12ð Þp! 0 as n!1
f1
2nþ 12
� �
p
!
6! 0
¼ f 0ð Þ:It follows that f is not continuous at 0; and so, by Corollary 1, f is not differentiable at 0.
7. For any non-zero h, the difference quotient Q(h) at 0 is
Q hð Þ ¼ f hð Þ � f 0ð Þh
¼h2 sin 1
h
� �
h
¼ h sin1
h
�
! 0 as h! 0:
It follows that f is differentiable at 0, and that f 0(0)¼ 0.
Now, for x 6¼ 0, f 0 xð Þ ¼ 2x sin 1x� cos 1
x. Hence, if n2N
f 01
2np
�
¼ 1
npsin 2npð Þ � cos 2npð Þ
¼ 0� 1
¼ �1; as n!1:Since f 0(0) 6¼�1, it follows that f 0 is not continuous at 0.
416 Appendix 4
8. Let c be any point in R . At c, the difference quotient for f is
Q hð Þ ¼ cos cþ hð Þ � cos c
h
¼ cos c cos h� sin c sin h� cos c
h
¼ cos c� cos h� 1
h� sin c� sin h
h
! cos c� 0� sin c� 1; as h! 0;
¼ � sin c:
It follows that f is differentiable at c, and f 0(c)¼� sin c. Since c is an arbitrary point
of R , it follows that f is differentiable on R .
9. For c> 0, the different quotient for f at c is
Q hð Þ ¼ cþ hð Þ3� c3
h
¼ 1
h3c2hþ 3ch2 þ h3� �
¼ 3c2 þ 3chþ h2 ! 3c2 as h! 0;
so that f is differentiable at c and f 0(c)¼ 3c2.
Next, for c< 0, the different quotient for f at c is
Q hð Þ ¼ cþ hð Þ2� c2
h
¼ 1
h2chþ h2� �
¼ 2cþ h! 2c as h! 0;
so that f is differentiable at c and f 0(c)¼ 2c.
Finally, the difference quotient for f at 0 is
Q hð Þ ¼ f hð Þ � f 0ð Þh
¼h2�0
h; h < 0,
h3�0h; h > 0,
(
¼h; h < 0,
h2; h > 0,
�
! 0; as h! 0:
It follows that f is differentiable at 0, and that f 0(0)¼ 0.
Therefore, the function f 0 is given by the following formula
f 0 xð Þ ¼2x; x < 0,
0; x ¼ 0,
3x2; x > 0.
(
It follows that f 0 is continuous at 0, since limx!0�
f 0 xð Þ ¼ limx!0þ
f 0 xð Þ ¼ f 0 0ð Þ ¼ 0:
Section 6.2
1. (a) f 0 xð Þ ¼ 7x6 � 8x3 þ 9x2 � 5; x 2 R :
(b) f 0 xð Þ ¼ x3 � 1ð Þ2x� x2 þ 1ð Þ3x2
x3 � 1ð Þ2
¼ �x4 � 3x2 � 2x
x3 � 1ð Þ2; x 2 R � 1f g:
We assume that h is sufficientlysmall that jhj< c, so thatcþ h> 0; and hence that we areusing the correct value for f atcþ h.
We assume that h is sufficientlysmall that jhj<� c, so thatcþ h< 0; and hence that we areusing the correct value for f atcþ h.
Solutions to the problems 417
(c) f 0 xð Þ ¼ 2 cos2 x� 2 sin2 x
¼ 2 cos 2x; x 2 R :
(d) f 0 xð Þ ¼ 3þ sin x� 2 cos xð Þex � ex cos xþ 2 sin xð Þ3þ sin x� 2 cos xð Þ2
¼ ex 3� sin x� 3 cos xð Þ3þ sin x� 2 cos xð Þ2
; x 2 R :
2. f 0 xð Þ ¼ e2x þ 2xe2x ¼ e2x 1þ 2xð Þ;f 00 xð Þ ¼ 2e2x 1þ 2xð Þ þ 2e2x ¼ e2x 4þ 4xð Þ;f 000 xð Þ ¼ 2e2x 4þ 4xð Þ þ 4e2x ¼ e2x 12þ 8xð Þ:
3. In each case, we use the Quotient Rule and the derivatives of sine and cosine.
(a) Here f xð Þ ¼ sin xcos x
, so that
f 0 xð Þ ¼ cos x cos x� sin x �sin xð Þcos2 x
¼ 1
cos2 x
¼ sec2 x;
on the domain of f.
(b) Here f xð Þ ¼ 1sin x
, so that
f 0 xð Þ ¼ � cos x
sin2 x
¼ �cosec x cot x;
on the domain of f.
(c) Here f xð Þ ¼ 1cos x
, so that
f 0 xð Þ ¼ sin x
cos2 x
¼ sec x tan x;
on the domain of f.
(d) Here f xð Þ ¼ cos xsin x
, so that
f 0 xð Þ ¼ sin x � sin xð Þ � cos x cos x
sin2 x
¼ �1
sin2 x
¼ �cosec2x;
on the domain of f.
4. (a) Here f xð Þ ¼ 12
ex � e�xð Þ; x 2 R ; so that
f 0 xð Þ ¼ 1
2ex þ e�xð Þ
¼ cosh x; x 2 R :
(b) Here f xð Þ ¼ 12
ex þ e�xð Þ; x 2 R ; so that
f 0 xð Þ ¼ 1
2ex � e�xð Þ
¼ sinh x; x 2 R :
(c) Here
f xð Þ ¼ sinh x
cosh x; x 2 R ;
418 Appendix 4
so that
f 0 xð Þ ¼ cosh x cosh x� sinh x sinh x
cosh2 x
¼ 1
cosh2 x
¼ sech2x; x 2 R :
5. (a) Here
f xð Þ ¼ sinh x2� �
; x 2 R ;
so that
f 0 xð Þ ¼ 2x cosh x2� �
; x 2 R :
(b) Here
f xð Þ ¼ sin sinh 2xð Þð Þ; x 2 R ;
so that
f 0 xð Þ ¼ cos sinh 2xð Þð Þ2 cosh 2xð Þ; x 2 R :
(c) Here
f xð Þ ¼ sincos 2x
x2
�
; x 2 0;1ð Þ;
so that
f 0 xð Þ ¼ coscos 2x
x2
�
� �x22 sin 2x� 2x cos 2x
x4
�
; x 2 0;1ð Þ:
6. (a) The function
f xð Þ ¼ cos x; x 2 0; pð Þ;is continuous and strictly decreasing on (0, p), and
f 0; pð Þð Þ ¼ �1; 1ð Þ:Also, f is differentiable on (0, p), and its derivative f 0(x)¼�sin x is non-zero
there.
So f satisfies the conditions of the Inverse Function Rule.
Hence f�1¼ cos�1 is differentiable on (�1, 1); and, if y¼ f (x), then
f�1� �0
yð Þ ¼ 1
f 0 xð Þ ¼ �1
sin x:
Since sin x> 0 on (0, p), and sin2 xþ cos2x¼ 1, it follows that
sin x ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1� cos2 xp
¼ffiffiffiffiffiffiffiffiffiffiffiffiffi
1� y2p
:
Hence
f�1� �0
yð Þ ¼ �1ffiffiffiffiffiffiffiffiffiffiffiffiffi
1� y2p :
If we now replace the domain variable y by x, we obtain
f�1� �0
xð Þ ¼ �1ffiffiffiffiffiffiffiffiffiffiffiffiffi
1� x2p ; x 2 �1; 1ð Þ:
(b) The function
f xð Þ ¼ sinh x; x 2 R ;
is continuous and strictly increasing on R , and
f Rð Þ ¼ R :
Solutions to the problems 419
Also, f is differentiable on R , and its derivative f 0(x)¼ cosh x is non-zero there.
So f satisfies the conditions of the Inverse Function Rule.
Hence f�1¼ sinh�1 is differentiable on R ; and, if y¼ f (x), then
f�1� �0
yð Þ ¼ 1
f 0 xð Þ ¼1
cosh x:
Since cosh x> 0 on R , and cosh2 x¼ 1þ sinh2 x, it follows that
cosh x ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ sinh2 xp
¼ffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ y2p
:
Hence
f�1� �0
yð Þ ¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ y2p :
If we now replace the domain variable y by x, we obtain
f�1� �0
xð Þ ¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ x2p ; x 2 R :
7. If x1< x2, then x15 < x2
5; it follows that f (x1)< f (x2), so that f is strictly increasing
on R .
Since f is a polynomial function, it is continuous and differentiable on R . Also
f 0 xð Þ ¼ 5x4 þ 1 6¼ 0 on R :
Thus, f satisfies the conditions of the Inverse Function Theorem.
Now, f (0)¼�1, f (1)¼ 1 and f (�1)¼�3. Hence, by the Inverse Function Rule
f�1� �0 �1ð Þ ¼ 1
f 0 0ð Þ ¼ 1;
f�1� �0
1ð Þ ¼ 1
f 0 1ð Þ ¼1
6;
and
f�1� �0 �3ð Þ ¼ 1
f 0 �1ð Þ ¼1
6:
8. By definition,
f xð Þ ¼ xx ¼ exp x loge xð Þ:The functions x 7! x and x 7! loge x are differentiable on (0, 1), and exp is differ-
entiable on R . It follows, by the Product Rule and the Composition Rule, that f is
differentiable on (0,1), and that
f 0 xð Þ ¼ exp x loge xð Þ � loge xþ x� 1
x
�
¼ xx loge xþ 1ð Þ:
Section 6.3
1. Since f is a polynomial function, f is continuous on [�1, 2] and differentiable on
(�1, 2); also
f 0 xð Þ ¼ x3 � x2 ¼ x2 x� 1ð Þ:Thus f 0 vanishes at 0 and 1.
First, we consider the behaviour of f near 0. For x2 (�1, 1), for example
f xð Þ ¼ 1
4x3 x� 4
3
�
has the opposite sign to that of x, since x� 43< 0; thus
420 Appendix 4
f xð Þ > 0 for x 2 �1; 0ð Þand
f xð Þ < 0 for x 2 0; 1ð Þ:
Since f (0)¼ 0, it follows that 0 is not a local extremum of f.
Next, we consider the behaviour of f near 1. Now
f xð Þ � f 1ð Þ ¼ 1
4x4 � 1
3x3
�
� 1
4� 1
3
�
¼ 1
4x4 � 1� �
� 1
3x3 � 1� �
¼ 1
4x� 1ð Þ x3 þ x2 þ xþ 1
� �
� 1
3x� 1ð Þ x2 þ xþ 1
� �
¼ 1
12x� 1ð Þ 3x3 � x2 � x� 1
� �
¼ 1
12x� 1ð Þ2 3x2 þ 2xþ 1
� �
:
It follows that, for x2 (0, 2), for example
f xð Þ � f 1ð Þ � 0;
so that f has a local minimum at 1, with value f 1ð Þ ¼ � 112
.
2. Since the sine and cosine functions are continuous and differentiable on R , so also is f.
Now
f 0 xð Þ ¼ 2 sin x cos x� sin x ¼ sin x 2 cos x� 1ð Þ;
thus f 0 vanishes in 0; 12p
� �
only when cos x ¼ 12; that is, when x ¼ 1
3p.
Since f (0)¼ 1, f 12p
� �
¼ 1 and
f1
3p
�
¼ 1
2
ffiffiffi
3p� 2
þ 1
2¼ 3
4þ 1
2¼ 5
4;
it follows that, on 0; 12p
� �
:
the minimum of f is 1, and occurs when x¼ 0 and x ¼ 12p;
the maximum of f is 54, and occurs when x ¼ 1
3p.
3. Since f is a polynomial function, f is continuous on [1, 3] and differentiable on (1, 3).
Also, f (1)¼ 2 and f (3)¼ 2, so that f (1)¼ f (3). Thus f satisfies the conditions of
Rolle’s Theorem on [1, 3].
Now
f 0 xð Þ ¼ 4x3 � 12x2 þ 6x ¼ 2x 2x2 � 6xþ 3� �
;
so that f 0(x)¼ 0 when x¼ 0 and x ¼ 12
3�ffiffiffi
3p� �
(the roots of the quadratic equation
2x2� 6xþ 3¼ 0). Thus the only value of x in (1, 3) such that f 0(x)¼ 0 is
x ¼ 12
3þffiffiffi
3p
� �
, so we must take c ¼ 12
3þffiffiffi
3p
� �
’ 2:37.
4. (a) NO: f is not defined at 12p.
(b) NO: f is not differentiable at 1, and f (0) 6¼ f (2).
(c) YES: All the conditions are satisfied.
(d) NO: f 0ð Þ 6¼ f 12p
� �
.
