Chapter 5: Limits, Continuity, and Differentiability · Chapter 5 Overview: Limits, Continuity and...

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Chapter 5:

Limits, Continuity, and Differentiability

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Chapter 5 Overview: Limits, Continuity and Differentiability Derivatives and Integrals are the core practical aspects of Calculus. They were the first things investigated by Archimedes and developed by Liebnitz and Newton. The process involved examining smaller and smaller pieces to get a sense of a progression toward a goal. This process was not formalized algebraically, though, at the time. The theoretical underpinnings of these operations were developed and formalized later by Bolzano, Weierstrauss and others. These core concepts in this area are Limits, Continuity and Differentiability. Derivatives and Integrals are defined in terms of limits. Continuity and Differentiability are important because almost every theorem in Calculus begins with the condition that the function is continuous and differentiable. The Limit of a function is the function value (y-value) expected by the trend (or sequence) of y-values yielded by a sequence of x-values that approach the x-value being investigated. In other words, the Limit is what the y-value should be for a given x-value, even if the actual y-value does not exist. The limit was created/defined as an operation that would deal with y-values that were of an indeterminate form. Indeterminate Form of a Number--Defn: "A number for which further analysis is necessary to determine its value."

Means: the number equals 00

, ∞∞ , 00 , 1∞ ,or other strange things.

The formal definition is rather unwieldy and we will not deal with it in this course other that to show the Formal Definition and translate it:

Formal Definition of a Limit

x→aLim f (x) = L if and only if for every ε > 0 , there exists δ > 0 such that

if 0 < x − a <δ , then f (x)− L < ε .

Formal Definition of a Limit means:

When x almost equals a , the limit almost equals y .

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There are four kinds of Limits:

• Two-sided Limits (most often just referred to as Limits) • One-sided Limits • Infinite Limits • Limits at Infinity

Continuity basically means a function’s graph has no breaks in it. The formal definition involved limits. Since all the families of functions investigated in PreCalculus are continuous in their domain, it is easier to look at when a curve is discontinuous rather than continuous. There are four kinds of discontinuity: Removable Discontinuity

(Limx→a

f (x) does exist) Essential Discontinuity (Limx→a

f (x) does not exist) f(a) does not exist

x

y

Point of Exclusion (POE)

x

y

Vertical asymptotes

f(a) exists

x

y

Point of Displacement (POD)

x

y

Jump Discontinuity

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5.1: Limits, L’Hopital’s Rule, and The Limit Definitions of a Derivative As mentioned in the intro to this chapter and last year, the limit was created/defined as an operation that would deal with y-values that were of an indeterminate form.

x→a lim f (x) is read "the limit, as x approaches a, of f of x." What the definition

means is, if x is almost equal to a (the difference is smaller than some small numberδ ), f(x) is almost equal to L (the difference is smaller than some small numberε ). In fact, they are so close, we could round off and consider them equal. In practice, usually . In other words, the limit is the y for a given

x = a--as long as y ≠ 0/0. If y=0/0, we are allowed to factor and cancel the terms that gave the zeros. No matter how small the factors get, they cancel to 1 as long as they do not quite equal 0/0. Ex 1 Find

x→5Lim (x + 2) and

x→−4Lim (x2 + 3)

x→5Lim (x + 2) = 5 + 2 = 7

x→−4Lim (x2 + 3) = (−4)2 + 3 = 19

Ex 2 Find x→5Lim

x2 − 25x − 5

If x = 5 here, x2 − 25x − 5

, which is (x − 5)(x + 5)

x − 5, would =

00

. But with a Limit, x is

only almost equal to 5, and, therefore, x − 5x − 5

=1. So

x→5Lim

x2 − 25x − 5

=x→5Lim

(x + 5)(x − 5)x − 5

=x→5Lim (x + 5)

= 5 + 5

=10

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As we can see from both the graph (in the given window) and the table, while no y-value exists for x=5, the y-values of the points on either side of x=5 show y should be 10.

Basically, the limit allows us to factor and cancel before we substitute the number "a" for x. OBJECTIVE Evaluate Limits algebraically. Evaluate Limits using L’Hopital’s Rule. Recognize and evaluate Limits which are derivatives. Use the nDeriv function on the calculator to find numerical derivatives.

Ex 3 Find

Since if x = 2, x2 + 4x −12x2 + 3x −10

= 00,we must be able to factor and cancel this

fraction. And, in fact, we know one of the factors in each must be x − 2( ) ,

otherwise the fraction would not yield zeros. So,

= 87

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Ex 4 Find

x→ −3lim 2x3 + x2 −13x + 6

x2 + x − 6=

x→ −3lim

x + 3( ) 2x2 −5x + 2( )x + 3( ) x − 2( )

=x→ −3lim 2x2 −5x + 2

x − 2

= 35−5

= −7

Ex 5 Find

Unlike the previous examples, this fraction does not factor. Yet it must simplify, somehow, to eliminate the Indeterminate Number. If we multiply by conjugates to eliminate the radicals from the numerator:

= x→2lim

4 − x − 2( ) 4 − x + 2( )x − 2( ) 4 − x + 2( )

= x→2Lim

4 − x − 2

x − 2( ) 4 − x + 2( )

= x→2Lim

2 − xx − 2( ) 4 − x + 2( )

= x→2Lim

−14 − x + 2

= −12 + 2

= −1

2 2 or

− 2

4

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One of the more powerful tools in Calculus for dealing with Indeterminate Forms and limits is called L’Hopital’s Rule.

L'Hopital's Rule

Iff a( )g a( ) =

00

or ∞∞

, then Limx→a

f x( )g x( ) = Limx→a

f ' x( )g ' x( ) .

This rule allows us to evaluate limits of functions that do not factor, such as those that involve transcendental functions.

Ex 5 (again) Find using L’Hopital’s Rule.

Since, at x = 0, x→2lim

4 − x − 2x − 2

= 4 − 2 − 2

2 − 2 = 0

0, the condition for

L’Hopital’s Rule is satisfied.

x→2lim

4 − x − 2x − 2

=x→2lim

Dx 4 − x − 2( )Dx x − 2( )

=x→2lim

14 − x

−1( )1

= −12 2

Obviously, this is a much faster process than the algebraic one.

Ex 6 Demonstrate that x→0lim

sin xx

=1.

Last year, we used this limit to prove the derivative of sin x. but we never proved that the limit actually equals 1. This limit is relatively difficult to evaluate without L’Hopital’s Rule (and requires a complicated theorem called the Squeeze Theorem), but it is very easy with L’H.

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At x = 0, sin xx

= 00

, L’H applies.

x→0Lim

sin xx

= x→0Lim

cos x1

=1

Obviously, this is not a proof, because of the circular reasoning. We used a derivative whose proof involved the limit that we were evaluating. But we are just interested in the process of applying L’H here. Ex 7 Evaluate

x→0Lim 1+ 2x( )csc x .

