Ch11 Limits Apreviewtocalculus

Limits: A Preview to Calculus

A112-acre parcel of land

in Pavo, Georgia, is

bounded by a street,

two other property lines,

and Little Creek, which

runs along the back of it.

How did the surveyors in Brooks County determine that the parcel is 112 acres? In this chapter, you will

use limits to find the area of a region with a curved boundary. If we consider Little Creek a function, then

the area below the curve of the function, above Old Pavo Road, and between the two side property

lines represents the parcel of land. Surveyors use limits (which are fundamental later in calculus) to

determine areas of irregular parcels.*

11

*See Section 11.5, Exercises 41 and 42.

OldOld PavoPavo RoadRoadOld Pavo Road

Little Creek

c11aLimitsAPreviewtoCalculus.qxd 6/10/13 3:55 PM Page 1076

1077

IN THIS CHAPTER we will first define what a limit of a function is and then discuss how to find limits of functions.

We will discuss finding limits numerically with tables, graphically, and algebraically. We will use limits to define tangent

lines to curves and then to define the slope of a function––the derivative. We will discuss limits at infinity and the limits of

sequences and summations, and then apply limits to application problems like finding the area below a curve.

• Limit Laws

• Finding Limits

Using Limit Laws

• Finding Limits

Using Direct

Substitution

• Finding Limits

Using Algebraic

Techniques

• Finding Limits

Using Left-Hand

and Right-Hand

Limits

• Definition of

a Limit

• Estimating Limits

Numerically and

Graphically

• Limits That Fail

to Exist

• One-Sided Limits

• Tangent Lines

• The Derivative

of a Function

• Instantaneous

Rates of Change

• Limits at Infinity

• Limits of

Sequences

• Limits of

Summations

• The Area Problem

LIMITS: A PREVIEW TO CALCULUS

L E A R N I N G O B J E C T I V E S

■ Understand the meaning of a limit and be able to estimate limits.

■ Apply limit laws and algebraic techniques to find exact values of limits and understand how

these techniques differ from estimating techniques.

■ Find the tangent line to an arbitrary point on a curve representing a function

and understand how the slope of that line corresponds to the derivative of that function.

■ Find limits at infinity.

■ Use the limits of summations to find the area under a curve.

11.1Introduction to

Limits:

Estimating Limits

Numerically and

Graphically

11.2Techniques for

Finding Limits

11.3Tangent Lines

and Derivatives

11.4Limits at Infinity;

Limits of

Sequences

11.5Finding the Area

Under a Curve


1078

Definition of a Limit

The notion of a limit is a fundamental concept in calculus. The question: “What happens

to the values f (x) of a function f as x approaches the real number a?” can be answered

with a limit.Let us consider the quadratic function If we rewrite this function

as we can see its graph is a parabola opening upward with a vertex at the

point (2, 0). Let’s investigate the behavior of this function f as x approaches 3. We start by

listing a table of values of f(x) for values of x near 3, but not equal to 3. It is important to

take values of x approaching from both the left (values less than 3) and the right (values

greater than 3).

f (x) = (x - 2)2,

f (x) = x2- 4x + 4.

y

1 2 3 4

x

1

2

3

4

As x approaches 3

f (x)approaches

1

f (x) = (x – 2)2x 2.9 2.99 2.999 3 3.001 3.01 3.1

f(x) 0.810 0.980 0.998 ? 1.002 1.020 1.21

x approaches 3 from the left x approaches 3 from the right

f(x) approaches 1 f (x) approaches 1

We see in both the table and the graph that when x is close to 3, the values of f (x) are

close to 1. In other words, as x approaches 3 from either side (left or right), the values of

f(x) approach 1.

WORDS MATH

The limit of the function

as xapproaches 3 is equal to 1.

You may be thinking that if we had evaluated the function at we would

have found it to be equal to 1. Although that is true, the concept of a limit is the behavior

of the function as x approaches a value. In fact, a function does not even have to be defined

at a value for a limit to exist at that value.

f (3) = 1,x = 3,

f (x) = x2- 4x + 4

limxS3

(x2- 4x + 4) = 1

CONCEPTUAL OBJECTIVES

■ Understand that the limit of a function at a point may

exist even though the function may not be defined at

that point.

■ Understand when a limit of a function fails to exist.

■ Understand the difference between a limit of a function

and a one-sided limit of a function.

INTRODUCTION TO LIMITS: ESTIMATING

LIMITS NUMERICALLY AND GRAPHICALLY

SKILLS OBJECTIVES

■ Use tables of values to estimate limits of

functions numerically.

■ Estimate limits of functions by inspecting graphs.

■ Determine whether limits of functions exist.

S E C T I O N

11.1


11.1 Introduction to Limits: Estimating Limits Numerically and Graphically 1079

In other words, the values of f(x) keep getting closer and closer to some real number L as xkeeps getting closer and closer to some real number a (from either side of a). An alternative

notation for is

as

This is read as “ f(x) approaches L as x approaches a.” This is the notation we used in

Section 2.6 when discussing asymptotes of rational functions.

It is important to note in defining the limit as that we do not set

regardless of whether x can equal a. This means that we are interested only in the values

of x close to a and we do not even consider when For all three figures below,

even though in part (b), and in part (c), f(a) is not defined.f (a) Z L,limxSa

f (x) = Lx = a.

x = a,x S a

x S af (x) S L

limxSa

f (x) = L

If the values of f(x) become arbitrarily close to L as x gets sufficiently close to a,

but not equal to a, then

WORDS MATH

The limit of f(x), as x approaches a, is L. limxSa

f (x) = L

The Limit of a FunctionDE F I N I T I O N Study Tip

Imagine a very small difference

between x and a, and imagine

making it smaller and smaller. In the

same way, the difference between

f(x) and L gets smaller and smaller.

y

a

L

x

y

a

L

x

y

a

L

x

f(a) is not definedf (a) Z Lf (a) = L

limxSa

f (x) = LlimxSa

f (x) = LlimxSa

f (x) = L

(a) (b) (c)

Estimating Limits Numerically and Graphically

In this section, we use calculators to make tables of values of functions and we inspect

graphs of functions to surmise whether a limit of a function exists and, if so, to estimate

limits of functions. It is important to note now (and we will summarize again at the end of

this section) that it is possible for calculators and graphing technologies to give incorrectvalues and pictures of behaviors. In the next section, however, we will discuss analyticmethods for calculating limits, which are foolproof.


x 0.9 0.99 0.999 1 1.001 1.01 1.1

f(x) 1.9 1.99 1.999 ? 2.001 2.01 2.1


f (x) approaches 2 f (x) approaches 2

EXAMPLE 1 Estimating a Limit Numerically and Graphically

Estimate the value of using a table of values and a graph.

Solution:

STEP 1 Make a table with values of x approaching 1 from both the left and the right.

limxS1

x2

- 1

x - 1

Technology Tip

Both the table and the graph indicate

that f(x) approaches 2 as xapproaches 1.

STEP 2 Draw the graph of

and inspect the behavior of f (x) as

x approaches 1 from both

the left and the right.

f (x) =

x2- 1

x - 1

Both the table and the graph indicate that our

estimate should be 2. limxS1

x2

- 1

x - 1= 2

■ Answer: - 4

Study Tip

Notice in Example 1 that the limit of

as exists even

though is not in the domain of f.x = 1

x S 1f (x) =

x2- 1

x - 1

y

1 2 3 4

x

1

2

3

4

As x approaches 1

f (x)approaches

2

f (x) = x2 – 1x – 1

■ YOUR TURN Estimate the value of using a table of values and a graph.limxS-2

x2

- 4

x + 2

The graph shows that is not in

the domain of f.x = 1

1080 CHAPTER 11 Limits: A Preview to Calculus

Classroom Example 11.1.1 Answer:Estimate the following limits using a table of values. a. �10 b.

a. b.* limxS

14

x3-

116 x

x -14

limxS5

25 - x2

x - 5

18


Technology Tip

Set the viewing rectangle as

by

Both the table and the

graph indicate that f(x) approaches

0 as x approaches 0.

[- 0.3, 0.4].

[- 0.000021, 0.000021]

Study Tip

The notation corresponds to

and is often used when

zooming in on graphs when values

are very small.

2 * 10-8

2E - 8

EXAMPLE 2 Tables and Graphing Technology PitfallsWhen Estimating Limits

Estimate the value of using a table of values and a graphing utility.

Solution:

If we zoom in closer and closer (x approaches 0 from both sides), one might be led to

believe from the table and the graph that the limit is equal to 0.

limxS0

2x2

+ 4 - 2

x2

x 0 0.00000001 0.0000001 0.00001

f(x) 0.25000 0.26645 0.00000 ? 0.00000 0.26645 0.25000

- 0.00000001- 0.0000001- 0.00001

–1E-6 –6E-7 –2E-7 2E-7 6E-7 1E-60.020.06

x

y

0.38

0.1

0.340.3

0.260.220.180.14

–1E-7 –6E-8 –2E-8 2E-8 6E-8 1E-70.020.06

x

y

0.38

0.1

0.340.3

0.260.220.180.14

f (x) =

2x2+ 4 - 2

x2

In the next section, we will show that this limit is equal to It is important to note that

calculators and graphing utilities can sometimes yield incorrect estimates of limits. In

Section 11.2, we will discuss analytic techniques to find limits that always yield

correct values.

14.

Limits That Fail to Exist

Limits do not necessarily exist. There are three classic examples (Examples 3–5)

illustrated here:

■ Piecewise-defined functions with a jump

■ Functions with oscillating behavior (they never approach a single value)

■ Functions with unbounded behavior (vertical asymptote)


Classroom Example 11.1.2 Answer:

Estimate using a table of values and a graph.limxS0

2x2+ 9 - 3

x2

16



Technology Tip

Both the table and the graph indicate

that f(x) approaches two different

values as x approaches 2.

EXAMPLE 3 A Limit That Fails to Exist Because of a Jump

Show that the following limit does not exist:

where

Solution:

METHOD 1 Make a table with values of x approaching 2 from both the left and the right.

f (x) = b x x 6 2

x2 x Ú 2limxS2

f (x),

x 1.5 1.9 1.99 2 2.01 2.1 2.5

f(x) 1.5 1.9 1.99 ? 4.0401 4.41 6.25


f(x) approaches 2 f(x) approaches 4

f (x) � x2f (x) � x

–2

1

1 2 3 4

2

3

4

5

6

7

8

9

10

x

y

As x approaches 2

f (x)approaches 2

f (x)approaches 4

f(x) =x x < 2x2 x ≥ 2METHOD 2 Draw the graph of

and inspect the behavior of f(x) as xapproaches 2 from both the left and

the right.

f (x) = b x x 6 2

x2 x Ú 2

Both the table and the graph indicate that as x approaches 2, there is a “jump” in the

values of the function f(x).

As x approaches 2 from the left, f(x) approaches 2. As x approaches 2 from the right, f(x)

approaches 4. Since f(x) does not approach a single value (it approaches two different

values), we say that does not exist.limxS2

f (x)

Classroom Example 11.1.3Show that where

does not exist.

Answer:

b1x + 2, x 7 0

- 1- x, x … 0,f (x) =

limxS0

f (x),

The graph approaches different

values from the left and right

of x � 0. So, the limit does

not exist.



EXAMPLE 4 A Limit That Fails to Exist Because ofOscillating Behavior


Solution:

METHOD 1 Make a table with values of x approaching 0 from the left and the right.

At first glance, it appears that may be zero:limxS0

sinapx b

limxS0

sinapx b .

Technology Tip


by Both the table

and the graph illustrate that f(x)

oscillates between and 1 as xapproaches 0.

- 1

[- 1.5, 1.5].

[- 1, 1]

However, selecting other values for x illustrates the continued oscillating behavior

between and 1.- 1

METHOD 2 Use a graphing utility to draw the graph

of on

and inspect the behavior of f(x)

as x approaches 0.

Since the value of f(x) does not approach a single (fixed) value as x approaches 0, we say

that .limxS0

sinapx b does not exist

[- 2, 2]f (x) = sinapx b

x 0 1

f(x) 0 0 0 ? 0 0 0

110

1100-

1100-

110- 1

x 0

f(x) 1 1 ? 1 1 - 1- 1- 1- 1

23

25

27

29-

29-

27-

25-

23

x

y

–1

–0.5

1 2–2 –1

0.5

1

f (x) = sin �x( )

The graph does not approach a

single (fixed) value as xapproaches 1. So, the limit

does not exist.

Classroom Example 11.1.4 Answer:

Show that does not exist.limxS1

cos°p

2

1 - x¢


METHOD 2 Graph and inspect the behavior of f(x) as x approaches 0.

Both the table and the graph indicate that as x approaches 0 from either side, the values

of f(x) continue to grow without bound. Since the values of f(x) do not approach a fixed

real number, does not exist. Even though the limit does not exist, we denote this limxS0

1

x2

f (x) =

1

x2


x

y

2–2

f (x) = 1x2

100

x 0 0.001 0.01 0.1

f(x) 100 10,000 1,000,000 ? 1,000,000 10,000 100

- 0.001- 0.01- 0.1

EXAMPLE 5 A Limit That Fails to Exist Because ofUnbounded Behavior


Solution:

METHOD 1 Make a table with values of x approaching 0 from the left and the right.

limxS0

1

x2.

Technology Tip


by Both the table and

the graph illustrate that the values

of f(x) grow without bound as xapproaches 0.

[- 20, 120].

[- 2, 2]

In Section 2.6, we discussed rational functions that often have vertical asymptotes,

which corresponded to an increasing (or decreasing) behavior without bound on either side

of a particular value of x. In Example 5, we will see that the function values increase

without bound as x approaches 0 from the left or right. Even though the behavior on both

sides of the asymptote is the same, growing positive without bound, we still say that the

limit does not exist—that is, that the limit is infinite—because the values continue to

increase, and therefore, do not approach a fixed real number.

Study Tip

Vertical asymptotes of graphs of

rational functions correspond to

limits that increase without bound.

special type of behavior (growing without bound) with the notation: .

It is important to note that does not represent a number. This notation is used

for the special case of a limit not existing due to growing without bound. In this case,

it indicates that continues to increase without bound as x gets closer and

closer to 0.

f (x) =

1

x2

ˆ

limxS0

1

x2= �

Classroom Example 11.1.5

Show that does not exist.

Answer:

limxS -4

- 1

(x + 4)2



One-Sided Limits

In Example 3, we discussed the limit of a piecewise-defined function:

In this case, we found that did not exist because the function approached two

different values as x approached 2. Recall that as x approached 2 from the left, f(x)

approached 2 and as x approached 2 from the right, f(x) approached 4. We use the following

notation to represent these one-sided limits:

WORDS MATH

f(x) approaches 2 as x approaches 2

from the left.

f(x) approaches 4 as x approaches 2

from the right.

The notation implies x approaching 2 from the left. In other words, we can consider

only those values of x that are less than 2. The notation implies x approaching 2

from the right. In other words, we can consider only those values of x that are greater than 2.

x S 2+

x S 2-

limxS2+

f (x) = 4

limxS2-

f (x) = 2

limxS2

f (x)

f (x) = b x x 6 2

x2 x Ú 2

Study Tip

For the standard two-sided limit to

exist, the left-hand and the right-hand

limits both must exist and both must

be equal.

Left-Hand Limit


by considering only values less than a, then

WORDS MATH

The left-hand limit of f(x) as

x approaches a is L.

In other words, the limit of f(x) as x approaches a from the left is L.

Right-Hand Limit


by considering only values greater than a, then

WORDS MATH

The right-hand limit of f(x) as

x approaches a is L.

In other words, the limit of f(x) as x approaches a from the right is L.

limxSa+

f (x) = L

limxSa-

f (x) = L

DE F I N I T I O N The One-Sided Limit of a Function



EXAMPLE 6 Using a Graph to Find Limits of a Piecewise-Defined Function

Find the indicated limits of the function, if they exist.

a. b.

c. d.

e. f.

Solution:

a. As x approaches 0 from the left, f(x) approaches 0.

b. As x approaches 0 from the right, f (x) approaches 0.

c. Since the left-hand and the right-hand limits are equal, the limit is 0.

d. As x approaches 1 from the left, f(x) approaches 0.

e. As x approaches 1 from the right, f (x) approaches 1.

f. Since the left-hand and the right-hand limits are not equal, does not exist.limxS1

f (x)

limxS1+

f (x) = 1

limxS1-

f (x) = 0

limxS0

f (x) = 0

limxS0+

f (x) = 0

limxS0-

f (x) = 0

limxS1

f (x)limxS1+

f (x)

limxS1-

f (x)limxS0

f (x)

limxS0+

f (x)limxS0-

f (x)

f (x) = L- x x 6 0

0 0 6 x 6 1

x2 x Ú 1

x

y

21–2 –1

4

3

2

1

0

For the standard (two-sided) limit to exist, the left-hand and the right-hand limits must exist and

must be equal. If the one-sided limits are not equal, then the (two-sided) limit does not exist.

if and only if and limxSa+

f (x) = LlimxSa-

f (x) = LlimxSa

f (x) = L

SUMMARY

SECTION

11.1

It is important to note that a function does not have to be defined at

the point where a limit exists. We discussed left-hand,

and right-hand, , limits, and if these exist and are equal,

then the (two-sided) limit exists. If a limit fails to exist due

to unbounded behavior, we use the notation or

limxSa

f (x) = - �.

limxSa

f (x) = �

limxSa+

f (x)

limxSa-

f (x),

In this section, we used tables and graphs to estimate limits. We

also revealed the fallibility of table and graphing methods. We found

that limits do not always exist. The special cases we discussed

where limits fail to exist are

■ Piecewise-defined functions with a jump

■ Some functions with oscillating behavior (never approach a

single value)

■ Functions with unbounded behavior (vertical asymptote)

Classroom Example 11.1.6Find the indicated limits, if they

exist:

a.

b.

c.

d.

e.

Answer:a. 0 b. 0 c. 2 d. 3 e. DNE

limxS -1

g(x)

limxS -1-

g(x)

limxS -1+

g(x)

limxS1-

g(x)

limxS1+

g(x)

g(x) = L 3, x … - 1

1 - x, - 1 6 x 6 1,

1x - 1, x Ú 1

c11aLimitsAPreviewtoCalculus.qxd 6/11/13 11:14 AM Page 1086


In Exercises 1–10, complete a table of values to four decimal places (according to the methods followed in Example 1 and elsewhere) and use the result to estimate the limit.

1. 2. 3. 4. 5.

6. 7. 8. 9. 10. limxS0

ln(x + 1)

xlim

xS0+ x ln xlim

xS0

1

e1/xlimxS0

ex

- 1

xlimxS1

cos(px) + 1

x - 1

limxS0

sin x

xlimxS9

1x - 3

x - 9lim

xS -3 11 - x - 2

x + 3lim

xS -2

x + 2

x2- 4

limxS1

x - 1

x2- 1

In Exercises 11–24, use the graph to estimate the limit, if it exists.

