580 ! Chapter 9 Derivatives

Limits

] Application PreviewAlthough everyone recognizes the value of eliminating any and all particulate pollution fromsmokestack emissions of factories, company owners are concerned about the cost of removing thispollution. Suppose that USA Steel has shown that the cost C of removing p percent of the par-ticulate pollution from the emissions at one of its plants is

To investigate the cost of removing as much of the pollution as possible, we can evaluate the limitas p (the percent) approaches 100 from values less than 100. (See Example 6.) Using a limit isimportant in this case, because this function is undefined at (it is impossible to remove100% of the pollution).

In various applications we have seen the importance of the slope of a line as a rate ofchange. In particular, the slope of a linear total cost, total revenue, or profit function for aproduct tells us the marginals or rates of change of these functions. When these functionsare not linear, how do we define marginals (and slope)?

We can get an idea about how to extend the notion of slope (and rate of change) tofunctions that are not linear. Observe that for many curves, if we take a very close (or“zoom-in”) view near a point, the curve appears straight. See Figure 9.1. We can think ofthe slope of the “straight” line as the slope of the curve. The mathematical process used toobtain this “zoom-in” view is the process of taking limits.

p ! 100

C ! C(p) !7300p

100 " p

We have used the notation f(c) to indicate the value of a function f(x) at If we needto discuss a value that f(x) approaches as x approaches c, we use the idea of a limit. Forexample, if

then we know that is not in the domain of f(x), so does not exist even thoughf(x) exists for every value of Figure 9.2 shows the graph of with an openy ! f(x)x " #2.

f(#2)x ! #2

f(x) !x2 # x # 6

x $ 2

x ! c.

9.1OBJECTIVES

! To use graphs and numericaltables to find limits offunctions, when they exist

! To find limits of polynomialfunctions

! To find limits of rationalfunctions

x

y

P

Zooming in near point P, the curve appears straight.

P

P

y = f (x)

Figure 9.1

Notion of a Limit

Limit

9.1 Limits ! 581

circle where The open circle indicates that does not exist but shows thatpoints near have functional values that lie on the line on either side of the opencircle. Even though is not defined, the figure shows that as x approaches fromeither side of the graph approaches the open circle at and the values of f(x)approach Thus is the limit of f(x) as x approaches and we write

This conclusion is fairly obvious from the graph, but it is not so obvious from the equationfor f(x).

We can use the values near in Table 9.1 to help verify that asNote that to the left of the values of x increase from to in

small increments, and in the corresponding column for f(x), the values of the function f(x)increase from to To the right of the values of x decrease from to while the corresponding values of f(x) decrease from to Hence,Table 9.1 and Figure 9.2 indicate that the value of f(x) approaches as x approachesfrom both sides of

From our discussion of the graph in Figure 9.2 and Table 9.1, we see that as xapproaches from either side of the limit of the function is the value L that the func-tion approaches. This limit L is not necessarily the value of the function at Thisleads to our intuitive definition of limit.

Let f(x) be a function defined on an open interval containing c, except perhaps at c.Then

is read “the limit of f(x) as x approaches c equals L.” The number L exists if we canmake values of f(x) as close to L as we desire by choosing values of x sufficiently closeto c. When the values of f(x) do not approach a single finite value L as x approaches c,we say the limit does not exist.

As the definition states, a limit as can exist only if the function approaches asingle finite value as x approaches c from both the left and right of c.

x S c

limxSc

f(x) ! L

x ! #2.#2,#2

x ! #2.#2#5

#4.999.#4.000#1.999#1.000#2,#5.001.#6.000

#2.001#3.000#2,x S #2.f(x) S #5x ! #2

limxS#2

f(x) ! #5, or f(x) S #5 as x S #2

#2,#5#5.(#2, #5)#2,

#2f(#2)x ! #2

f(#2)x ! #2.