Section 6.4
1. (a) Since f is a polynomial function, f is continuous on [�2, 2] and differentiable on
(�2, 2). Thus f satisfies the conditions of the Mean Value Theorem on [�2, 2].
Solutions to the problems 421
Nowf 0 xð Þ ¼ 3x2 þ 2;
so that c satisfies the conclusion of the theorem when
3c2 þ 2 ¼ f 2ð Þ � f �2ð Þ2� �2ð Þ ¼
12� �12ð Þ2� �2ð Þ
¼ 6;
that is, when 3c2¼ 4 so that c ¼ �ffiffi
43
q
’ �1:15.
Thus there are two possible values of c.
(b) The function exp is continuous on [0, 3] and differentiable on (0, 3). Thus f
satisfies the conditions of the Mean Value Theorem on [0, 3].
Now, f 0(x)¼ ex, so that c satisfies the conclusion of the theorem when
ec ¼ f 3ð Þ � f 0ð Þ3� 0
¼ 1
3e3 � 1� �
;
that is, when c ¼ loge13
e3 � 1ð Þ� �
’ 1:85.
2. (a) Here f 0 xð Þ ¼ 4x13 � 4 ¼ 4 x
13 � 1
� �
, so that f 0(x)> 0 on (1, 1). Hence, by the
Increasing-Decreasing Theorem, f is strictly increasing on [1,1).
(b) Here f 0 xð Þ ¼ 1� 1x, so that f 0(x)< 0 on (0, 1). Hence, by the Increasing-
Decreasing Theorem, f is strictly decreasing on (0, 1].
3. (a) Let
f xð Þ ¼ sin�1 xþ cos�1 x; x 2 �1; 1½ �:
Then f is continuous on [�1, 1] and differentiable on (�1, 1).
Now
f 0 xð Þ ¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffi
1� x2p � 1
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1� x2p ¼ 0; for x 2 �1; 1ð Þ:
It follows, from Corollary 1, that
sin�1 xþ cos�1 x ¼ c; for x 2 �1; 1½ �;for some constant c.
Putting x¼ 0, we obtain 0þ 12p ¼ c, so that c ¼ 1
2p. Hence
sin�1 xþ cos�1 x ¼ 1
2p; x 2 �1; 1½ �:
(b) Let
f xð Þ ¼ tan�1 xþ tan�1 1
x; x 2 0;1ð Þ:
Then f is continuous and differentiable on (0,1).
Now
f 0 xð Þ ¼ 1
1þ x2�
1x2
1þ 1x
� �2¼ 0; for x 2 0;1ð Þ:
It follows, from Corollary 1, that
tan�1 xþ tan�1 1
x¼ c; for x 2 0; 1ð Þ;
for some constant c.
Putting x¼ 1, we obtain 14pþ 1
4p ¼ c, so that c ¼ 1
2p. Hence
tan�1 xþ tan�1 1
x¼ 1
2p; x 2 0; 1ð Þ:
422 Appendix 4
4. We have
f 0 xð Þ ¼ 3x2 � 6x ¼ 3x x� 2ð Þ; for x 2 R :
Thus f 0(x)¼ 0 when x¼ 0 and x¼ 2, so that c¼ 0, 2.
Now
f 00 xð Þ ¼ 6x� 6; for x 2 R ;
so that f 0 0(0)¼�6< 0 and f 0 0(2)¼ 6> 0. It follows, from the Second Derivative
Test, that f has a local maximum that occurs at 0 and a local minimum that occurs at 2.
The value of the local maximum is f (0)¼ 1, and the value of the local minimum is
f (2)¼�3.
5. (a) Let
f xð Þ ¼ x� sin x; x 2 0;1
2p
� �
:
Then f is continuous on 0; 12p
� �
and differentiable on 0; 12p
� �
.
Now
f 0 xð Þ ¼ 1� cos x > 0; for x 2 0;1
2p
�
;
and f (0)¼ 0.
It follows that f (x)� 0 for x 2 0; 12p
� �
, so that
x � sin x; for x 2 0;1
2p
� �
:
(b) Let
f xð Þ ¼ 2
3xþ 1
3� x
23; x 2 0; 1½ �:
Then f is continuous on [0, 1] and differentiable on (0, 1).
Now
f 0 xð Þ ¼ 2
3� 2
3x�
13
¼ 2
31� x�
13
� �
< 0; for x 2 0; 1ð Þ;
and f 1ð Þ ¼ 23þ 1
3� 1 ¼ 0.
It follows that f (x)� 0 for x2 [0, 1], so that
2
3xþ 1
3� x
23; for x 2 0; 1½ �:
Section 6.5
1. Here
f xð Þ ¼ x3 þ x2 sin x and g xð Þ ¼ x cos x� sin x on 0; p½ �:Since polynomial functions and the sine and cosine functions are continuous and
differentiable on R , it follows, by the Combination Rules, that f and g are continuous
on [0, p] and differentiable on (0, p). It follows that Cauchy’s Mean Value Theorem
applies to the functions f and g on [0, p].
Now, g(0)¼ 0� 1� 0¼ 0 and g(p)¼ pcosp� sinp¼ � p, so that g(p) 6¼ g(0).
Also
f 0 xð Þ ¼ 3x2 þ 2x sin xþ x2 cos x
¼ x2 3þ cos xð Þ þ 2x sin x
Solutions to the problems 423
and
g0 xð Þ ¼ cos x� x sin x� cos x
¼ �x sin x
on (0, p). In particular, g0(x) 6¼ 0 on (0, p).
It therefore follows from Cauchy’s Mean Value Theorem that there exists at
least one point c in (0, p) for which
c2 3þ cos cð Þ þ 2c sin c
�c sin c¼ p3ð Þ � 0ð Þ�pð Þ � 0ð Þ ;
so thatc 3þ cos cð Þ þ 2 sin c
� sin c¼ �p2:
By cross-multiplying, we obtain
c 3þ cos cð Þ þ 2 sin c ¼ p2 sin c;
so that
3c ¼ p2 � 2� �
sin c� c cos c:
Hence the equation 3x¼ (p2� 2)sinx� xcosx has at least one root in (0, p).
2. (a) Letf xð Þ ¼ sinh 2x and g xð Þ ¼ sin 3x; for x 2 R :
Then f and g are differentiable on R , and
f 0ð Þ ¼ g 0ð Þ ¼ 0;
so that f and g satisfy the conditions of l’Hopital’s Rule at 0.
Now
f 0 xð Þ ¼ 2 cosh 2x and g0 xð Þ ¼ 3 cos 3x;
so that
f 0 xð Þg0 xð Þ ¼
2 cosh 2x
3 cos 3x:
It follows, from l’Hopital’s Rule, that the desired limit limx!0
f xð Þg xð Þ exists and equals
limx!0
f 0 xð Þg0 xð Þ ¼ lim
x!0
2 cosh 2x
3 cos 3x;
provided that this last limit exists.
But, by the Combination Rules for continuous functions, the function
x 7! 2 cosh 2x
3 cos 3x
is continuous at 0, so that
limx!0
f 0 xð Þg0 xð Þ ¼
f 0 0ð Þg0 0ð Þ
¼ 2
3:
It follows, from l’Hopital’s Rule, that the original limit exists, and that its
value is 23.
(b) Let
f xð Þ ¼ 1þ xð Þ15� 1� xð Þ
15
and
g xð Þ ¼ 1þ 2xð Þ25� 1� 2xð Þ
25;
for x 2 � 12; 1
2
� �
:
424 Appendix 4
Then f and g are differentiable on � 12; 1
2
� �
, and
f 0ð Þ ¼ g 0ð Þ ¼ 0;
so that f and g satisfy the conditions of l’Hopital’s Rule at 0.
Now
f 0 xð Þ ¼ 1
51þ xð Þ�
45 þ 1
51� xð Þ�
45
and
g0 xð Þ ¼ 4
51þ 2xð Þ�
35 þ 4
51� 2xð Þ�
35;
so that
f 0 xð Þg0 xð Þ ¼
15
1þ xð Þ�45 þ 1
51� xð Þ�
45
45
1þ 2xð Þ�35 þ 4
51� 2xð Þ�
35
:
It follows, from l’Hopital’s Rule, that the desired limit limx!0
f xð Þg xð Þ exists and equals
limx!0
f 0 xð Þg0 xð Þ ¼ lim
x!0
15
1þ xð Þ�45 þ 1
51� xð Þ�
45
45
1þ 2xð Þ�35 þ 4
51� 2xð Þ�
35
;
provided that this last limit exists.
But, by the Combination Rules and the Power Rule for continuous functions,
the functions f 0 and g0 are continuous at 0; hence, by the Quotient Rule
limx!0
f 0 xð Þg0 xð Þ ¼
f 0 0ð Þg0 0ð Þ
¼15þ 1
545þ 4
5
¼ 1
4:
It follows, from l’Hopital’s Rule, that the original limit exists, and equals 14.
(c) Let
f xð Þ ¼ sin x2 þ sin x2� �� �
and g xð Þ ¼ 1� cos 4x; for x 2 R :
Then f and g are differentiable on R , and
f 0ð Þ ¼ g 0ð Þ ¼ 0;
so that f and g satisfy the conditions of l’Hopital’s Rule at 0.
Now
f 0 xð Þ ¼ cos x2 þ sin x2� �� �
� 2xþ 2x cos x2� �� �
and
g0 xð Þ ¼ 4 sin 4x;
so that
f 0 xð Þg0 xð Þ ¼
cos x2 þ sin x2ð Þð Þ � 2xþ 2x cos x2ð Þð Þ4 sin 4x
:
It follows, from l’Hopital’s Rule, that the limit limx!0
f xð Þg xð Þ exists and equals
limx!0
f 0 xð Þg0 xð Þ ; ( )
provided that this last limit () exists.
Next, f 0(0)¼ 0 and g0(0)¼ 0, and f 0 and g0 are differentiable on R . Thus f 0 and
g0 satisfy the conditions of l’Hopital’s Rule at 0.
It follows, from l’Hopital’s Rule, that the limit limx!0
f 0 xð Þg0 xð Þ exists and equals
Solutions to the problems 425
limx!0
f 00 xð Þg00 xð Þ ; ()
provided that this last limit () exists.
Now
f 00 xð Þ ¼ �sin x2 þ sin x2� �� �
� 2xþ 2x cos x2� �� �2
þ cos x2 þ sin x2� �� �
� 2þ 2 cos x2� �
� 4x2 sin x2� �� �
and
g00 xð Þ ¼ 16 cos 4x:
But, by the Combination Rules for continuous functions, the functions f 0 0 and g0 0
are continuous at 0, so that
limx!0
f 00 xð Þg00 xð Þ ¼
f 00 0ð Þg00 0ð Þ
¼ 4
16¼ 1
4:
It follows, from l’Hopital’s Rule, that the limit () exists, and that its value is 14.
It then follows, from a second application of l’Hopital’s Rule, that the original
limit exists, and that its value is 14.
(d) Let
f xð Þ ¼ sin x� x cos x and g xð Þ ¼ x3; for x 2 R :
Then f and g are differentiable on R , and
f 0ð Þ ¼ g 0ð Þ ¼ 0;
so that f and g satisfy the conditions of l’Hopital’s Rule at 0.
Now
f 0 xð Þ ¼ x sin x and g0 xð Þ ¼ 3x2;
so that
f 0 xð Þg0 xð Þ ¼
x sin x
3x2
¼ 1
3� sin x
x:
It follows, from l’Hopital’s Rule, that the desired limit limx!0
f xð Þg xð Þ exists and equals
limx!0
f 0 xð Þg0 xð Þ ¼ lim
x!0
1
3� sin x
x
�
¼ 1
3� lim
x!0
sin x
x
�
;
provided that this last limit exists.
But limx!0
sin xx
� �
¼ 1, so that the limit limx!0
f 0 xð Þg0 xð Þ does exist and equals 1
3. It follows,
from l’Hopital’s Rule, that the original limit also exists, and that its value is 13.
Chapter 7
Section 7.1
1. In this case, the function f is continuous, except at 12
and 1.