11. limxS1

(x3- 1) 12. lim

xS1(1 - x4 ) 13. lim

xS0

ƒx ƒ

x

14. limxS2

ƒx - 2 ƒ

x - 215. lim

xS0+ ln x 16. lim

xS0 e

-ƒ x ƒ

x

y

–10–8–6–4–2

1 2–2 –1

108642 x

y

–10–8–6–4–2

1 2–2 –1

108642 x

y

–2

–1

1 2–2 –1

2

1

x

y

–2

–1

42 6–2

2

1

x

y

–2

–1

42 31 65–2 –1

2

1

x

y

42 31 5–5 –4 –3 –2 –1

1

0.2

0.4

0.6

0.8

EXERCISES

SECTION

11.1

■ SKILLS



17. where f (x) = e - x2+ 1 x Z 1

2 x = 1limxS1

f (x), 18. where f (x) = e - x3 x Z 0

- 2 x = 0limxS0

f (x),

x

y

–3

–2

–11 2–2 –1

5

4

3

2

1x

y

–4

–3

–2

–1 1 2–2 –1

4

3

2

1

19. limxS0

1

x20. lim

xS0 ln ƒ x ƒ

x

y

–2

–1 1 2–1

4

3

2

1

–4

–3

–2

y

–3

–2

2

1

–4

x

1–1–1

21. limxS0

tan x

3�2

�2

x

y

–3

–4 8–1

–2

1

2

3

2– �

22. limxSp/2

tan x 23. limxS0

cos

p

x24. lim

xS0 sin

1

x2

3�2

�2

x

y

–3

–4 8–1

–2

1

2

3

2– � 4 8–8 –4

1

x

y

2

–2

–1

3 4.5–4.5 –3

x

y

2

–2

1.5–1.5

–1

1

In Exercises 25–32, for the graph of the function f shown, state the value of the given quantity.

x

y

–12 431–4 –2

10987654321

–3 –1

25. limxS -2-

f (x) 26. limxS -2+

f (x) 27. limxS -2

f (x)

28. f (- 2) 29. limxS2-

f (x) 30. limxS2+

f (x)

31. limxS2

f (x) 32. f(2)



In Exercises 33–44, for the graph of the function f shown, state the value of the given quantity.

33. limxS -1-

f (x) 34. lim

xS -1+

f (x) 35. limxS -1

f (x)

36. f (- 1) 37. limxS1-

f (x) 38. limxS1+

f (x)

39. limxS1

f (x) 40. f(1) 41. limxS2+

f (x)

42. limxS2-

f (x) 43. limxS2

f (x) 44. f(2)

x

y

–2

–143–4 –2

4

3

1

–3 –1 2

2

1

In Exercises 45–48, graph the piecewise-defined function and use that graph to estimate the limits, if they exist.

45. 46.

47. 48. limxS0

f (x)f (x) = e ƒ cos x ƒ x Z 0

0 x = 0limxS0

f (x)f (x) = e sin x x 6 0

cos x x 7 0

limxS1

f (x)f (x) = L1

(x - 1)2x Z 1

0 x = 1limxS0

f (x)f (x) = e - x x … 0

x + 1 x 7 0

50. Greatest-Integer Function. The greatest-integer function

is a step function defined by

-

integer n such

Find the following

values, if they exist:

a.

b.

c.

d.

e. 331 � x244limxS0

[ [1] ]

limxS1

[ [x] ]

limxS1+

[ [x] ]

limxS1-

[ [x] ]

that n … x.

[ [x] ] = the greatest

[ [x] ]

49. Heaviside Function. The Heaviside function H, also

called the unit step function, is a discontinuous function

whose value is 0 for negative arguments and 1 for

nonnegative arguments. The function is named after the

mathematician/engineer Oliver Heaviside and is used in

signal processing to represent a signal that switches on at

some time, typically taken to be and is never

turned off.

Find the following

values, if they exist:

a.

b.

c.

d. H(0)

limtS0

H(t)

limtS0+

H(t)

limtS0-

H(t)t

y

–2

–11 3–2

3

2

1

H(t)

–3

–3 2–1

H(t) = e0 t 6 0

1 t Ú 0

(t = 0),

x

y

–2

1 3–2

3

2[[x]]

1

–3

–3 2–1

■ A P P L I C AT I O N S



x 0

f(x) 0 0 0 ? 0 0 0

23

25

29-

29-

25-

23

In Exercises 51 and 52, explain the mistake that is made.

51. Find the limit, if it exists:

Solution:

Make a table of values.

limxS0

cos apx b . 52. Find the limit, if it exists: where

Solution:

Evaluate f (0).

When

This is incorrect. What mistake was made?

limxS0

f (x) = - 1

f (0) = 0 - 1 = - 1f (x) = x - 1.x = 0,

f (x) = e x - 1 x … 0

x + 1 x 70

limxS0

f (x),


limxS0

cos apx b = 0

In Exercises 53–56, determine whether each statement is true or false.

53. If then

54. If then limxSa

f (x) = L.f (a) = L,

f (a) = L.limxSa

f (x) = L,

In Exercises 57 and 58, determine the value for c so that exists.

57. f (x) = e x - 1 x 6 c

1 - x x 7 c

limxSc

f (x)

55. If then exists

and is equal to L.

56. If both the right-hand and left-hand limits exist, then the

two-sided limit exists.

limxSa

f (x)limxSa-

f (x) = limxSa+

f (x) = L,

58. f (x) = d 0 x 6 0

sin(px) 0 6 x 6 c

cos(px) c 6 x 6 1

- 1 x 7 1

In Exercises 59–64, use a graphing utility to determine whether the limit exists. Estimate the limit to three decimal places, if it exists.

59. 61. 63.

60. 62. 64. limxS1

cos(2px) - 1

cos(px) + 1limxS0

x3

- 4

xlim

xS -1 x3

+ x2+ 5x + 5

x + 1

limxS1

sinap2

xb - 1

cos(px) + 1limxS0

x3

+ x2- 3

xlimxS1

x3- x2

+ 7x - 7

x - 1

■ C AT C H T H E M I S TA K E

■ C O N C E P T UA L

■ CHALLENGE

■ T E C H N O L O G Y

c11aLimitsAPreviewtoCalculus.qxd 6/11/13 11:14 AM Page 1090

1091

In the last section, we estimated limits of functions using tables of values and inspecting

graphs. However, we found that such methods can sometimes incorrectly describe behavior.

In this section, we discuss calculating limits exactly using limit laws together with visual

inspections of graphs or algebraic techniques.

Limit Laws

The following properties can be used to calculate limits.

Let a and c be real numbers and let f and g be functions with the following limits:

and

Then

limxSa

g(x) = MlimxSa

f (x) = L

LIMIT LAWS

LAW

NO.

1

2

3

4

5

LIMIT OF

A . . .

Sum

Difference

Constant

(Scalar)

Multiple

Product

Quotient

WORDS

The limit of a sum is the sum

of the limits.

The limit of a difference is the

difference of the limits.

The limit of a constant times a

function is the constant times the

limit of the function.

The limit of a product is the

product of the limits.

The limit of a quotient is the

quotient of the limits (provided

the limit of the denominator is

not equal to 0).

CONCEPTUAL OBJECTIVE

■ Understand when limit laws can and cannot

be applied.

TECHNIQUES FOR FINDING LIMITS

SKILLS OBJECTIVES

■ Find limits of functions using limit laws.

■ Find limits of functions using direct substitution.

■ Find limits of functions using algebra.

■ Use left-hand and right-hand limits to find the limit of

a function.

SECTION

11.2

MATH

limxSa

[ f (x) + g(x)] = limxSa

f (x) + limxSa

g(x) = L + M

limxSa

[ f (x) - g(x)] = limxSa

f (x) - limxSa

g(x) = L - M

limxSa

[cf (x)] = c limxSa

f (x) = cL

limxSa

[ f (x)g(x)] = limxSa

f (x) # limxSa

g(x) = LM

limxSa

c f (x)

g(x)d =

limxSa

f (x)

limxSa

g(x)=

L

M if M Z 0

c11bLimitsAPreviewtoCalculus.qxd 6/10/13 3:53 PM Page 1091


These laws agree with our intuition. Looking at Law 1, if f(x) is close to L and g(x) is close

to M, then it makes sense that is close to Similarly, looking at Law 4,

if f(x) is close to L and g(x) is close to M, then it makes sense that is close to LM.

These five laws will be proved in calculus once we have a precise definition of a limit.

Here are two additional limit laws:

f (x)g(x)

L + M.f (x) + g(x)

Let a be a real number and n be a positive integer and let f be a function with

the following limit:

Then

limxSa

f (x) = L

LIMIT LAWS

Let a and c be real numbers and n be a positive integer. Then

SPECIAL

LIMIT NO. WORDS MATH

1 The limit of a constant function

2 The limit of the identity function

3 The limit of a power function

4 The limit of a radical function limxSa

1n

x = 1n

a, where a 7 0

limxSa

xn= an

limxSa

x = a

limxSa

c = c

SPECIAL LIMITS

LAW

NO.

6

7

LIMIT OF

A . . .

Power

Root

WORDS

The limit of a power is

the power of the limit.

The limit of a root is

the root of the limit.

MATH

limxSa

[ f (x)]n= [ lim

xSa f (x)]n

= Ln

Note: If n is even, we assume L 7 0.

limxSa

2n

f (x) = 2n

limxSa

f (x) = 2n

L

Law 6 can be shown by repeating Law 4 with

Before we find limits using the limit laws, let us first mention four special limits.

g(x) = f (x).

Special Limits 1 and 2 can be found by inspecting the graphs of the constant and identity

functions (Section 1.2). Special Limits 3 and 4 are special cases of Limit Laws 6 and 7

when f (x) = x.


Classroom Example 11.2.1Given the graphs of the functions

f and g, find the following limits:

a.

b.* , where

a and b are real numbers.

c.

d.

e.*

Answer:a. 5 b. a � 4b c. 0

d. e. -7216

limxS -3

f (x) - g(x)

g(x)

limxS5-

1g(x)

limxS -4+

f(x)g(x)

limxS0

[af(x) + bg(x)]

limxS0

[ f (x) + g(x)]

f(x)

g(x)

11.2 Techniques for Finding Limits 1093

Finding Limits Using Limit Laws

Let us now use these limit laws and special limits.

EXAMPLE 1 Finding Limits Using Limit Laws and Graphs

Given the graphs of functions f and g, find the following limits:

V V

2 0

V V

6 -2

x

y

–3 0 32

8

6

4

–2

–4

1–1–2g (x)

f (x)

2

a. b.

c. d.

e. f.

g. h.

Solution (a):

Use the limit of a sum (Law 1).

Inspect the graphs of f and gto determine the limits.

Simplify.

Solution (b):

Use the limit of a difference (Law 2).

Inspect the graphs of f and gto determine the limits.

Simplify.

limxS2

[ f (x) - g(x)] = 8

= 6 - (- 2) = 8

= limxS2

f (x) - limxS2

g(x)

limxS2

[ f (x) - g(x)] = limxS2

f (x) - limxS2

g(x)

limxS0

[ f (x) + g(x)] = 2

= 2 + 0 = 2

= limxS0

f (x) + limxS0

g(x)

limxS0

[ f (x) + g(x)] = limxS0

f (x) + limxS0

g(x)

limxS2

1f (x) + g(x)limxS0

[ f (x)]2

limxS -1

f (x)

g(x)limxS0

f (x)

g(x)

limxS -1

[ f (x)g(x)]limxS2

[ f (x) - 2g(x)]

limxS2

[ f (x) - g(x)]limxS0

[ f (x) + g(x)]



V6

Solution (c):

Use the limit of a difference (Law 2).

Use the limit of a constant

multiple (Law 3).

Inspect the graphs of f and g to

determine the limits.

Simplify.

Solution (d):

Use the limit of a product (Law 4).



Simplify.

Solution (e):

Use the limit of a quotient (Law 5).


determine the limits. =

limxS0

f (x)

limxS0

g(x)=

2

0

lim

xS0 f (x)

g(x)=

limxS0

f (x)

limxS0

g(x)

limxS -1

[ f (x)g(x)] = 0

= (0)(- 1) = 0

= limxS -1

f (x) #

limxS -1

g(x)

limxS -1

[ f (x)g(x)] = lim

xS -1 f (x) # lim

xS -1 g(x)

limxS2

[ f (x) - 2g(x)] = 10

= 6 - 2(- 2) = 10

= limxS2

f (x) - 2 lim xS2

g(x)

= limxS2

f (x) - 2 limxS2

g(x)

limxS2

[ f (x) - 2g(x)] = lim

xS2 f (x) - lim

xS2[2g(x)]

V V

0 -1

V

0

V2

V

-1

V0

Study Tip

If are both

equal to zero, the might

exist. If is nonzero and

is equal to zero, then

does not exist.limxSa

f (x)

g(x)

limxSa

g(x)

limxSa

f (x)

limxSa

f (x)

g(x)

limxSa

f (x) and limxSa

g(x)

limxSa

f (x)

g(x)=

limxSa

f (x)

limxSa

g(x)

V

-2

The limit, does not exist because the limit in the numerator is nonzero and the

limit in the denominator is equal to zero. Hence, Limit Law 5 cannot be used. And

looking at a table of values would indicate unbounded behavior of at x � 0, where f (x)

g(x)

lim

xS0

f (x)

g(x),

but the therefore, the limit does not exist.

Solution (f):

Use the limit of a quotient (Law 5).



Simplify.

limxS -1

f (x)

g(x)= 0

=

0

- 1= 0

=

limxS-1

f (x)

limxS-1

g(x)

limxS -1

f (x)

g(x)=

limxS -1

f (x)

limxS -1

g(x)

lim

xS0-

f (x)

g(x)= - �;lim

xS0+

f (x)

g(x)= �,



V

2

Solution (g):

Use the limit of a power (Law 6).

Inspect the graph of f to

determine the limit.

Simplify.

Solution (h):

Use the limit of a root (Law 7).

Use the limit of a sum (Law 1).



Simplify.

■ YOUR TURN Given the graphs of functions f and g in Example 1, find the following

limits, if they exist:

a. b. c. limxS0

[2 f (x) - g(x)]limxS -1

g(x)

f (x)limxS2

f (x)

g(x)

limxS2

1f (x) + g(x) = 2

= 16 - 2 = 14 = 2

= 1 limxS2

f (x) + limxS2

g(x)

= 1 limxS2

f (x) + limxS2

g(x)

limxS2

1f (x) + g(x) = 1 limxS2

[ f (x) + g(x)]

limxS0

[ f (x)]2

= 4

= 22= 4

= [ limxS0

f(x)]2

limxS0

[ f (x)]2= [ lim

xS0 f (x)]2

EXAMPLE 2 Finding Limits Using Limit Laws and Special Limits

Find the following limits:

a. b.

Solution (a):

Use the limit of a sum

(Law 1).


multiple (Law 3) and the

limit of a power (Law 6).

Use the special limits

1, 2, and 3.

Simplify.

limxS -1

(x3+ 2x + 5) = 2

= 2

= (- 1)3+ 2(- 1) + 5

= limxS -1

x 3+ 2 lim

xS -1 x + lim

xS -1 5

limxS -1

(x3+ 2x + 5) = lim

xS -1 x

3+ lim

xS -1 2x + lim

xS -1 5

limxS1

x2- 3x + 2

x2- 4

limxS -1

(x3+ 2x + 5)

V V

6 -2

■ Answer: a.b. does not exist

c. 4

- 3

Technology Tip


by The table indicates that

f(x) approaches 2 as x approaches - 1.

[- 4, 8].

[- 2, 2]

a.

1 1


Classroom Example 11.2.2Let a, b, and c be real numbers.

Compute the following limits:

a.

b.*

Answer:a. a � b � cb. 3

limxS2a

(x + a)2

x2- a2

limxS -1

(ax2+ bx + c )


Solution (b):

Use the limit of a quotient

(Law 5).

Use the limit of a sum/difference

(Laws 1 and 2).


multiple (Law 3).

Use the special limits 1, 2, and 3.

Simplify.

■ YOUR TURN Find the following limits:

a. b. limxS -1

x2

+ 1

x3+ 2x

limxS1

(x2- x + 2)

limxS1

x2

- 3x + 2

x2- 4

= 0

=

12- 3(1) + 2

12- 4

=

0

- 3= 0

=

limxS1

x2

- 3 limxS1

x + lim

xS1 2

limxS1

x2

- limxS1

4

=

limxS1

x2

- 3 limxS1

x + lim

xS1 2

limxS1

x2

- limxS1

4

=

limxS1

x2

- limxS1

3x + limxS1

2

limxS1

x2

- limxS1

4

limxS1

x2

- 3x + 2

x2- 4

=

limxS1

(x2- 3x + 2)

limxS1

(x2- 4)

V(1)2 V(1) V(2)

V

(1)2

V

(4)

■ Answer: a. 2 b. -23

The table illustrates that f(x)

approaches 0 as x approaches 1.

b.

Finding Limits Using Direct Substitution

Recall from the previous section that the limit of a function can exist even if the function

is not defined [part (c) below], or has another value at that point [part (b) below]. However,

in some cases [part (a) below] the limit is equal to the function value.

y

a

L

x

y

a

L

x

y

a

L

x

f(a) is not definedf (a) Z Lf (a) = L

limxSa

f (x) = LlimxSa

f (x) = LlimxSa

f (x) = L

(a) (b) (c)



It is important to note that direct substitution can be used to find the limits of any

continuous function as long as the value that x is approaching is in the domain of the

function. Sinusoidal functions and exponential functions are examples of continuous

functions for which direct substitution can be used for all real values. Radical functions,

logarithmic functions, and other trigonometric functions have domain restrictions. As long

as the value x is approaching is in the domain, then direct substitution can be used.

Recall from Section 1.2 that a function is continuous if you can draw it without picking up

your pencil (no holes or jumps). All polynomial functions are continuous. In fact, the sine

and cosine functions are also continuous functions.

When the limit of a function at a point is equal to the function value at that point

then the function is said to be continuous at the point a. Limits of continuous functions

can be evaluated using direct substitution.

Look again at Example 2(a), which is the limit of a polynomial function.

where

If we directly substitute into the function to get

we see that we indeed get the same result as the limit:

Look again at Example 2(b), which is the limit of a rational function.

where

If we directly substitute to get we see that we

also get the same limit:

Special attention must be given to the domain of the rational function. Remember that

values of x that make the denominator equal to zero, or must be excluded from

the domain.

d(x) = 0,

limxS1

f (x) = f (1).

f (1) =

12- 3(1) + 2

12- 4

=

0

- 3= 0,x = 1

f (x) =

n(x)

d(x)=

x2- 3x + 2

x2- 4

limxS1

x2- 3x + 2

x2- 4

= limxS1

f (x),

limxS -1

f (x) = f (- 1).

f (- 1) = (- 1)3+ 2(- 1) + 5 = 2,x = - 1

f (x) = x3+ 2x + 5lim

xS -1(x3

+ 2x + 5) = limxS -1

f (x),

limxSa

f (x) = f (a)

Study Tip

Rational functions are continuous

everywhere except where they are

not defined (denominator . = 0)

FUNCTION DIRECT SUBSTITUTION RESTRICTIONS ON a

Polynomial: a is any real number.

