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Transcript of Introduction to Calculus Set
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Introduction to Calculus
Bhaskara Aacharya (1114 1185)
Legendary Indian Mathematician who gave preliminary concepts of infinitesimal
calculus and mathematical analysis , in his work , Siddhanta Shiromani.
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Calculus is the mathematics of change.
It has two major branches, differential calculus and integral calculus, which arerelated by the Fundamental Theorem of Calculus.
Differential calculus determines varying rates of change. It helps solve
problems involving force
Integration is the "inverse" (or opposite) of differentiation or a sum of series. It
measures accumulations over periods of change. Integration can find volumes
and lengths of curves, measure forces and work, etc.
Calculus has widespread applications in science, economics, finance and
engineering.
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2.3 Introduction to Functions
A is any set of ordered pairs.The set of all first components of the
Definition of a Relation
relation
domainordered pairs is called the of
the relation, and the set of all second
components is called the of the
relat
range
ion.
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Ex 1: Determine whether each
relation is a function.
a. {(4,5), (6,7), (8,8)}
b. {(5,6), (4,7), (6,6), (6,7)}
We begin by making a igure
or each relation that sho s set , the
domain, and set , t
So
he
luti
ra
o
e.
n
ng
X
Y
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Solution for part (a)
4
6
8
5
7
8
X Y
Domain ange
The figure shows that every
element in the domaincorresponds to exactly one
element in the range.
No two ordered pairs in the given relation have
the same first component different second
components. Thus, the relation is a function
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Solution for part (b)
4
5
6
6
7
X Y
Domain ange
The igure sho s that 6corresponds to both 6 and 7.
I any element in the domain corresponds tomore than one element in the range, the
relation is not a unction, Thus, the relation
is not a unction.
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Function Notation
( )
The variable is called the, because it can be assigned any
of the permissible numbers from the domain.
The variable is called the
independentvariable
dependent
, becausevar itsiable value depe
y f x
x
y
!
nds on .x
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Function Notation
The special notation ( ), read " of "
or " at ," represents the value of the
function at the number .
The notation ( ) does not mean
" times ."
f x f x
f x
x
f x
f x
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Ex2: Determine whether each equation
defines yas a function ofx.
2
1. 25
2. 25
x y
x y
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Solution continued
From this last equation we can see that
for each value of , there is one and
only one value of . Thus, the equation
defines y as a function of .
x
y
x
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Solution of22
25x y !2 25y x!
25y x! s The in this last equation sho s that
there are t o values o or in thedomain o
i.e the interval ( ,5). For this reason
the equation does not de ine as a unction o .
y x
f
y x
s
g
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Finding a Functions Domain
I a unction does not model data or
verbal conditions, its domain is the
largest set o real numbers or hich the
value o ( ) is a real number.
f
f x
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Finding a Functions Domain
Exclude rom a unction's domain real
numbers that cause division by zero
and real numbers that result in an even
root o a negative number.
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Ex 5: Find the domain of each
function2a. ( ) 8 5 2
2 b. ( )
5
c. ( ) 2
f x x x
g xx
h x x
!
!
!
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Solution part a
2The unction ( ) 8 5 2 contains
neither division nor an even root.
The domain o is the set o all real numbers.
f x x x
f
!
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Solution part b
2
The function ( ) contains division.5
Because division by zero is undefined, we
must exclude from the domain values of
that cause 5 to be 0. Thus cannot equal
to 5. The domain of function is
g x x
x
x x
g
!
{ | 5}.x x {
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Solution part c
The function ( ) 2 contains an even
root. Because only non-negative numbershave real square roots, the quantity under the
radical sign, 2 must be greater than or
equal to 0. Thus, 2 0 or 2
Th
h x x
x
x x
!
u u
erefore the domain of is { | 2}
or the interval [ 2, ).
h x x u
g
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Limit
We say that the limit o ( ) as approaches is and rite f x x a L
lim ( )x a
f x Lp
!
if the values of ( ) approach as approaches . f x x a
a
( ) y f x!
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c) Find2
3 if 2lim ( ) where ( )
1 if 2x
x x f x f x
xp
{ !