Figure 9.2

TABLE 9.1

Left of

x

#3.000 #6.000#2.500 #5.500#2.100 #5.100#2.010 #5.010#2.001 #5.001

Right of

x

#1.000 #4.000#1.500 #4.500#1.900 #4.900#1.990 #4.990#1.999 #4.999

f (x) !x2 " x " 6

x # 2

"2

f (x) !x2 " x " 6

x # 2

"2

x

y

2

−6

−4

−2

(−2, −5)

f (x) = x2 − x − 6x + 2

−2

582 ! Chapter 9 Derivatives

! EXAMPLE 1 Limits

Figure 9.3 shows three functions for which the limit exists as x approaches 2. Use this figureto find the following.

(a) and f(2) (if it exists)

(b) and g(2) (if it exists)

(c) and h(2) (if it exists)limxS2

h(x)

limxS2

g(x)

limxS2

f(x)

Solution(a) From the graph in Figure 9.3(a), we see that as x approaches 2 from both the left

and the right, the graph approaches the point (2, 3). Thus f(x) approaches the singlevalue 3. That is,

The value of f(2) is the y-coordinate of the point on the graph at Thus (b) Figure 9.3(b) shows that as x approaches 2 from both the left and the right, the graph

approaches the open circle at Thus

The figure also shows that at there is no point on the graph. Thus g(2) isundefined.

(c) Figure 9.3(c) shows that

The figure also shows that at there is a point on the graph at (2, 4). Thusand we see that

As Example 1 shows, the limit of the function as x approaches c may or may not bethe same as the value of the function at

In Example 1 we saw that the limit as x approaches 2 meant the limit as x approaches2 from both the left and the right. We can also consider limits only from the left or onlyfrom the right; these are called one-sided limits.

x ! c.

limxS2

h(x) " h(2).h(2) ! 4,x ! 2

limxS2

h(x) ! 1

x ! 2

limxS2

g(x) ! #1

(2, #1).

f(2) ! 3.x ! 2.

limxS2

f(x) ! 3

−2 2 4−1

2

4

y

x

(a)

(2, 3)

y = f (x)

−2 1 4

2

4

y

x

(b)

−2(2, −1)

y = g(x)

−2 2 4

−2

2

4

y

x

(c)

(2, 4)

(2, 1)

y = h(x)

Figure 9.3

One-Sided Limits

9.1 Limits ! 583

Solution(a) As from the left side and the right side of f(x) increases without bound,

which we denote by saying that f(x) approaches as In this case,

does not exist [denoted by DNE] because f(x) does not approach a finite

value as In this case, we write

The graph has a vertical asymptote at (b) As from the left, g(x) approaches and as from the right, g(x)

approaches so g(x) does not approach a finite value as Therefore, thelimit does not exist. The graph of has a vertical asymptote at x ! 2.y ! g(x)

x S 2.$%,x S 2#%,x S 2

x ! 2.

f(x) S $% as x S 2

x S 2.

limxS2

f(x)

limxS2

f(x)x S 2.$%x ! 2,x S 2

Figure 9.4 (a)

1 2 3 4

1

2

3

4

5

6

x

y

f (x) = 1(x − 2)2

-3

-2

-1

1

2

3

x

y

g(x) = 1x − 2

(b)

1 2 3 4

-1 1 2 3 4-1

1

2

3

45

6

7

x

y

(c)

h(x) = 4 − x2 if x ≤ 22x − 1 if x > 2

Limit from the Right:

means the values of f(x) approach the value L as but

Limit from the Left:

means the values of f(x) approach the value M as but

Note that when one or both one-sided limits fail to exist, then the limit does notexist. Also, when the one-sided limits differ, such as if above, then the values off(x) do not approach a single value as x approaches c, and does not exist. So far,

the examples have illustrated limits that existed and for which the two one-sided limits werethe same. In the next example, we use one-sided limits and consider cases where a limitdoes not exist.