Rational: a is any real number

such that d(a) Z 0.

limxSa

f (x) = f (a) =

n(a)

d(a)f (x) =

n(x)

d(x)

limxSa

f (x) = f (a)f (x)

DIRECT SUBSTITUTION: LIMITS OF POLYNOMIAL

AND RATIONAL FUNCTIONS

Study Tip

Direct substitution can be used

on any continuous function.

Radical functions, exponential

functions, logarithmic functions,

and trigonometric functions all

are continuous over some domain.

In that domain, direct substitution

can be used.


Classroom Example 11.2.3Let a, b, and c be real numbers.

Compute the following limits:

a.

b.

c.

Answer:a. c b. 3 c. 0

limxS3p

tan(2x)

limxS -1-

(4x + 1)2

4x2- 1

limxSb

[a(x - b)2+ c]


EXAMPLE 3 Finding Limits Using Direct Substitution

Use direct substitution to find the following limits:

a. b. c.

Solution (a):

The function is a polynomial function with a domain of all real numbers.

Use direct substitution.

Evaluate f(3).

Solution (b):

The function is a rational function.

The domain is the set of all real numbers except 1 and .

The value x is approaching in this limit is 2, which is in the domain.


Evaluate f(2).

Solution (c):

The function is a function with a domain of all real numbers.


Evaluate

■ YOUR TURN Use direct substitution to find the following limits, if possible:

a. b. c. limxSp/2

(x sin x)limxS1

1

x2+ 2

limxS -1

(x3+ 2)

limxSp

(x cos x) = -p

= p cos p = -pf (p).

limxSp

f (x) = f (p)

f (x) = x # cos x

limxS2

x2+ 1

x2- 1

=

5

3

=

22+ 1

22- 1

=

5

3

limxS2

f (x) = f (2)

(- �, - 1) � (- 1, 1) � (1, �)- 1:

f (x) =

x2+ 1

x2- 1

limxS3

(x2- 7) = 2

= 32- 7 = 2

limxS3

f (x) = f (3)

f (x) = x2- 7

limxSp

(x cos x)limxS2

x2+ 1

x2- 1

limxS3

(x2- 7)

Technology Tip


by The graph shows that

f(x) approaches as x approaches 2.53

[- 6, 8].

[- 1, 4]

b.


by The graph shows that f(x)

approaches as x approaches p.-p

[- 5, 5].

[0, 2p]

c.

■ Answer: a. 1 b. c.p

213

V

-1

Finding Limits Using Algebraic Techniques

Of the techniques for finding limits we have looked at thus far, direct substitution is the easiest.

However, there are many times when direct substitution (or limit laws) cannot be used.

For example, to find direct substitution is not permitted because is not x = 2limxS2

x - 2

x2- 4

,



Compute .

Answer: -148

limxS4

4 - x

x3- 64

in the domain of the rational function Furthermore, Limit Law 5 (limit of a

quotient) cannot be used because the limit of the denominator is zero. We can, however,

use algebra to simplify the expression, first so we can then apply direct substitution.

There are three algebraic techniques we will discuss: dividing out a common factor,

rationalizing, and general simplification, which are illustrated in Examples 4 to 6.

x - 2

x2- 4

,

f (x) =

x - 2

x2- 4

.

EXAMPLE 5 Finding a Limit by Rationalizing

Find

Solution:

Note that we cannot use direct substitution, because zero is not in the domain of the function

nor can we use Limit Law 5, because the limit of the denominator is

zero. Instead, the following algebraic steps enable the expression to be simplified

first and then the limit can be found.

f (x) =

2x2+ 4 - 2

x2,

limxS0

2x2

+ 4 - 2

x2.

Technology Tip


by [- 2, 2].

[- 4, 4]

After dividing out the common

factor, direct substitution

works and f(x) approaches as xapproaches 2.

14

x - 2,

Technology Tip


by [- 0.5, 0.5].

[- 2, 2]


EXAMPLE 4 Finding a Limit by Dividing Out a Common Factor

Find

Solution:

Note that we cannot use direct substitution, because is not in the domain of the

function nor can we use Limit Law 5 because the limit of the denominator

would be zero. Instead, the following algebraic steps enable the expression to be simplified

first and then the limit can be found.

Factor the denominator.

Divide out the common factor.

Simplify.

Use direct substitution (let

■ YOUR TURN Find limxS -1

x + 1

x2- 1

.

limxS2

x - 2

x2- 4

=

1

4

=

1

2 + 2=

1

4x S 2).

= limxS2

1

x + 2

= limxS2

(x - 2)

(x - 2)(x + 2)

limxS2

x - 2

x2- 4

= limxS2

(x - 2)

(x - 2)(x + 2)

f (x) =

x - 2

x2- 4

,

x = 2

limxS2

x - 2

x2- 4

.

■ Answer: -12


Classroom Example 11.2.5Let a be a positive real number.

Compute .

Answer: -

1a

2a

limxSa

1a - 1x

x - a

Classroom Example 11.2.6Let a be a nonzero real number

and b be any real number.

Compute

for

Answer: 3ax2

f (x) = ax3+ b.

limhS0

f (x + h) - f (x)

h


Recall from Section 1.2 the difference quotient, An important limit in

calculus is the limit of the difference quotient as the denominator goes to zero:

Notice that the denominator goes to zero, so neither Limit Law 5 nor direct substitution

can be used. Instead, simplifying allows us to find the limit.

limhS0

f (x + h) - f (x)

h

f (x + h) - f (x)

h.

EXAMPLE 6 Finding a Limit by Simplifying

Find given that

Solution:

Let

Expand the numerator.

Simplify the numerator.

Factor the numerator. = limhS0

h(2x + h)

h

= limhS0

2xh + h2

h

= limhS0

x2

+ 2xh + h2- x2

h

limhS0

f (x + h) - f (x)

h= lim

hS0 (x + h)2

- x2

hf (x) = x2.

f (x) = x2.limhS0

f (x + h) - f (x)

h,

Rationalize the

numerator.


Divide out the common

factor.


This demonstrates algebraically the limit we found in Example 2 of Section 11.1. Recall that

this is the one for which technology gave us misleading information.

■ YOUR TURN Find limxS0

2x2

+ 1 - 1

x2.

limxS0

2x2

+ 4 - 2

x2=

1

4

=

1

( 202+ 4 + 2)

=

1

24 + 2=

1

4

= limxS0

1

( 2x2+ 4 + 2)x2

= limxS0

x2

x2 ( 2x2+ 4 + 2)

= limxS0

(x2+ 4) - 4

x2 ( 2x2+ 4 + 2)

limxS0

2x2

+ 4 - 2

x2= lim

xS0 ( 2x2

+ 4 - 2)x2

#( 2x2

+ 4 + 2)

( 2x2+ 4 + 2)

After rationalizing the numerator,

direct substitution works and f(x)

approaches as x approaches 0.14

■ Answer: 12

Divide out the common factor h.


For the function � 2x .limhS0

f (x + h) - f (x)

hf (x) = x2,

= 2x + 0 = 2x

= limhS0

(2x + h)


Classroom Example 11.2.7Find

Answer: 2

f (x) = b 12(x + 3)2

+ 2 x … - 3

�x + 1� x 7 - 3

limxS -3

f (x), where

Classroom Example 11.2.8Compute the following limits,

if they exist:

a.

b.

Answer: a. 1 b. DNE

limxS0

g(x)

limxS1

g(x)

g(x) =

1

x x 6 0

1 0 … x 6 1

�x � x 7 1

μ


Finding Limits Using Left-Hand and Right-Hand Limits

Recall from Section 11.1 that for the standard (two-sided) limit to exist, then the left-hand

and right-hand limits must exist and must be equal. If the one-sided limits are not equal,

then the (two-sided) limit does not exist.

if and only if and

We can now use the techniques from this section (limit laws, direct substitution, and

algebraic techniques) to find the one-sided limits, and if those exist and are equal, then the

result is the standard two-sided limit.

limxSa+

f (x) = LlimxSa-

f (x) = LlimxSa

f (x) = L

EXAMPLE 7 Finding Limits by Evaluating One-Sided Limits

Find where

Solution:

Find the left-hand limit.

Find the right-hand limit.

Since the left-hand and right-hand limits both exist and are equal: .limxS1

f (x) = 1

limxS1+

f (x) = limxS1+

x2

= 12= 1

limxS1-

f (x) = limxS1-

(2 - x) = 2 - 1 = 1

f (x) = e2 - x x 6 1

x2 x 7 1.lim

xS1 f (x),

EXAMPLE 8 Finding Limits by Evaluating One-Sided Limits

Find if it exists, where

Solution:

Find the left-hand limit.

Find the right-hand limit.

Both limits exist, but they are not equal; therefore, does not exist.

This was Example 3 of Section 11.1, where we found the same result by inspecting the graph.

■ YOUR TURN For find the following limits, if they exist:

a. b. limxS1

f (x)limxS0

f (x)

f (x) = L- x x 6 0

0 0 6 x 6 1

x2 x Ú 1

,

limxS2

f (x)

limxS2+

f (x) = limxS2+

x2

= 22= 4

limxS2-

f (x) = limxS2-

x = 2

f (x) = e x x 6 2

x2 x Ú 2.lim

xS2 f (x),

■ Answer: a. 0

b. does not exist



In Exercises 1–12, given the graphs of functions f and g, find the following limits, if they exist.

In Exercises 13–20, given the graphs of functions f and g, find the following limits, if they exist.

In Exercises 21–28, find the indicated limit by using limit laws and special limits.

21. 22. 23. 24.

25. 26. 27. 28. limxS -1

ax2+ 2x - 1

x3+ 2

b 2

limxS1

[(x - 3)(x + 2)]2limxS1

2x2+ 8lim

xS0 x2

+ 2

x2- 1

limxS -1

(x4

- x + 3)limxS9

1xlimxS -2

x3lim

xS5 17

14. limxS -2

[g(x) - f (x)]13. limxS0

[ f (x) + g(x)] 15. limxS -1

f (x)

g(x)

17. limxS1

[ f (x)]216. limxS2

[ f (x)g(x)] 18. limxS 0

[g(x)]2

20. limxS -1

1g(x) - f (x)19. limxS2

1f (x) + g(x)

x

y

–4–3–2–1

1 32–4 –2 –1–3

43

65

21

g (x)

f (x)

x

y

4321

–4–3–2–1

g (x)

f (x)

–3 21–2 –1

1. limxS0

[3f (x) + g(x)] 2. limxS0

[ f (x) - 3g(x)] 3. limxS -2

[ f (x)g(x)]

4. limxS1

[ f (x)g(x)] 5. limxS1

f (x)

g(x)6. lim

xS -1 f (x)

g(x)

7. limxS -2

g(x)

f (x)8. lim

xS0 g(x)

f (x)9. lim

xS -1 [ f (x)]2

10. limxS1

[g(x)]2 11. limxS -2

1f (x) - g(x) 12. limxS0

13f (x) + g(x)

SUMMARY

SECTION

11.2

Exact methods of finding limits include using limit laws (sum,

difference, scalar multiples, product, quotient, powers, and roots

of limits), direct substitution, and algebraic techniques. Special

limits are another aid: constant functions, identity function,

power functions, and radical functions. For all functions, direct

substitution is the simplest method, but it sometimes cannot be

used. For example, if a denominator is equal to zero, then in that

case, the algebraic techniques might help. The algebraic techniques

(simplifying, dividing out a common factor, and rationalizing)

can yield exact values for limits in some cases. The methods

discussed in the previous section (tables and graphs produced by

calculators) yield estimates, and these methods can sometimes

give incorrect behavior.

We also discussed finding limits by first finding the one-sided

limits using the techniques discussed in this section, and if both

one-sided limits exist and are equal, then the traditional two-sided

limit exists.

■ SKILLS

EXERCISES

SECTION

11.2

c11bLimitsAPreviewtoCalculus.qxd 6/11/13 11:33 AM Page 1102


In Exercises 29–50, find the limit, if it exists.

29. 30. 31. 32.

33. 34. 35. 36.

37. 38. 39. 40.

41. 42. 43. 44.

45. 46. 47. 48.

49. 50.

In Exercises 51–62, find

51. 52. 53. 54.

55. 56. 57. 58.

59. 60. 61. 62.

In Exercises 63–70, evaluate the one-sided limits in order to find the limit, if it exists.

63. where 64. where

65. where 66. where

67. where 68. where

69. 70. limxS -4

ƒx + 4 ƒ

x + 4limxS3

ƒx - 3 ƒ

x - 3

f (x) = μsin x x 6

p

2

cos x x 7

p

2

limxSp/2

f (x),f (x) = e sin x x 6 0

cos x x 7 0limxS0

f (x),

f (x) = e - 2x + 1 x 6 1

3x - 1 x Ú 1limxS1

f (x),f (x) = e - x + 1 x … 1

2x + 1 x 7 1limxS1

f (x),

f (x) = e - x + 1 x 6 0

x + 1 x 7 0limxS0

f (x),f (x) = e - x2 x 6 0

x x 7 0limxS0

f (x),

f (x) = - x2- 3x + 1f (x) = - x2

+ 2x + 3f (x) = 1xf (x) =

1

x

f (x) = - 3x2+ 2f (x) = - 2x2

+ 1f (x) = x2- 3f (x) = x2

+ 2

f (x) = - 3x - 2f (x) = - 2x + 3f (x) = 3x + 1f (x) = 5x + 2

limhS0

f (x � h) � f (x)

h.

limxS0

1

x - 1+ 1

xlimxS0

1

x + 2-

1

2

x

limtS -1

1

t+ 1

t + 1limtS2

1

t-

1

2

t - 2limxS1

1x + 8 - 3

x - 1limxS4

2 - 1x

x - 4

limxS0

1x + 4 - 2

xlimxS0

1x + 1 - 1

xlimxS0

e2x

- 1

ex+ 1

limxS0

e2x

- 1

ex- 1

limxS0

sec x

csc xlimxS0

tan x

sec xlim

xSp/2 1 - sin x

cos xlimxS0

1 - cos x

sin x

limxS -2

x4

- 16

x + 2limxS1

x4

- 1

x - 1limxS2

x2

- x - 2

x - 2lim

xS -2 x2

- x - 6

x + 2

limxS0

- 3x2

- x + 5

4x2+ 2x + 3

limxS2

5x2

+ 2x + 7

x2+ x + 6

limxS -5

x2

+ 25

x - 5limxS1

x + 1

x2+ 1

71. Gravity. A person standing near the edge of a cliff

100 feet high throws a rock upward with an initial speed

of 32 feet per second. The height of the rock above the

lake at the bottom of the cliff is a function of time:

We found in Section 2.1,

Exercise 75, that the rock would hit the lake at

approximately 3.59 seconds. Find the velocity of the rock

when seconds (shortly before it hits the lake).t = 3

h(t) = - 16t2+ 32t + 100.

Exercises 71 and 72 involve gravity on a falling object. The height function h(t) is given in terms of time t. The velocity at time

is given by limtSa

h(t) � h(a)

t � a.t � a


72. Gravity. A person holds a pistol straight upward and

fires. The initial velocity of most bullets is around

1200 feet/second. The height of the bullet is a function of

time: We found in Section 2.1,

Exercise 76, that the bullet would hit the ground in

75 seconds. Find the velocity when seconds

(shortly before it hits the ground).

t = 75

h(t) = - 16 t 2

+ 1200t.



In Exercises 73–76, explain the mistake that is made.

73. Find

Solution:

Divide out the x.



74. Find

Solution:

Divide out the x.



= (0 - 1) = - 1

limxS0

x2

- 1

x= lim

xS0 (x - 1)

limxS0

x2

- 1

x.

= (0 - 8) = - 8

limxS0

x3

- 8

x= lim

xS0 (x2

- 8)

limxS0

x3

- 8

x. 75. Find

Solution:

Use direct

substitution.


76. Find

Solution:

Use Limit Law 5 (limit of a quotient).

Use Limit Law 2 (limit

of a difference).

Use special limits 1, 2, and 3.

Simplify.


=

0

0= 1

=

23- 8

2 - 2

=

limxS2

x3

- limxS2

8

limxS2

x - limxS2

2

limxS2

x3

- 8

x - 2=

limxS2

(x3

- 8)

limxS2

(x - 2)

limxS2

x3

- 8

x - 2.

limxS1

x2

- 1

x - 1=

12- 1

1 - 1=

0

0= 1

limxS1

x2

- 1

x - 1.

In Exercises 77–80, determine whether each statement is true or false. Assume d(x) and n(x) are polynomials.

79. If exists and exists, then exists.

80. If then and limxSa+

f (x) = L.limxSa-

f (x) = LlimxSa

f (x) = L,

limxSa

f (x)limxSa+

f (x)limxSa-

f (x)

82. f (x) = L0 x 6 0

sin x 0 6 x 6 c

sin(2x) c 6 x 6 p

In Exercises 81 and 82, determine the value(s) for c so that exists.

81. f (x) = e2x + 6 x 6 c

x2- 3x x 7 c

limxSc

f (x)

77. If and then is either equal to

or (does not exist).

78. If and then limxSa

n(x)

d(x)= 1.n(a) = 0,d(a) = 0

+�- �

limxSa

n(x)

d(x)n(a) Z 0,d(a) = 0



■ CHALLENGE


In Exercises 83 and 84, use a graphing utility to estimate thelimit, if it exists.

83. where

84. where f (x) =

19 - x - 3

1x + 9 - 3limxS0

f (x),

f (x) = x cos xlimxS0

f (x),

In Exercises 85 and 86, use a graphing utility to estimate thelimit, if it exists. Confirm by finding the exact limit using thelimit laws and algebraic techniques.

85. 86. limxS2

114 + x - 4

16 - x - 2limxS3

11 + x - 2

17 - x - 2

SECTION

11.3


■ Understand that the slope of a tangent line to a graph

of a function at a point is equal to the value of the

derivative of that function at that point.

■ Understand that the derivative of a function at a point

is its instantaneous rate of change at that point.

SKILLS OBJECTIVES

■ Use limits to find the slope of a tangent line to a graph

of a function at a point.

■ Use limits to find the derivative of a function.

■ Find the instantaneous rate of change of a function,

which is the value of the derivative of that function at

a point.