!
-2
62 2
lim ( ) = lim 3
x x
f x xp
p
Note: f(-2) 1
is not involved
23 lim
3( 2) 6
xx
p!
! !
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The Definition ofLimit-I H
lim ( )We say if and only ifx a
f xp
!
given a positive number , there exists a positive such thatI H
if 0 | | , then | ( ) | . x a f xH I
( ) y f x!a
LL I
L I
a H a H
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such that or all in ( , ), x a a aH H{
then we can find a (small) interval ( , )a aH H
( ) is in ( , ). f x L LI I
This means that i e are given a
small interval ( , ) centered at , L L LI I
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Examples2
1. Sho that lim(3 4) 10.x
xp
!
Let 0 be given.I " We need to find a 0 such thatH "if - 2 | ,x H then | (3 4) 10 | .x I
But | (3 4) 10 | | 3 6 | 3 | 2 | x x x I ! !
i | 2 |3
x I So we choose .3
IH !
1
12. Show that lim 1.
x xp!
Let 0 be given. We need to find a 0 such thatI H" "1if | 1 | , then | 1 | .xx
H I
1 11But | 1 | | | | 1 | .x
xx x x
! ! What do we do with the
x?
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1 31I e decide | 1| , then .2 22
x x
1And so
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The right-hand limit off(x), asx approaches a,
equals L
written:
if we can make the valuef(x) arbitrarily close
to Lby takingx to be sufficiently close to the
right ofa.
lim ( )x a
f x Lp
!
a
L
( ) y f x!
One-Sided LimitOne-Sided Limits
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The left-hand limit off(x), asx approaches a,
equals M
written:
if we can make the valuef(x) arbitrarily close
to Lby takingx to be sufficiently close to theleft ofa.
lim ( )x a
f x Mp
!
a
M
( ) y f x!
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2 i 3( )
2 i 3
x xf x
x x
e! "
1. Given
3
lim ( )x
f x
p
3 3lim ( ) lim 2 6
x x f x x
p p! !
2
3 3lim ( ) lim 9
x x f x x
p p! !
Find
Find3
lim ( )x
f xp
Examples
Examples ofOne-Sided Limit
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1, i 0
2. Let ( ) 1, i 0.
x x
f x x x
"
! e Find the limits:
0lim( 1)
xx
p! 0 1 1! !
0a) lim ( )
xf x
p
0 b) lim ( )
xf x
p 0lim( 1)
xx
p! 0 1 1! !
1
c) lim ( )x
f xp 1
lim( 1)x
xp
! 1 1 2! !
1d) lim ( )
xf x
p 1lim( 1)
xx
p! 1 1 2! !
More Examples
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lim ( ) i and only i lim ( ) and lim ( ) .x a x a x a f x L f x L f x L p p p! ! !
For the function
1 1 1lim ( ) 2 because lim ( ) 2 and lim ( ) 2.x x x
f x f x f x p p p
! ! !But
0 0 0lim ( ) does not exist because lim ( ) 1 and lim ( ) 1.x x x f x f x f x p p p! !
1, i 0( )
1, i 0.
x xf x
x x
"!
e
This theorem is used to show a limit does not
exist.
ATheorem
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Examples Using Limit Rule
Ex.
2
3lim 1x xp
2
3 3lim lim1x xxp p!
2
3 3
2
lim lim1
3 1 10
x xx
p p!
! !
Ex.1
2 1lim
3 5x
x
xp
1
1
lim 2 1
lim 3 5
x
x
x
x
p
p
!
1 1
1 1
2 lim lim1
3lim lim5
x x
x x
x
x
p p
p p
!
2 1 1
3 5 8
! !
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More Examples
3 31. Suppose lim ( ) 4 and lim ( ) 2. Findx x f x g xp p! !
3
a) lim ( ) ( )x
f x g xp
3 3
lim ( ) lim ( )x x
f x g xp p
!
4 ( 2) 2! !
3
b) lim ( ) ( )x
f x g xp
3 3
lim ( ) lim ( )x x
f x g xp p
!
4 ( 2) 6! !