On the three subintervals 0; 13
� �
, 13; 3
4
� �
and 34; 1
� �
in P, we have
426 Appendix 4
m1 ¼ f 0ð Þ ¼ 0; M1 ¼ f1
3
�
¼ 2
3; �x1 ¼
1
3� 0 ¼ 1
3;
m2 ¼ f1
2
�
¼ 0; M2 ¼ f3
4
�
¼ 3
2; �x2 ¼
3
4� 1
3¼ 5
12;
m3 ¼ f 1ð Þ ¼ 1; M3 ¼ 2�
¼ limx!1�
f xð Þ�
; �x3 ¼ 1� 3
4¼ 1
4:
It then follows, from the definitions of L( f, P) and U( f, P), that
L f ;Pð Þ ¼X
3
i¼1
mi�xi ¼ m1�x1 þ m2�x2 þ m3�x3
¼ 0� 1
3þ 0� 5
12þ 1� 1
4
¼ 0þ 0þ 1
4¼ 1
4;
U f ;Pð Þ ¼X
3
i¼1
Mi�xi ¼ M1�x1 þM2�x2 þM3�x3
¼ 2
3� 1
3þ 3
2� 5
12þ 2� 1
4
¼ 2
9þ 5
8þ 1
2¼ 97
72:
2. In this case, the function f is increasing and continuous on [0, 1]. Thus, on each
subinterval in [0, 1], the infimum of f is the value of f at the left end-point of
the subinterval and the supremum of f is the value of f at the right end-point of the
subinterval.
Hence, on the ith subinterval i�1n; i
n
� �
in Pn, for 1� i� n, we have
mi ¼ fi� 1
n
�
¼ i� 1
n
� 2
; Mi ¼ fi
n
�
¼ i
n
� 2
; and
�xi ¼i
n� i� 1
n¼ 1
n:
It then follows, from the definitions of L( f, Pn) and U( f, Pn), that
L f ;Pnð Þ ¼X
n
i¼1
mi�xi ¼X
n
i¼1
i� 1
n
� 2
� 1
n
¼ 1
n3
X
n
i¼1
i2 � 2iþ 1� �
¼ 1
n3
n nþ 1ð Þ 2nþ 1ð Þ6
� 2n nþ 1ð Þ
2þ n
� �
¼ 1
n2
nþ 1ð Þ 2nþ 1ð Þ6
� nþ 1ð Þ þ 1
� �
¼ 1
6n22n2 þ 3nþ 1� �
� 6 nþ 1ð Þ þ 6 �
¼ 1
6n22n2 � 3nþ 1 �
¼ 1
3� 1
2nþ 1
6n2;
U f ;Pnð Þ ¼X
n
i¼1
Mi�xi ¼X
n
i¼1
i
n
� 2
� 1
n
¼ 1
n3
X
n
i¼1
i2
¼ 1
n3� n nþ 1ð Þ 2nþ 1ð Þ
6
¼ 2n2 þ 3nþ 1
6n2¼ 1
3þ 1
2nþ 1
6n2:
x134
12
0
1
2
y = f (x)
y
13
x
1
1
y
y = f (x)
Solutions to the problems 427
It follows that
limn!1
L f ; Pnð Þ ¼ limn!1
1
3 � 1
2n þ 1
6n2
�
¼ 1
3
and
limn!1
U f ; Pnð Þ ¼ limn!1
1
3 þ 1
2n þ 1
6n2
�
¼ 1
3:
3. By definition of least upper bound, we know that f xð Þ � supx 2J
f , for x 2 J . It follows,
from the fact that I J , that
f xð Þ � supx 2J
f ; for x 2 I :
Hence supx2 J
f is an upper bound for f on I, so that supx2 I
f � supx 2J
f .
4. In Problem 2, we showed that for the function f ( x ) ¼ x 2 , x 2 [0, 1], and the partition
Pn ¼ 0 ; 1n
� �
; 1n ; 2
n
� �
; . . .; i�1n; i
n
� �
; . . .; 1 � 1n ; 1
� � �
of [0, 1]
limn!1
L f ; Pnð Þ ¼ 1
3and lim
n!1U f ; Pnð Þ ¼ 1
3:
It follows thatZ
�
1
0
f � 1
3and
Z
�1
0
f � 1
3:
ButZ
�
1
0
f �Z
�1
0
f ; by part ðbÞ of Theorem 4:
It follows that we must haveR
�
1
0 f ¼
R
�1
0f ¼ 1
3, so that f is integrable on [0, 1] and
R 1
0 f ¼ 1
3 .
5. Here the function f ( x ) = k , x 2 [0, 1], is the constant function on [0, 1].
Hence, on the i th subinterval i�1n; i
n
� �
in Pn , for 1 � i � n, we have
mi ¼ k ; M i ¼ k ; and � x i ¼i
n � i � 1
n¼ 1
n:
It then follows, from the definitions of L( f , Pn) and U ( f , Pn ), that
L f ; Pnð Þ ¼X
n
i¼1
mi � x i ¼X
n
i ¼1
k � 1
n
¼ kX
n
i¼1
1
n ¼ k ;
U f ; Pnð Þ ¼X
n
i¼ 1
Mi � x i ¼X
n
i¼1
k � 1
n
¼ kX
n
i ¼1
1
n ¼ k :
It follows thatZ
�
1
0
f � k and
Z
�1
0
f � k :
ButZ
�
1
0
f �Z
�1
0
f ; by part ðbÞ of Theorem 4:
It follows that we must haveR
�1
0f ¼
R
�1
0f ¼ k, so that f is integrable on [0, 1] and
R 1
0f ¼ k:
428 Appendix 4
6. (a) Let Pn ¼ x0; x1½ �; x1; x2½ �; . . .; xi�1; xi½ �; . . .; xn�1; xn½ �f g be a standard partition of
[0, 1].
The function f is constant on the interval [0, 1), so that, for this partition Pn,
we have
mi ¼ �2 for all i;
Mi ¼�2; 1 � i � n� 1;
3; i ¼ n;
�
�xi ¼1
n:
It follows that
L f ;Pnð Þ ¼X
n
i¼1
mi�xi ¼X
n
i¼1
�2ð Þ � 1
n
¼ �2X
n
i¼1
1
n¼ �2;
and
U f ;Pnð Þ ¼X
n
i¼1
Mi�xi ¼X
n�1
i¼1
Mi�xi þMn�xn
¼X
n�1
i¼1
�2ð Þ� 1
nþ 3� 1
n
¼ �2X
n�1
i¼1
1
nþ 3
n
¼ �2n� 1
nþ 3
n¼ �2þ 5
n:
Thus, in particular
U f ;Pnð Þ � L f ;Pnð Þ ¼ 5
n:
Now let " be any given positive number. It follows that, if we choose n such that5n< " (that is, if we choose n > 5
" ), then
U f ;Pnð Þ � L f ;Pnð Þ ¼ 5
n
�
< ":
It follows, from Riemann’s Criterion for integrability, that f is integrable on [0, 1].
Alternatively, let P¼ {[x0, x1], [x1, x2], . . ., [xi�1, xi], . . ., [xn�1, xn]} be any
partition of [0, 1] . Then, since Mi ¼ mi for 1� i� n�1, we have
U f ;Pð Þ � L f ;Pð Þ ¼X
n
i¼1
Mi � mið Þ�xi ¼ Mn � mnð Þ�xn
¼ 3� �2ð Þð Þ�xn ¼ 5�xn: ()
Now let " be any given positive number, and choose P to be any partition of [0, 1]
such its mesh, jjPjj, is less than "5; then, in particular, we have �xn <
"5. It then
follows, from the inequality (), that
U f ;Pð Þ � L f ;Pð Þ ¼ 5�xn
< 5� "5¼ ":
It follows, from Riemann’s Criterion for integrability, that f is integrable
on [0, 1].
3
2
1
1 x
y
–1
–2
Solutions to the problems 429
(b) Let P¼ {[x0, x1], [x1, x2], . . ., [xi�1, xi], . . ., [xn�1, xn]} be any partition of [0, 1].
xi–1 xi
1
y
10 x
y = 1, x ∈
y = 0, x ∉
Then, for the partition P, we have
mi ¼ 0 and Mi ¼ 1;
for all i.
It follows that
L f ;Pð Þ ¼X
n
i¼1
mi�xi ¼X
n
i¼1
0� �xi ¼ 0;
and
U f ;Pð Þ ¼X
n
i¼1
Mi�xi ¼X
n
i¼1
1� �xi
¼X
n
i¼1
�xi ¼ 1:
Thus, for ALL partitions P of [0, 1], we have
U f ;Pð Þ � L f ;Pð Þ ¼ 1:
In particular, it follows that, for any positive " for which 0<"< 1, there is NO
partition P of [0, 1] for which
U f ;Pð Þ � L f ;Pð Þ < ":
It follows, from Riemann’s Criterion for integrability, that f is not integrable
on [0, 1].
Section 7.2
1. First, we want to use the Strategy to prove that
infx2I
k þ f xð Þf g ¼ k þ infx2I
f xð Þf g:
Since f is bounded on I, infx2I
f xð Þf g exists. In particular
f xð Þ � infx2I
f xð Þf g; for all x 2 I:
It follows that
k þ f xð Þ � k þ infx2I
f xð Þf g; for all x 2 I;
so that k þ infx2I
f xð Þf g is a lower bound for kþ f (x) on I.
Next, let m0 be any number greater than k þ infx2I
f xð Þf g. It follows that
m0 � k > infx2I
f xð Þf g;
430 Appendix 4
so that, in particular, there is some number x in I such that
m0 � k > f xð Þ:We may reformulate this fact as: there is some number x in I such that
m0 > k þ f xð Þ;so that m0 is not a lower bound for kþ f (x) on I.
It follows that the greatest lower bound for kþ f (x) on I is k þ infx2I
f xð Þf g; in other
wordsinfx2I
k þ f xð Þf g ¼ k þ infx2I
f xð Þf g:
Next, we want to use the Strategy to prove that
supx2I
k þ f xð Þf g ¼ k þ supx2I
f xð Þf g:
Since f is bounded on I, supx2I
f xð Þf g exists. In particular
f xð Þ � supx2I
f xð Þf g; for all x 2 I:
It follows that
k þ f xð Þ � k þ supx2I
f xð Þf g; for all x 2 I;
so that k þ supx2I
f xð Þf g is an upper bound for kþ f (x) on I.
Next, let M0 be any number less than k þ supx2I
f xð Þf g. It follows that
M0 � k < supx2I
f xð Þf g;
so that, in particular, there is some number x in I such that
M0 � k < f xð Þ:We may reformulate this fact as: there is some number x in I such that
M0 < k þ f xð Þ;so that M0 is not an upper bound for kþ f (x) on I.
It follows that the least upper bound for kþ f (x) on I is k þ supx2I
f xð Þf g; in other
wordssupx2I
k þ f xð Þf g ¼ k þ supx2I
f xð Þf g:
2. The function f is decreasing on the interval (�2, 0], so that
infx2ð�2;0�
f ¼ f 0ð Þ ¼ 0 and supx2ð�2;0�
f ¼ limx!�2
þf xð Þ ¼ 4;
f is increasing on the interval [0, 3], so that
infx2 0;3½ �
f ¼ f 0ð Þ ¼ 0 and supx2 0;3½ �
f ¼ f 3ð Þ ¼ 9:
Since I¼ (�2, 0] [ [0, 3], it follows that
infI
f ¼ f 0ð Þ ¼ 0 and supI
f ¼ maxn
f 3ð Þ; limx!�2
þf xð Þ
o
¼ 9:
Since 0� f (x)� 9 and 0� f (y)� 9, so that�9 �� f (y)� 0, we have
� 9 � f xð Þ � f yð Þ � 9:
We will prove that infx;y2I
f xð Þ � f yð Þf g ¼ �9 and supx;y2I
f xð Þ � f yð Þf g ¼ 9.
First, since f (x)� f (y) ��9, �9 is a lower bound for f (x)� f (y), for x, y2 I. To
prove that �9 is the greatest lower bound, we now need to prove that:
for each positive number ", there are X and Y in I = (�2, 3] for which
f Xð Þ � f Yð Þ < �9þ ":
Solutions to the problems 431
Now, by the definition of infimum and supremum on I, we know that, since 12" > 0,
there exist X and Y in (�2, 3] such that
f Xð Þ < 12" and f Yð Þ > 9� 1
2":
It follows from these two inequalities that
f Xð Þ � f Yð Þ < 12"
� �
þ �9þ 12"
� �
¼ �9þ ":
It follows that infx;y2I
f xð Þ � f yð Þf g ¼ �9.
Finally, since f (x)� f (y)� 9, 9 is an upper bound for f (x)� f (y), for x, y2 I. To
prove that 9 is the least upper bound, we now need to prove that:
for each positive number ", there are X and Y in I¼ (�2, 3] for which
f Xð Þ � f Yð Þ > 9� ":
Now, by the definition of infimum and supremum on I, we know that, since 12" > 0,
there exist X and Y in (�2, 3] such that
f Xð Þ > 9� 12" and f Yð Þ < 1
2":
It follows from these two inequalities that
f Xð Þ � f Yð Þ > 9� 12"
� �
þ �12"
� �
¼ 9� ":
It follows that supx;y2I
f xð Þ � f yð Þf g ¼ 9.