TANGENT LINES AND DERIVATIVES

In this section, we will use limits to define tangent lines to curves and then to define a derivativeof a function.As you will see in this section, a derivative of a function defines the rate of change

of that function, and the instantaneous rate of change of a function is the value of the derivative

at a specific point. As you proceed to calculus, you will see many applications of rates of

change, such as housing prices, manufacturing costs, gas mileage of a car as a function of speed,

global warming temperatures, populations, and instantaneous velocity ( just to name a few).

Tangent Lines

In Section 1.2, we discussed functions increasing, decreasing, and being constant on certain

intervals. The parabola shown on the right is increasing on and decreasing on .

Now we want to investigate this behavior in more detail. How quickly is it increasing and how

quickly is it decreasing?

The rate at which the graph is increasing (rising) or decreasing (falling) depends on

which point along the graph we are talking about. Therefore, we classify the rate of change

of a function at a specific point.The slope of a line corresponds to the rate at which the line rises or falls. For lines, the

slope (rate of change) is the same at every point along the line. For graphs other than lines

(like a parabola), the rate at which the graph rises (increases) or falls (decreases) is

different from point to point.

(3, �)(- �, 3)

RisingLess

Quickly

RisingQuickly

x

y

Level

Falling

3FallingMore

Quickly

A tangent line to a curve at a point is a line that just touches that curve at a single point

(as long as you stay in the vicinity of the point).


11.3 Tangent Lines and Derivatives 1105



It is important to note that when we say “tangent line to a curve” we mean at a

specific point. For example, in the figure below, we see that the tangent line in the vicinity

of x � a only touches (intersects but does not cross) the curve at one point, x � a. It does

not matter that the line also intersects the curve at the point x � b. The tangent line may

intersect the curve at other points away from the specified point.

ba

y

x

ba

y

x

Realize that the tangent line to the curve at x � b is a different tangent line.

If we look at the tangent lines to the parabola on the left, we see that they approximate

the behavior of the graph of the function around each point. Specifically, we say that the

slope of the tangent line is equal to the slope of the graph at the point. In other words,

the rate of change (slope) of the tangent line is equal to the rate of change (slope) of the

graph of the function at that point.

Recall that in Section 1.2, we discussed the average rate of change of a function in

terms of the slope of the secant line, msecant:

msecant = average rage of change =

f (x2) - f (x1)

x2 - x1

RisingLess

Quickly

x

y

Level

Falling

FallingMore

Quickly

RisingQuickly

Secant

(x2, f (x2))

(x1, f (x1))

x

y

x1 x2

x2 – x1

f (x2) – f (x1)



The average rate of change is a “global” or macro level description of how the function

changes over some specified segment of the curve. In this section, we seek the rate of

change of the function at a single point, which is a more micro-level description of how

the function is changing. Let us look at the graph of some function f. If we are interested

in the slope of the curve at the point (a, f(a)), then we first consider a nearby point (x, f(x)),

and then we let x approach a.

WORDS MATH

Start with the slope of the secant line to the

graph of f between points (a, f(a)) and (x, f(x)).

Take the limit as .

The result is the slope of the tangent line m to

the graph of the function f at the point (a, f(a)). m = limxSa

f (x) - f (a)

x - a

limxSa

msecant = limxSa

f (x) - f (a)

x - ax S a

msecant =

f (x) - f (a)

x - a

The tangent line to the graph of f at the point (a, f(a)) is the line that

■ passes through the point (a, f(a)) and

■ has slope m:

m is also called the slope of the graph of f at the point (a, f(a)).

m = limxSa

f (x) - f (a)

x - a

A Tangent LineDE F I N I T I O N

Secant Line Secant Lines

Tangent Line

x

y

a xx

y

ax

y

ax

x – a

f (x) – f (a)

(a, f (a)) (a, f (a))(a, f (a))

(x, f (x))

Recall from Section 1.2 that if we let x � a � h, then we can rewrite this slope in terms

of the limit of the difference quotient.

m = limhS0

f (a + h) - f (a)

h

x

y

a a + h

Secant Line

h

f (a + h) – f (a)

(a, f (a))

(a + h, f (a + h))



EXAMPLE 1 Finding the Equation of a Tangent Line to a Curve

Find the equation of the tangent line to the graph of at the point (2, 4).

Solution:

STEP 1 Find the slope of the tangent line.

Write the formula for slope of the

tangent line to a curve.

Let f(x) � x2.

Identify a: (2, 4) � (a, f(a)).

Let a � 2.

Factor the numerator.

Divide out the common factor x � 2.

Let . m � 4

STEP 2 Find the equation of the tangent line.

Write the equation of a line. y � mx � b

Let m � 4. y � 4x � b

The line passes through the point (2, 4). 4 � 4(2) � b

Solve for b. b � �4

The equation for the tangent line is .

STEP 3 Check with a graph.

The graph is a good check because

we see that the tangent line indeed

“touches” the graph of the parabola

at the point (2, 4).

■ YOUR TURN Find the tangent line to the graph of f(x) � x2 at the point (3, 9).

y = 4x - 4

x S 2

m = limxS2

(x + 2)

m = limxS2

(x - 2)(x + 2)

(x - 2)

m = limxS2

x2- 4

x - 2

a = 2

m = limxSa

x2- a2

x - a

m = limxSa

f (x) - f (a)

x - a

f (x) = x2

x

y

–3–2

–4

–11 432–1

43

10

5

21

87

9

6

(2, 4)

y = 4x – 4

f (x) = x2

–5

f(x) f(a)

⎫ ⎬ ⎭ ⎫ ⎬ ⎭

Study Tip

In Step 2 of Example 1, we could

have used the point–slope form to

find the equation of the line.

■ Answer: y � 6x � 9

Technology Tip

To confirm the equation of the

tangent line, enter Y1 � x2 and

Y2 � 4x � 4, and graph both equations.

Classroom Example 11.3.1Find the tangent line to the

graph of

Answer: y = - 3x + 8

at (3, - 1).

f (x) = - (x - 2)3

c11cLimitsAPreviewtoCalculus.qxd 6/10/13 4:39 PM Page 1108

EXAMPLE 2 Finding the Equation of a Tangent Line to a Curve (Using the Difference Quotient)

Find the equation of the tangent line to the graph of f(x) � x2 � 2 at the point (1, 3).

Solution:

STEP 1 Find the slope of the tangent line.

Write the formula for

the slope of the tangent

line to a curve.

Let f(x) � x2 � 2.

Identify a: (1, 3) � (a, f(a)). a � 1

Let a � 1.

Eliminate the parentheses

in the numerator.

Eliminate the brackets in

the numerator.

Simplify.



factor h.

Let m � 2

STEP 2 Find the equation of the tangent line.

Write the equation of a line. y � mx � b

Let m � 2. y � 2x � b

The line passes through the point (1, 3). 3 � 2(1) � b

Solve for b. b � 1

The equation for the tangent line is y � 2x � 1 .

STEP 3 Check with a graph.

h S 0.

m = limhS0

(2 + h)

m = limhS0

h(2 + h)

h

m = limhS0

2h + h2

h

m = limhS0

3 + 2h + h2- 3

h

m = limhS0

(1 + 2h + h2+ 2) - (3)

h

m = limhS0

[(1 + h)2+ 2] - (12

+ 2)h

m = limhS0

[(a + h)2+ 2] - (a2

+ 2)h

m = limhS0

f (a + h) - f (a)

h


Technology Tip

To draw the tangent line to the graph

of f(x) � x2 � 2 at (1, 3), enter

Y1 � x2 � 2. Set the viewing window

as [�2, 5] by [�2, 10]. Enter

2nd DRAW 5:Tangent( ENTER

VARS � Y-VARS 1:Function...

ENTER 1:Y1 ENTER , 1 )

ENTER .

To highlight the tangent line at the

point (1, 3), type

TRACE 1 ENTER .

To confirm the equation of the

tangent line, enter Y1 � x2 � 2 and

Y2 � 2x � 1, and graph both equations.

f(a)⎫ ⎬ ⎭f(a � h)⎫ ⎪ ⎬ ⎪ ⎭

x

1 32–1

y

–2–1

43

10

5

21

87

9

6

y = 2x + 1

f (x) = x2 + 2

(1, 3)

■ YOUR TURN Find the tangent line to the graph of f(x) � �x2 � 1 at the point (�1, �2).■ Answer: y � 2x

Classroom Example 11.3.2Find the tangent line to the

graph of

Answer: y = - 3x + 10

at (3, 1).

f (x) = 2 - (x - 2)3



The Derivative of a Function

We have seen that the slope of the tangent line to a curve at a point is the rate of change

of the curve at that point. In general, the slope of the tangent line to a curve at the

point (x, f(x)) is given by

This limit is a function of x and is called the derivative of f at x. It is denoted and we

say “f prime of x.”

The derivative is an instantaneous rate of change.

f ¿(x)

m = limhS0

f (x + h) - f (x)

h

The derivative of a function f at x, denoted , is

provided this limit exists.

f ¿(x) = limhS0

f (x + h) - f (x)

h

f ¿(x)

Derivative of a FunctionDE F I N I T I O N

Notice that if we let x � a, we get

which is a limit of our difference quotient form of the slope of the tangent line to the graph

of the function f at the point (a, f(a)).

We may want to calculate the derivative (rate of change) of a function at several points.

Therefore, we calculate the derivative first as a function of x and then allow x to take on

certain values.

f ¿(a) = limhS0

f (a + h) - f (a)

h

EXAMPLE 3 Finding the Derivative of a Function

Let f (x) � �x2 � 6x � 3.

a. Find .

b. Find , and .

Solution (a):

Write the formula for the

derivative of a function.

Let f (x) � �x2 � 6x � 3.

Eliminate the parentheses in the numerator.

f ¿(x) = limhS0

( - x2

- 2xh - h2+ 6x + 6h - 3) - ( - x2

+ 6x - 3)h

f ¿(x) = limhS0

[ - (x + h)2

+ 6(x + h) - 3] - ( - x2+ 6x - 3)

h

f ¿(x) = limhS0

f (x + h) - f (x)

h

f ¿(5)f ¿(1), f ¿(2), f ¿(3), f ¿(4)

f ¿(x)

f(x)⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭f(x � h)⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

Study Tip

The derivative of a function

is the limit as of the

difference quotient.

h S 0


Eliminate the brackets in the numerator.

Simplify.


Divide out the common factor h.

Let

Solution (b):

Start with the derivative of

Let

Let

Let

Let

Let

The derivative of a function at a

point is equal to the slope of the

tangent line at that point.

f ¿(3) = 0

f ¿(5) = - 4f ¿(2) = 2

f ¿(4) = - 2f ¿(1) = 4

f ¿(5) = - 2(5) + 6 = - 4x = 5.

f ¿(4) = - 2(4) + 6 = - 2x = 4.

f ¿(3) = - 2(3) + 6 = 0x = 3.

f ¿(2) = - 2(2) + 6 = 2x = 2.

f ¿(1) = - 2(1) + 6 = 4x = 1.

f ¿(x) = - 2x + 6f (x) = - x2+ 6x - 3.

f ¿(x) = - 2x + 6h S 0.

f ¿(x) = limhS0

(- h - 2x + 6)

f ¿(x) = limhS0

h(- h - 2x + 6)

h

f ¿(x) = limhS0

- h2

- 2xh + 6h

h

f ¿(x) = limhS0

- x2

- 2xh - h2+ 6x + 6h - 3 + x2

- 6x + 3

h


■ YOUR TURN Let

a. Find b. Find .f ¿(0), f ¿(2), and f ¿(4)f ¿(x).

f (x) = x2- 4x + 3.

■ Answer:a.b. and

f ¿(4) = 4

f ¿(0) = - 4, f ¿(2) = 0,

f ¿(x) = 2x - 4

FallingRisingLess

Quickly

x

y

–4

–3

–2

–14 651 32

4

8

Level

5

6

7

Zero Slope

MoreNegative

Slope

f �(5) = –4

f �(4) = –2f �(3) = 0f �(2) = 2

f �(1) = 4

LessPositiveSlope

NegativeSlope

RisingQuickly

FallingMore

Quickly1PositiveSlope

EXAMPLE 4 Finding the Derivative of a Function

Let

a. Find .

b. Find , and .

Solution (a):

Write the formula for the

derivative of a function.f ¿(x) = lim

hS0 f (x + h) - f (x)

h

f ¿(9)f ¿(1), f ¿(4)

f ¿(x)

f (x) = 1x.

Technology Tip

To find the derivative of

f (x) � �x2 � 6x �3 at x � 1, enter

MATH � 8:nDeriv( ENTER (�)

X, T, , n x2 � 6 X, T, , n � 3

, X, T, , n , 1 ) ENTER .u

uu

Classroom Example 11.3.3*Let a, b, c, and d be real

numbers, where and

f(x) � ax3 � bx2 � cx � d.

a. Find .

b. Compute

Answer:a. 3ax2 � 2bx � cb. f ¿(1) = 3a + 2b + c

f ¿(2) = 12a + 4b + c

f ¿(1) and f ¿(2).

f ¿(x)

a Z 0,



Let

Rationalize the numerator.

Simplify the numerator

(difference of two squares).



factor h.

Let

Solution (b):

Start with the derivative of

Let

Let

Let

The derivative of a function at a point

is equal to the slope of the tangent

line at that point. Here, we find that

the derivatives are all positive; hence,

the slopes are all positive, but they

become less positive as x increases.

f ¿(9) =

1

6f ¿(4) =

1

4f ¿(1) =

1

2

f ¿(9) =

1

219=

1

6x = 9.

f ¿(4) =

1

214=

1

4x = 4.

f ¿(1) =

1

211=

1

2x = 1.

f ¿(x) =

1

21xf (x) = 1x.

f ¿(x) =

1

1x + 1x=

1

21xh S 0.

f ¿(x) = limhS0

1

1x + h + 1x

f ¿(x) = limhS0

h

h (1x + h + 1x )

f ¿(x) = limhS0

(x + h) - x

h (1x + h + 1x )

f ¿(x) = limhS0

( 1x + h - 1x )h

#( 1x + h + 1x )

1x + h + 1x

f ¿(x) = limhS0

1x + h - 1x

hf (x) = 1x.

RisingLess

Quickly

RisingEven LessQuickly

y

4

3

2

5

6

0

1

x

5 106 7 8 91 432

f �(9) =LessPositiveSlope

LesserPositiveSlope

RisingQuickly

PositiveSlope

16

f �(4) = 14

f �(1) = 12

■ YOUR TURN Let .

a. Find b. Find and .f ¿(2)f ¿(12

), f ¿(1),f ¿(x).

f (x) =

1

x

Instantaneous Rates of Change

In developing the forms of the slope of the tangent line and the derivative of a function at

a point, we mentioned our Section 1.2 definition of average rate of change of a function f :

Average rate of change =

change in y

change in x=

f (x) - f (a)

x - a

Technology Tip

To find the derivative of f(x) �

at x � �1, enter

MATH � 8:nDeriv( ENTER 2nd

X, T, , n ) , X, T, , n , 1

) ENTER .

uu1

1x

The TI will give the approximated

values of the derivatives.

■ Answer:

a.

b. , and

f ¿(2) = -14

f ¿ (12 ) = - 4, f ¿(1) = - 1

f ¿(x) = -

1

x2

f (x � h)

⎫ ⎬ ⎭

f (x)r

Classroom Example 11.3.4Let .

a. Find

b. Compute

Answer:

a.

b. g¿(4) =

1a

4g¿(1) =

1a

2,

1ax

2x

g¿(1) and g¿(4).

g¿(x).

g(x) = 1ax, where a 7 0



EXAMPLE 5 Instantaneous Velocity

A person standing near the edge of a cliff 100 feet high throws a rock upward with an

initial speed of 32 feet per second. The height of the rock above the lake at the bottom of

the cliff is a function of time: We found in Section 2.1,

Exercise 75 that the rock would hit the ground at approximately 3.59 seconds. Find the

average velocity between 2 and 3 seconds and the instantaneous velocity of the rock

when seconds (shortly before it hits the lake).

Solution (a):

Average velocity:

Let

Simplify.

The average velocity of the rock between and seconds is �48 ft/sec. The

negative implies falling.

Solution (b):

Instantaneous velocity:

Let

and

Eliminate the brackets in

the numerator.



factor

Let

The instantaneous velocity at seconds is ft/sec.- 64t = 3

= - 64t S 3.

= limtS3

[- 16(t + 1)]t - 3.

= limtS3

- 16(t - 3)(t + 1)

(t - 3)

= limtS3

- 16t2

+ 32t + 48

t - 3

a = 3.

= limtS3

[ - 16t2

+ 32t + 100] - [ - 16(3)2+ 32(3) + 100]

t - 3h(t) = - 16t2

+ 32t + 100

limtSa

h(t) - h(a)

t - a

t = 3t = 2

= - 48

=

[ - 16(3)2+ 32(3) + 100] - [ - 16(222 + 32(2) + 100]

3 - 2

h(t) = - 16t2+ 32t + 100.

h(3) - h(2)

3 - 2

t = 3

h(t) = - 16t2+ 32t + 100.

As mentioned earlier, this gives us a global- or macro-level rate of change of the function.

To see how the function is changing near a point, a, we let x approach a.

The instantaneous rate of change of a function f, at a point is the

limit of the average rate of change as x approaches a.

Instantaneous rate of change = limxSa

f (x) - f (a)

x - a= f ¿(a)

x = a,

INSTANTANEOUS RATE OF CHANGE

Technology Tip

a. To find the average velocity

between and seconds,

enter On

the home screen, enter

( VARS � Y-VARS ENTER

1:Y1 ENTER ( 3 ) – VARS

� Y-VARS ENTER 1:Y1 ENTER

( 2 ) ) � ( 3 – 2 ) ENTER .

Y1 = - 16x2+ 32x + 100.

t = 3t = 2

b. Finding the instantaneous velocity

at is the same as finding the

derivative of the function h(t) at

, which is

MATH � 8:nDeriv( ENTER

(�) 16 X, T, n x2 32

X, T, n 100 , X, T, n ,

3 ) ENTER

u,+u,

+u,

h¿(3).t = 3

t = 3

Study Tip

The derivative is an instantaneous

rate of change.

Classroom Example 11.3.5Replace the function in

Example 5 by

h(t) � �16t2 � v0 t � s0.

a. Determine the average

velocity between 0 and

2 seconds.

b. Determine the instantaneous

velocity at t � 2 seconds.

Answer: a. �32 � v0 ft/sec

b. �64 � v0 ft/sec


SECTION

11.3


In Exercises 1–10, find the slope of the tangent line to the graph of f at the given point.

1. at the point 2. at the point


5. at the point 6. at the point (1, 7)

7. at the point 8. at the point (�1, 2)

9. at the point (4, 1) 10. at the point (5, 2)

In Exercises 11–20, find the equation of the tangent line to the graph of f at the given point.