This limit yields a different indeterminate form from before, namely, 1∞ . Because the variable is in the base and the exponent, we might apply the logarithm rules before applying L’Hopital’s Rule.

y =x→0lim 1+ 2x( )csc x

ln y =x→0lim csc x ln 1+ 2x( )

=x→0lim

ln 1+ 2x( )sinx

=L 'H

x→0lim

21+ 2xcosx

ln y = 2

y = e2

The Limit Definitions of a Derivative As we recall from last year, the derivative was initially created as a function that would yield the slope of the tangent line. The slope formula for any line though

two points is m = y2 − y1x2 − x1

. When considering a tangent line, though, that equation

has too many variables. We can simplify this some by realizing y = f x( ) . If we consider h to represent the horizontal distance between the points and realize that y = f x( ) , then the two points that form the secant line would be x, f x( )( )and

x + h, f x + h( )( ) .

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(x , f(x))

(x+h , f(x+h))

Then the slope formula becomes m = f (x + h)− f (x)x + h − x

or m = f (x + h)− f (x)h

.

When h = 0, the points would merge and we would have the tangent line. h = 0

gives m = 00

, therefore, we can use the h→0Lim and

will represent the slope of the tangent line.

h→0Lim

f (x + h)− f (x)h

is the derivative.

Since we already know the derivative rules from the last chapter, we will not be using this very often. What we might see instead is the Numerical Derivative, which yields the slope of the tangent line at a specific point. There are two versions of the Numerical Derivative formula:

The Numerical Derivative

f ' a( ) =x→aLim

f (x) − f (a)x − a

or f ' a( ) =h→0Lim

f (a + h)− f (a)h

Often, there are questions on the AP test that look like limit questions, but which are really questions about recognition of these formulas.

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Ex 8 Evaluate h→0Lim

(2 + h)3 − (2)3h

.

We could FOIL the numerator out (using Pascal’s Triangle for ease) and solve this limit algebraically:

h→0Lim

(2 + h)3 − (2)3h

=h→0Lim

8 +12h + 3h2 + h3 − 8h

=h→0Lim

12h + 3h2 + h3h

=h→0Lim 12 + 2h + h2( )

=12

or we could apply L’Hopital’s Rule:

h→0Lim

(2 + h)3 − (2)3h

=h→0Lim

3(2 + h)21

= 3 2( )2=12

But the quickest way is to recognize that h→0Lim

(2 + h)3 − (2)3h

is really f ' 2( ) ,

where f x( ) = x3 . Of course, f ' x( ) = 3x2 and f ' 2( ) = 3 2( )2 =12 .

Ex 9 Evaluate

h→0Lim

Ln(e+ h)−1h

= d dx x=e

Ln x( ) = 1x x=e

= 1e

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Ex 10 Evaluate

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5.1 Homework Set A Evaluate the following Limits.

1. x→−3Lim

x2 + 7x +12x2 − 9

12. x→9Lim

3− xx2 − 81

2. x→0Lim

1− cosxx2 13.

x→πLim cos2 x −1

2cos2 x − 5cos x − 7

3. x→0Lim

x2

tan x 14.

x→1Lim

x2 −1ln x

4. x→2Lim

4x −12x2 − 3x −10

15. x→−1Lim

1− x2

ex+1 −1

5.

x→0Lim x cot x( ) 16.

x→πLim csc x 1+ sec x( )

6. x→1Lim

x2 − xex − 1

17. x→−1Lim

x +1ex+1

7. x→1Lim

x3 −1x2 + 3x − 4

18. x→5Lim

1− x − 5x − 6

8. x→4Lim

ln x4

⎛⎝⎜

⎞⎠⎟

x − 2 19.

x→0Lim 1+ x

2⎛⎝⎜

⎞⎠⎟

cot x

9. x→1Lim

ln xtan x

20. x→π

2

Lim 1− sin2 x

3sin2 x − sin x − 2

10. x→−3Lim

x2 + 4x −1x3 + 3x

21. x→5Lim csc x − 5( )ln x − 4( )

11. x→ 2Lim

x4 + 4x2 −12x3 + x2 − 2x − 2

22. x→πLim

1+ cosxex −1

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Evaluate the following Limits.

23. h→0Lim

3+ h( )3 − 27h

24. x→πLim

sin xx − π

25. h→0Lim

e3+h − e3

h 26.

x→π

2

Lim 1+ sin x

x + π2

27. h→0Lim

cos π + h( ) +1h

28. x→2Lim

ln x − ln2x − 2

29. x→4Lim

sin x − sin4x − 3

30. x→aLim

x3 − a3

x − a

31. x→aLim

x3 − a3

x2 − a2 32. h→0Lim

2 3+ h( )4 −162h

Solve the following multiple choice problems 33. Given the graph of f x( ) below, tell which of the following are TRUE.

A. I only B. I and II C. II and III D. I, II, and III E. III only

I. does not exist.

II. III.

Graph of f

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34. Given the graph of f x( ) below, tell which of the following are TRUE.


35. If a ≠ 0 , then Limx→a

x4 − a4

x − a is

A.a B. 3a2 C. 4a3 D. 0 E. nonexistent

36. If the average rate of change of a function f over the interval from x = 2 to x = 2+ h is given by 4eh − 4sinh , then f ' 2( ) =

A. 0 B. 1 C. 2 D. 3 E. 4 5.1 Homework Set B

1.

limt→2

ln t −1( )tan t −1( ) 12.

limx→π

tan2 x +1cos x

2. limx→e

ln x2( )− 2x − e 13.

limx→4

x +5 − 3x − 4

3. limx→π

tan x1+ cos x 14.

limx→e

2+ln x( )

I. ( )0

2 12

1 Limh

hfh→

+ −=−

II. ( )2 1f = III. ( )

21 Lim

xxf

→=

Graph of f

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4. limx→0

ln 1−15x( )tan x 15.

limt→3

et−3 −13− t

5. limx→0

x2 − 9x2 +16

16. limy→π

y +π cos ysin y

6. limx→0

x12 −5x6 + 6ex 17.

limx→e

1− ln x

sin πxe

⎛

⎝⎜⎞

⎠⎟

7. limx→2


18. limx→9

x3 − 9x2 −5x + 45x2 −11x +18

8. limx→3

x2 − 9x + 4

19.

limx→π

2

cos x

x − π2 sin x

9. limx→0

ex − cos xtan x 20.

limx→4

x2 −16x + 4

10. limx→e

1− ln xx − e 21.

limx→0

csc x ln 1+ 2x( )

11. limt→3

t3 − 9tt2 −5t + 6

22.

limt→2

tan−1 t −1( )ln 2e

t⎛

⎝⎜⎞

⎠⎟

23. limx→e

ln x4( )− 4x − e 24.

limx→2


25. limx→π

tan xx −π

26. limh→0

e5+h − e5

h 26.

limx→ e

ln x2( )−1x2 − e

28. limh→0

2 x + h( )2 − 7 x + h( ) + 6− 2x2 − 7x + 6( )h

29. limh→0

2+ h( )5 − 32h

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30. limx→e

ln x( )−1x − e 31.

limh→0

4+ h( )3 − 64h

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5.2: Continuity, One-sided Limits, and Infinite Limits In 2-1, we said that Lim

x→a f (x) is basically what the y-value should be when x = a,

even if a is not in the domain. For all of the families of functions that we studied last year, this was enough. But what if that y-value might be two different numbers? In a graph of a piece-wise defined function like this,

It is not clear whether the y-value for x=1 is 1 or 2. As we can see from the table of values, x-values less than 1 have y-values that approach 1 while x-values greater than 1 have y-values that approach 2:

The algebraic ways to describe theses differences are one-sided limits. The symbols we use are:

Limx→a-

f (x)or Limx→a+

f (x)

Limx→a-

f (x)reads "the limit as x approaches a from the left," while Limx→a+

f (x) reads

"the limit as x approaches a from the right."