3. (a) For any x in I, f xð Þ � supI
f , so that, since l> 0
lf xð Þ � l supI
f ; for all x 2 I:
Thus
supx2I
lf xð Þð Þ � l supI
f :
To prove that supx2I
lf xð Þð Þ ¼ l�
supI
f�
, we now need to prove that:
for each positive number ", there is some X in I for which
lf Xð Þ > l�
supI
f�
� ": ( )
Now, since "> 0 and l> 0, we have "l > 0. It follows, from the definition of
supremum of f on I, that there exists an X in I for which
f Xð Þ > supI
f � "l:
Multiplying both sides by the positive number l, we obtain the desired result ().(b) For any x in I, f xð Þ � inf
If so that
� f xð Þ � � infI
f ; for all x 2 I:
Thus
supx2I
�f xð Þð Þ � � infI
f :
To prove that supx2I
�f xð Þð Þ ¼ � infI
f , we now need to prove that:
for each positive number ", there is some X in I for which
� f Xð Þ > � infI
f � ": ( )
For example, for any number Xin (�2, 3] with Xj j � 1
2
ffiffiffi
"p
, wehave
f Xð Þ ¼ X2 � 1
4";
we could, for instance, chooseX ¼ 1
2
ffiffiffi
"p
.
432 Appendix 4
Now, since "> 0 it follows, from the definition of infimum of f on I, that there
exists an X in I for which
f Xð Þ < infI
f þ ";so that
� f Xð Þ > � infI
f � ":
This is precisely the desired result ().
Section 7.3
1. (a) F0 xð Þ ¼ xþffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � 4p� ��1
� 1þ 1
2x2 � 4� ��1
2 2xð Þ�
¼ xþffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 � 4p� ��1
� x2 � 4� ��1
2� x2 � 4� �
12þx
� �
¼ x2 � 4� ��1
2:
(b) F0 xð Þ ¼ 1
2
ffiffiffiffiffiffiffiffiffiffiffiffiffi
4� x2p
þ 1
2x� 1
24� x2� ��1
2 �2xð Þ þ 2 1� 1
4x2
� �121
2
¼ 4� x2� ��1
21
24� x2� �
� 1
2x2 þ 2
�
¼ 4� x2� ��1
2 4� x2� �
¼ 4� x2� �
12:
2. (a) From the Fundamental Theorem of Calculus and the Table of Standard
Primitives, we obtainZ 4
0
x2 þ 9� �
12dx ¼ 1
2x x2 þ 9� �
12 þ 9
2loge xþ x2 þ 9
� �
12
� �
� �4
0
¼ 10þ 9
2loge 9� 9
2loge 3
¼ 10þ 9
2loge 3:
(b) From the Fundamental Theorem of Calculus and the Table of Standard
Primitives, we obtainZ e
1
loge xdx ¼ x loge x� x½ �e1
¼ e� eð Þ � 0� 1ð Þ¼ 1:
3. Using the Table of Standard Primitives and the Combination Rules, we obtain the
following primitives:
(a) F xð Þ ¼ 4 x loge x� xð Þ � tan�1 x
2
� �
;
(b) F xð Þ ¼ 23
loge sec 3xð Þ þ 15e2x 2 sin x� cos xð Þ:
4. (a) Here we use integration by parts. Let
f xð Þ ¼ loge x and g0 xð Þ ¼ x13;
so that
f 0 xð Þ ¼ 1
xand g xð Þ ¼ 3
4x
43:
Solutions to the problems 433
It follows thatZ
x13 loge x ¼ 3
4x
43 loge x �
Z
x �13
4 x
43dx
¼ 3
4x
43 loge x � 3
4
Z
x13dx
¼ 3
4x
43 loge x � 9
16x
43:
(b) Here we use integration by parts, twice. On each occasion we differentiate
the power function and integrate the trigonometric function.
HenceZ p
2
0
x 2 cos xdx ¼ x 2 sin x� �
p2
0�Z p
2
0
2x sin xdx
¼ 1
4p 2 � 2
Z p2
0
x sin xdx ;
andZ p
2
0
x sin xdx ¼ x � �cos xð Þ½ �p2
0 �Z p
2
0
� cos xð Þdx
¼ 0 þ sin x½ �p2
0
¼ 1:
It follows thatZ p
2
0
x 2 cos xdx ¼ 1
4p 2 � 2:
5. (a) We follow the strategy just before Example 3, with x in place of t and u in
place of x .
Let u ¼ g( x ) ¼ 2sin 3x , x 2R . Then
du
dx ¼ 6 cos 3 x; so that du ¼ 6 cos 3xdx:
Hence Z
sin 2 sin 3xð Þ cos 3xdx ¼Z
1
6sin udu
¼ �1
6cos u
¼ �1
6cos 2 sin 3xð Þ:
(b) We follow the strategy just before Example 3, with x in place of t and u in place of x.
Let u ¼ g xð Þ ¼ e x , x 2R . The function g is one–one on R . Then
du
dx ¼ ex ; so that du ¼ exdx;
also
when x ¼ 0; then u ¼ 1;
when x ¼ 1; then u ¼ e:
HenceZ 1
0
ex
1þ exð Þ2dx ¼
Z e
1
du
1þ uð Þ2
¼ �1
1þ u
� �e
1
¼ � 1
1þ eþ 1
2
¼ e� 1
2 1þ eð Þ :
434 Appendix 4
6. (a) We follow the strategy just before Example 4.
Let t ¼ g ðxÞ ¼ ðx � 1Þ12; x 2 ð1;1Þ. The function g is one–one on (1, 1).
Then t2 ¼ x � 1, so that x ¼ t2 þ 1. It follows that
dx
dt¼ 2t ; so that dx ¼ 2 tdt :
HenceZ
dx
3 x � 1ð Þ32þ x x� 1ð Þ
12
¼Z
2 tdt
3t 3 þ t 2 þ 1ð Þt
¼Z
2dt
4t 2 þ 1
¼ tan�1 2tð Þ
¼ tan�1 2 x � 1ð Þ12
� �
:
(b) We follow the strategy just before Example 4.
Let t ¼ g xð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ exp
, x2 [0,1). The function g is one–one on [0,1).
Then t2¼ 1þ ex, so that ex¼ t2�1 and x¼ loge (t2�1). It follows that:
dx
dt¼ 2t
t2 � 1; so that dx ¼ 2t
t2 � 1dt;
also:
when x¼ 0, then t ¼ffiffiffi
2p
,
when x¼ loge 3, then t¼ 2.
HenceZ loge 3
0
exffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ exp
dx ¼Z 2
ffiffi
2p t2 � 1� �
� t � 2t
t2 � 1dt
¼Z 2
ffiffi
2p 2t2dt
¼ 2
3t3
� �2
ffiffi
2p
¼ 16� 4ffiffiffi
2p
3:
Section 7.4
1. Since
x sin1
x10
�
� x; for x 2 1; 3½ �;
it follows, from part (a) of Theorem 2, thatZ 3
1
x sin1
x10
�
dx �Z 3
1
xdx
¼ 1
2x2
� �3
1
¼ 1
29� 1ð Þ¼ 4:
2. Since
ex2� 1
1� x2
� 1
1� 1
4
¼ 4
3; for x 2 0;
1
2
� �
;
Solutions to the problems 435
it follows, from part (b) of Theorem 2, thatZ 1
2
0
ex2
dx � 4
3
1
2� 0
�
¼ 2
3:
(Remark We could obtain a smaller upper estimate for the integral by applying
part (a) of Theorem 2 to the inequality ex2� 11�x2.)
Next, since
ex2�1þ x2 � 1; for x 2 0;1
2
� �
;
it follows, from part (b) of Theorem 2, thatZ 1
2
0
ex2
dx � 1� 1
2� 0
�
¼ 1
2:
3. (a) Since
sin1
x
�
� 1; for x 2 1; 4½ �;
and
2þ cos1
x
�
� 1; for x 2 1; 4½ �;
it follows that
sin 1x
� �
2þ cos 1x
� �
� 1; for x 2 1; 4½ �:
It follows, from Theorem 3, thatZ 4
1
sin 1x
� �
2þ cos 1x
� �
� 1� 4� 1ð Þ
¼ 3:
(b) Since
tan x � 0 for x 2 0;p4
h i
;
and
3� sin x2� �
� 2; for x 2 0;p4
h i
;
it follows that
tan x
3� sin x2ð Þ �1
2tan x; for x 2 0;
p4
h i
:
It follows, from Theorems 2 and 3, thatZ p
4
0
tan x
3� sin x2ð Þ dx
�Z p
4
0
tan x
3� sin x2ð Þ
dx
�Z p
4
0
1
2tan xdx
¼ 1
2loge sec xð Þ
� �p4
0
¼ 1
2loge sec
p4
� �
� loge 1� �
¼ 1
2loge
ffiffiffi
2p¼ 1
4loge 2:
436 Appendix 4
4. (a) I0 ¼Z 1
0
exdx
¼ ex½ �10 ¼ e� 1:
(b) Using integration by parts, we obtain
In ¼Z 1
0
exxndx
¼ exxn½ �10 �Z 1
0
exnxn�1dx
¼ e� nIn�1:
(c) Using the result of part (b), with n¼ 1, 2, 3 and 4 in turn, we obtain
I1 ¼ e� I0 ¼ e� e� 1ð Þ ¼ 1;
I2 ¼ e� 2I1 ¼ e� 2;
I3 ¼ e� 3I2 ¼ e� 3 e� 2ð Þ ¼ 6� 2e;
I4 ¼ e� 4I3 ¼ e� 4 6� 2eð Þ ¼ 9e� 24:
5. (a) a1 ¼2
1� 23¼ 4
3;
a2 ¼2
1� 23� 43� 45¼ 64
45;
a3 ¼2
1� 23� 43� 45� 65� 67¼ 256
175;
b1 ¼1!ð Þ222
2!ffiffiffi
1p ¼ 2;
b2 ¼2!ð Þ224
4!ffiffiffi
2p ¼ 4
3
ffiffiffi
2p
;
b3 ¼3!ð Þ226
6!ffiffiffi
3p ¼ 16
15
ffiffiffi
3p
:
(b) b21 ¼ 4 and 3a1 ¼ 4;
b22 ¼
32
9and
5
2a2 ¼
32
9;
b23 ¼
256
75and
7
3a3 ¼
256
75:
(c) We know that
b2n ¼
n!ð Þ424n
2nð Þ!ð Þ2n:
We now express an in a similar way, tackling the numerator and denominator
separately.
The numerator is a product of 2n even numbers; so, taking a factor 2 from
each term, we obtain
2:2:4:4: . . . : 2nð Þ: 2nð Þ ¼ 22n 1:1:2:2: . . . :n:nð Þ¼ 22n n!ð Þ2:
The denominator of an cannot be tackled in quite the same way, as all its factors
are odd. However, we can relate it to factorials by introducing the missing
even factors:
Solutions to the problems 437
1:3:3:5:5: . . . : 2n� 1ð Þ: 2nþ 1ð Þ ¼ 1:2:2:3:3:4:4: . . . : 2n� 1ð Þ: 2nð Þ: 2nð Þ: 2nþ 1ð Þ2:2: 4:4: . . . : 2nð Þ: 2nð Þ
¼ 2nð Þ!ð Þ2 2nþ 1ð Þ22n n!ð Þ2
:
It follows that
an ¼22n n!ð Þ22nð Þ!ð Þ2 2nþ1ð Þ
22n n!ð Þ2
¼ 24n n!ð Þ4
2nð Þ!ð Þ2 2nþ 1ð Þ¼ b2
n
n
2nþ 1;
so that
b2n ¼
2nþ 1
nan:
6. Let
f xð Þ ¼ 1
xp; for x 2 1;1½ Þ;
where p> 0 and p 6¼ 1.
Then f is positive and decreasing, and
f xð Þ ! 0 as x!1:Also,
Z n
1
f ¼Z n
1
dx
xp
¼ x1�p
1� p
� �n
1
¼ n1�p � 1
1� p: ( )
Now, if p> 1, then 1� p< 0, so thatZ n
1
f ¼ 1
p� 1� n1�p
p� 1<
1
p� 1:
Since the setR n
1f : n 2 N
�
is bounded above, it follows, from the Maclaurin
Integral Test, that the series converges.
Finally, if 0< p< 1, then 1� p> 0, so that
n1�p !1 as n!1:It follows, from the Maclaurin Integral Test, that the series diverges.