11. at the point (2, 17) 12. at the point

13. at the point 14. at the point (4, 10)

15. at the point (�1, 6) 16. at the point


19. at the point (6, 2) 20. at the point (4, 0)

In Exercises 21–30, find the derivative of the function at the specified value of x.

21. where 22. where

23. where 24. where

25. where 26. where

27. where 28. where

29. where 30. where f (x) =

3

x + 2f ¿(- 3),f (x) =

2

x - 5f ¿(2),

f (x) = -

1

1xf ¿(1),f (x) =

1

1xf ¿(1),

f (x) = 1x - 1 + 2f ¿(5),f (x) = 1 - 1x - 3f ¿(12),

f (x) = x4f ¿(- 1),f (x) = x3f ¿(1),

f (x) = 2 - x2f ¿(3),f (x) = 1 - 2x2f ¿(2),

f (x) = 2 - 1xf (x) = 1x - 2

(- 1, - 1)f (x) = -

1

x2(1, 12 )f (x) =

x

x + 1

(2,- 1)f (x) = - 3x2+ 2x + 7f (x) = 2x2

- 3x + 1

f (x) = 3x - 2(- 3, 7)f (x) = - 2x + 1

(5, - 6)f (x) = - 6f (x) = 17

f (x) = 1x - 1f (x) = 1x - 1

f (x) =

2

x2(3, 14 )f (x) =

1

x + 1

f (x) = 7x2(1,- 3)f (x) = - 3x2

(- 3, 18)f (x) = - 5x + 3(0, - 1)f (x) = 4x - 1

(- 1, 2)f (x) = 2(5, - 3)f (x) = - 3

The derivative of a function at a point is equal to the slope

of the tangent line to the curve at that point (provided the

limit exists):

The average rate of change is a global- or macro-level behavior,

whereas the instantaneous rate of change (derivative) represents

how a function is changing at a single point (instant) of time.

f ¿(x) = lim hS0

f (x + h) - f (x)

h

In this section, we defined a tangent line to a curve at a specific

point. The tangent line to the graph of a function f, at the point

(a, f (a)), is the line that

■ touches the graph of f at the point (a, f (a)) and

■ has slope m: or

m = lim hS0

f (a + h) - f (a)

h

m = limxSa

f (x) - f (a)

x - a

SUMMARY

■ SKILLS

EXERCISES

SECTION

11.3

c11cLimitsAPreviewtoCalculus.qxd 6/11/13 11:35 AM Page 1114


In Exercises 31–40, find the derivative

31. 32. 33.

34. 35. 36.

37. Also note any domain restrictions on 38. Also note any domain restrictions on

39. Also note any domain restrictions on 40. Also note any domain restrictions on f ¿(x).f (x) =

1

x3f ¿(x).f (x) =

1

x4

f ¿(x).f (x) =

1

1x - 1f ¿(x).f (x) =

1

1 + 1x

f (x) = 2x2- xf (x) = x - x2f (x) = 9x + 2

f (x) = - 7x + 1f (x) = pf (x) = 2

f ¿(x).

45. Gravity. A person holds a pistol straight upward and fires.

The initial velocity of most bullets is around 1200 feet per

second. The height of the bullet t seconds after it is fired is

a function described by

a. What is the average velocity of the bullet between 30

and 40 seconds after it is fired?

b. What is the instantaneous velocity of the bullet when

sec. (shortly before it hits the ground)?

46. Gravity. For the height function given in Exercise 45:

a. What is the average velocity of the bullet between 37

and 38 seconds after it is fired?

b. What is the instantaneous velocity of the bullet when

sec.?t = 37.5

t = 70

h(t) = - 16t2+ 1200t

41. Gas Mileage. The gas mileage (miles per gallon, mpg) for a

typical sedan can be approximated by the quadratic function

where x represents the

speed of the car in miles per hour (mph) and f represents the

gas mileage in mpg. Find the instantaneous rate of change

of the gas mileage of this sedan when the speed is 70 mph.

42. Gas Mileage. Find the instantaneous rate of change of the

gas mileage of the sedan in Exercise 41 when the speed is

55 miles per hour.

43. Profit. If the profit function P(x) in dollars corresponding to

producing x units is given by

Find the instantaneous rate of change of the profit when

units. Is the profit increasing or decreasing at

40 units? What is the meaning of and what are its units?

44. Profit. If the profit function P(x) in dollars corresponding to

producing x units is given by

Find the instantaneous rate of change of the profit when

units. Is the profit increasing or decreasing at

100 units? What is the meaning of and what are

its units?

P¿(x)

x = 100

P(x) = - x2+ 80x - 1000,

P¿(x)

x = 40

P(x) = - x2+ 80x - 1000,

f (x) = - 0.008(x - 50)2+ 30,

48. Find where

Solution:

Write the formula for the derivative of a function.

Let

Simplify.

Let


f ¿(x) = 2h S 0.

f ¿(x) = limhS0

h2

+ 2h

h= lim

hS0(h + 2)

f ¿(x) = limhS0

[(x + h)2+ 2(x + h) - 1] - (x2

+ 2x - 1)h

f (x) = x2+ 2x - 1.

f ¿(x) = limhS0

f (x + h) - f (x)

h

f (x) = x2+ 2x - 1.f ¿(x),


47. Find where

Solution:

Write the formula for the derivative of a function.

Let

Simplify.

Let


f ¿(x) = 0h S 0.

f ¿(x) = lim hS0

- 3h2

h= lim

hS0(- 3h)

f ¿(x) = lim hS0

[ - 3(x + h)2+ 1] - ( - 3x2

+ 1)h

f (x) = - 3x2+ 1.

f ¿(x) = lim hS0

f (x + h) - f (x)

h

f (x) = - 3x2+ 1.f ¿(x),





For Exercises 49–52, determine whether each statement is true or false.

52. A tangent line to the graph of a function can never be

a vertical line.

53. Find the derivative of the function

54. Find the derivative of the function f (x) = ax2+ b.f ¿(x)

f (x) = ax + b.f ¿(x)

49. The derivative of a constant function is 0.

50. The instantaneous rate of change can never equal the

average rate of change.

51. A tangent line to the graph of a function can never be

a horizontal line.

57. Given find

58. Given find f ¿(x).f (x) =

a

x2,

f ¿(x).f (x) = ax2+ bx + c,55. Explain why does not exist for .

56. The function is defined at but the

derivative is not. Why?f ¿(x)

x = 0f (x) = 1x

f (x) = ƒ x ƒf ¿(0)

In Exercises 59–62, use a graphing utility.

59. Graph and graph where h � 0.1,

h � 0.01, and Based on what you see,

what would you guess to be?

60. Graph and graph where

h � 0.1, h � 0.01, and Based on what you

see, what would you guess to be?f ¿(x)

h = ;0.001.;;

sin (x + h) - sin x

h,f (x) = sin x,

f ¿(x)

h = ;0.001.;

;

ex + h- ex

h,f (x) = ex, 61. Graph and graph

where h � 0.1,

h � 0.01, and Based on what you see,

what would you guess to be?

62. Graph and graph

where h � 0.1, h � 0.01, and Based on

what you see, what would you guess to be?f ¿(x)

h = ;0.001.;;

cos [2(x + h)] - cos(2x)

h,f (x) = cos(2x),

f ¿(x)

h = ;0.001.;

;

(x + h)[ln(x + h) - 1] - x(ln x - 1)

h,

f (x) = x (lnx - 1),


■ Understand that if a limit of a function at infinity

exists, that corresponds to a horizontal asymptote.

■ Understand that if a limit of a sequence at infinity

exists, then the sequence is convergent. If the limit

does not exist, it is a divergent sequence.

LIMITS AT INFINITY;

L IMITS OF SEQUENCES

SKILLS OBJECTIVES

■ Evaluate limits of functions at infinity.

■ Find limits of sequences.

SECTION

11.4

All of the limits we have discussed so far have been where x approaches some constant.

The result was one of two things: Either the limit existed (some real number) or the limit

did not exist. Now we turn our attention to another type of limit called a limit at infinity.

limxS�

f (x)

limxSc

f (x)


■ CHALLENGE



11.4 Limits at Infinity; Limits of Sequences 1117

This examines the behavior of some function f as x gets large (or approaches infinity). We

will also examine the limits of sequences, as n gets large, which will be useful to us in

the last section when we find the area under a curve (graph of a function).

Limits at Infinity

We actually have already found limits at infinity in Section 2.6 when we found horizontal

asymptotes. In Example 5(b) from Section 2.6, we found that the rational function

has a horizontal asymptote and the notation we used in that section was

as

We now use the limit notation from this chapter.

WORDS MATH

The limit of f (x) as x approaches infinity is 2.

The limit of f(x) as x approaches negative infinity is 2.

It is important to note that we use the word “infinity” and the symbol to represent

growing without bound in the positive direction, and we use the words “negative infinity”

and the symbol to represent growing without bound in the negative direction. In the

graph of the rational function on the right, we see that in this case both the right (positive

infinity) and left (negative infinity) limits are equal, at 2. In fact, if any rational function

has a horizontal asymptote, then both the right and left horizontal asymptotes must be the

same. There are, however, other functions such as exponential functions when only one

limit as or exists and the other does not.x S - �x S �

- �

�

limxS -�

a8x2+ 3

4x2+ 1b = 2

limxS�a8x2

+ 3

4x2+ 1b = 2

f (x) S 2x S �,

y = 2,

f (x) =

8x2+ 3

4x2+ 1

an,

Let f be a function; then we use the following notation to represent limits at infinity:

Note the following:

■ Infinity and negative infinity do not represent actual numbers, but

instead indicate growth without bound.

■ The above limits do not have to exist. However, if they do exist, they correspond

to horizontal asymptotes.

(- �)(�)

DE F I N I T I O N Limits at Infinity

HORIZONTAL

LIMIT AT . . . MATH WORDS ASYMPTOTE

Infinity The limit of f(x) as xapproaches infinity

is L.

Negative Infinity The limit of f(x) as xapproaches negative

infinity is M.

y = MlimxS -�

f (x) = M

y = LlimxS�

f (x) = L

8–8 0

y

5

4

3

1

2y = 2

f (x) = 8x2 + 34x2 + 1

x �x –�

x

c11dLimitsAPreviewtoCalculus.qxd 6/10/13 5:38 PM Page 1117

x f(x)

-

1

1000= - 0.001- 1000

-

1

100= - 0.01- 100

-

1

10= - 0.1- 10

x f(x)

1

1000= 0.0011000

1

100= 0.01100

1

10= 0.110

54321–5 –4 –3 –2–2–4–6–8

–10

–1

y

10

68

42

y = 0 is a horizontalasymptote

y = 0 is a horizontalasymptote

f (x) = 1x

x

We see that as x gets large either in the positive or negative direction, approaches 0.

and

Using these limits with the limit of a power (Law 6 from Section 11.2) yields the following

special limits.

limxS -�

1

x= 0lim

xS� 1

x= 0

1

x

HORIZONTAL

LIMIT AT . . . MATH WORDS ASYMPTOTE

Infinity The limit of as

x approaches infinity

is 0.

Negative Infinity The limit of as

x approaches negative

infinity is 0.

y = 01

x nlimxS -�

1

xn = 0

y = 01

x nlimxS�

1

x n = 0

Let n be any positive integer. Then

SPECIAL LIMITS

Before reading Example 1, reread Section 2.6, Example 5. The graphs of all three

functions are shown there, and the horizontal asymptotes are found. In Section 2.6, we

stated a rule for finding horizontal asymptotes of rational functions by comparing degrees of

The limit laws from Section 11.2 hold for limits at infinity. There are two special limits at infinity that we will use in combination with the limit laws to evaluate limits at

infinity. Special cases of these special limits are

and

Recall the reciprocal function, which was discussed in Section 1.2. Inspecting

a table of values as x gets large in both directions and the graph of the function, we see that

is the horizontal asymptote.

As AS x S ˆx S �ˆ

y = 0

f (x) =

1

x,

limxS�

1

xlim

xS -� 1

x




the numerator and denominator. Here, we present an algebraic technique to determine

limits at infinity for rational functions. The first step is to divide the numerator and

denominator by where n is the degree of the denominator.xn,

EXAMPLE 1 Finding Limits at Infinity for Rational Functions

Find the following limits, if they exist:

a. b. c.

Solution (a):

Since the degree of the denominator

is 2, divide both the numerator and

denominator by


Use Limit Law 1 (limit of a sum).

Use the special limits.

Simplify.

A similar calculation can be done to find that the limit as is also 0. Looking

back at Example 5(a) in Section 2.6, we see both algebraically and graphically that the

horizontal asymptote of is

Solution (b):



denominator by

Use Limit Law 5 (limit of a quotient). =

limxS�a8 +

3

x2 blimxS�a4 +

1

x2 b

limxS�

8x2

+ 3

4x2+ 1

= limxS�

8 +

3

x2

4 +

1

x2

x2.

y = 0.f (x) =

8x + 3

4x2+ 1

x S - �

limxS�

8x + 3

4x2+ 1

= 0

=

0

4= 0

=

8 limxS�a1

x b + 3 limxS�a 1

x2 blimxS�

4 + limxS�a 1

x2 b

=

limxS�a8

x b + limxS�a 3

x2 blimxS�

(4) + limxS�a 1

x2 b

=

limxS�a8

x+

3

x2 blimxS�a4 +

1

x2 b

limxS�

8x + 3

4x2+ 1

= limxS�

8

x+

3

x2

4 +

1

x2

x2.

limxS�

8x3

+ 3

4x2+ 1

limxS�

8x2

+ 3

4x2+ 1

limxS�

8x + 3

4x2+ 1

0 r 0 r0

r

4

rTechnology Tip


by .[- 2, 4][- 10, 10]

a.

The graph indicates that the limiting

value of y is 0 as and as

. That is, the horizontal

asymptote is

b.

y = 0.

x S �

x S - �


by .[- 2, 4][- 10, 10]

The graph indicates that the limiting

value of y is 2 as and as

. That is, the horizontal

asymptote is y = 2.

x S �

x S - �





Simplify.

A similar calculation can be done to find that the limit as is also 2. Looking back

at Example 5(b) in Section 2.6, we see both algebraically and graphically that the horizontal

asymptote of is

Solution (c):



denominator by




We conclude that (does not exist) because the numerator grows

without bound while the denominator approaches 4. A similar calculation can be done to

find that the limit as also does not exist. Looking back at Example 5(c) in Section 2.6,

we see both algebraically and graphically that has no horizontal asymptote.

■ YOUR TURN Find the following limits, if they exist:

a. b. c. limxS�

2x3

- 1

x2+ 2

limxS�

2x - 1

x2+ 2

limxS�

2x2- 1

x2+ 2

h(x) =

8x3+ 3

4x2+ 1

x S - �

limxS�

8x3+ 3

4x2+ 1

= �

=

limxS�

(8x) + limxS�a 3

x2 blimxS�

(4) + limxS�a 1

x2 b

=

limxS�

(8x) + limxS�a 3

x2 blimxS�

(4) + limxS�a 1

x2 b

=

limxS�a8x +

3

x2 blimxS�a4 +

1

x2 b

limxS�

8x3

+ 3

4x2+ 1

= limxS�

8x +

3

x2

4 +

1

x2x2.

y = 2.g(x) =

8x2+ 3

4x2+ 1

x S - �

limxS�

8x2

+ 3

4x2+ 1

= 2

=

8

4= 2

=

limxS�

(8) + limxS�a 3

x2 blimxS�

(4) + limxS�a 1

x2 b

=

limxS�

(8) + limxS�a 3

x2 blimxS�

(4) + limxS�a 1

x2 b8 r 0 r

0

r4

r

� r 0 r0

r

4

rSet the viewing rectangle as

by .[- 20, 20][- 10, 10]

The graph indicate that the limiting

value of y approaches as and

approaches as . There is

no horizontal asymptote.

x S - �- �

x S ��

■ Answer:a. 2

b. 0

c. (does not exist)�

c.

Classroom Example 11.4.1Find the following limits, if they

exist:

a.

b.

c.

Answer:a. 2 b. DNE c. 0

limxS�

2x + 1

2x2+ 1

limxS�

(2x + 1)4

2x3+ 1

limxS�

(2x + 1)2

2x2+ 1



EXAMPLE 2 Finding Limits at Infinity for Trigonometric Functions

Use the graph of the trigonometric functions to determine the following limits, if they exist:

a.

b. and

Solution (a):

The sine function continues to oscillate between

the y-values of and 1. Therefore, the sine

function does not approach a single value as xgets large.

Solution (b):

limxS -�

arctan x = -

p

2

limxS�

arctan x =

p

2

limxS�

sin x does not exist.

- 1

limxS -�

arctan xlimxS�

arctan x

limxS�

sin x

EXAMPLE 3 Finding Limits at Infinity for Exponential Functions

Use graphs to find the following limits at

infinity of exponential functions:

a. and

b. and

Solution (a):

■ (does not exist) because

the function continues to grow without

bound to the right.

■

Solution (b):

■

■ (does not exist) because the

function continues to grow without

bound to the left.

limxS -�

e-x

= �

limxS�

e- x

= 0

limxS -�

ex

= 0

limxS�

ex

= �

limxS -�

e-xlimxS�

e-x

limxS -�

e xlimxS�

e x

–1

–2

y

x

2

1

100806040200

y

x2010–10–20

�2

3�4

�4

�

�2

3�4

–

–

–

4

lim arctan xx → �

lim arctan xx → ��

–3 –2–4 –1 10 2 3 4

x

y

10

5

y = ex

y = e–x

–3 –2–4 –1 10 2 3 4

x

y

10

5

Classroom Example 11.4.2*Determine the following limits,

if they exist:

a.

b.

c.

Answer:a. DNE b. DNE c. 1

limxS�

cosa 1

2x blim

xS -� cos(2x)

limxS�

1

cos x



Limits of Sequences

Recall that in Section 10.1 we discussed the idea of a sequence of numbers

that can be represented by a general term: where Here, we want to

investigate the behavior of the sequence as n gets large:

For example, take the sequence for If we write out the terms

. . . , we see that the terms of this sequence get closer and closer to 0. We

say that this sequence converges to 0 and we use the following limit notation:

Let us graph the sequence for . . . and the function

for Notice that the sequence has the positive integers as its domain, whereas the

function is defined for all positive real numbers and zero. We see the same behavior in that

both the function and the sequence approach the number 2 as we move to the right.

x Ú 0.

f (x) =

2x

x + 1n = 1, 2, 3,an =

2n

n + 1

limnS�

1

n2= 0

1, 14, 19, 116, 1

25, 136,

n = 1, 2, 3, . . . .an =

1

n2

n S �.

n = 1, 2, 3, . . . .an,

(a1, a2, a3, . . .)

Let be a sequence where represents the nth term of the sequence.

If the nth term gets closer and closer to L as n gets larger and larger, then we

define the limit of a sequence as:

an

ana1, a2, a3, . . .