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In this example, Limx→1- f (x) =1and Lim

x→1+ f (x) = 2 . The Lim

x→1 f (x) does not exist,

because the one-sided limits do not equal each other. By “the Limx→1

f (x) does not exist,” we mean there is not one REAL number that the limit equals. OBJECTIVE Evaluate one-sided limits graphically, numerically, and algebraically. Evaluate two-sided limits in terms of one-sided limits. Prove continuity or discontinuity of a given function. Interpret Vertical Asymptotes in terms of one-sided limits.

Ex 1 Does Limx→-1

f (x) exist for f x( ) = x2 +1, if x ≥ −13− x, if x < −1

⎧⎨⎪

⎩⎪

For Lim

x→-1 f (x) to exist, Lim

x→-1− f (x)must equal Lim

x→-1+ f (x) . The domain states

that any number less than -1 goes into 3 - x. Therefore,

Limx→-1−

f (x) = Limx→-1

3− x( ) = 3− −1( ) = 4

Similarly, numbers greater than -1 go into x2 +1, and

Limx→-1+

f (x) = Lim (x→-1

x2 +1)

= (−1)2 +1 = 2

You can see that the two one-sided limits are not equal. Therefore, Limx→-1

f (x) does not exist. We see in the graph that the two parts do not come together:

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There are two main topics where one-sided limits most often come into play: Continuity and Infinite Limits. A limit for a function only exists if the left and right limits are equal. Since this was not much of a concern for the functions in Precalculus, we pretty much ignored the fact, but in Calculus (both here and in college) teachers love to deal with this fact.

exists if and only if

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Ex 2 Evaluate each of the limits given the function pictured below.

a) b) c) d)

e) f) g) h)

i) j) k) l)

a) 1 e) 0 i) -1 b) -1 f) 0 j) -1 c) DNE g) 0 k) -1 d) 1 h) DNE l) -1

y = g (x)

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CONTINUITY One of the main topics early in Calculus is CONTINUITY. It really is a simple concept, which, like the Limit, is made complicated by its mathematical definition. Let us take a look at the formal definition of continuous: Continuous--Defn: "A function f(x) is continuous at x = a if and only if: i. f(a) exists*, ii. Lim

x→a f (x) exists*,

and iii. Limx→a

f (x) = f (a) ." *By “exists,” we mean that it equals a real number.

i) "f(a) exists" means a must be in the domain. ii) " Lim

x→a f (x) exists" means Lim

x→a- f (x) = Limx→a+

f (x) .

iii) " Limx→a

f (x) = f (a) " should be self explanatory.

NB. All the families of functions which were explored in PreCalculus are continuous in their domain.

Ex 3 g x( ) =x2 − 2x +1, if x > −12 , if x = −13− x, if x < −1

⎧

⎨⎪⎪

⎩⎪⎪

Is g(x) continuous at x = -1?

To answer this question, we must check each part of the definition. i) Does g(-1) exist? Yes, the middle line says that -1 is in the domain and it tells us that y = 2 if x = -1. ii) Does the Lim

x→-1 g(x) exist? Yes, the two one-sided limits are equal.

iii) Does Limx→-1

g(x) = g(-1)? No. Limx→-1

g(x)=4, while g(-1) = 2

So g(x) is not continuous at x = -1 because the limit does not equal the function.

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Ex 4 F x( ) = x2 − 5, if x > 0x + 2, if x < 0

⎧⎨⎪

⎩⎪ Is F(x) continuous at x = 0? Why not?

F(x) is not continuous at x = 0 because 0 is not in the domain. Notice neither inequality includes an equals sign.

Ex 5 IfG x( ) =x + 2, if x > −1 3, if x = −1

x2 − 5, if x < −1

⎧

⎨⎪⎪

⎩⎪⎪

, is G(x) continuous at x = 1? Why not?

i) Does G(1) exist? Yes, the second line says that 1 is in the domain

(and it tells us that y = 3, if x = 1). ii) Does the Lim

x→-1 g(x) exist? No. The two-sided limit only exists if the

two one-sided limits are equal. But,

Limx→1- G(x) = Lim

x→1- x2 − 5( ) = −4 and Limx→1+

G(x) = Limx→1+

x + 2( ) = 3

The two one-sided limits are not equal. Therefore,

G(x) is not continuous at x = 1, because Limx→1

G(x)does not exist.

INFINITE LIMITS The other situation where one-sided limits come into play is at vertical asymptotes. Here, the y-value goes to infinity (or negative infinity), which is why these limits are also called.

Ex 6 Evaluate a) Limx→2-

2

x − 2and b) Lim

x→2+ 1− x2 − x

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In both these cases, we are considering a vertical asymptote. The limits, being y values, would be either positive or negative infinity, depending on if the curve went up or down on that side of the asymptote. We can look at the limit algebraically (vs. graphically) though:

a) Limx→2-

2

x − 2= 2

0- = −∞

Note that the 0- is not “from the left” because the 2 is from the left, but rather that x-2 is negative for any x values less than 2.

b) Limx→2+

1− x2 − x

= −10− = +∞

Note in this case that the numerator’s sign effects the outcome (two negatives make a positive).

Note: It is debatable whether the Infinite Limits exist or not. It depends on whether “exist” is defined as equal to a real number or not. Some books would say that Lim

x→2-

2x − 2

exists because there is one answer. It just happens to be a

transfinite number. Other books would say Limx→2-

2

x − 2 does not exist (DNE)

because the answer is not a Real number. There are certain Infinite Limits that we just need to know.

x→0+Lim

1x=∞

x→0−Lim

1x= −∞

x→∞Limxa=∞

x→0+Lim Ln x( ) = −∞

x→π2

+Lim tan x( ) = −∞ x→π

2

−Lim tan x( ) =∞

The first two rules ( and ) reverse signs

if a is negative. All these rules are based on the asymptotic behavior of the functions.

273

Here is the graph of .

As we trace along the graph approaching 0 from the right side, the graph heads up (to ). As we trace along the graph approaching 0 from the left side, the graph heads down (to – ).


As we trace along the graph approaching 0 from the right side, the graph heads down (to – ).