7. Let t¼ g(x)¼ loge x, x2 (1,1). The function g is one–one on (1,1).
Then x¼ et; also
dt
dx¼ 1
xand so dt ¼ dx
x:
HenceZ
dx
x loge xð Þ2¼Z
dt
t2
¼ � 1
t¼ � 1
loge x:
438 Appendix 4
Now let
f xð Þ ¼ 1
x logexð Þ2; x 2 2;1½ Þ:
Then f is positive and decreasing on [2,1), and
f xð Þ ! 0 as x!1:Also
Z n
2
f ¼Z n
2
dx
x loge xð Þ2
¼ � 1
loge x
� �n
2
¼ 1
loge 2� 1
loge n
� 1
loge 2:
Since the setR n
2f : n 2N
�
is bounded above, it follows, from the Maclaurin
Integral Test, that the series converges.
8. Let t¼ g(x)¼ loge x, x2 (1,1). The function g is one–one on (1,1).
Then x¼ et; also
dt
dx¼ 1
xand so dt ¼ dx
x:
HenceZ
dx
x loge x¼Z
dt
t
¼ loge t ¼ loge loge xð Þ:Now let
f xð Þ ¼ 1
x loge x; x 2 2;1½ Þ:
Then f is positive and decreasing on [2,1), and
f xð Þ ! 0 as x!1:Also
Z n
2
f ¼Z n
2
dx
x loge x
¼ loge loge xð Þ½ �n2¼ loge loge nð Þ � loge loge 2ð Þ! 1 as n!1:
Hence, by the Maclaurin Integral Test, the series diverges.
9. Let f xð Þ ¼ 11þx2, x2 [0, 1].
Then f is positive and decreasing on [0, 1]; it follows, from the strategy, that
1
n
X
n
i¼1
fi
n
�
!Z 1
0
f :
Now
1
n
X
n
i¼1
fi
n
�
¼ 1
n
X
n
i¼1
1
1þ in
� �2¼ n
X
n
i¼1
1
n2 þ i2
¼ n1
n2 þ 12þ 1
n2 þ 22þ � � � þ 1
n2 þ n2
�
;
Solutions to the problems 439
andZ 1
0
f ¼Z 1
0
1
1þ x2dx
¼ tan�1 x� �1
0
¼ tan�1 1� tan�1 0 ¼ 1
4p:
It follows that
n1
n2 þ 12þ 1
n2 þ 22þ � � � þ 1
n2 þ n2
�
! 1
4p as n!1:
10. Let f xð Þ ¼ 1ffiffi
xp , x2 [1,1).
Then f is positive and decreasing on [1,1), and f (x)! 0 as x!1.
Now
X
n
i¼1
f ið Þ ¼X
n
i¼1
1ffiffi
ip
¼ 1þ 1ffiffiffi
2p þ 1
ffiffiffi
3p þ � � � þ 1
ffiffiffi
np ;
andR n
1f ¼
R n
1x�
12dx
¼ 2x12
h in
1
¼ 2n12 � 2!1 as n!1:
It follows, from the strategy, that
1þ 1ffiffiffi
2p þ 1
ffiffiffi
3p þ � � � þ 1
ffiffiffi
np � 2n
12 � 2 as n !1;
and so, using the remark before the problem, that
1þ 1ffiffiffi
2p þ 1
ffiffiffi
3p þ � � � þ 1
ffiffiffi
np � 2n
12 as n!1:
Section 7.5
1. 20! ’ 2:4329� 1018;
30! ’ 2:6525� 1032;
40! ’ 8:1592� 1047;
50! ’ 3:0414� 1064;
60! ’ 8:3210� 1081:
2. First, f2(n) � f5(n); that is, sin 1n2
� �
� 1n2 as n!1.
We know that
sin x
x! 1 as x! 0;
since the sequence 1n2
�
tends to 0 as n!1, it follows that
sin 1n2
� �
1n2
! 1 as n!1:
In other words, sin 1n2
� �
� 1n2 as n!1.
Second, f3(n) � f7(n); that is, 1� cos 1n
� �
� 12n2 as n!1.
We can use the formula cos x ¼ 1� 2 sin2 12
x� �
to obtain
440 Appendix 4
1 � cos x
x2¼
2 sin2 12 x
� �
x 2
¼ 1
2 �
sin 12
x� �
12
x�
sin 12
x� �
12
x:
Then, since 12
x ! 0 as x ! 0, we deduce, from the fact that sin xx! 1 as x ! 0, that
1 � cos x
x2! 1
2as x ! 0 :
Then, since the sequence 1n
�
tends to 0 as n !1, it follows that
1 � cos 1n
� �
1n 2
! 1
2as n !1;
so that
1 � cos 1n
� �
12 n2
! 1 as n !1;
in other words, 1 � cos 1n
� �
� 12n 2 as n !1.
3. Using the Hint, let f (n) ¼ n2 and g( n) ¼ n.
Then
f nð Þð Þ1n ¼ n2
� �
1n
¼ n1n
� �2
! 1ð Þ2 ¼ 1 as n !1;and
g nð Þð Þ1n ¼ nð Þ
1n
! 1 as n !1;
so thatf nð Þð Þ
1n
g nð Þð Þ1n! 1 as n !1. In other words, f nð Þð Þ
1n� g nð Þð Þ
1n as n !1.
However
f nð Þg nð Þ ¼
n2
n
¼ n 6! 1 as n !1;so that f ( n) 6� g(n) as n !1.
4. First, using a calculator, we find that the value of the expressionffiffiffiffiffiffiffiffi
2p np
ne
� �nwhen
n ¼ 5 is approximately 118.019.
Now 5! ¼ 120, so that the error in the Stirling’s Formula approximation is about
120 � 118.019 ¼ 1.981. The percentage error in this approximation is thus about
1 :981
120� 100 ’ 1: 651%:
This approximation is therefore not within 1% of the exact value.
5. Using Stirling’s Formula, we obtain
nn
n!� nn
ffiffiffiffiffiffiffiffi
2pnp
ne
� �n
¼ en
ffiffiffiffiffiffi
2pnp ;
it follows from Example 1 that
nn
n!
� 1n
� e
p1
2n 2nð Þ1
2n
as n!1:
Solutions to the problems 441
We have seen previously that, for any positive number a
a1n ! 1 and n
1n ! 1; as n!1:
It follows that
p1
2n ! 1 and 2nð Þ1
2n ! 1; as n!1:It follows that
nn
n!
� 1n
� e as n!1;
that is
nn
n!
� 1n
! e as n!1:
6. (a) 300
150
!
1
2300¼ 300!
150!� 150!� 2300
’ffiffiffiffiffiffiffiffiffiffi
600pp
300e
� �300
300p 150e
� �3002300
¼ 1
30
ffiffiffi
6
p
r
¼ 0:046 ðto two significant figuresÞ:(b)
300!
100!ð Þ3� 1
3300’
ffiffiffiffiffiffiffiffiffiffi
600pp
300e
� �300
200pð Þ32 100
e
� �3003300
¼ffiffiffi
3p
200p
¼ 0:0028 ðto two significant figuresÞ:
7. Now
4n
2n
�
¼ 4nð Þ!2nð Þ!ð Þ2
and2n
n
�
¼ 2nð Þ!n!ð Þ2
;
so that
4n
2n
�
2n
n
� ¼ 4nð Þ! n!ð Þ2
2nð Þ!ð Þ3:
It follows, from Stirling’s Formula, that
�
4n
2n
�
2n
n
�ffiffiffiffiffiffiffiffi
8pnp
4ne
� �4n2pn n
e
� �2n
4pnð Þ32 2n
e
� �6n
¼ 44n
ffiffiffi
2p� 26n
¼ 22n
ffiffiffi
2p :
Hence l ¼ 1�ffiffiffi
2p
.
442 Appendix 4
Chapter 8
Section 8.1
1. The tangent approximation to f at a is
f xð Þ ’ f að Þ þ f 0 að Þ x� að Þ:
(a) f xð Þ ¼ ex; f 2ð Þ ¼ e2;
f 0 xð Þ ¼ ex; f 0 2ð Þ ¼ e2:
Hence the tangent approximation to f at 2 is
ex ’ e2 þ e2 x� 2ð Þ:
(b) f xð Þ ¼ cos x; f 0ð Þ ¼ 1;
f 0 xð Þ ¼ �sin x; f 0 0ð Þ ¼ 0:
Hence the tangent approximation to f at 0 is
cos x ’ 1þ 0 x� 0ð Þ ¼ 1:
2. (a) f xð Þ ¼ ex; f 2ð Þ ¼ e2;
f 0 xð Þ ¼ ex; f 0 2ð Þ ¼ e2;
f 00 xð Þ ¼ ex; f 00 2ð Þ ¼ e2;
f 000 xð Þ ¼ ex; f 000 2ð Þ ¼ e2:
Hence
T1 xð Þ ¼ f 2ð Þ þ f 0 2ð Þ1!
x� 2ð Þ
¼ e2 þ e2 x� 2ð Þ;
T2 xð Þ ¼ f 2ð Þ þ f 0 2ð Þ1!
x� 2ð Þ þ f 00 2ð Þ2!
x� 2ð Þ2
¼ e2 þ e2 x� 2ð Þ þ 1
2e2 x� 2ð Þ2;
T3 xð Þ ¼ f 2ð Þ þ f 0 2ð Þ1!
x� 2ð Þ þ f 00 2ð Þ2!
x� 2ð Þ2þ f 000 2ð Þ3!
x� 2ð Þ3
¼ e2 þ e2 x� 2ð Þ þ 1
2e2 x� 2ð Þ2þ 1
6e2 x� 2ð Þ3:
(b) f xð Þ ¼ cos x; f 0ð Þ ¼ 1;
f 0 xð Þ ¼ �sin x; f 0 0ð Þ ¼ 0;
f 00 xð Þ ¼ �cos x; f 00 0ð Þ ¼ �1;
f 000 xð Þ ¼ sin x; f 000 0ð Þ ¼ 0:
Hence
T1 xð Þ ¼ f 0ð Þ þ f 0 0ð Þ1!
x ¼ 1;
T2 xð Þ ¼ f 0ð Þ þ f 0 0ð Þ1!
x þ f 00 0ð Þ2!
x2 ¼ 1� 1
2x2;
T3 xð Þ ¼ f 0ð Þ þ f 0 0ð Þ1!
x þ f 00 0ð Þ2!
x2 þ f 000 0ð Þ3!
x3 ¼ 1� 1
2x2:
Solutions to the problems 443
3. The Taylor polynomial of degree 4 for f at a is
T4 xð Þ ¼ f að Þ þ f 0 að Þ1!
x� að Þ þ f 00 að Þ2!
x� að Þ2þ f 000 að Þ3!
x� að Þ3þ f 4ð Þ að Þ4!