If exists, then we say that the sequence converges (or is convergent),

and if the limit does not exist, we say that the sequence diverges (or is divergent).

limnS�

an = L

DE F I N I T I O N Limit of a Sequence

WORDS MATH

The limit of as n gets sufficiently large is L. limnS�

an = Lan

610 7 8 9 102 3 4 5

n

an

3

1

2

an = 2nn + 1

610 7 8 9 102 3 4 5

x

y

3

1

2

f (x) = 2xx + 1

We can define a limit of a sequence similarly to how we defined a limit of a function

at infinity.



If and for every positive integer n, then limnS�

an = L.f (n) = anlimxS�

f (x) = L

The limit laws and special limits also hold for sequences.

EXAMPLE 4 Finding the Limit of a Convergent Sequence

Determine whether the sequence is convergent. If it is, state the limit.

Solution:

Simplify an.

Take the limit of an as

Use Limit Law 1 (Limit of a Sum).

Evaluate the limits.

Simplify.

The sequence converges to 3.

■ YOUR TURN Find the limit of the sequence an =

5

n2 cn(n + 1)

2d .

limnS�

an = 3

limnS�

an = 3 + 0 + 0

limnS�

(3) + limnS�a 9

2n b + limnS�a 3

2n2 blimnS�

an =

limnS�a3 +

9

2n+

3

2n2 blimnS�

an =n S �.

= 3 +

9

2n+

3

2n2

=

6n3

2n3+

9n2

2n3+

3n

2n3

=

6n3+ 9n2

+ 3n

2n3

an =

3(2n3+ 3n2

+ n )2n3

an =

9

n3 cn(n + 1)(2n + 1)

6d

■ Answer: 52

Technology Tip

To enter the sequence

you

need to set the MODE to Seq and

use

9 X, T, n ^ 3 (

X, T, n ( X, T, n 1 ) (

2 X, T, n � 1 ) 6 ),u,

+u,u,

u,,Y =

Y = .

an =

9

n3cn(n + 1)(2n + 1)

6d ,

Set the viewing window with

and

[0, 50] by [0, 5].

nMin = 1, nMax = 50,

The graph shows that the sequence

converges to 3 as n approaches �.

Classroom Example 11.4.4Determine whether the sequence is convergent. If it is, state its limit.

Answer:Convergent with limit .1

3

an =

(n + 1)2(n - 1)2

3n4



EXAMPLE 5 Finding That a Sequence Diverges

Show that the limit of diverges (does not converge).

Solution:

Make a table and graph of the sequence.

an = (- 1)n

We see that the terms in this sequence are

either 1 or and they continue to

oscillate as n grows.

Therefore, this sequence does not converge

to a single value. Hence, we say that this

sequence diverges.

- 1

n 1 2 3 4 5 6 7

1 1 1 - 1- 1- 1- 1an

610 7 8 9 102 3 4 5

x

y

2

–1

–2

1

Technology Tip

To enter the sequence

you need to set the MODE to Seq

and use

( 1 ) ^ X, T, n

Set the viewing window with

and [0, 50]

by [- 2, 2].

nMax = 50,nMin = 1,

u,(- )Y =

Y = .

an = (- 1)n,

The graph shows that the sequence

does not converge to a single value.

Recall that in Section 10.1, we said that a sequence whose general term alternated between

positive and negative, usually through behavior, is said to be an alternating sequence.

It is important to note that just because it is oscillating does not mean it does not converge.

For example, is an alternating sequence, but as n increases, it is oscillating

(positive to negative) around zero. Therefore, we say that limnS�

an = limnS�

(- 1)n

n= 0.

an =

(- 1)n

n

(- 1)n

4321

–1

1

n

n

an

an = , n ≥ 1(–1)n

function f has a horizontal asymptote, We also discussed

limits of sequences. If the limit of a sequence exists, we say that

the sequence converges, and if the limit of the sequence does not

exist, we say that the sequence diverges.

y = L.In this section, we defined limits at infinity. Specifically, we

allowed x to increase without bound and decrease without

bound If the limit at infinity of a function exists,

that is, if or then the graph of thelimxS -�

f (x) = L,limxS�

f (x) = L(x S - �).

(x S �)

SUMMARY

SECTION

11.4

Classroom Example 11.4.5 Answer:Show that the sequence defined The odd-indexed terms are all 1, while the even-indexed terms

by diverges. are all 3. Hence, the terms oscillate back and forth between 1

and 3 without ever settling down to a single value.

bn = 2 + (- 1)n



In Exercises 1–16, find the limit, if it exists.

1. 2. 3. 4.

5. 6. 7. 8.

9. 10. 11. 12.

13. 14. 15. 16.

In Exercises 17–28, determine whether the sequence converges. If it does converge, state the limit.

17. 18. 19. 20.

21. 22. 23. 24.

25. 26. 27. 28. an =

12

n3 cn(n + 1)(2n + 1)

6dan =

12

n4 cn2(n + 1)2

4dan = (- 1)n + 1n!an = (- 1)nn2

an = cos (np)an = sin anp

2ban =

(n - 1)!

(n + 1)!an =

n!

(n + 1)!

an =

(n - 1)2

(n + 1)2an =

2n2- 1

n2an =

2n + 1

nan =

n

n + 1

limxS�

ax -

1

x blimxS�

a3x +

4

x blimxS�

a 1

x2+

3x + 4

2x - 1blim

xS� a 1

x3-

2x + 1

x - 5b

limxS -�

5exlimxS�

4e- xlimxS�

tan xlimxS�

cos x

limxS -�

6x2+ 6x + 1

3x2- 5x - 2

limxS -�

x2

- 4x + 5

2x2+ 6x - 4

limxS�

2 - x3

2x - 7limxS�

7x3+ 1

x + 5

limxS�a-

2x

x2+ 9blim

xS�

7x

x2+ 16

limxS�a- 4

x blimxS�

3

x

29. Medicine. The concentration C of a particular drug in a

person’s bloodstream t minutes after injection is given by

What do you expect the concentration to

be after several days?

30. Medicine. The concentration C of aspirin in the bloodstream

t hours after consumption is given by What

do you expect the concentration to be after several days?

31. Keyboarding. An administrative assistant is hired by a law

firm upon graduation from high school and learns to

keyboard (type) on the job. The number of words he can

enter per minute is given by where n

is the number of months he has been on the job. How

many words per minute would you expect him to enter if

he worked at the same firm until he retired?

32. Memorization. A professor teaching a large lecture course

tries to learn students’ names. The number of names she can

remember increases with each week in the semester n,

and is given by According to this relation,

what is the greatest number of names she can remember?

N(n) =

600n

n + 20.

N(n)

an =

130n + 260

n + 5,

C(t) =

t

t2+ 40

.

C(t) =

2t

t2+ 100

.

33. Weight. The amount of food that cats typically eat increases

as their weight increases. A function that describes this is

where the amount of food F(x) is given in

ounces and the weight of the cat x is given in pounds.

How many ounces of food will most adult cats eat?

6 8 102 4

x

y

10

8

6

4

2

F(x) =

10x2

x2+ 4

,

■ SKILLS

EXERCISES

SECTION

11.4




34. Memorization. The 2004 Guinness Book of World Recordsstates that Dominic O’Brien (from the U.K.) memorized on

a single sighting a random sequence of 54 separate packs

of cards all shuffled together (2808 cards in total) at

Simpson’s-In-The-Strand, London, on May 1, 2002. He

memorized the cards in 11 hours 42 minutes, and then

recited them in exact sequence in a time of 3 hours

30 minutes. With only a 0.5% margin of error allowed

(no more than 14 errors), he broke the record with just

8 errors. If we let t represent the time (hours) it takes to

memorize the cards and y represent the number of cards

memorized, then a rational function that models this

event is given by . According to this model,

what is the greatest number of cards that can be memorized?

y =

2800t2+ t

t2+ 2

35. Carrying Capacity. In biology, the term carrying capacityrefers to the maximum number of animals that a system can

sustain due to food sources and other environmental factors. If

the number of golden lancehead vipers that live on an island

off the coast of Brazil is given by

where S is the number of vipers and t is the number of years

(assume corresponds to the year 2000), what is the

carrying capacity of the island for this pit viper?

36. Automotive Sales. A new model of an automobile is being

released. The number N of this model on the roads in the

United States is given by where t represents the

number of weeks after its release. What is the greatest

number of these automobiles that will ever be expected on

the road in the United States according to the formula?

100,000

1 + 10e-2t ,

t = 0

S = 3000(1 - e-0.02t ),

38. Find

Solution:

Rewrite secant as the

reciprocal of cosine.

Use Limit Law 5

(limit of a quotient).

Evaluate the limits.


=

1

1= 1

=

limnS�

1

limnS�

cos (np)

limnS�

sec (np) = limnS�

1

cos (np)

limnS�

sec (np).


37. Find

Solution:


Use Limit Law 4

(limit of a product).

Evaluate using

special limits.

Simplify.


= 2 + 0 = 0

= [ lim2xS�

] [ limxxS�

] + limxS�

1

x

= [ lim2xS�

] [ limxxS�

] + limxS�

1

x

limxS�

a2x +

1

x b = limxS�

2x + limxS�

1

x

limxS�

a2x +

1

x b .

2⎫ ⎬ ⎭

does notexist⎫ ⎬ ⎭

0⎫ ⎬ ⎭

In Exercises 39–42, determine whether each statement is true or false.

41. If the limit at infinity of a function exists, then the function

has a horizontal asymptote.

42. If the terms of a sequence continue to increase without

bound as n increases, we say that the sequence is

convergent.

39. When the degree of the numerator is less than the degree

of the denominator for a rational function, then the limit at

infinity does not exist.

40. When the degree of the numerator is equal to the degree of

the denominator for a rational function, then the limit at

infinity exists.

44. Find .limxS�

x – sinx43. Find .limxS�

sin x

x



■ CHALLENGE


11.5 Finding the Area Under a Curve 1127

Prior to Exercises 45 and 46, use a graphing utility to showthat the following limits equal zero:

45. Which of these three limits most quickly converges to zero?

46. Which of these three limits most slowly converges to zero?

limnS ˆ

1n!

limnS ˆ

1

2nlimnS ˆ

1

n2

For Exercises 47 and 48, use a graphing utility to determinewhether the sequence converges. If it does converge, find thelimit to four decimal places.

47.

48. an = a1 -

2

n bn

an = n ln a1 +

12

n b

One of the fundamental problems in calculus is the area problem.

FINDING THE AREA

UNDER A CURVE

SECTION

11.5

CONCEPTUAL OBJECTIVE

■ Understand that the more rectangles used to

approximate the area under a curve, the better

the approximation.

SKILLS OBJECTIVES

■ Find limits of summations.

■ Find the area under a curve using rectangles.

■ Find the area under a curve by using limits of

summations.

Find the area of the region S that lies

under the curve f(x) from to x = b.x = a

THE AREA PROBLEM

y

x

a

f (x)

S

b

The challenge here is that the region S has a curved boundary. Notice that three of the

boundaries are straight lines: and but the boundary, which is the graph

of f(x), is curved. If we had wanted the area of a region such as a rectangle, triangle, or

circle, we would already know the formula. But what happens when a boundary is defined

by a specified function?

Recall that in Section 11.3, we used the limit of the slopes of secant lines to a curve

(average rate of change) to approximate the slope of a tangent line to a curve (instantaneous

rate of change) as the two points defining the secant line got closer and closer together. In

x = b,x = a,y = 0,




For example, let’s take a finite geometric series (Section 10.3) with

Let

Below are summation properties and summation formulas. Many of the summation

formulas were derived in Section 10.4 (Mathematical Induction), and they will be useful

in the area problem later in this section.

limnS�

Bank = 1

xk - 1R = limnS�

B (1 - xn )(1 - x)

R = L1

1 - x�x � 6 1

does not exist �x � Ú 1

n S �:

an

k = 1

xk - 1= 1 + x + x2

+ x3+ Á + xn-1

=

(1 - xn )(1 - x)

a1 = 1:

limnS�

an

k = 1 ak = a

�

k = 1

ak = a1 + a2 + a3 + Á

slimit of a summation winfinite series

1. c is a constant

2.

3. an

k = 1

(ak - bk) = an

k = 1

ak - an

k = 1

bk

an

k = 1

(ak + bk) = an

k = 1

ak + an

k = 1

bk

an

k = 1

cak = can

k = 1

ak,

SUMMATION PROPERTIES

1. c is a constant 2.

3. 4. an

k = 1

k3=

n2(n + 1)2

4an

k = 1

k2=

n(n + 1)(2n + 1)

6

an

k = 1

k =

n(n + 1)

2an

k = 1

c = cn,

SUMMATION FORMULAS

this section, we will use limits again; only this time we will approximate the area of S using

rectangles (which we know the area of) and let the width of each rectangle approach 0. This

will require us to take limits of summations. So we will first discuss limits of summations,

and then we will develop an approach to solving the area problem.

Limits of Summations

In Section 10.1, we used sigma notation to represent a finite series, or summation:

Recall that the sum of a finite series can always be found. When we allow n to grow without

bound, that is, allow the result is an infinite series, and the sum may or may not

exist (depending on whether the limit exists).

n S �,

an

k = 1

ak = a1 + a2 + a3 + Á + an



}

40

}

40

}

40

EXAMPLE 1 Evaluate a Summation

Evaluate the summation:

Solution:

Use summation property (1).

Use summation formula (3) with

Simplify.

■ YOUR TURN Evaluate the summation:

a25

k = 1

4k3= 4 + 4(2)3

+ 4(3)3+

Á+ 4(25)3

a40

k = 1

3k2= 66,420

= 3 c40(41)(81)

6d = 66,420

= 3 sn(n + 1)(2 # n + 1)

6tn = 40.

a40

k = 1

3k2= 3a

40

k = 1

k2

a40

k = 1

3k2= 3 + 3(2)2

+ 3(3)2+

. . .+ 3(40)2.

EXAMPLE 2 Finding the Limit of a Summation

Find

Solution:

STEP 1 Find the sum.

Use summation property (1)

(n is a constant).


Use summation formulas (1) and (2).

Simplify.

Combine the expression inside the

brackets into one fraction.

Eliminate the parentheses inside

the brackets.


Eliminate the brackets.

STEP 2 Find the limit of the sum.

Let

in the limit.

an

k = 1

k - 1

n2=

n2- n

2n2

=

n2- n

2n2

=

1

n2 cn2

- n

2d

=

1

n2 cn2

+ n - 2n

2d

=

1

n2 cn(n + 1) - 2n

2d

=

1

n2 cn(n + 1)

2- nd

=

1

n2 Ban

k = 1

k - an

k = 1

1R =

1

n2 Ban

k = 1

k - an

k = 1

1R a

n

k = 1

k - 1

n2=

1

n2 a

n

k = 1

k - 1

limnS�

an

k = 1

k - 1

n2.

rn(n + 1)2 rn

limnS�

an

k = 1

k - 1

n2= lim

nS� n2

- n

2n2

■ Answer: 22,100

Technology Tip

To find the sum of the finite series

you need to use the SUMa40k = 13k2,

Technology Tip

To find the limit of the sum of the

series you need to use

the command. We shall put in

100, 800. Enter

.ENTER)1,50

,1,X, T, �, n,x250,

)1-X, T, �, n(ENTER

5:seq(�OPS�LIST

ENTER5:sum(�MATH

�LISTn = 50,

SUM

limnS�a

n

k = 1

k - 1

n2,

command. Enter

. ENTER)1,40,1

,X, T, �, n,x2X, T, �, n

3ENTER5:seq(�OPS

�LISTENTER5:sum(�

MATH�LIST

Classroom Example 11.5.1Evaluate the summations:

a. *b.

Answer:a. �2550 b. �2544

a50

k = 3

- 2ka50

k = 1

- 2k



Divide the numerator and

denominator by

Simplify.

Let

■ YOUR TURN Find limnS�

an

k = 1

2k + 1

n2.

limnS�

an

k = 1

k - 1

n2 = 1

2

= limnS�

1 -

1

n

2=

1 - 0

2=

1

2n S �.

= limnS�

1 -

1

n

2

n2.

The Area Problem

Now that we can find the limits of summations, we are ready to solve the area problem.

The goal is to find the area bounded by the graph of a function f, which is above the x-axis,

and three straight boundaries: and This bounded region S is often

called the area under the curve (where “the curve” is the graph of f ). We will assume that

the function f is continuous and is nonnegative (lies entirely on or above the x-axis). First

we will start by approximating the area with a finite number of rectangles and then we will

increase the number of rectangles and see that, as the number of rectangles increases, the

approximation gets better. Ultimately, we will define the exact area of the region as a limit

of the sum of the areas of infinitely many rectangles.

x = b.y = 0, x = a,

■ Answer: 1

y

x

a

f (x)

S

b

Technology Tip

EXAMPLE 3 Approximating the Area Under a Curve with Rectangles

Use the area of four rectangles to

approximate the area of the region

bounded by the graph of

the x-axis, and the lines

and

Solution:

STEP 1 Label four rectangles.

Divide the x-interval [0, 1] into

four equal subintervals.

c0, 1

4d , c1

4, 1

2d , c1

2, 3

4d , c3

4, 1d

x = 1.x = 0

f (x) = x2,

x

y

14

1

12

12

34

10

f (x) = x2

Label points along the curve that

correspond to the right endpoints

of the subintervals. x

y

14

1

12

916

14116

34

10

f (x) = x2

= limnS�

n2

n2-

n

n2

2n2

n2


Find

Answer: 23

limnS�a

n

k = 1

2k2+ n

n3.



STEP 2 Find the areas of the four rectangles.

Write the formula for area of a rectangle.

Each of the four rectangles has width

The length (or height) of each rectangle is equal to the function value f(x).

Area of Rectangle 1:




STEP 3 Sum the areas of the four rectangles.

R4 = (1)2 a1

4b = (1)a1

4b =

1

4

R3 = a3

4b 2a1

4b = a 9

16b a1

4b =

9

64

R2 = a1

2b 2a1

4b = a1

4b a1

4b =

1

16

R1 = a1

4b 2a1

4b = a 1

16b a1

4b =

1

64

R = l #1

414.

R = l # w

f (1) = (1)2= 1

f a3

4b = a3

4b 2

=

916

f a1

2b = a1

2b 2

=

14

f a1

4b = a1

4b 2

=

116

x

y

14

1

12

916

141

1634

10

f (x) = x2

R1 R2

R3

R4

The area is equal to the

sum of the four rectangles. R1 + R2 + R3 + R4 =

1

64+

1

16+

9

64+

1

4

Combine the four fractions

into a single fraction.

square units

■ YOUR TURN Use the area of four rectangles

(using left endpoints) to

approximate the area of the

region bounded by the graph

of the x-axis, and

the lines and x = 1.x = 0

f (x) = x2,

R1 + R2 + R3 + R4 = 0.46875

=

1 + 4 + 9 + 16

64=

30

64=

15

32

x

y

14

1

12

916

14116

34

1

f (x) = x2

R1 R2R3

R4

Draw the four rectangles that have width and height equal to the function

values corresponding to the right endpoints of the subintervals.