274


As we trace along the

graph approaching from

the right side, the graph heads down (to – ). As we trace along the

graph approaching from

the left side, the graph heads up (to ).

275

5.2 Homework 1. The graph of a function, f, is shown below. Which of the following

statements are TRUE?

A. I only B. II only C. I, II, and III D. I and III E. II and III

2. The graph of a function, f , is shown below. Which of the following



I. ( ) ( )2' 2 Lim

xf f x

→>

II. ( ) ( )2

' 1 Limx

f f x→

<

III. ( )

2 does not existLim

xxf

→

I. ( ) ( )22

LimLim xxf x f x

−+ →→=

II.

f 2( ) = 2

III. ( )

2 does not existLim

xxf

→

Graph of f

Graph of f

276



A. I only B. II only C. I, II, and III D. I and II E. II and III



1 2 3 4

1

2

3

4

x

y


I. ( ) ( ) ( )2 2

2Lim Lim

2x x

f x ff x

x→ →

−>

−

II. ( ) ( )2 3

Lim Limx x

f x f x→ →

=

III. ( ) ( ) ( ) ( )2 3

2 3Lim Lim

2 3x x

f x f f x fx x→ →

− −=

− −

I. f is continuous at x = 3 II. ( ) ( )

2 2Lim Limx x

f x f x+ −→ →

=

III. ( )

2 does not existLim

xxf

→

Graph of f

Graph of f

277

5. The graph of a function, f , is shown below. Which of the following statements are TRUE?


6. The graph of a function, f , is shown below. Which of the following statements are TRUE?

A. I only B. I and III C. I, II, and III D. II only E. II and III

Determine if each of the following functions is continuous at x = a and use the definition to prove it.

I. ( ) ( )1Lim 1x

f x f→

= II. ( ) ( )

1 1Lim Limx x

f x f x+ −→ →

≠

III. ( ) ( ) ( ) ( )2 4

<2 4

Lim Lim2 4x x

f x f f x fx x→ →

− −− −

I. ( ) ( )2Lim 2x

f x f→

= II. ( ) ( )

2 2Lim Limx x

f x f x+ −→ →

=

III. does not exist

Graph of f

Graph of f

278

7. f x( ) =x2 −1, if x > −10 , if x = −14 − x, if x < −1

⎧

⎨⎪⎪

⎩⎪⎪

; a = -1

8. g x( ) = x2 − 4x − 5x2 −1

; a = -1

9. h x( ) =4, if x = 2

x2 − 4x − 2

, if x ≠ 2⎧

⎨⎪⎪

⎩⎪⎪

; a = 2

10. k x( ) = 2x +1, if x ≤ −1

3− x2, if x > −1

⎧⎨⎪

⎩⎪; a = 3

11. k x( ) = 2x +1, if x ≤ −1

3− x2, if x > −1

⎧⎨⎪

⎩⎪; a = -1

12. F x( ) = 2x +1, if x ≤ −1

−x2, if x > −1

⎧⎨⎪

⎩⎪; a = -1

13. H x( ) =2x +1, if x > −10 , if x = −1

−x2, if x < −1

⎧

⎨⎪⎪

⎩⎪⎪

; a = -1

14. G x( ) = Cos x, if x ≤πx2 −πx −1, if x >π

⎧⎨⎪

⎩⎪; a = π

15.

f x( ) = tan−1 x − 3( ) if x ≤ 3

1− cos x − 3( ) if x > 3

⎧⎨⎪

⎩⎪; a = 3

279

16.

f x( ) = ln 1+ x( ) if x < 0x2 +5x if x > 0

⎧⎨⎪

⎩⎪; a = 0

Evaluate the following Limits.

17. Limx→0+

ln xsin x

18. Limx→2−

7xx2 − 4

19. Limx→0−

cot x2 − 5x( )ln x2

Solve the following multiple choice problems. 20. Given the graph of the function,

f x( ) , below, which of the following



21. Given the graph of the function,

f x( ) , below, which of the following


I. ( )

1 Lim

xxf

→ does not exist.

II. ( )

1 1Lim

xxf

+→=

III. ( )

1 3Lim

xxf

−→=

Graph of f

280

A. I only B. I and II C. II and III D. I, II, and III E. III only 22. Evaluate each of the limits given the function pictured below.

a) b) c)

I. ( )

2 Lim

xxf

→ does not exist.

II. ( ) ( )

2 2 Lim Lim

x xx xf f

+ −→ →=

III. ( )

2 3Lim

xxf

→=

Graph of f

y = g (x)

281

d) e) f)

g) h) i)

j) k) l)

5.2 Homework Set B Evaluate the following Limits.

1. lim

x→−4+x2 +16x + 4

2.

limx→1+

ln x −1( )tan πx

2⎛

⎝⎜⎞

⎠⎟

3. limx→0−

ln 1− x( )

4. limt→5−

ln 5− t( )t − 3

5. limv→e−

v2 +1v − e 6.

limx→0+

tan−1 ex − 3x

⎛

⎝⎜

⎞

⎠⎟

7. limx→2−

ln 2− x( )x + 4

8.

limx→4+

x +16ln x − 4( ) 9.

limt→2−

tan−1 −2ln 2− t( )( )

10. limx→5−

tan−1 15− x

⎛

⎝⎜⎞

⎠⎟ln x − 3( ) 11.

limx→1−

ln 1− x( ) 12. limx→0+

5− ln x( )

13. limx→2+

ln x2 − 3 14. limx→3+

x2 −169− x2

15. Evaluate each of the following for the graph of f (x), shown below.

282

a.

lim

x→−2−f x( ) = b.

lim

x→−2+f x( ) = c.

limx→−2

f x( ) =

d.

f −2( ) = e.

limx→0+

f x( ) = f. limx→0−

f x( ) =

g

limx→0

f x( ) = h. f 0( ) = i.

limx→3+

f x( ) =

j.

limx→3−

f x( ) = k. limx→3

f x( ) = l. f 3( ) =

m.

limx→4+

f x( ) = n. limx→4−

f x( ) =

16. Given the function f x( ) = −2x +1

x2 −5x − 6 find each of the following:

a. limx→∞

f x( ) b. lim

x→−∞f x( ) c.

limx→6+

f x( ) d.

limx→6−

f x( ) e. lim

x→−1+f x( )

283

5.3: Differentiability and Smoothness

The second (after Continuity) important underlying concept to Calculus is differentiability. Almost all theorems in Calculus begin with “If a function is continuous and differentiable…” It is actually a very simple idea. Differentiability—Defn: the derivative exists. F(x) is differentiable at a if and

only if F’(a) exists and is differentiable on [a, b] if and only F(x) is differentiable at every point in the interval.

F(x) is differentiable at x=a if and only if i. F(x) is continuous at x=a, ii. Lim

x→a- F '(x) and Limx→a+

F '(x) both exist,

and iii. Limx→a- F '(x) = Lim

x→a+ F '(x)

NB. All the families of functions studied in PreCaculus are Continuous and differentiable in their domains.