x� að Þ4:
(a) f xð Þ ¼ 7� 6xþ 5x2 þ x3; f 1ð Þ ¼ 7;
f 0 xð Þ ¼ �6þ 10xþ 3x2; f 0 1ð Þ ¼ 7;
f 00 xð Þ ¼ 10þ 6x; f 00 1ð Þ ¼ 16;
f 000 xð Þ ¼ 6; f 000 1ð Þ ¼ 6;
f 4ð Þ xð Þ ¼ 0; f 4ð Þ 1ð Þ ¼ 0:
Hence
T4 xð Þ ¼ 7þ 7 x� 1ð Þ þ 8 x� 1ð Þ2þ x� 1ð Þ3:
(b) f xð Þ ¼ 1� xð Þ�1; f 0ð Þ ¼ 1;
f 0 xð Þ ¼ 1� xð Þ�2; f 0 0ð Þ ¼ 1;
f 00 xð Þ ¼ 2 1� xð Þ�3; f 00 0ð Þ ¼ 2;
f 000 xð Þ ¼ 3! 1� xð Þ�4; f 000 0ð Þ ¼ 3!;
f 4ð Þ xð Þ ¼ 4! 1� xð Þ�5; f 4ð Þ 0ð Þ ¼ 4!:
Hence
T4 xð Þ ¼ 1þ xþ x2 þ x3 þ x4:
(c) f xð Þ ¼ loge 1þ xð Þ; f 0ð Þ ¼ 0;
f 0 xð Þ ¼ 1þ xð Þ�1; f 0 0ð Þ ¼ 1;
f 00 xð Þ ¼ � 1þ xð Þ�2; f 00 0ð Þ ¼ �1;
f 000 xð Þ ¼ 2 1þ xð Þ�3; f 000 0ð Þ ¼ 2;
f 4ð Þ xð Þ ¼ �3! 1þ xð Þ�4; f 4ð Þ 0ð Þ ¼ �3!:
Hence
T4 xð Þ ¼ x� 1
2x2 þ 1
3x3 � 1
4x4:
(d) f xð Þ ¼ sin x; fp4
� �
¼ 1ffiffiffi
2p ;
f 0 xð Þ ¼ cos x; f 0p4
� �
¼ 1ffiffiffi
2p ;
f 00 xð Þ ¼ � sin x; f 00p4
� �
¼ � 1ffiffiffi
2p ;
f 000 xð Þ ¼ � cos x; f 000p4
� �
¼ � 1ffiffiffi
2p ;
f 4ð Þ xð Þ ¼ sin x; f 4ð Þ p4
� �
¼ 1ffiffiffi
2p :
Hence
T4 xð Þ ¼ 1ffiffiffi
2p 1þ x� p
4
� �
� 1
2x� p
4
� �2
� 1
6x� p
4
� �3
þ 1
24x� p
4
� �4�
:
444 Appendix 4
(e)f xð Þ ¼ 1þ 1
2x� 1
2x2 � 1
6x3 þ 1
4x4; f 0ð Þ ¼ 1;
f 0 xð Þ ¼ 1
2� x� 1
2x2 þ x3; f 0 0ð Þ ¼ 1
2;
f 00 xð Þ ¼ �1� xþ 3x2; f 00 0ð Þ ¼ �1;
f 000 xð Þ ¼ �1þ 3!x; f 000 0ð Þ ¼ �1;
f 4ð Þ xð Þ ¼ 3!; f 4ð Þ 0ð Þ ¼ 3!:
Hence
T4 xð Þ ¼ 1þ 1
2x� 1
2x2 � 1
6x3 þ 1
4x4:
4. f xð Þ ¼ tan x; f 0ð Þ ¼ 0;
f 0 xð Þ ¼ sec2 x; f 0 0ð Þ ¼ 1;
f 00 xð Þ ¼ 2 sec2 x tan x; f 00 0ð Þ ¼ 0;
f 000 xð Þ ¼ 4 sec2 x tan2 xþ 2 sec4 x; f 000 0ð Þ ¼ 2:
Hence
T3 xð Þ ¼ xþ 1
3x3;
so that
T3 0:1ð Þ ¼ 0:1þ 1
3� 0:001 ¼ 0:1003:
Since (by calculator) tan 0.1¼ 0.10033467. . ., the percentage error involved is
about
0:10033467� 0:10033333
0:10033467� 100 ’ 0:001%:
5. (a) f xð Þ ¼ 1� xð Þ�1; f 0ð Þ ¼ 1;
f 0 xð Þ ¼ 1� xð Þ�2; f 0 0ð Þ ¼ 1;
f 00 xð Þ ¼ 2� 1� xð Þ�3; f 00 0ð Þ ¼ 2;
..
.
f nð Þ xð Þ ¼ n!ð Þ � 1� xð Þ�n�1; f nð Þ 0ð Þ ¼ n!:
Hence
Tn xð Þ ¼ f 0ð Þ þ f 0 0ð Þxþ 1
2!f 00 0ð Þx2 þ � � � þ 1
n!f nð Þ 0ð Þxn
¼ 1þ xþ x2 þ � � � þ xn:
(b) f xð Þ ¼ loge 1þ xð Þ; f 0ð Þ ¼ 0;
f 0 xð Þ ¼ 1þ xð Þ�1; f 0 0ð Þ ¼ 1;
f 00 xð Þ ¼ �1ð Þ � 1þ xð Þ�2; f 00 0ð Þ ¼ �1;
f 000 xð Þ ¼ �1ð Þ �2ð Þ � 1þ xð Þ�3; f 000 0ð Þ ¼ 2;
..
.
f nð Þ xð Þ ¼ �1ð Þnþ1n� 1ð Þ!� 1þ xð Þ�n; f nð Þ 0ð Þ ¼ �1ð Þnþ1
n� 1ð Þ!:Hence
Tn xð Þ ¼ f 0ð Þ þ f 0 0ð Þxþ 1
2!f 00 0ð Þx2 þ � � � þ 1
n!f nð Þ 0ð Þxn
¼ x� 1
2x2 þ 1
3x3 � � � � þ �1ð Þnþ1
nxn:
Solutions to the problems 445
(c) f xð Þ ¼ ex; f 0ð Þ ¼ 1;
and, for each positive integer k
f kð Þ xð Þ ¼ ex; f kð Þ 0ð Þ ¼ 1:
Hence
Tn xð Þ ¼ f 0ð Þ þ f 0 0ð Þxþ 1
2!f 00 0ð Þx2 þ � � � þ 1
n!f nð Þ 0ð Þxn
¼ 1þ xþ x2
2!þ x3
3!þ � � � þ xn
n!:
(d) f xð Þ ¼ sin x; f 0ð Þ ¼ 0;
f 0 xð Þ ¼ cos x; f 0 0ð Þ ¼ 1;
f 00 xð Þ ¼ �sin x; f 00 0ð Þ ¼ 0;
f 000 xð Þ ¼ �cos x; f 000 0ð Þ ¼ �1;
f 4ð Þ xð Þ ¼ sin x; f 4ð Þ 0ð Þ ¼ 0;
and, in general, for each positive integer k
f ðkÞð0Þ ¼0; if k is even,
1; if k � 1 (mod 4),
�1; if k � 3 (mod 4).
8
<
:
Hence
Tn xð Þ ¼ x� x3
3!þ x5
5!þ � � � þ 0 or 1 or�1ð Þ x
n
n!;
where the coefficient of xk isf kð Þ 0ð Þ
k! , and the value of f (k)(0) is as specified above.
(e) f xð Þ ¼ cos x; f 0ð Þ ¼ 1;
f 0 xð Þ ¼ �sin x; f 0 0ð Þ ¼ 0;
f 00 xð Þ ¼ �cos x; f 00 0ð Þ ¼ �1;
f 000 xð Þ ¼ sin x; f 000 0ð Þ ¼ 0;
f 4ð Þ xð Þ ¼ cos x; f 4ð Þ 0ð Þ ¼ 1;
and, in general, for each positive integer k
f kð Þ 0ð Þ ¼0; if k is odd;1; if k � 0 mod 4ð Þ;�1; if k � 2 mod 4ð Þ:
8
<
:
Hence
Tn xð Þ ¼ 1� x2
2!þ x4
4!� � � � þ 0 or 1 or� 1ð Þ x
n
n!;
where the coefficient of xk isf kð Þ 0ð Þ
k! , and the value of f (k)(0) is as specified
above.
Section 8.2
1. Here
f xð Þ ¼ 1
1� x; f 0 xð Þ ¼ 1
1� xð Þ2and f 00 xð Þ ¼ 2
1� xð Þ3:
Hence
R1 xð Þ ¼ f 00 cð Þ2!
x2 ¼ x2
1� cð Þ3:
446 Appendix 4
To calculate the value of c, we use Taylor’s Theorem, with n¼ 1
f xð Þ ¼ f að Þ þ f 0 að Þ x� að Þ þ R1 xð Þ:This gives
f3
4
�
¼ f 0ð Þ þ f 0 0ð Þ 34þ R1
3
4
�
;
that is
4 ¼ 1þ 3
4þ
34
� �2
1� cð Þ3;
and so
9
4¼
34
� �2
1� cð Þ3:
It follows that 1� cð Þ3¼ 14; so that 1� c ¼ 1
4
� �13 ’ 0:630. Thus c ’ 0:370:
2. When f is a polynomial of degree n or less
f nþ1ð Þ cð Þ ¼ 0; so that Rn xð Þ ¼ 0:
3. f xð Þ ¼ cos x; f 0ð Þ ¼ 1;
f 0 xð Þ ¼ �sin x; f 0 0ð Þ ¼ 0;
f 00 xð Þ ¼ �cos x; f 00 0ð Þ ¼ �1;
f 000 xð Þ ¼ sin x; f 000 0ð Þ ¼ 0;
f 4ð Þ xð Þ ¼ cos x:
Hence, by Taylor’s Theorem with a¼ 0, f (x)¼ cos x and n¼ 3, we have
cos x ¼ 1� 1
2x2 þ R3 xð Þ;
where
R3 xð Þ ¼ f 4ð Þ cð Þ4!
x4:
Thus
R3 xð Þj j � cos cj j24
x4
� 1
24x4:
4. f xð Þ ¼ loge 1þ xð Þ; f 0ð Þ ¼ 0;
f 0 xð Þ ¼ 1
1þ x; f 0 0ð Þ ¼ 1;
f 00 xð Þ ¼ �1
1þ xð Þ2; f 00 0ð Þ ¼ �1;
f 000 xð Þ ¼ 2
1þ xð Þ3:
Hence, by Taylor’s Theorem with x¼ 0.02
loge 1:02 ¼ 0þ 1� 0:02ð Þ � 1
2� 0:022
�
þ R2 0:02ð Þ
¼ 0:0198þ R2 0:02ð Þ:Now, for c2 (0, 0.02)
f 000 cð Þj j ¼ 2
1þ cð Þ3
� 2:
Solutions to the problems 447
Hence, by the Remainder Estimate with M¼ 2, we obtain
jR2 0:02ð Þj � 2
3!� 0:023 ¼ 0:0000026:
It follows that
loge 1:02 ¼ 0:0198 ðto four decimal placesÞ:5. For any positive integer n
f nð Þ xð Þ ¼ � cos x or � sin x;
hence
f nð Þ cð Þ
� 1; for all c 2 R ;
so that M¼ 1.
It follows that the remainder term in the Remainder Estimate satisfies the
inequality
Rn 0:2ð Þj j � 1
nþ 1ð Þ! 0:2ð Þnþ1:
Now
0:2ð Þ4
4!¼ 0:00006 and
0:2ð Þ5
5!¼ 0:0000026 . . .;
so we should try n¼ 4 in the Remainder Estimate, to be safe.
Here
f xð Þ ¼ cos x; f 0ð Þ ¼ 1;
f 0 xð Þ ¼ �sin x; f 0 0ð Þ ¼ 0;
f 00 xð Þ ¼ �cos x; f 00 0ð Þ ¼ �1;
f 000 xð Þ ¼ sin x; f 000 0ð Þ ¼ 0;
f 4ð Þ xð Þ ¼ cos x; f 4ð Þ 0ð Þ ¼ 1
f 5ð Þ xð Þ ¼ �sin x:
Hence, by Taylor’s Theorem with f (x)¼ cos x, a¼ 0, x¼ 0.2 and n¼ 4, we have
f xð Þ ¼ 1� 1
2x2 þ 1
24x4 þ R4 xð Þ;
in other words
cos 0:2 ¼ 1� 1
20:2ð Þ2þ 1
240:2ð Þ4þR4 0:2ð Þ
¼ 1� 0:02þ 0:00006þ R4 0:2ð Þ¼ 0:9801 ðrounded to four decimal placesÞ:
6. Using the same function f as in Problem 5, we find that
f pð Þ ¼ �1; f 0 pð Þ ¼ 0; f 00 pð Þ ¼ 1;
f 000 pð Þ ¼ 0; f 4ð Þ pð Þ ¼ �1:
It follows that
T4 xð Þ ¼ �1þ 1
2x� pð Þ2� 1
24x� pð Þ4:
The difference between T4(x) and f (x) is
R4 xð Þ ¼ f 5ð Þ cð Þ5!
x� pð Þ5
¼ � sin c
120x� pð Þ5;
448 Appendix 4
for some number c between p and x. Hence, for x 2 34p; 5
4p
� �
, we have
R4 xð Þj j � 1
120
p4
� �5
¼ 0:00249 . . . < 3� 10�3:
Section 8.3
1. (a) Applying the Ratio Test with an¼ 2nþ 4n, we obtain
anþ1
an
¼ 2nþ1 þ 4nþ1
2n þ 4n¼ 2þ 2nþ2
1þ 2n
¼ 212
� �nþ212
� �nþ1
! 2� 2 ¼ 4 as n!1:Hence the series has radius of convergence 1
4:
(b) Applying the Ratio Test with an ¼ ðn!Þ2ð2nÞ!, we obtain
anþ1
an
¼ nþ 1ð Þ nþ 1ð Þ2nþ 1ð Þ 2nþ 2ð Þ ¼
1þ 1n
� �
1þ 1n
� �
2þ 1n
� �
2þ 2n
� �
! 1
4as n!1:
Hence the series has radius of convergence 4.
(c) Applying the Ratio Test with an¼ nþ 2�n, we obtain
anþ1
an
¼ nþ 1þ 2�n�1
nþ 2�n¼
1þ 1nþ 1
n2nþ1
1þ 1n2n
! 1 as n!1:Hence the series has radius of convergence 1.