=14

■ Answer: 0.21875 sq units


It seems that the upper and lower estimates are converging to . We will now let n be the

number of intervals and then allow n to approach infinity. In Example 4, we show that the

exact area under the curve is .13

13

EXAMPLE 4 Finding the Exact Area Under a Curve

Find the area of the region bounded

by the graph of the x-axis,

and the lines and

Solution:

STEP 1 Let n rectangles approximate the area under the curve.

Note that if n is the number of

intervals, then each of the equal

subintervals has width

Write the subintervals in terms of n.

Identify the right endpoints.1

n, 2

n, 3

n, Á ,

n

n

c0, 1

nd , c1

n, 2

nd , c2

n, 3

nd , Á , cn - 1

n, n

nd

1

n.

x = 1.x = 0

f (x) = x2,


In Example 3, by choosing the rectangles using the right endpoints of the intervals, we

calculated an approximate area that overpredicts the actual area under the curve. In the Your

Turn following Example 3, by choosing the rectangles using the left endpoints of the

intervals, we calculated an approximate area that underpredicts the actual area under the

curve. The actual area lies somewhere in between:

Furthermore, had we chosen more than four rectangles, we would have calculated better

estimates.

0.21875 6 A 6 0.46875

NUMBER OF RECTANGLES UPPER AND LOWER ESTIMATES

10

100

1000 0.33283 6 A 6 0.33383

0.32835 6 A 6 0.33835

0.28500 6 A 6 0.38500

x

y

14

1

12

12

34

10

f (x) = x2

x

y

1

0

f (x) = x2

Rk

1n

kn

nn

k – 1n



STEP 2 Find a general formula for the area of each of the rectangles.

Rectangle 1 has width and length

(height) equal to


(height) equal to


(height) equal to

Rectangle k has width and length

(height) equal to

The general area of each rectangle is

STEP 3 Sum the area of the n rectangles.

Sum the areas of the nrectangles.

Write using sigma notation.

Use summation property

(1) (n is a constant).

Use summation formula (3).

STEP 4 Use the limit to get the exact area under the curve as .

As n gets large, approximates

the area under the curve, A.

Let

Divide the numerator and

denominator by = limnS�

2 +

3n

+

1

n2

6n3.

A = limnS�

2n3

+ 3n2+ n

6n3An =

2n3+ 3n2

+ n

6n3 .

A = limnS�

An

An

n S �

=

n(n + 1)(2n + 1)

6n3 =

2n3+ 3n2

+ n

6n3

=

1

n3 # n(n + 1)(2n + 1)

6

=

1

n3an

k = 1k2

= an

k = 1 Rk = a

n

k = 1 1n

a k

n b2

= an

k = 1

k2

n3

=

1n

a1n b

2

+

1n

a2n b

2

+

1n

a3n b

2

+. . .

+

1n

an

n b2

An = R1 + R2 + R3 +. . .

+ Rn

Rk =

1n

a k

n b2

.

Rk =

1n

a k

n b2

f a k

n b = a k

n b2

:

1n

R3 =

1n

a3n b

2

f a3n b = a3

n b2

:

1n

R2 =

1n

a2n b

2

f a2n b = a2

n b2

:

1n

R1 =

1n

a1n b

2

f a1n b = a1

n b2

:

1n

Technology Tip


series you need to use

the command. We shall

put in 800, 900. Enter

. ENTER)1,100,1,

X, T, �, n,3^100,x2

X, T, �, nENTER5:seq(

�OPS�LISTENTER

5:sum(�MATH�LIST

n = 100,SUM

limnS�

ank=1

k2

n3,

c11eLimitsAPreviewtoCalculus.qxd 6/10/13 6:15 PM Page 1133


Use Limit Laws 1 and 5

(limit of a sum/quotient).

Let

Simplify.

square unitsA = 1

3

=

2

6=

1

3

=

2 + 0 + 0

6n S �.

=

limnS�

2 + limnS�

3

n+ lim

nS� 1

n2

limnS�

6

We can generalize the procedure outlined in Example 4 to find the exact area under a

curve by taking the limit of the sum of the areas of n rectangles as n approaches In

calculus this limit of sums represents an integral.�.

EXAMPLE 5 Finding the Area Under a Curve

Find the area bounded by the

curve and

the x-axis.

Solution:

STEP 1 Find the x-intercepts of f.

Let f(x) � 0.

Solve for x.

or x = 2x = 0

x(2 - x) = 0

2x - x2= 0

f (x) = 2x - x2

Let f be a continuous nonnegative function on the interval [a, b]. The area of the

region bounded by the graph of f , the x-axis and the vertical

lines and is given by

where:

Width of each rectangle

Right endpoint of the kth rectangle

Height of the kth rectangle

A = limnS�

an

k = 1

f ca +

(b - a)k

nd ab - a

n b

f (xk) = f (a + k¢x)

xk = a + k¢x

¢x =

b - a

n

A = limnS�

an

k = 1 f (xk)¢x

x = bx = a(y = 0),( f (x) 7 0)

Area Under a CurveDEFINITION

dr r

area of rectangle, Rk

height width

10 32

x

y

2

1f (x) = 2x – x2

Classroom Example 11.5.4a. Find the area under the curve

.

b. Find the area under the curve

Answer:a. 21 sq units

b. 150 sq units

f (x) = 2x2+ 1 on [0, 6].

f (x) = 2x2+ 1 on [0, 3]



STEP 2 Find the dimensions of the rectangles.

Find the width.

Find the right endpoints.

Find the height.

STEP 3 Find the sum of the areas of the n rectangles.

Areas of the rectangles.

Let and

Eliminate the parentheses.



(n is a constant).

Use summation formulas (2)

and (3).

Simplify


Combine into a single fraction.

Simplify.

STEP 4 Find the area under the curve by taking the limit as n

Let

Use Limit Law 2

(limit of a difference).

Divide out n2 in the first limit.

Find the individual limits.

sq units

■ YOUR TURN Find the area bounded by the curve and the x-axis.f (x) = 3x - x2

A = 4

3

=

4

3- 0 =

4

3

= limnS� a4

3b - lim

nS�a 4

3n2 b

= limnS�

a4n2

3n2 b - limnS�

a 4

3n2 b

A = limnS� a4n2

- 4

3n2 bn S �.

S �.

=

4n2- 4

3n2

=

12n2+ 12n - 8n2

- 12n - 4

3n2

=

4n + 4

n-

8n2+ 12n + 4

3n2

=

4(n + 1)

n-

4(n + 1)(2n + 1)

3n2

=

8

n2#n(n + 1)

2-

8

n3#n(n + 1)(2n + 1)

6

=

8

n2an

k = 1

k -

8

n3an

k = 1

k2

= an

k = 1

8k

n2- a

n

k = 1

8k2

n3

= an

k = 1

a8k

n2 -

8k2

n3 b

= an

k = 1

ca4k

n-

4k2

n2 b a2

n bdf (xk) =

4k

n-

4k2

n2.¢x =

2

n

An = an

k = 1

f (xk)¢x

=

4k

n-

4k2

n2f (xk) = f a2k

n b = 2 a2k

n b - a2k

n b2

xk = a + k¢x = 0 + k 2

n=

2k

n

¢x =

b - a

n=

2 - 0

n=

2

n

df (x) � 2x � x2

Technology Tip


series

you need to

use the command. We shall

put in and 800. Enter

8

. ENTER)

1,100,1,X, T, �, n,

3^100,(X, T, �, n�100

(X, T, �, nENTER5:seq(

�OPS�LISTENTER


n = 100

SUM

limnS�a

nk=1

8k(n - k)

n3,

limnS�a

nk=1a4k

n-

4k2

n2 b2

n=

■ Answer: sq units92

Similarly:

Classroom Example 11.5.5Find the area under the curve

above the x-axis.

Answer: sq units1253

g(x) = - 2x2+ 2x + 12



EXAMPLE 6 Finding the Area Under a Curve

Find the area bounded by the graph of

and the x-axis, between

and

Solution:

STEP 1 Find the dimensions of the rectangles.

Find the width.

Find the right endpoints.

Find the height.

STEP 2 Find the sum of the areas of the n rectangles.

Areas of the rectangles.

Let and




(n is a constant).

Use summation formulas (2) and (3).

Simplify.

Divide out common factors.

Eliminate parentheses. = 8 -

n + 1

n-

2n2+ 3n + 1

6n2

= 8 -

(n + 1)

n-

(n + 1)(2n + 1)

6n2

=

8

nn -

2

n2#n(n + 1)

2-

1

n3#n(n + 1)(2n + 1)

6

=

8

n a

n

k = 1

1 -

2

n2 a

n

k = 1

k -

1

n3 a

n

k = 1

k2

=

8

n a

n

k = 1

1 -

2

n2 a

n

k = 1

k -

1

n3 a

n

k = 1

k2

= an

k = 1

8

n- a

n

k = 1

2k

n2- a

n

k = 1

k2

n3

= an

k = 1

c8n

-

2k

n2-

k2

n3d

An = an

k = 1

ca8 -

2k

n-

k2

n2 ba1

n bd

f (xk) = 8 -

2k

n-

k2

n2.¢x =

1

n

An = an

k = 1

f (xk)¢x

f (xk) = f a1 +

k

n b = - a1 +

k

n b2

+ 9 = 8 - 2k

n-

k2

n2

xk = a + k¢x = 1 + k 1

n= 1 +

k

n

¢x =

b - a

n=

2 - 1

n=

1

n

x = 2.x = 1

f (x) = - x2+ 9

10 32

x

y

10987654321

f (x) = –x2 + 9

x = 2x = 1

d d

n n(n + 1)

2

d

n(n + 1)(2n + 1)

6

df(x) � �x2 � 9

Technology Tip


series

you need

to use the command. We

shall put in Enter

. ENTER)1,100,

1,X, T, �, n,3^100,

)x2X, T, �, n-X, T, �, n

)100(2-x2)

100(8(ENTER5:seq(

�OPS�LISTENTER


n = 100, 800, 900.

SUM

limnS�a

nk=1

8n2- 2nk - k2

n3,

limnS�a

nk=1a8 -

2k

n-

k2

n2 b1

n =

Classroom Example 11.5.6Find the area bounded by the

curve

between

Answer: sq units1123

x = - 1 and x = 3.

g(x) = - 2x2+ 2x + 12



STEP 3 Find the area under the curve by taking the limit as n

Let

Use Limit Law 2

(limit of a

difference).

Divide out the

common factors:

n and

Use Limit Law 1

(limit of a sum).

Find the individual

limits.

square units

■ YOUR TURN Find the area bounded by the graph of and the

x-axis, between and x = 2.x = 1

f (x) = - x2+ 4

A = 20

3

= 8 - 1 - 0 -

1

3- 0 - 0 =

20

3

= limnS�

8 - limnS�

1 - limnS�

1

n- lim

nS� 1

3- lim

nS�

1

2n - lim

nS�

1

6n2

= 8 - limnS�

a1 +

1

n b - limnS�

a1

3+

1

2n+

1

6n2 bn2.

= limnS�

8 - limnS� an + 1

n b - limnS�

a2n2+ 3n + 1

6n2 b

A = limnS� a8 -

n + 1

n-

2n2+ 3n + 1

6n2 bn S �.

S �.

■ Answer: sq unitsA =53

In Exercises 1–10, evaluate the summations. Note: In Exercise 9 the summation starts at 0, and in Exercise 10 the summation starts at 2.

1. 2. 3. 4. 5.

6. 7. 8. 9. 10. an

k = 2

(k2+ 2)a

n

k = 0

(k2+ 2)a

n

k = 1

(2k3- k)a

n

k = 1

(2k3+ k2)a

n

k = 1

(2k - 5)

n2

an

k = 1

(3k - 4)

na17

k = 1

2k2a25

k = 1

3ka30

k = 1

8a20

k = 1

5

SUMMARY

SECTION

11.5

In this section, we first discussed summation properties and

formulas.

1.

2.

3.

4. an

k = 1

k3=

n2(n + 1)2

4

an

k = 1

k2=

n(n + 1)(2n + 1)

6

an

k = 1

k =

n(n + 1)

2

an

k = 1

c = cn, c is a constant

Then we discussed how to find the area under a specified curve.

We first started by approximating the area by sums of

rectangles. Either right or left endpoints of the intervals can be

used to calculate the height of the rectangles. We found that for

increasing functions, right endpoints resulted in an overestimate

and left endpoints resulted in an underestimate. For decreasing

functions the opposite is true. Then we defined the exact area in

terms of the limit of the sums of areas of rectangles as the number

of rectangles, n, approaches .�

■ SKILLS

EXERCISES

SECTION

11.5



In Exercises 11–20, find the limit of the summations.

11. 12. 13. 14.

15. 16. 17. 18.

19. 20.

In Exercises 21–23, approximate the area under the curve (between the graph of f, the x-axis, and ) using thegiven number of rectangles of equal width.

21. Approximate with four rectangles and use left endpoints.

22. Approximate with four rectangles and use right endpoints.

23. Based on Exercises 21 and 22 the actual area A is

In Exercises 24–26, approximate the area under the curve (between the graph of f, the x-axis, and ) using thegiven number of rectangles of equal width.

24. Approximate with two rectangles and use left endpoints.

25. Approximate with two rectangles and use right endpoints.

26. Based on Exercises 24 and 25 the actual area A is

In Exercises 27–30, approximate the area under the curve using the given number of rectangles of equal width.

27. on [0, 8]; eight rectangles with right endpoints

28. on [0, 2]; four rectangles with right endpoints

29. on [0, 4]; four rectangles with left endpoints

30. on [1, 2]; eight rectangles with left endpoints

In Exercises 31–40, find the area under the curve on the specified interval.

31. 32. 33.

34. 35. 36.

37. 38. 39.

40. 3 … x … 4f (x) = - x2+ 7x - 10,

2 … x … 3f (x) = - x2+ 5x - 4,1 … x … 2f (x) = x2

- 3x,1 … x … 3f (x) = x2- 4x,

0 … x … 1f (x) = x3+ 1,0 … x … 2f (x) = 8 - x3,0 … x … 2f (x) = 5 - x2,

0 … x … 1f (x) = 2 - x2,1 … x … 3f (x) = x - 1,0 … x … 1f (x) = - 2x + 3,

f (x) =

1

x

f (x) = 1x

f (x) = x2+ 1

f (x) = 4 -

1

2x

____ 6 A 6 ____.

x � 2x � 0

____ 6 A 6 ____.

x � 4x � 0

limnS�

an

k = 1a k2

5n3-

2

n+

3k

n2 blimnS�

an

k = 1a k2

2n3+

2

n-

4k

n2 b

limnS�

an

k = 1a2k3

3n5-

k

4n2 blimnS�

an

k = 1a2k3

n4+

k2

3n3 blimnS�

an

k = 1a5

n-

k

n2 blimnS�

an

k = 1a2k

n2-

3

n b

limnS�

an

k = 1

2k2

n3lim

nS� a

n

k = 1

3k

n2lim

nS� a

n

k = 1

8

nlim

nS� a

n

k = 1

5

n

x

y

54321

10

5

f (x) = x212

x

y

321

10

5

f (x) = 8 – x3



42. Surveying. Steve and Peggy bought a parcel of land so their

dogs, Cisco, Cali, Cruz, and Auti, could have more room to

train for field trials. One side of their property is bounded by

a road, and two sides of their property are bounded by their

two neighbors’ properties. The back of the property is

bordered by a creek whose path can be approximately

modeled by the function where xis given in feet. Calculate the area of their parcel of land to

the nearest hundredth of an acre (1 acre = 43,560 ft2).

f (x) = - 0.006x2+ 5000,

41. Surveying. Colby and Michelle bought a parcel of land so

their dog Chief could have more room to train for field

trials. One side of their property is bounded by a road, and

two sides of their property are bounded by their two

neighbors’ properties. The back of the property is bordered

by a creek whose path can be approximately modeled by

the function

where x is given in feet. Calculate the area of their

parcel of land to the nearest hundredth of an acre

See graph below:(1 acre = 43,560 ft2).

f (x) = 200 +

1

100,000 (- x3

+ 600x2)

For Exercises 43 and 44, refer to the following:

The work that it takes to move an object that weighs F pounds

a distance of d feet is the product, On the Cartesian

graph, work in foot pounds is an area (in this case, the force

is constant).

W = Fd.

43. Work. A man tries to move a large object. In the

beginning, he will push with more force, and as he gets

tired, he will push with less and less force until he can

push no more. Calculate the work exerted when a man

pushes with a force of pounds from

to

44. Work. Calculate the work exerted when a man pushes with

a force of pounds from to x = 5 ft.x = 0f (x) = 400 - x2

x = 10 ft.

x = 0f (x) = 200 - x2

1000 200 300 400 500 600

100

200

300

Colby and Michelle’sProperty

400

500

600f (x) = 200 + (–x3 + 600x2)

x

y

CREEK

1100,000

0 1000 2000

1000

2000

3000

4000

5000 f (x) = –0.0006x2 + 5000

x

y

Little Creek

d (ft)

F (lb)

W (ft-lb)

d

F

W


When the force is not constant (varying), the work is still the

area—only now we must use a method similar to the area

problem for computing work.



When speed is varying (nonconstant), we can still think of distance

as the area under the speed curve over a given time interval.

For Exercises 45 and 46, refer to the following:

Recall the distance–rate–time formula, If a car travels at

a constant speed of, say, 60 miles per hour, then it will take

hour to drive 30 miles. The distance can be thought of as the

area under the speed function (curve) over a given time interval.

12

d = rt.


47. Find

Solution:

Use summation

property (1).

Use Limit Law 1.

Let


= 2an

k = 1

0 = 0

= 2an

k = 1

limnS�

1

nn S �.

= 2 limnS�

an

k = 1

1

n

limnS�

an

k = 1

2

n= lim

nS� 2 a

n

k = 1

1

n

limnS�

an

k = 1

2

n. 48. Find

Solution:

Use summation

property (1).

Use Limit Law 4

(limit of product).

Let


= limnS�

1

n2 lim nS�a

n

k = 1

k = 0n S �.

= limnS�

1

n2 limnS�

an

k = 1

k

limnS�

an

k = 1

k

n2= lim

nS� 1

n2 a

n

k = 1

k

limnS�

an

k = 1

k

n2.

For Exercises 49 and 50, determine whether each statement istrue or false.

49. 50.

51. Find the area bounded by the graph of and

the x-axis.

f (x) = - x2+ 1

limnS�

an

k = 1

1

k= 0a

n

k = 1

k2= ca

n

k = 1

kd 2

r0

t (s)

r (ft/s)

d (ft)

■ C AT C H T H E M I S TA K E K


52. Find the area bounded by the graph of and

the x-axis.