3
2 ( ) ( )c) lim
( ) ( )x
f x g x
f x g xp
3 3
3 3
lim 2 ( ) lim ( )
lim ( ) lim ( )
x x
x x
f x g x
f x g x
p p
p p
!
2 4 ( 2) 5
4 ( 2) 4
! !
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Indeterminate forms occur when substitution in the limit
results in 0/0. In such cases either factor or rationalize the
expressions.
Ex.25
5lim
25x
x
xp
Notice form
5
5lim
5 5x
x
x xp
!
Factor and cancel
common factors
51 1
lim5 10x xp
! !
Indeterminate Forms
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The Squeezing Theorem
I ( ) ( ) ( ) hen is near , and if x g x h x x ae e, thenlim ( ) lim ( )
x a x a f x h x L
p p! ! lim ( )
x ag x L
p!
2
0
Show that liExampl m 0e: .x
x sin
x
T
p
!
0Note that e cannot use product rule because limx sin xTp DNE! But 1 sin 1xT e e 2 2 2and so sin . x x xxT e e
2 2
0 0Since lim lim( ) 0,x xx xp p! ! we use the Squeezing Theorem to conclude
20
lim 0.x
x sinx
T
p!
See Graph
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Continuity
A functionfis continuous at the pointx =a if
the following are true:
) ( ) is definedi f a
) lim ( ) existsx a
ii f xp
a
f(a)
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A functionfis continuous at the pointx =a if
the following are true:
) ( ) is definedi f a
) lim ( ) existsx a
ii f xp
) lim ( ) ( )x a
iii f x f ap
!
a
f(a)
xamp e
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At which value(s) of x is the given function
discontinuous?
1. ( ) 2 f x x! 2
92 . ( )3
xg xx
!
Continuous everywhereContinuous everywhere
except at 3x !
( 3) is undefinedg
lim( 2) 2x a x ap !
and so lim ( ) ( )x a
f x f ap
!
-4 -2 2 4
-2
2
4
6
-6 -4 -2 2 4
-10
-8
-6
-4
-2
2
4
xamp e
s
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2, if 13. ( )
1, if 1
x xh x
x
"!
e
1lim ( )
xh x
pand
Thus h is not cont. atx=1.
1!1
lim ( )x
h xp
3!
h is continuous everywhere else
1, if 04. ( )
1, if 0
xF x
x
e!
"
0lim ( )x F xp 1! and 0lim ( )x xp 1!
ThusFis not cont. at 0.x !
F is continuous everywhere else
-2 2 4
-3
-2
-1
1
2
3
4
5
-1
-5 5 1
-3
-2
-1
1
2
3
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Continuous Functions
Apolynomial function y =P(x) is continuous at
every pointx.
A rational function is continuous
at every pointx in its domain.
( )( )
( )p x
R xq x
!
Iffandgare continuous atx =a, then
, , and ( ) 0 are continuous
at
f f g fg g a
g
x a
s {
!
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Example
? A
2Given ( ) 3 2 5,
Show that ( ) 0 has a solution on 1, 2 .
f x x x
f x
!
!
(1) 4 0
(2) 3 0
f
f
! ! "
f(x) is continuous (polynomial) and sincef(1) < 0andf(2) > 0, by the Intermediate Value Theorem
there exists a c on [1, 2] such thatf(c) = 0.
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Limits at Infinity
For all n > 0,1 1lim lim 0
n nx xx xpg pg! !
provided that is defined.
1
nx
Ex.2
2
3 5 1lim
2 4x
x x
xpg
2
2
5 13lim
2 4x
x x
xpg
!
3 0 0 3
0 4 4
! !
Divide
by2
x
2
2
5 1lim 3 lim lim
2lim lim 4
x x x
x x
x x
x
pg pg pg
pg pg
!
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More Examples
3 2
3 2
2 3 21. lim 100 1x
x x
x x xp g
3 2
3 3 3
3 2
3 3 3 3
2 3 2
lim100 1x
x x
x x x x x x
x x x x
pg
!
3
2 3
3 22
lim1 100 1
1x
x x
x x x
p g
!
22
1! !
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0!