(d) Applying the Ratio Test with an ¼ 1
n!ð Þ1n; we obtain
anþ1
an
¼ n!ð Þ1n
nþ 1ð Þ!ð Þ1
nþ1
�ffiffiffiffiffiffiffiffi
2pnp� �
1n n
e
� �
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2p nþ 1ð Þp� �
1nþ1 nþ1
e
� �
¼ffiffiffiffiffiffi
2pp� �
1n
ffiffiffiffiffiffi
2pp� �
1nþ1
��
n1n
�12
ð nþ 1ð Þ1
nþ1Þ12
� n
nþ 1
! 1 as n!1:(Here we have used Stirling’s Formula to estimate n! and (nþ 1)!.)
Hence the series has radius of convergence 1.
2. (a) Applying the Ratio Test with an¼ n, we obtain
anþ1
an
¼ nþ 1
n! 1 as n!1:
Hence the series has radius of convergence 1.
Thus the series converges if x2 (�1, 1), and diverges if jx j> 1.
When x¼�1, the sequence {nxn} is non-null; hence, by the Non-null Test, the
series diverges.
Hence the interval of convergence of the series is (�1, 1).
(b) Applying the Ratio Test with an ¼ 1n3n, we obtain
Solutions to the problems 449
anþ1
an
¼ n3n
nþ 1ð Þ3nþ1¼ 1
1þ 1n
� �
3
! 1
3as n!1:
Thus the series converges if x2 (�3, 3), and diverges if jx j> 3.
When x¼ 3, the series isP
1
n¼1
1n, which is divergent.
When x¼�3, the series isP
1
n¼1
�1ð Þnn
, which is convergent.
Hence the interval of convergence of the series is [�3, 3).
3. Applying the Ratio Test with an ¼ � ��1ð Þ: ... : ��nþ1ð Þn! , we obtain
anþ1
an
¼ �� n
nþ 1
¼�n� 1
1þ 1n
! 1 as n!1:
Hence the series has radius of convergence 1.
4. From the Hint, we know that tan�1 x is the sum function of the series
X
1
n¼0
�1ð Þn
2nþ 1x2nþ1 ¼ x� x3
3þ x5
5� x7
7þ � � �; for xj j < 1: ( )
From the Hint, we also know that the radius of convergence of this power series is 1.
It follows that, by applying Abel’s Limit Theorem to the power series (), we may
obtain the following
limx!1�
tan�1 x� �
¼X
1
n¼0
�1ð Þn
2nþ 1;
provided that this last series is convergent; it is indeed convergent, by the Alternating
Test for series.
Since the function x 7! tan�1 x is continuous at 1, it follows that
X
1
n¼0
�1ð Þn
2nþ 1¼ lim
x!1�tan�1 x� �
¼ tan�1 1 ¼ p4:
Section 8.4
1. (a) We know that
ex ¼ 1þ xþ x2
2!þ x3
3!þ � � � þ xn
n!þ � � �; for x 2 R ;
and so
e�x ¼ 1� xþ x2
2!� x3
3!þ � � � þ �1ð Þnxn
n!þ � � �; for x 2 R :
Hence, by the Sum and Multiple Rules for power series
sinh x ¼ 1
2ex � e�xð Þ
¼ xþ x3
3!þ � � � þ x2nþ1
2nþ 1ð Þ!þ � � �; for x 2 R :
This series converges for all x.
For x 7! tan�1 x is continuouson its domain R .
450 Appendix 4
(b) We know that
loge 1� xð Þ ¼ �x� x2
2� x3
3� � � � � xn
n� � � �; for xj j < 1;
and1
1� x¼ 1þ xþ x2 þ x3 þ � � � þ xn þ � � �; for xj j < 1:
Hence, by the Sum and Multiple Rules for power series
loge 1� xð Þþ 2
1� x
¼ �x� x2
2� x3
3� � � � � xn
n� � � �
�
þ 2þ 2xþ 2x2 þ 2x3 þ � � � þ 2xn þ � � �� �
¼ 2þ xþ 3
2x2 þ 5
3x3 þ � � � þ 2n� 1
nxn þ � � �; for xj j < 1:
The radius of convergence of this series is 1.
2. (a) We know that
sinh x ¼ xþ x3
3!þ � � � þ x2nþ1
2nþ 1ð Þ! þ � � �; for x 2 R ;
and that
sin x ¼ x� x3
3!þ � � � þ �1ð Þnx2nþ1
2nþ 1ð Þ! þ � � �; for x 2 R :
Hence, by the Sum and Multiple Rules for power series
sinh xþ sin x
¼ xþ x3
3!þ � � � þ x2nþ1
2nþ 1ð Þ!þ � � ��
þ x� x3
3!þ � � � þ �1ð Þnx2nþ1
2nþ 1ð Þ! þ � � ��
¼ 2 xþ x5
5!þ � � � þ x4nþ1
4nþ 1ð Þ!þ � � ��
; for x 2 R :
This series converges for all x.
(b) We know that
loge 1þ xð Þ ¼ x� x2
2þ x3
3� � � � þ �1ð Þnþ1xn
nþ � � �; for xj j < 1;
and
loge 1� xð Þ ¼ �x� x2
2� x3
3� � � � � xn
n� � � �; for xj j < 1:
Hence, by the Sum and Multiple Rules for power series
loge
1þ x
1� x
�
¼ loge 1þ xð Þ � loge 1� xð Þ
¼ x� x2
2þ x3
3� � � � þ �1ð Þnþ1xn
nþ � � �
�
� �x� x2
2� x3
3� � � � � xn
n� � � �
�
¼ 2 xþ x3
3þ � � � þ x2nþ1
2nþ 1þ � � �
�
; for xj j < 1:
The radius of convergence of this series is 1.
(c) We know that
1
1� x¼ 1þ xþ x2 þ x3 þ � � � þ xn þ � � � ; for xj j < 1:
Solutions to the problems 451
It follows, by replacing x by � 2x 2, that
1
1 þ 2x 2 ¼ 1 þ �2x 2
� �
þ �2x2� �2þ �2x 2
� �3þ � � � þ �2x 2� �nþ � � �
¼ 1 � 2x 2 þ 22 x 4 � 23 x 6 þ � � � þ �1ð Þn2n x 2 n þ � � �;
for j2 x2 j< 1; that is, for xj j < 1ffiffi
2p .
The radius of convergence of this series is 1ffiffi
2p .
3. (a) We know that
loge 1 þ xð Þ ¼ x � x2
2þ x3
3� � � � þ �1ð Þnþ 1x n
nþ � � � ; for xj j < 1:
Hence, by the Product Rule
1 þ xð Þ loge 1 þ xð Þ
¼ x � x 2
2þ x 3
3� � � � þ �1ð Þn þ1x n
nþ � � �
�
þ x 2 � x 3
2þ x 4
3� � � � þ �1ð Þn xn
n � 1 þ � � �
�
¼ x þ x2
2� x 3
6� � � � þ �1ð Þn x n
n n� 1ð Þ þ � � � ; for xj j < 1:
The radius of convergence of this series is 1.
(b) We know that
1
1� x¼ 1þ xþ x2 þ x3 þ � � � þ xn þ � � � ; for xj j < 1;
and (from Example 1) that
1þ x
1� xð Þ2¼ 1þ 3xþ 5x2 þ � � � þ 2nþ 1ð Þxn þ � � � ; for xj j < 1:
Hence, by the Product Rule, we obtain
1þ x
1� xð Þ3
¼ 1
1� x� 1þ x
1� xð Þ2
¼ 1þ xþ x2 þ x3 þ � � � þ xn þ � � �� �
� 1þ 3xþ 5x2 þ � � � þ 2nþ 1ð Þxn þ � � �� �
¼ 1þ 3þ 1ð Þxþ 5þ 3þ 1ð Þx2 þ 7þ 5þ 3þ 1ð Þx3 þ � � �þ 2nþ 1ð Þ þ 2n� 1ð Þ þ � � � þ 1ð Þxn þ � � �
¼ 1þ 4xþ 9x2 þ 16x3 þ � � � þ nþ 1ð Þ2xn þ � � � ; for xj j < 1:
(We can prove, by Mathematical Induction, that
1þ 3þ 5þ � � � þ 2nþ 1ð Þ ¼X
n
k¼0
2k þ 1ð Þ ¼ nþ 1ð Þ2; for n 2 N :Þ
The radius of convergence of this series is 1.
4. (a) We know that
1� xð Þ�1¼ 1þ xþ x2 þ x3 þ � � � þ xn þ � � � ; for xj j < 1:
Hence, by the Differentiation Rule, we obtain
1� xð Þ�2¼ 1þ 2xþ 3x2 þ 4x3 þ � � � þ nxn�1 þ � � � ; for xj j < 1:
The radius of convergence of this series is 1.
452 Appendix 4
(b) If we differentiate the series in part (a ), by the Differentiation Rule for power
series, we obtain
2 1� xð Þ�3 ¼ 2 þ 6x þ 12x 2 þ � � � þ n n� 1ð Þxn �2 þ � � �; for xj j < 1:
Hence, by the Multiple Rule, we have
1 � xð Þ�3 ¼ 1 þ 3x þ 6x 2 þ � � � þ n n� 1ð Þ2
x n� 2 þ � � �; for xj j < 1:
The radius of convergence of this series is 1.
(c) Notice that, for the function f ( x ) ¼ tanh� 1 x, j x j< 1 , we have
f 0 xð Þ ¼ 1
1 � x 2 ¼ 1 þ x 2 þ x 4 þ � � � þ x 2n þ � � �:
Hence, by the Integration Rule for power series, the Taylor series for f at 0 is
tanh �1 x ¼ c þ x þ x 3
3þ x 5
5þ � � � þ x 2n þ1
2n þ 1 þ � � �; for xj j < 1;
since f (0) ¼ 0, it follows that c ¼ 0. Hence
tanh �1 x ¼ x þ x 3
3þ x 5
5þ � � � þ x 2n þ1
2n þ 1 þ � � �; for xj j < 1:
The radius of convergence of this series is 1.
5. We know that
e x ¼ 1 þ x þ x 2
2!þ x 3
3!þ � � � þ x n
n !þ � � �; for x 2 R ;
and we know, from part (a ) of Problem 4, that
1 � xð Þ� 2¼ 1 þ 2x þ 3x 2 þ 4x 3 þ � � � þ n þ 1ð Þx n þ � � �; for xj j < 1:
Hence, by the Product Rule for power series, we have
e x 1 � xð Þ� 2 ¼ 1 þ x þ x 2
2þ x 3
6þ � � �
�
� 1 þ 2x þ 3x 2 þ 4x 3 þ � � �� �
¼ 1 þ 2 þ 1ð Þx þ 3 þ 2 þ 1
2
�
x 2 þ � � �
¼ 1 þ 3x þ 11
2x 2 þ � � � ; for xj j < 1:
The radius of convergence of this series is 1.
6. We are given that
f xð Þ ¼ x þ x3
1:3 þ x5
1: 3:5 þ � � � þ x 2n þ1
1: 3: ::: : 2n þ 1ð Þ þ � � �; for x 2 R :
(a) By the Differentiation Rule, we can differentiate the power series term-by-term;
this gives
f 0 xð Þ ¼ 1þ x2
1þ x4
1:3þ � � � þ x2n
1:3: ::: : 2n� 1ð Þ þ � � �; for x 2 R :
(b) It follows, from the definition of f, that
xf xð Þ ¼ x2 þ x4
1:3þ x6
1:3:5þ � � � þ x2nþ2
1:3: ::: : 2nþ 1ð Þ þ � � �; for x 2 R :
Hence, by the Combination Rules for power series,
f 0 xð Þ � xf xð Þ ¼ 1;
since all the other terms cancel out.
Solutions to the problems 453
7. Since
ex ¼ 1þ xþ x2
2!þ x3
3!þ � � � þ xn
n!þ � � �; for x 2 R ;
we may deduce that
e�x2 ¼ 1� x2 þ x4
2!� x6
3!þ � � � þ �1ð Þn x2n
n!þ � � �; for x 2 R :
It follows, from the Integration Rule for power series, that
Z 1
0
e�x2
dx ¼ x� x3
3þ x5
5� 2!� x7
7� 3!þ � � � þ �1ð Þn x2nþ1
2nþ 1ð Þ � n!þ � � �
� �1
0
¼ 1� 1
3þ 1
10� 1
42þ � � � þ �1ð Þn 1
2nþ 1ð Þ � n!þ � � �:
8. By the General Binomial Theorem
1þ 6xð Þ14 ¼X
1
n¼0
14
n
�
xn; for 6xj j < 1;
where
14
n
�
¼14
� �
� 34
� �
� 74
� �
. . . 14� nþ 1
� �
n!:
It follows that
1þ 6xð Þ14 ¼ 1þ
14
� �
16xð Þ þ
14
� �
� 34
� �
26xð Þ2þ
14
� �
� 34
� �
� 74
� �
66xð Þ3þ � � �
¼ 1þ 3
2x� 27
8x2 þ 189
16x3 � � � �; for xj j < 1
6:
The radius of convergence of this series is 16.