! EXAMPLE 2 One-Sided Limits

Using the functions graphed in Figure 9.4, determine why the limit as does notexist for

(a) f(x)(b) g(x)(c) h(x)

x S 2

limxSc

f(x)L " M

x & c.x S c

limxSc#

f(x) ! M

x ' c.x S c

limxSc$

f(x) ! L

584 ! Chapter 9 Derivatives

In this case we summarize by writing

and

(c) As from the left, the graph approaches the point at (2, 0), so As

from the right, the graph approaches the open circle at (2, 3), so

Because these one-sided limits differ, does not exist.

Examples 1 and 2 illustrate the following two important facts regarding limits.

1. The limit of a function as x approaches c is independent of the value of thefunction at c. When exists, the value of the function at c may be (i) the

same as the limit, (ii) undefined, or (iii) defined but different from the limit (seeFigure 9.3 and Example 1).

2. The limit is said to exist only if the following conditions are satisfied:(a) The limit L is a finite value (real number).(b) The limit as x approaches c from the left equals the limit as x approaches c

from the right. That is, we must have

Figure 9.4 and Example 2 illustrate cases where does not exist.

1. Can exist if f (c) is undefined?

2. Does exist if

3. Does if

4. If does exist?

Fact 1 regarding limits tells us that the value of the limit of a function as will notalways be the same as the value of the function at However, there are many func-tions for which the limit and the functional value agree [see Figure 9.3(a)], and for thesefunctions we can easily evaluate limits. The following properties of limits allow us to iden-tify certain classes or types of functions for which equals f(c).lim

xScf(x)

x ! c.x S c

limxSc

f(x)limxSc#

f(x) ! 0,

limxSc

f(x) ! 1?f(c) ! 1

f(c) ! 0?limxSc

f(x)

limxSc#

f(x)

limxSc

f(x)

limxSc#

f(x) ! limxSc$

f(x)

limxSc

f(x)

limxS2

h(x)

limxS2$

h(x) ! 3.x S 2

limxS2#

h(x) ! 0.x S 2

limxS2

g(x) DNE

limxS2$

g(x) DNE or g(x) S $% as x S 2$

limxS2#

g(x) DNE or g(x) S #% as x S 2#

! Checkpoint

Properties of Limits,Algebraic Evaluation

Properties of Limits

If k is a constant, and then the following are true.

I. IV.

II. V.

III. VI.

provided that when n is even.L ' 0

limxSc1n f(x) ! 1n lim

xScf (x) !1n L,lim

xSc3 f(x) ( g(x) 4 ! L ( M

limxSc

f(x)g(x)

!LM if M " 0lim

xScx ! c

limxSc3 f(x) $ g(x) 4 ! LMlim

xSck ! k

limxSc

g(x) ! M,limxSc

f(x) ! L,

9.1 Limits ! 585

If f is a polynomial function, then Properties I–IV imply that can be found

by evaluating f(c). Moreover, if h is a rational function whose denominator is not zero atthen Property V implies that can be found by evaluating h(c). The follow-

ing summarizes these observations and recalls the definitions of polynomial and rationalfunctions.

limxSc

h(x)x ! c,

limxSc

f(x)

Function

Polynomial function

Rational function

Definition

The function

where and n is a positive integer, is called apolynomial function of degree n.

The function

where both f(x) and g(x) are polynomial functions, iscalled a rational function.

h(x) !f(x)g(x)

an " 0$ p $ a1x $ a0,f(x) ! anxn $ an#1xn#1

Limit

for all values c (by Properties I–IV)

when (by Property V)g(c) " 0

limxSc

h(x) ! limxSc

f(x)g(x)

!f (c)g(c)

limxSc

f(x) ! f (c)

! EXAMPLE 3 Limits

Find the following limits, if they exist.

(a)

(b)

Solution(a) Note that is a polynomial, so

Figure 9.5(a) shows the graph of (b) Note that this limit has the form

where f(x) and g(x) are polynomials and Therefore, we have

Figure 9.5(b) shows the graph of g(x) !x2 # 4xx # 2

.

limxS4

x2 # 4xx # 2

!42 # 4(4)

4 # 2!