OBJECTIVE Determine if a function is differentiable or not. Demonstrate understanding of the connections and differences between differentiability and continuity. There are two ways that a function would not be differentiable: I. The tangent line could be vertical, causing the slope to be infinite. Ex 1 Is y = x3 differentiable at x = 0?

If y = x3 , then dydx

= 1

3x23

. At x = 0, dydx

= 10= dne .

284

II. Just as the Limit does not exist if the two one-sided limits are not equal, a derivative would not exist if the two one-sided derivatives are not equal. That is, the slopes of the tangent lines to the left of a point are not equal to the tangent slopes to the right. A curve like this is called non-smooth. Ex 2 Is the function represented by this curve differentiable at x=1?

We can see that the slope to the left of x=1 is 1, while the slope to the right of x=1 is 0. So this function is not differentiable at x=1. This is an example of a curve that is not smooth.

285

Ex 3 Is G x( ) = Cos x, if x ≤πx2 −πx −1, if x >π

⎧⎨⎪

⎩⎪ differentiable at x = π ?

G ' x( ) = −Sin x, if x ≤π2x −π , if x >π

⎧⎨⎪

⎩⎪

Limx→π- G '(x) = Lim

x→π −Sin x( ) = 0

Limx→π+

G '(x) = Limx→π

2x −πx( ) = 2π −π = π

Clearly, these two one-sided derivatives are not equal. Therefore, G(x) is not differentiable at x = π .

All three of these examples show functions that are continuous, but not differentiable. Continuity does not insure differentiability. But the converse is true.

If f(x) is differentiable, it MUST BE continuous

and

If f(x) is not continuous, it CANNOT BE differentiable

286

Ex 4 Given that f x( ) = k x +1, if x ≤ 32 +mx, if x > 3

⎧⎨⎪

⎩⎪ is differentiable at x=3, find m and k.

f ' x( ) = k1

2 x +1, if x ≤ 3

m, if x > 3

⎧

⎨⎪

⎩⎪⎪

If f(x) is differentiable at x=3, then k 12 3+1

= m→ 14k = m .

If f(x) is differentiable at x=3, it is also continuous. Therefore, at x=3, k x +1 = 2 +mx

k 3+1 = 2 +m32k = 3m + 2

Since both 14k = m and 2k = 3m + 2 must be true, we can use linear

combination or substitution to solve for k and m.

2k = 3 14k

⎛⎝⎜

⎞⎠⎟+ 2

2k = 34k + 2

54k = 2

k = 85→ m = 2

5

287

5.3 Homework 1. Determine if

f x( ) = x23 is differentiable or not.

2. Determine if f x( ) = sin−1 x, if −1≤ x <1

ln x, if x ≥1⎧⎨⎪

⎩⎪ is differentiable or not.

3. Determine if f x( ) = x2 − 2x +1, if x ≤1

ln x, if x >1⎧⎨⎪


4. Determine if f x( ) = x2 + 2x −5, if x ≤1

x3 + x − 4, if x >1⎧⎨⎪


5. Given that f x( ) = mx + 2, if x ≤1

k ln x, if x >1⎧⎨⎪

⎩⎪ is differentiable at x = 1, find m and k.

6. Given that f x( ) = mx −5, if x ≤ −2

kx2 +1, if x > −2⎧⎨⎪

⎩⎪ is differentiable at x = –2 , find m and k.

7. Given that f x( ) = ke2x , if x ≤ 0

3− mx, if x > 0⎧⎨⎪


8. Given that f x( ) = mx − 2, if x ≤ 2

k x2 − 3, if x > 2

⎧⎨⎪


288

9. At what x values is f (x) not differentiable.



Graph of f

Graph of f

289

12. At what x value is function g not differentiable.

13. Use the graph of f below to select the correct answer from the choices below.

Graph of g

Graph of f

Graph of g

290

A.

limx→3

f x( ) = f 3( ) B. f is not continuous at x = 3 C. f is differentiable at x = 3 D.

f ' 1( ) < f ' 4( )

E. limx→3

f x( ) does not exist 14. Use the graph of f below to select the correct answer from the choices below.

A. f has no extremes B. f is continuous at x = 2

C. f is differentiable for x ∈ 0,4( ) D. f has a relative maximum at 2

E. f is concave up for x ∈ 0,4( )

Graph of f

291

5.4: Limits at Infinity and End Behavior

Vocabulary Limit at Infinity—Defn: the y-value when x approaches infinity or

negative infinity Infinite Limit— Defn: a limit where y approaches infinity End Behavior—Defn: The graphical interpretation of a limit at infinity OBJECTIVE Evaluate Limits at infinity. Interpret Limits at infinity in terms of end behavior of the graph. Evaluating limits at infinity for algebraic functions relies on one fact:

x→±∞Lim 1x

= 0

This fact is what led to our general rule about horizontal asymptotes.

Ex 1 Evaluatex→−∞Lim

1− 3x2

4x2 + 3.

We already know, from last year, that this function, y = 1− 3x2

4x2 + 3, has a

horizontal asymptote at y = − 34

. Why is this true? Because

292

x→−∞Lim1− 3x2

4x2 + 3=x→−∞Lim

1− 3x2x2

4x2 + 3x2

=x→−∞Lim

1x2

− 3

4 + 3x2

= x→−∞Lim1x2

− 3

4 +x→−∞Lim

3x2

= 0 − 34 + 0

= − 34

Ex 2 Find

x→∞Lim ex and x→−∞Lim ex and interpret these limits in terms of the end

behavior ofy=ex .

x→∞Lim ex = e∞ =∞ . This means that the right end (+∞ ) goes up.

x→−∞Lim ex = e−∞ = 1e∞

= 1∞= 0 . This means that, on the left (-∞ ), y = ex has a

horizontal asymptote at y = 0. **NB. The syntax we used here—treating ∞ as if it were a real number and “plugging” it in—is incorrect. Though it works as a practical approach, writing this on the AP test will lose you points for process. In particular, this will come up with Improper Integrals. We will discuss it again in Chapter 7.

There are certain Limits at Infinity (which come from our study of end behavior) that we just need to know.

x→∞Lim Tan−1 x = π2

or x→−∞Lim Tan−1 x = −π

2

x→∞Lim ex =∞ and x→−∞Lim ex = 0

For y ' = ky A − y( ) and its solution y = A1 + Be−kt

, x→∞Lim f x( ) = A

293

These last two equations are the Logistic Growth equations. We will discuss them further in a later chapter.

Ex 3 Evaluate

x→∞Lim Tan−1 x − x3( )

x→∞Lim Tan−1 x − x3( ) = Tan−1 x→∞Lim x − x3( )⎛

⎝⎞⎠

= Tan−1 −∞( )= −π

2

Limits at Infinity often involve ratios and L’Hospital’s Rule can be applied. But some math teachers consider this to be “using a cannon to kill a fly.” It is easier to just know the relative growth rates of functions—i.e., which families of functions grow faster than which. The End Behavior studied in PreCalculus gives a good sense of order.