9. (a) By the General Binomial Theorem
1� xð Þ�12 ¼X
1
n¼0
� 12
n
�
�xð Þn; for xj j < 1;
where
� 12
n
�
¼� 1
2
� �
� 32
� �
� 52
� �
. . . � 12� nþ 1
� �
n!:
It follows that
1� xð Þ�12¼ 1þ 1
2xþ 3
8x2 þ � � � þ �1ð Þn � 1
2
n
�
xn þ � � �; for xj j < 1:
The radius of convergence of this series is 1.
(b) We know that
d
dxsin�1 x ¼ 1
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1� x2p ; for x 2 �1; 1ð Þ:
Then, by the result of part (a), with x replaced by x2, we obtain
1ffiffiffiffiffiffiffiffiffiffiffiffiffi
1� x2p ¼ 1þ 1
2x2 þ 3
8x4 þ � � � þ �1ð Þn � 1
2
n
�
x2n þ � � �; for xj j < 1:
Hence, by the Integration Rule for power series
sin�1 x ¼ cþ xþ 1
6x3 þ 3
40x5 þ � � �
þ �1ð Þn � 12
n
�
x2nþ1
2nþ 1þ � � �; for xj j < 1:
454 Appendix 4
Putting x¼ 0 into this equation, we see that c¼ 0. It follows that
sin�1 x ¼ xþ 1
6x3 þ 3
40x5 þ � � �
þ �1ð Þn � 12
n
�
x2nþ1
2nþ 1þ � � �; for xj j < 1:
The radius of convergence of this series is 1.
Section 8.5
1. We use the Addition Formula
tan�1 xþ tan�1 y ¼ tan�1 xþ y
1� xy
�
; ( )
which holds provided that tan�1 xþ tan�1 y lies in the interval � p2; p
2
� �
.
First, we deduce, from the Addition Formula (), that
tan�1 1
3
�
þ tan�1 1
4
�
¼ tan�113þ 1
4
1� 13� 1
4
!
¼ tan�1 4þ 3
12� 1
�
¼ tan�1 7
11
�
:
This equation holds, since
tan�1 1
3
�
þ tan�1 1
4
�
’ 0:3218þ 0:2450 ¼ 0:5668;
and 0:5668 2 � p2; p
2
� �
’ �1:5708; 1:5708ð Þ:Next, we deduce, from the Addition Formula (), that
tan�1 1
3
�
þ tan�1 1
4
�
þ tan�1 2
9
�
¼ tan�1 7
11
�
þ tan�1 2
9
�
¼ tan�17
11þ 2
9
1� 711� 2
9
!
¼ tan�1 63þ 22
99� 14
�
¼ tan�1 1ð Þ ¼ p4:
This equation holds, since
tan�1 7
11
�
þ tan�1 2
9
�
’ 0:5667þ 0:2187 ¼ 0:7854;
and 0:7854 2 � p2; p
2
� �
’ �1:5708; 1:5708ð Þ:2. First
I0 ¼Z 1
�1
cos1
2px
�
dx
¼ 2
psin
1
2px
� � �1
�1
¼ 4
p:
Solutions to the problems 455
It follows that p I0¼ 4.
Next, using integration by parts twice, we obtain
I1 ¼Z 1
�1
1� x2� �
cos1
2px
�
dx
¼ 1� x2� � 2
psin
1
2px
� � �1
�1
�Z 1
�1
�2xð Þ 2
psin
1
2px
�
dx
¼ 4
p
Z 1
�1
x sin1
2px
�
dx
¼ 4
px � 2
p
�
cos1
2px
� � �1
�1
� 4
p
Z 1
�1
� 2
p
�
cos1
2px
�
dx
¼ 8
p2
2
psin
1
2px
� � �1
�1
¼ 32
p3:
It follows that p 3I1 ¼ 32.
456 Appendix 4
Index
ax, definition and continuity, 161
Abel’s limit theorem, 333
absolute convergence test, 104
absolute convergence theorem, 332, 337
absolute value, 12
absolutely convergent series, 104
addition formula for tan�1, 166
alternating test, 107
antipodal points theorem, 145
approximation by Taylor
polynomials, 317
Archimedean property of R , 7
arithmetic in R , 8, 30
arithmetic mean – geometric mean
inequality, 18, 21
asymptotic behaviour of functions, 176
basic
continuous functions, 142
differentiable functions, 224
null sequences, 48
power series, 325
series, 97
Bernoulli’s inequality, 20, 237
Bessel function, 329
Bijection, 355
binomial theorem, 358, 342
blancmange function, 244
Bolzano–Weierstrass theorem, 70
bound, greatest lower, 26
bound, least upper, 25
boundedness theorem, 62, 92, 149
Cauchy condensation test, 97
Cauchy’s mean value theorem, 239
Cauchy–Schwarz inequality, 20, 311
Chain Rule, 218
combination rules
continuous functions, 136
convergent sequences, 55
convergent series, 89
differentiable functions, 216
functions which tend to1, 177
inequalities, 14
infimum, 275
integrable functions, 278
limits of functions, 172
null sequences, 46
power series, 338
primitives, 284
sequences which tend to1, 65
series, 89
supremum, 275
tilda, 305
common limit criterion, 268
common refinement, 262
comparison test, 94
composition rule
asymptotic behaviour of functions, 180
continuous functions, 136
differentiable functions, 218
limits of functions, 173
conditionally convergent series, 104
continuity
continuity, 132, 194
limits of functions, 171
one-sided, 134, 194
uniform, 201
continuous functions
basic, 142
boundedness theorem, 149
combination rules, 136
composition rule, 136
extreme values theorem, 169
integrability, 277
intermediate value theorem, 143
inverse function rule, 153
squeeze rule, 137
convergence, interval of, 330
radius of, 330
convergent sequence, 53
combination rules, 55
quotient rule, 55
squeeze rule, 58
convergent series, 85
combination rules, 89
cos�1, 156
cosh, power series, 339
cosh�1, 158
cosine function
derived function, 214
power series, 326
Darboux’s theorem, 254
decimal representation of numbers, 3, 87
density property of R , 7
derivatives, 207
higher order, 215
one-sided, 210
standard, 359
difference quotient, 206
differentiability of
f(x)¼ ex, 214
f(x)¼ ax, 223
f(x)¼ x�, 223
f(x)¼ xx, 223
trigonometrical functions, 214
differentiable functions
combination rules, 216
composition rules, 218
inequalities involving, 237
inverse function rule, 221
differentiation, 210
rule for power series, 341
Dirichlet’s function, 197, 310
discontinuity, removable, 172
divergent sequence, 61
series, 85
domination hierarchy, 56
e, definition, 74
irrationality, 175
"� � game, 43, 187
Euclidean algorithm, 78
even subsequence, 66
ex, definition, 75, 122
fundamental property, 127
inverse property, 75
exponent laws, 163
exponential function
continuity, 141
derived function, 214
differentiability, 223
inequalities, 141
power series, 326
extreme values theorem, 149
extremum, 228
field, 9
first subsequence rule, 67
function, nowhere differentiable, 244
fundamental inequality for integrals, 288
fundamental theorem of Algebra, 146
of Calculus, 283, 292
general binomial theorem, 342
geometric series, 87
457
Goethe, ix
greatest lower bound, 26, 272
property of R , 29
Gregory’s series for estimating p, 346
harmonic series, 93, 109
higher-order derivatives, 215
hyperbolic functions, differentiability, 224
hypergeometric series, 352
increasing–decreasing theorem, 234
inequalities, power rule, 10
rules for integrals, 289
infimum, 26, 272
infinite series, 85
inheritance property of subsequences, 66
Int I, 234
Integrability, 264
combination rules, 278
continuous functions, 277
integral test, 297
modulus rule, 278
monotonic functions, 276
Riemann’s criterion, 267
integral, 264
additivity, 280
inequality rules, 289
lower, 264
upper, 264
integration by parts, 285
by substitution, 286
integration rule for power series, 341
integration, reduction of order
method, 293
intermediate value theorem, 143
interval image theorem, 149
interval of convergence, 230
inverse function rule for continuous
functions, 153
differentiable functions, 221
inverse hyperbolic functions, 158
inverse trigonometric functions, 156
irrational numbers, 5
K" lemma, 49, 193
least upper bound, 25, 262
property of R , 29
Leibniz notation, 207
Leibniz test, 107
Leibniz’s series for estimating p, 346
l‘Hopital’s rule, 241
limit comparison test, 95
limit inequality rule for limits, 175
sequences, 60
limit of a function, 169, 177, 185
as x!1, 177
one-sided, 175
limit of a sequence, 53, 182
limits of functions and continuity, 171
limits of functions, combination
rules, 172
composition rule, 173
one-sided, 175
squeeze rule, 174
local extremum theorem, 229
loge 1þ xð Þ, power series for, 325
lower integral, 264
lower Riemann sum, 259
Maclaurin integral test, 297
mathematical induction, 357
maximum, 23
mean value theorem, 233
minimum, 24
modulus function, continuity, 135
modulus rule for integrable
functions, 278
modulus, 12
monotone convergence theorem, 68
monotonic function, 153
integrability, 276
monotonic sequence theorem, 69
multiplication of series, 114
n!, Stirling’s formula for, 306
neighbourhood, 169
non-null test, 91
nth partial sum of series, 85
nth root function, continuity, 155
nth root of a positive real number,
32, 155
null partition criterion, 269
null sequence, 43
odd subsequence, 66
one-one function, 152, 355
one-sided derivative, 210
one-sided limit, 175
onto function, 355
order properties of R , 7
p, 76
irrationality of, 348
numerical estimates, 346
partial sum of series, 85
partition of [a, b], 258
standard, 258
power series
absolute convergence, 332
basic, 325
combination rules, 338
differentiation rule, 341
integration rule, 341
uniqueness theorem, 342
primitives, 282
combination rules, 278
scaling rule, 278
standard, 360
uniqueness theorem, 284
radius of convergence theorem, 329
ratio test, 96
for radius of convergence, 330
rational function, continuity, 136
differentiability, 217
rational power, 34
rearrangement of series, 109
reciprocal rule for
functions which tend to1, 177
sequences which tend to1, 64
tilda, 305
recursion formula, 71
reduction of order method, 293
refinement, 262
remainder estimate, 322
removable discontinuity, 172
Riemann’s rearrangement theorem, 112
Riemann’s function, 198
Riemann’s-criterion for integrability, 267
Rolle’s theorem, 231
scaling rule for primitives, 284
second derivative test, 236
second subsequence rule, 67
sequences, 38
basic null, 48
combination rules, 55
convergent, 53
divergent, 61
limit inequality rule, 60
monotonic, 40
null, 43
unbounded, 62
which tend to1, 63
which tend to �1, 65
squeeze rule, 58
series, 85
absolutely convergent, 104
basic, 97
convergent, 85
divergent, 85
geometric, 87
harmonic, 93, 109
hypergeometric, 352
integral test, 297
multiplication, 114
product rule, 114
Taylor, 325
telescoping, 88
sine function
derived function, 214
power series, 326
sine inequality, 140
sin�1, 156
sinh, 158
power series, 339
square root function, continuity, 134
squeeze rule
as x!1, 178
continuous functions, 137
458 Index
convergent sequences, 58
limits of functions, 174
null sequences, 47
sequences which tend to1, 179
standard derivatives, 359
standard partition, 258
standard primitives, 360
Stirling’s formula, 306
strategy for testing for convergence, 116
sub-interval theorem, 280
subsequence, 66
sum function, 325
supremum, 35
tan�1, 157
power series, 341
tanh�1, 159
power series, 344
tangent approximation, 314
Taylor polynomial, 316
Taylor series, 325
Taylor’s theorem, 320
tilda notation, 300, 304
combination rules, 305
transitive property of R , 7
transitive rule for inequalities, 291
triangle inequality, 15, 16
backwards form, 15
infinite form, 105
integrals, 291
trichotomy property of R , 7
trigonometric functions
continuity, 139
differentiability, 213
unbounded sequence, 62
uniform continuity, 201
uniqueness theorem, power series, 342
primitives, 284
upper integral, 264
upper Riemann sum, 259
Wallis’s formula, 293
Weierstrass, K., x
zero derivative theorem, 235
zero of polynomial, 295
zeros localisation theorem, 147
Index 459