02

! 0

g(c) " 0.

limxSc

f(x)g(x)

f (x) ! x3 # 2x.

limxS#1

f(x) ! f (#1) ! (#1)3 # 2(#1) ! 1

f (x) ! x3 # 2x

limxS4

x2 # 4xx # 2

limxS#1

(x3 # 2x)

586 ! Chapter 9 Derivatives

We have seen that we can use Property V to find the limit of a rational functionas long as the denominator is not zero. If the limit of the denominator of

is zero, then there are two possible cases.

I. Both and or

II. and

In case I we say that has the form at We call this the 0!0 inde-terminate form; the limit cannot be evaluated until is factored from both f(x) andg(x) and the fraction is reduced. Example 4 will illustrate this case.

In case II, the limit has the form where a is a constant, This expression isundefined, and the limit does not exist. Example 5 will illustrate this case.

! EXAMPLE 4 0!0 Indeterminate Form

Evaluate the following limits, if they exist.

(a)

(b)

Solution(a) We cannot find the limit by using Property V because the denominator is zero at

The numerator is also zero at so the expression

has the indeterminate form at Thus we can factor from both thenumerator and the denominator and reduce the fraction. (We can divide by because while )

limxS2

x2 # 4x # 2

! limxS2

(x # 2)(x $ 2)x # 2

! limxS2

(x $ 2) ! 4

x S 2.x # 2 " 0x # 2

x # 2x ! 2.0!0

x2 # 4x # 2

x ! 2,x ! 2.

limxS1

x2 # 3x $ 2x2 # 1

limxS2

x2 # 4x # 2

a " 0.a!0,

x # cx ! c.0!0f(x)!g(x)

limxSc

f(x) " 0.limxSc

g(x) ! 0

limxSc

f(x) ! 0,limxSc

g(x) ! 0

f(x)!g(x)f(x)!g(x)

x

y

−2 2

−4

−2

2

4

(−1, 1)

f (x) = x3 − 2x

(a)

x

y

−2 4 6

-4

−2

2

4

(4, 0)

2

(b)

g(x) = x2 − 4xx − 2

Figure 9.5

9.1 Limits ! 587

Figure 9.6(a) shows the graph of Note the open circleat (2, 4).

(b) By substituting 1 for x in we see that the expression has theindeterminate form at so is a factor of both the numerator and the

denominator. (We can then reduce the fraction because while )

Figure 9.6(b) shows the graph of Note the opencircle at (1, #1

2).g(x) ! (x2 # 3x $ 2)!(x2 # 1).

!1 # 21 $ 1

!#12

(by Property V)

! limxS1

x # 2x $ 1

limxS1

x2 # 3x $ 2x2 # 1

! limxS1

(x # 1)(x # 2)(x # 1)(x $ 1)

x S 1.x # 1 " 0x # 1x ! 1,0!0

(x2 # 3x $ 2)!(x2 # 1),

f(x) ! (x2 # 4)!(x # 2).

x

y

−2 2 4

2

4

6

(2, 4)

(a)

f (x) = x2 − 4x − 2

-2 2 4

-2

2

4

(b)

g(x) = x2 − 3x + 2x2 − 1

(1, – )12

x

y

Figure 9.6

Note that although both problems in Example 4 had the indeterminate form, theyhad different answers.

! EXAMPLE 5 Limit with a!0 Form

Find if it exists.

SolutionSubstituting 1 for x in the function results in so this limit has the form with and is like case II discussed previously. Hence the limit does not exist. Because the numer-ator is not zero when we know that is not a factor of the numerator, and wecannot divide numerator and denominator as we did in Example 4. Table 9.2 confirms thatthis limit does not exist, because the values of the expression are unbounded near x ! 1.

x # 1x ! 1,

a " 0,a!0,6!0,

limxS1

x2 $ 3x $ 2x # 1

,

0!0

Rational Functions:Evaluating Limits of the

Form where

limxSc

g(x) ! 0

limxSc

f(x)g(x)

588 ! Chapter 9 Derivatives

The left-hand and right-hand limits do not exist. Thus does not exist.