The Hierarchy of Functions 1. Logs grow the slowest. 2. Polynomials, Rationals and Radicals grow faster than logs and

the degree of the EBM determines which algebraic function grows fastest. For example, y = x12grows more slowly thany = x2 .

3. The trig inverses fall in between the algebraic functions at the value of their respective horizontal asymptotes.

4. Exponential functions grow faster than the others. (In BC Calculus, we will see the factorial function, y = n !, grows the fastest.)

5. The fastest growing function in the combination determines the end behavior, just as the highest degree term did among the algebraics.

294

Ex 4 Evaluate (a) x→∞Lim

x − x3

ex − 95 and (b)

x→∞Lim x − x3

x2 − ln x

(a) Repeated iterations of L’Hospital’s Rule will give the same result, but,

because exponential grow faster than polynomials, x→∞Lim

x − x3

ex − 95 = 0

(b) Since x − x3 essentially has a degree of 32

, the denominator has a

higher degree and x→∞Lim

x − x3

x2 − ln x = 0 .

EX 6 Use the concept of the Hierarchy of Functions to find the end behavior of: (a) y = 2xex and (b) y = x Ln x .

(a) For y = 2xex , the exponential y = ex determines the end behavior. So the right end goes up and the left end has a horizontal asymptote at y = 0 . (b) We still must consider the domain, first. There is no end behavior on left, because the domain is y∈ 0, ∞( ) . For y = x Ln x , y = x dominates, so, on the right, the curve goes up.

295

5.4 Homework Set A Evaluate the Limit.

1. Limx→∞

1−15x +12x2

16x2 −1 2. Lim

x→−∞xex

3. Limx→−∞

2+ ex( )Tan−1 x2 +1( ) 4. Limx→∞

ln Tan−1xex

+1⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎞

⎠⎟

5. Limx→∞

cosx18− 4x + x2 6.

Limx→∞

2x +1x48 − 48x

7, Limx→−∞

sin xln −x( ) 8.

Limx→−∞

Tan−1x2 + 5x +1x − 2

⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜

⎞

⎠⎟

9.

Limx→−∞

Tan−1x3 + 7x2 +10x

x + 5⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜

⎞

⎠⎟ 10.

Limx→−∞

Tan−1 x3

x3 − 3x⎛

⎝⎜⎞

⎠⎟

ex +1( )

11. Limx→∞

Cot−1xe−x +1

Use limits to determine the end behavior of the following functions. 12.

f x( ) = ex +1( ) Tan−1x3( )

13.

f x( ) = cosx

x2

14. f x( ) = ln x

x − 4

15. f x( ) = ln x2

4 − x

296

5.4 Homework Set B

1. lim

x→−∞2xex 2.

limx→∞

x2 − 9x2 +16

3. lim

x→−∞tan−1 1+ ex( )

4. limx→∞

x12 −5x6 + 6ex 5.

lim

x→−∞

ex

x12 −5x6 + 6 6.

lim

x→−∞tan−1 x2 −1( )

7. limx→∞

tan−1 11− x

⎛

⎝⎜⎞

⎠⎟ 8.

lim

x→−∞sin tan−1 5x +1( )( ) 9.

limt→∞

t2 + 4et

10. limy→∞

y +5y3 −12y

11. limx→∞

5x2 −15x + 27x2 −16

12. lim

x→−∞5+ e3x

13.

limx→∞

5x2 −15x + 27ex x2 −16( ) 14.

limx→∞

tan−1 1− x2( ) 15. lim

x→−∞

x2 −169− x2

16. limy→∞

16−5y4

y4 −8y2 +12 17.

limt→∞

sin tan−1 x5

5x +1⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

18. lim

x→−∞

ex x2 − 25( )x2 −5x + 4

19. lim

t→−∞e2x − tan−1 x2 + x +1( )( )

20. limt→∞

cos tan−1 et

te +1⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

21. lim

y→−∞

16−5y4( )ex3

y4 −8y2 +12

22. lim

x→−∞tan−1 1− x3( )

23. lim

t→−∞e2x − tan−1 x2 + x +1

5+ x2

⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

297

5.5 Type I Improper Integrals We have only looked into definite integrals over finite regions or integrals on bounded functions. Integrals over an infinite region are called Improper Integrals. Since an integral is a sum, it should make sense that unbounded integrals (integrals over an infinite area) might still be finite. That is, Improper Integrals can be Convergent or Divergent. There are two types of improper integrals: Type I: Definite integrals over an infinite interval (there’s an infinity in one of

the limits of integration) --

1x2 dx

1

∞

∫

Type II: Integrals over an unbounded function (i.e., integrating over a vertical

asymptote) --

1x2 dx

−1

1

∫

In this section, we will only look at Type I Improper Integrals. Though the region is infinite in the x-direction, the area might still be finite. Vocabulary Convergent Integral--Defn: an improper integral that has a total. Divergent Integral --Defn: an improper integral that does not have a total. Objectives Evaluate Type I Improper Integrals. Type I Improper Integral of come in three forms, based on the boundaries. Either the upper boundary, the lower boundary, or both boundaries are infinite.

(a) If

f x( )dxa

b

∫ exists for every number b ≥ a , then

f x( )dxa

∞

∫ = limb→∞

f x( )dxa

b

∫ , provided this limit exists as a finite number.

298

(b) If

f x( )dxa

b

∫ exists for every number a ≤ b , then

f x( )dx

−∞

b

∫ = lima→−∞

f x( )dxa

b

∫ , provided this limit exists as a finite number.

(c) If both

f x( )dxc

∞

∫ and

f x( )dx−∞

c

∫ are convergent, then

f x( )dx

−∞

∞

∫ = f x( )dx−∞

c

∫ + f x( )dxc

∞

∫ , where c is in a,b( ) .

You can choose any c you want – so choose an “easy” c, like 0 or 1. If either of the two integrals on the right diverge, then the entire integral will diverge. The improper integrals are said to be convergent if these limits exist as a finite number and divergent if ANY of the limits do not exist.

Steps to Solving an Improper Integral:

1. Recognize the integral as improper. 2. Rewrite the improper integral using limits. 3. Evaluate the integral, saving the limit until the very end. 4. When you are finished with the integral, you will have an expression

containing the variable you are taking the limit of – evaluate this limit. 5. If your limit gives you a number, the integral converges to that number. If

your limit is undefined or infinite, the integral diverges.

Ex 1 Determine whether

1x2 dx

1

∞⌠⌡⎮

is convergent or divergent.

1x2 dx

1

∞⌠⌡⎮

= lim

b→∞

1x2 dx

1

b⌠⌡⎮

Write the improper integral using limits

= lim

b→∞x−2 dx

1

b

∫ Convert to a power so we can integrate

= lim

b→∞− x−1

1

b Use the FTC

= lim

b→∞−b−1( )− −1−1( )⎡

⎣⎢⎤⎦⎥ Keep using the FTC

= lim

b→∞− 1

b +1⎛

⎝⎜⎞

⎠⎟ Even more of the FTC

=1 Because we got a number we would say our integral converges.