2
3 2
4 5 212. lim
7 5 10 1x
x x
x x xpg
2
3 3 3
3 2
3 3 3 3
4 5 21
lim7 5 10 1x
x x
x x x
x x x
x x x x
pg
!
2 3
2 3
4 5 2 1
lim5 10 1
7x
x x x
x x x
p g
!
07
!
2 2 43. lim
12 31x
x x
xpg
2 2 4
lim12 31x
x x
x x xx
x x
pg
!
42
lim31
12x
xx
x
pg
!
21 2
g !
! g
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24. lim 1x
x xpg
2 22
1 1lim
1 1x
x x x x
x xpg
!
2 2
2
1lim
1x
x x
x xpg
!
2
1lim
1x x xpg
! 1 1
0! ! !
g
g g
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Infinite LimitsFor all n > 0,
1
limn
x a x ap
! g
1
lim if is evenn
x an
x ap
! g
1
lim i is oddn
x an
x ap
! g
-8 -6 -4 -2 2
-20
-15
-10
-5
5
10
15
20
-2 2 4 6
-20
-10
10
20
30
40
ore Graphs
-8 -6 -4 -2 2
-20
-15
-10
-5
5
10
15
20
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Examples
Find the limits2
20
3 2 11. lim
2x
x x
xp
2
0
2 13= lim
2x
x xp
3
2
g g! ! g
3
2 12. lim
2 6x
x
xp
3
2 1= lim
2( 3)x
x
xp
! g
-8 -6 -4 -2 2
-20
20
40
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Limit and Trig Functions
From the graph of trigs functions
( ) sin and ( ) cos f x x g x x! !
we conclude that they are continuous everywhere
-10 -5 5 10
-1
-0.5
0.5
1
-10 -5 5 10
-1
-0.5
0.5
1
limsin sin and lim cos cos x c x c
x c x cp p
! !
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Examples
2a) lim sec
x
xT
p
! g 2
b) lim secx
xT
p
! g
3 2c) lim tanx xTp
! 3 2
d) lim tanx
xT p
!
e) lim cotx
xT
p! g
g
3 2g) lim cot
x
xTp
! 3 2
cos 0lim 0
sin 1x
x
xTp! !
4
) lim tanx
xTp
!g
1
Li it d E ti l
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Limit and Exponential
Functions
-6 -4 -2 2 4 6
-2
2
4
6
8
10
, 1x
y a a! "
-6 -4 -2 2 4 6
-2
2
4
6
8
10 , 0 1xy a a!
The above graph confirm that exponential
functions are continuous everywhere.
lim x cx c
a ap
!
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Asymptotes
horizontal asymptotThe line is called a
of the curve ( ) if eihter
ey L
y f x
!
!
lim ( ) or lim ( ) .x x
f x L f x Lpg pg! !
vertical asymptoteThe line is called a
o the curve ( ) i eihter
x c
y f x
!
!
lim ( ) or lim ( ) . x c x c
f x f x p p
! sg ! sg
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Examples
Find the asymptotes of the graphs of the functions
2
2
11. ( )
1
xf x
x
!
1(i) lim ( )
xf x
p! g
There ore the line 1
is a vertical asymptote.
x !
1.(iii) lim ( )x
f xpg
!
1(ii) lim ( )
xf x
p! .g
Therefore the line 1
is a vertical asymptote.
x !
Therefore the line 1
is a horizonatl asymptote.
y !
-4 -2 2 4
-10
-7.5
-5
-2.5
2.5
5
7.5
10
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2
12. ( )
1
xf x
x
!
21 1
1(i) lim ( ) lim
1x x
xf x
xp p !
1 1
1 1 1= lim lim .
( 1)( 1) 1 2x x
x
x x xp p
! !
There ore the line 1
is a vertical asympNO t eT ot .
x !
1(ii) lim ( ) .
xf xp ! g
Therefore the line 1
is a vertical asymptote.
x !
(iii) lim ( ) 0.x
f xpg
!
Therefore the line 0
is a horizonatl asymptote.
y !
-4 -2 2 4
-10
-7.5
-5
-2.5
2.5
5
7.5
10
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