53. Find the area bounded by the graphs of and

54. Find the area bounded by the graphs of and

g (x) = x.

f (x) = x2

g (x) = - x2+ 2.

f (x) = x2

f (x) = - x2+ 4

t (h)

r (mi/h)

d (mi) 45. Distance–Rate–Time. Assume the speed of a smart car

(driven on a college campus) is given by

where v is the speed of the car in feet per second and t is the

time in seconds after the smart car starts moving. Find the

distance traveled by the smart car from to

46. Distance–Rate–Time. Repeat Exercise 45 for to

t = 6 s.

t = 4

t = 10 s.t = 0

v(t) = 10t - t2,


57. Approximate the area bounded by the graph of

the x-axis, and the vertical lines and using

four rectangles and the right endpoints. Repeat for ten

rectangles. Round to four decimal places.

58. Approximate the area bounded by the graph of

the x-axis, and the vertical lines and using four

rectangles and the right endpoints. Repeat for ten

rectangles. Round to four decimal places.

x =

p

2x = 0

y = cos x,

x = px = 0

y = sin x, 59. Use a graphing utility to approximate the area bounded by

the graph of the x-axis, and the vertical lines

and using 4, 100, and 500 rectangles. Round to four

decimal places.

60. Use a graphing utility to approximate the area bounded by

the graph of the x-axis, and the vertical

lines and using 4, 100, and 500 rectangles.

Round to four decimal places.

x = 2x = 0

y = ln(x + 1),

x = 2

x = 0y = ex,



56. Find the area bounded by the x-axis and the graph of the

function:

f (x) = Lx3 0 … x … 1

x2 1 6 x … 2

- 2x + 8 2 6 x … 4

55. Find the area bounded by the x-axis and the graph of the

function:

f (x) = Lx2 0 … x … 1

1 1 6 x … 2

3 - x 2 6 x … 3

■ CHALLENGE


1. Consider the graph of the function y � f (x) shown at the right, along with its tangentline at x � 1.

a. Determine the derivative of f at x � 1.

Hint: Recall from Section 11.3 that thederivative of a function at a point isdefined to be the slope of the tangentline at that point.

b. Next, use a ruler or straight edge tosketch the tangent line to the function, y � f (x), through each of the indicatedpoints (at x � �3, �2, �1, 0, 2, and 3).

c. Determine the derivative of f at x � �3,�2, �1, 0, 2, and 3 and record these in the table below.

1142

CHAPTER 11 INQUIRY-BASED LEARNING PROJECT

Graph of y � f(x)

x �3 �2 �1 0 1 2 3

y � f �(x)

d. Graph the derivative of f by plotting the points from your table in part (c). Do you notice anything interesting about this graph?

e. The graph of f � has an x-intercept at x � 0. What does this mean about the graph of f ?

f. The graph of f � is below the x-axis for x � 0 and above the x-axis for x � 0. What does this mean about the graph of f ?

2. The graph of y � g(x) is shown below.

a. Sketch the graph of its derivative, y � g�(x), by plotting points.

Graph of y � f �(x)

x �3 �2 �1 0 1 2 3

g�(x)


b. What can you say about the derivative of a linear function?

3. Lastly, suppose y � h(x) is a constant function. Try to predict what thegraph of its derivative, y � h�(x), will look like. Explain.

Graph of y � g�(x)Graph of y � g(x)

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MODELING OUR WORLD

The seven-wedges model is a proposed plan to keep the world fromdoubling our current CO2 emissions. In 2005, the world produced roughly 7 billion tons, or 7 gigatons of carbon (GtC) per year and is currently on apath to double that to 14 GtCs per year by 2055. Each of the seven wedgesrepresents a cumulative savings of 25 GtCs (the area of each wedge). In thisproject, we are going to assume a nonlinear model for carbon emissions anddetermine the total amount of carbon emission reduction between the models.

1. Develop two quadratic models of the form C(t ) � a(t � h)2 � k, where Cis the number of gigatons of carbon, t represents the year, t � 0corresponds to the year 2005, and the peak (vertex) of our carbonemissions occurs in the year 2055.

a. Model A: Maximum carbon emissions in 2055 is 9 GtC.

b. Model B: Maximum carbon emissions in 2055 is 10 GtC.

2. Calculate the total amount of GtCs that are emitted over the 50-year periodfrom 2005 to 2055. (The area under the curve of C represents the totalamount of carbon emitted over the time period.)

a. Model A

b. Model B

3. What is the difference in total carbon emissions between Models A and Bover the 50-year period?

4. Research ways to reducing carbon emissions (e.g., driving a 60-mpg carinstead of a 30-mpg car) and suggest ways to meet the goals of the twomodels.

Year

1975 2015 2055 2095 2135

Tons

of

Car

bon E

mit

ted /

Yea

r

(in b

illi

ons)

20

18

16

14

12

10

8

6

4

Wedges

FlatPath

7 GtC/yr

14 GtC/yr

7 “Wedges”HistoricalEmissions

Currently

Projected

Path

1.9

Total =25 Gigatons Carbon

50 years

1 GtC/yr

What is a “Wedge”?

A “solution” to the CO2 problem should provide at least one wedge.

Cumulatively, a wedge redirects the flow of 25 GtCs in its first 50 years.

A “wedge” is a strategy to reduce carbon emissions thatgrow in 50 years from 0 to 1.0 GtC/yr. The strategyhas already been commercialized at scale somewhere.


SECTION CONCEPT KEY IDEAS/FORMULAS

11.1 Introduction to limits: Estimating limits numericallyand graphically

Definition of a limit

The limit of f(x) as x approaches a is L.

Estimating limits Inspection of graphs and tables allows us to estimate

numerically and limits. Be careful: Technology sometimes indicates

graphically incorrect behavior.

Note: even if or f(a) is not defined.

Limits that fail to exist ■ Piecewise-defined functions with jumps

■ Functions with oscillating behavior

■ Functions with unbounded behavior:

or

One-sided limits The left-hand limit of f (x) as x approaches a is L:

The right-hand limit of f(x) as x approaches a is L:

If the left-hand and right-hand limits both exist and are

equal, then the two-sided limit exists.

if and only if

and

11.2 Techniques for finding limits

Limit laws Sum:

Difference:

Constant multiple:

Product:

Quotient:

Power: if exists

Root: if exists

Special limits:

limxSa1nx = 1nalim

xSa x

n= an

limxSa

x = alimxSa

c = c

limxSa

f (x)limxSa1n

f (x) = 1n

limxSa

f (x)

limxSa

f (x)limxSa

[ f (x)]n= c limxSa

f (x) d n

limxSac f (x)

g(x)d =

limxSa

f (x)

limxSa

g(x) if lim

xSag(x) Z 0

limxSa

[ f (x)g(x)] = lim

xSa f (x) # lim

xSa g(x)

lim

xSa[cf (x)] = c lim

xSa f (x)

limxSa

[ f (x) - g(x)] = limxSa

f (x) - lim

xSa g(x)

limxSa

[ f (x) + g(x)] = lim

xSa f (x) + lim

xSa g(x)

limxSa+

f (x) = LlimxSa-

f (x) = L

limxSa

f (x) = L

limxSa+

f (x) = L

limxSa-

f (x) = L

limxSa

f (x) = -�limxSa

f (x) = �

f (a) Z LlimxSa

f (x) can equal L

limxSa

f (x) = L

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Finding limits using Use limit laws and special limits with

limit laws information from graphs.

Finding limits using If f(x) is continuous at x � a, then .

direct substitution

Finding limits using ■ Simplifying and dividing out common factors

algebraic techniques ■ Rationalizing denominators or numerators

Finding limits using left-hand If then .

and right-hand limits

11.3 Tangent lines and derivatives

Tangent lines The tangent line to the graph of f at the point (a, f (a))

is the line that

■ passes through the point (a, f (a)) and

■ has slope m:

The derivative of a function The derivative of a function f at x, denoted is

provided this limit exists.

Instantaneous rates The instantaneous rate of change of a function fof change at a point is the limit of the average rate of change

as x approaches a.

11.4 Limits at infinity; limits of sequences

Limits at infinity

The limit of f (x) as x approaches is L.

f (x) has a horizontal asymptote:

Special limit at infinity:

has a horizontal asymptote:

Limits of sequences Convergent sequence:

Divergent sequence: does not exist.limnS�

an

limnS�

an = L

y = 0 for n 7 0.

f (x) =

1

xn

limxS�

1

xn = 0

y = L.�

aor limxS-�

f (x) = LblimxS�

f (x) = L

limxSa

f (x) - f (a)

x - a= f ¿(a)

x = a

f ¿(x) = limhS0

f (x + h) - f (x)

h

f ¿(x),

m = limxSa

f (x) - f (a)

x - a

limxSa

f (x) = LlimxSa-

f (x) = limxSa+

f (x) = L,

limxSa

f (x) = f (a)


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11.5 Finding the area Find the area of the region S that lies under the curve f (x)

under a curve and above the x-axis from to

Limits of summations Summation properties:

Summation formulas:

The area problem A = limnS�

an

k = 1 f (xk)¢x

an

k = 1k3

=

n2(n + 1)2

4an

k = 1k2

=

n(n + 1)(2n + 1)

6

an

k = 1k =

n(n + 1)

2an

k = 1c = cn

an

k = 1(ak ; bk) = a

n

k = 1ak ; a

n

k = 1bk

an

k = 1cak = ca

n

k = 1ak

y

x

a

f (x)

S

b

x = b.x = a


r }

height width

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11.1 Introduction to Limits: Estimating Limits Numerically and Graphically

Complete a table of values to four decimal places and usethe result to estimate each limit.

1. 2.

Use the graph to estimate the limit, if it exists.

3.

4.

5.

6. where

f (x) = e x2- 1 x Z 0

1 x = 0

limxS0

f (x),

limxSp/2

csc x

limxS -1

ƒ x + 1 ƒ

x + 1

limxS0

1

x2

limxS16

1x - 4

x - 16limxS3

x - 3

x2- 9

11.2 Techniques for Finding Limits

Find each limit, if it exists.

7. 8.

9. 10.

11. 12.

13. 14.

Find .

15. 16.

Evaluate the one-sided limits in order to find the limit, if it exists.

17. where

18. where

19. where

20. where

11.3 Tangent Lines and Derivatives

Find the slope of the tangent line to the graph of f at the given point.

21. at the point (0, 3)

22. at the point (�1, �3)

23. at the point (�1, 1)


Find the equation of the tangent line to the graph of f at thegiven point.


26. at the point (1, 2)f (x) = -3x + 5

f (x) = 12

f (x) = 1x + 1

f (x) =

1

x2

f (x) = -3x2

f (x) = 6x + 3

f (x) = e cos x x 6 p

sec x x 7 plimxSp

f (x),

f (x) = d

sin x x 6

p

4

cos x x 7

p

4

limxSp/4

f (x),

f (x) = e -x + 1 x 6 0

x2- 1 x 7 0

limxS0

f (x),

f (x) = e -x x 6 0

x2 x 7 0limxS0

f (x),

f (x) = -7x + 9f (x) = x2- 2x + 3

limhS0

f (x � h) � f (x)

h

limxSp

tan x

sec xlim

xS2p

1 - cos x

sin x

limxS9

1x - 3

x - 9lim

xS -1

x2- 1

x + 1

limxS -2

2x2+ 5lim

xS0

2x2- 3x + 5

-x2+ 10

limxS -1

x5limxS3

8

–5 –3 –1 21 3 4 5

x

y

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10

–5 –3 –1 21 3 4 5

x

y

–5

–3–2

–4

12345

� � 2�2

3�2

–4

–2

–3

–1

2

4

1

3

x

y

–5 –3 –1 21 3 4 5

x

y

–5

–3–2

–4

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CHAPTER 11 REVIEW EXERCISES



28. at the point

Find the derivative of the function at the specified value of x.

29. where

30. where

31. where

32. where

Find the derivative

33. 34.

35. 36.

11.4 Limits at Infinity; Limits of Sequences

Find each limit, if it exists.

37. 38.

39. 40.

41. 42.

Determine whether the sequence converges. If it does converge, state the limit.

43. 44.

45. 46.

11.5 Finding the Area Under a Curve

Evaluate the summations.

47. 48.

Find the area under the curve on the specified interval.

49.

50.

51.

52. 1 … x … 2f (x) = x2+ 3x + 1,

1 … x … 3f (x) = 5x + 3x2,

0 … x … 6f (x) = x2,

1 … x … 4f (x) = 5 - x,

a12

k = 1(5 + k2)a

30

k = 13k2

an = (-1)n n!an = sin (np)

an =

(n + 1)!

(n - 1)!n (n + 1)an =

3n - 4n

limxS�

a 5x

x2- 8 blim

xS� a1

x- 3xb

limxS�

ln xlimxS-�

e2x

limxS�

a4x2- 2x

3x2+ 9 blim

xS� a- 3

x b

f (x) = 1x - 1f (x) = 3x2- 2x + 1

f (x) = 6x + 7f (x) = e

f œ(x).

f (x) =

x

1 + xf ¿(0),

f (x) = 1 - x2f ¿(5),

f (x) = -x4f ¿(-1),

f (x) = 2 - 5x2f ¿(2),

a1

2, 2bf (x) =

1x

f (x) = 5x2+ 2 Technology Exercises

Section 11.1

Use a graphing utility to determine whether the limit exists.Estimate the limit to three decimal places, if it exists.

53. 54.

Section 11.2

Use a graphing utility to estimate the limit, if it exists. Confirm by finding the exact limit using the limit laws andalgebraic techniques.

55. where

56. where

Section 11.3

Use a graphing utility.


and Based on what you

see, what would you guess to be?


and Based on

what you see, what would you guess to be?

Section 11.4

Use a graphing utility to determine whether the sequence converges. If it does converge, find the limit to four decimal places.

59. 60.

Section 11.5

61. Use a graphing utility to approximate the area bounded

by the graph of the x-axis, and the vertical

lines and using 4, 100, and 500 rectangles.

Round to four decimal places.


the graph of the x-axis, and the vertical lines

and using 4, 100, and 500 rectangles. Round

to four decimal places.

x =

p

3x = 0

y = tan x,

x = 1x = 0y = 12x + 1,

an = n cos a 1n b - 1an = n sin

15n

f ¿(x)

h = ;0.001.h = ;0.01,h = ;0.1,

cos (x + h) - cos x

h,f (x) = cos x

f ¿(x)

h = ;0.001.h = ;0.01,h = ;0.1,

ln (x + h) - ln x

h,f (x) = ln x

f (x) =

e2 (x - 1)- 1

ex - 1- 1

limxS1

f (x),

f (x) =

x - 3

212 - x - 3limxS3

f (x),

limxS-2

x3+ 2x2

- x - 2

x + 2limxS1

tan ap3 xb - 13

tan ap4 xb - 1

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Use the graph to estimate each limit, if it exists.

1.

2. where

In Exercises 3–12, find each limit, if it exists.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12. limxS�

a 2x

x2- 1 blim

xS� a1

x- xb

limxS�

a 9x2+ x

3x2+ 7 blim

xS� a-

5

x - 1 b

limxSp

csc x

cot xlimxSp

1 - sin x

cos x

limxS -5

x2- 25

x + 5limxS0

-x2+ 4x - 2

x2+ 10

limxS -1

(x + 1)5limxS5

6

f (x) = e x2 x Z 0

2 x = 0limxS0

f (x),

limxS1

1

(x - 1)2

13. Find for

14. Find where .

15. Find the equation of the tangent line to the graph of

at the point ( ).

16. Find where .

In Exercises 17 and 18, find the derivative

17.

18.

19. Determine whether the sequence given by

converges. If it does converge, state the limit.

20. Evaluate the summation

In Exercises 21 and 22, find the area under the curve on thespecified interval.

21.

22.

23. Use a graphing utility to determine whether the sequence

converges. If it does converge, find the

limit to four decimal places.


the graph of the x-axis, and the vertical lines,

and , using 4 and 100 rectangles. Round each

to four decimal places.

x =

p

2x = 0

y = cos x,

an = n sin 13n

-1 … x … 1f (x) = -x2+ 1,

-2 … x … 2f (x) = x2,

a20k=1(2k + 5).

an =

(2n - 1)!

(2n + 1)!

f (x) = -2x2- 3x + 7

f (x) = 17x + 5

fœ(x).

f (x) = -x4+ 2f ¿(1),

1, -1f (x) = -3x2+ 2

f (x) = e -x + 2 x 6 0

x2+ 2 x 7 0

limxS0

f (x),

f (x) = -2 x2+ 3x - 7.lim

hS0

f (x + h) - f (x)

h

–5 –3 –1 21 3 4 5

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y

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CHAPTER 11 PRACTICE TEST

1150


13. Evaluate exactly cos .

14. Solve the triangle and ft.

15. Solve the system of linear equations: .

16. Find the determinant of .

17. For the matrices and , find ABand BA.

18. Identify the center and graph the ellipse

19. Solve the system of nonlinear equations

20. Find , if it exists.

21. Find , if it exists.

22. Find the area bounded by the curves

and x = 3.x = 0,y = 0,y = x2,

limxS�

5x2+ 2

10x2+ 1

limxS3

x2- 2x - 3

x - 3

(x - 1)2+ (y + 1)2

= 9

x + y = 3

9x2- 18x + 4y2

+ 16y = 11

J5 1

0 4

3 2KA = c2 3 1

0 1 2 d

†

3 0 1

2 -5 -1

1 2 7

†

4x + 3y - z = - 4

- x + y - z = 0

3x + 2y + 5z = 25

c = 3.2b = 3.7 ft,� = 103°,

c tan- 1 a- 13

3bd1. For the function find the difference

quotient .

2. Find the inverse of the function . State the

domain and range of both f and its inverse

3. Find the composite function where ,

and for and state its domain and range.

4. Find the vertex and graph the parabola given by

5. Write as a product of

linear factors.

6. State the vertical and horizontal asymptotes and graph the

rational function .

7. Solve

8. Solve . Round your answer to three decimal

places.

9. How many years will it take your initial principal to double if

invested in an account earning 4% annually and compounded

continuously?

10. Evaluate exactly .

11. State the domain and range of the function

12. Solve -sin2 x + cos x - 1 = 0.

f (x) = 2 sin (3x).

tan a7p

6 b

e2 x - 1= 15.7

log3 (2x + 7) - log3 (x - 1) = 1.

f (x) =

3x2+ x - 2

x2- 4

P(x) = x4- x3

+ 7x2- 9x - 18

f (x) = x2- 4x + 7.

x Ú 0g(x) = x2+ 1

f (x) = 1x - 1f (g(x)),

f -1.

f (x) =

2

5 + x

f (x + h) - f (x)

h

f (x) = -7x2+ 3x,

CHAPTERS 1–11 CUMULATIVE TEST

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Ch11 Limits Apreviewtocalculus