299

1x2 dx

1

∞⌠⌡⎮

converges to 1.


1x dx

1

∞⌠⌡⎮


1x dx

1

∞⌠⌡⎮

= lim

b→∞

1x dx

1

b⌠⌡⎮


= lim

b→∞x−1 dx

1

b

∫ Evaluate the integral, leaving the limit alone

= lim

b→∞ln x

1

b Why don’t we need the absolute values?

= lim

b→∞lnb( )− ln1( )⎡

⎣⎤⎦ Evaluate the limit

=∞ So,

1x dx

1

∞⌠⌡⎮

is divergent.

Note: Tuck this fact away–

1x dx

1

∞⌠⌡⎮

diverges but

1x2 dx

1

∞⌠⌡⎮

converges.


x1+ x2 dx

−∞

∞⌠⌡⎮


x1+ x2 dx

−∞

∞⌠⌡⎮

=

x1+ x2 dx

−∞

0⌠⌡⎮

+ x1+ x2 dx

0

∞⌠⌡⎮

= lim

a→−∞

x1+ x2 dx

a

0⌠⌡⎮

+ limb→∞

x1+ x2 dx

0

b⌠⌡⎮

Let u =1+ x2

du = 2xdx

=

12

lima→−∞

duux=a

x=0

∫ + 12

limb→∞

duux=0

x=b

∫

= 12

lima→−∞

ln 1+ x2( )a

0+ 1

2limb→∞

ln 1+ x2( )0

b

300

= 12

lima→−∞

ln1− ln 1+ a2( )⎡⎣⎢

⎤⎦⎥+ 1

2limb→∞

ln 1+ b2( )− ln1⎡⎣⎢

⎤⎦⎥

=−∞ + ∞

−∞+∞ is another one of those indeterminate forms in Calculus. And in terms of improper integrals, it means our integral is divergent because one of the limits is divergent. It does not matter that both are divergent. It only takes one. Really, you only would need to evaluate one integral at a time. If the first integral divergent, then the entire integral will diverge. If the first integral is convergent, evaluate the second one. The second integral also must converge for the entire integral to do converge.

301

5.5 Homework

1. 1

3x +1( )2 dx

1

∞⌠

⌡⎮⎮ 2. 1

3−w dw

−∞

−1⌠⌡⎮

3. e-x dx

0

∞∫ 4. y3 dy

−∞∞∫

5. xe-x2 dx

−∞∞∫ 6. cosθ dθ

0

∞∫

7. Lnxx

dx1

∞⌠⌡⎮

8. 1t2 + 4

dt−∞

∞⌠⌡⎮

9. x5−x2 dx−∞0∫ 10. 1

x2 + 9 dx

3

∞⌠⌡⎮

11. 1x ln2 x

dxe

∞⌠⌡⎮

12. Which of the following improper integrals converge?

I.

11+ x2

0

∞⌠⌡⎮

dx II.

1x ln x2

∞⌠⌡⎮

dx III.

1x1

∞⌠⌡⎮

dx

(a) I only (b) II only (c) I and II only (d) II and III only (e) I and III only

13.

u− 34 du

1

∞∫

(a) 4 (b)

54

(c)

14

(d) -4 (e) DNE

302

5.6 Type II Improper Integrals Like Type I, Type II Improper Integrals occur in three situations, depending on the vertical asymptote. Either the Asymptote is at the beginning, the end or in the middle of the interval defined by the boundaries. If f is a continuous function on

a,b⎡⎣ ) and is discontinuous at b, then

f x( )dxa

b

∫ = limt→b−

f x( )dxa

t

∫ , if this limit exists as a finite number.

If f is a continuous function on a,b( ⎤⎦ and is discontinuous at a, then

f x( )dxa

b

∫ = limt→a+

f x( )dxt

b

∫ , if this limit exists as a finite number.

If f has a discontinuity at c, where a < c < b , and both

f x( )dxa

c

∫ and

f x( )dxc

b

∫ are

convergent, then

f x( )dxa

b

∫ = f x( )dxa

c

∫ + f x( )dxc

b

∫ .

Objectives Determine convergence or divergence of a Type II Improper Integral.


1x2 dx

−1

0⌠⌡⎮


1x2 dx

−1

0⌠⌡⎮

= limb→0−

1x2 dx

−1

b⌠⌡⎮


= limb→0−

x−2 dx−1

b


= limb→0−

− x−1−1

b

= limb→0−

−b−1( )− −1( )−1⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

= limb→0−

− 1b −1

⎛

⎝⎜⎞

⎠⎟ Evaluate the limit

= dne This integral diverges

303


1x

dx0

4⌠

⌡⎮ is convergent or divergent.

1x

dx0

4⌠

⌡⎮ =

lim

a→0+1x

dxa

4⌠

⌡⎮ Write the improper integral using limits

= lim

a→0+x− 1

2 dxa

4


= lim

a→0+2x

12

a

4

= lim

a→0+4( )− 2a

12⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ Evaluate the limit

=4 This integral converges to 4


14x −5

dx0

2⌠⌡⎮


One might think this is an easy problem:

14x −5

dx0

2⌠⌡⎮

= 14

14x −5

4dx0

2⌠⌡⎮

= 14

Ln 4x −5⎤

⎦⎥

0

2

= 14

Ln 3 − Ln −5( )= 1

4Ln 3

5

This is a classic example of what goes wrong if the integral is not recognized

as improper. The answer 14

Ln 35

is wrong!!!

We need to split this into two integrals – split the integral where the function has a discontinuity, in this case at

x = 5

4 .

304

14x −5

dx0

2⌠⌡⎮

=1

4x − 5dx

0

54∫ +

14x − 5

dx5

4

2

∫

= limb→5

4−

14x −5

dx0

b

∫ + lima→5

4+

14x −5

dxa

2

∫

= limb→5

4−

14

ln 4x −50

b+ lim

a→54+

14

ln 4x −5a

2

= limb→5

4−

14

ln 4b−5 − ln5⎡⎣

⎤⎦ + lim

a→54+

14

ln3− ln 4a −5⎡⎣

⎤⎦

limb→5

4−

14

ln 4b−5⎡⎣

⎤⎦ = dne, therefore,

14x −5

dx0

2⌠⌡⎮

is divergent.

305

5.6 Homework Determine whether each integral is convergent or divergent. Evaluate those that are convergent.

1. 1x5 dx

0

5⌠⌡⎮

2. 1x +1

dx−1

−3⌠⌡⎮

3. csc2 t dt

0

π4∫

4. 1x4 dx

−2

3⌠⌡⎮

5. sec x dx

0

π∫

6. 32x − 3

dx4

0⌠⌡⎮


I.

11+ x2

0

−∞⌠⌡⎮

dx II.

1x ln x1

e⌠⌡⎮

dx III.

1x0

4⌠⌡⎮

dx

(a) I only (b) II only (c) I and II only (d) II and III only (e) I, II, and III only

8.

u−2du

−1

1

∫ (a) 1 (b) 2 (c)

12

(d)

14

(e) DNE

306

9. Which of the following improper integrals diverge?

I.

2x1− x2

0

e⌠⌡⎮

dx II.

1x2

−1

∞⌠⌡⎮

dx III.

1x0

1⌠⌡⎮

dx

(a) I only (b) II only (c) III only (d) II and III only (e) I, II, and III

307

Limit and Continuity Test 1. The function f is differentiable at x = b. Which of the following statements could be false?

(a) limx→b

f x( ) exists (b) limx→b

f x( ) = f b( ) (c) limx→b−

f x( ) = limx→b+

f x( )

(d) limx→b−

f ' x( ) = limx→b+

f ' x( ) (e) None of these

2. The function f is defined for all Reals such that

f x( ) =x2 + kx for x < 5

5sinπ2

x for x ≥ 5

⎧⎨⎪

⎩⎪.

For which value of k will the function be continuous throughout its domain?

(a) −2 (b) −1 (c)

23

(d) 1 (e) None of these


II.

e−x0

∞∫ dx II.

1x2

0

1⌠⌡⎮

dx III.

1x0

1⌠⌡⎮

dx

(a) I only (b) III only (c) I and II only (d) II and III only (e) I and III only

4. If f x( ) = x + 2 for x ≤ 3

4x − 7 for x > 3⎧⎨⎪

⎩⎪, which of the following statements are true?

I. limx→3

f x( ) exists II. f is continuous at x = 3 III. f is differentiable at x = 3 (a) None (b) I only (c) II only (d) I and II only (e) I, II, and III

308

5. Which of the following functions is differentiable at x = 0 ?

(a) f x( ) = 1+ x (b)

f x( ) = x (c)

f x( ) = x2 sin 1x for x ≠ 0

0 for x = 0

⎧⎨⎪

⎩⎪

(d)

f x( ) =1x for x ≠ 0

0 for x = 0

⎧⎨⎪

⎩⎪ (e)

f x( ) = cos x for x < 0

sin x for x ≥ 0⎧⎨⎪

⎩⎪

6. limh→0

sin π2 + h

⎛

⎝⎜⎞

⎠⎟−1

h =

(a)

π2

(b)

π4

(c) 0 (d) −π4

(e) DNE

7. limx→0

sint2dt0

x

∫x3

(a) 0 (b) 1 (c)

13

(d) 3 (e) DNE

8. limx→∞

4x5 + 3x4 + 2x3 + x2 +13x5 − 9x4 + 4x3 +15

=

(a) 0 (b)

34

(c)

43

(d) 3 (e) DNE

9.

u−54du

1

∞∫

(a) 4 (b)

54

(c)

14

(d) -4 (e) DNE

309

9. The graph of a function f is given below. Which of the following statements are true?

(A) I only (B) II only (C) I and II (D) I, II, and III (E) II and III

10. The graph of a function is shown below.

6

4

2

-2

5

Which of the following statement(s) is (are) true? I. lim

x→2−f x( ) exists

II. limx→2+

f x( ) exists

III. limx→2

f x( ) exists (a) I only (b) II only (c) I and II only

I. limx→2

f x( )− f 2( )x − 2

dne

II. limx→2

f x( ) = 4

III. limx→2

f x( ) dne

f ( )

1

1

x

y

310

(d) I and III only (e) I, II, and I

11. At which x-value is f (graphed above) differentiable but not continuous? (a) 0 (b) 1 (c) 2 (d) 4 (e) nowhere

12. f x( ) = sin−1 π x −1( )⎡⎣ ⎤⎦, if x <1

0 , if x =1

Ln x2, if x >1

⎧

⎨⎪⎪

⎩⎪⎪

a) Is f x( ) continuous? Why/Why not? b) Is f x( ) differentiable? Why/Why not?

311

13. 3x

3x2 + 2( )3−∞

∞⌠

⌡⎮⎮ dx 14. 1

4 − x2−2

0⌠

⌡

⎮⎮⎮

dx

15. Given the graph of f (x) below, find the values of the following:

a. Lim

x→−3− f (x) b. Lim

x→−3+ f (x) c. Lim

x→−3 f (x) d. f −3( )

e. Lim

x→−1− f (x) f. Lim

x→−1+ f (x) g. Lim

x→−1 f (x) h. f −1( )

i. Lim

x→1− f (x) j. Lim

x→1+ f (x) k. Lim

x→1 f (x) l. f 1( )

m. Lim

x→3− f (x) n. Lim

x→3+ f (x) o. Lim

x→3 f (x) p. f 3( )

312

5.1 Homework 1.

− 1

6 2.

12

3. 0

4. 13 5. 1 6. 0

7. 35 8. 1 9. 0

10. 19 11.

8 22+ 2

12. −1108

13. 29

14. 2 15. 2

16. 0 17. 0 18. -1 19. e

12 20.

− 2

5 21. 1

22. 0 23. 27 24. -1 25. e3 26.

271

27. 0 28. 12

20. 4

30. 3a

2 31.

32a 32. 216 33. C

34. D 35. C 36. E 5.2 Homework 1. E 2. B 3. D 4. D 5. C 6. B 7. Lim DNE 8. G(-1) DNE

313

9. Continuous 10. Continuous 11. Lim DNE 12. Continuous 13. H(-1) not equal to the LIM 14. Continuous 15. Continuous 16. dne 17. dne 18. −∞ 19. ∞

20. A 21. C 22a) ; b) c) d) = 1 e) f) g) h) i) j) k) l) 5.3 Homework

1. Not differentiable, dydx dne . 2. Not continuous.

3. Not differentiable,

f ' 1−( ) ≠ f ' 1+( ) . 4. Differentiable

5. m = −2, k = −2 6. m = −6, k = 3

2 7. m = −6, k = 3

8. m = 4

3, k = 2

3 9. Not differentiable 10. Not differentiable

11. Not differentiable 12. Differentiable 13. B 14. B

314

5.4 Homework

1.

34 2. 0 3. π 4.

Ln

π4

⎛⎝⎜

⎞⎠⎟

5. 0 6. ∞ 7. 0 8.

−π2

9.

π2

10. π4

11. 0

12. HA

y = −

π2

on Left, Up on the left

13. HA at y = 0 14. HA at y = 0 on Right 15. HA at y = 0 on Left 5.5 Homework

1. 112

2. Diverges 3. 1 4. Diverges

5. 0 6. Diverges 7. Diverges 8. π

2

9. −1

2ln5 10. π

3 11. 1 12. A 13. E

5.6 Homework 1. 2 3 2. Diverges 3. Diverges 4. Diverges 5. Diverges 6. Diverges 7. E 8. E 9. E

Chapter 5: Limits, Continuity, and Differentiability · Chapter 5 Overview: Limits, Continuity and...

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