In Example 5, even though the left-hand and right-hand limits do not exist (see Table9.2), knowledge that the functional values are unbounded (that is, that they become infinite)is helpful in graphing. The graph is shown in Figure 9.7. We see that is a verticalasymptote.

x ! 1

limxS1

x2 $ 3x $ 2x # 1

TABLE 9.2

Left of x ! 1 Right of x ! 1

x x

0 #2 2 120.5 #7.5 1.5 17.50.7 #15.3 1.2 35.20.9 #55.1 1.1 65.10.99 #595.01 1.01 605.010.999 #5,995.001 1.001 6,005.0010.9999 #59,995.0001 1.0001 60,005.0001

(f (x) S #% as x S 1#) (f (x) S $% as x S 1$)

x2 $ 3x $ 2x # 1

DNElimxS1$

x2 $ 3x $ 2x # 1

DNElimxS1#

x2 # 3x # 2x " 1

x2 # 3x # 2x " 1

x

y

−12 −8 4 8 12 16

−8

−4

4

8

12

16

20

f (x) unbounded ( f (x) → + %)as x → 1+

f (x) unbounded ( f (x) → − %)as x → 1−

y = x2 + 3x + 2x − 1

Figure 9.7

The results of Examples 4 and 5 can be summarized as follows.

Type I. If and then the fractional expression has the 0!0indeterminate form at We can factor from f(x) and g(x), reduce thefraction, and then find the limit of the resulting expression, if it exists.

Type II. If and then does not exist. In this case,

the values of are unbounded near the line is a vertical asymptote.x ! cx ! c;f(x)!g(x)

limxSc

f(x)g(x)

limxSc

g(x) ! 0,limxSc

f(x) " 0

x # cx ! c.

limxSc

g(x) ! 0,limxSc

f(x) ! 0

9.1 Limits ! 589

] EXAMPLE 6 Cost-Benefit (Application Preview)

USA Steel has shown that the cost C of removing p percent of the particulate pollutionfrom the smokestack emissions at one of its plants is

To investigate the cost of removing as much of the pollution as possible, find:

(a) the cost of removing 50% of the pollution.(b) the cost of removing 90% of the pollution.(c) the cost of removing 99% of the pollution.(d) the cost of removing 100% of the pollution.

Solution(a) The cost of removing 50% of the pollution is $7300 because

(b) The cost of removing 90% of the pollution is $65,700 because

(c) The cost of removing 99% of the pollution is $722,700 because

(d) The cost of removing 100% of the pollution is undefined because the denominator ofthe function is 0 when To see what the cost approaches as p approaches

100 from values smaller than 100, we evaluate This limit has the

Type II form for rational functions. Thus as which means

that as the amount of pollution that is removed approaches 100%, the cost increaseswithout bound. (That is, it is impossible to remove 100% of the pollution.)

5. Evaluate the following limits (if they exist).

(a) (b) (c)

In Problems 6–9, assume that f, g, and h are polynomials.

6. Does

7. Does

8. If and can we be certain that

(a) (b) exists?

9. If and what can be said about and limxSc

h(x)g(x)

?limxSc

g(x)h(x)

h(c) ! 0,g(c) " 0

limxSc

g(x)h(x)

limxSc

g(x)h(x)

! 0?

h(c) ! 0,g(c) ! 0

limxSc

g(x)h(x)

!g(c)h(c)

?

limxSc

f (x) ! f (c)?

limxS#3!4

4x4x $ 3

limxS5

x2 # 3x # 3x2 # 8x $ 1

limxS#3

2x2 $ 5x # 3x2 # 9

x S 100#,7300

100 # pS $%

limxS100#

7300p

100 # p.

p ! 100.

C(99) !7300(99)100 # 99

!722,700

1! 722,700

C(90) !7300(90)100 # 90

!657,000

10! 65,700

C(50) !7300(50)100 # 50

!365,000

50! 7300

C ! C(p) !7300p

100 # p

! Checkpoint

590 ! Chapter 9 Derivatives

As we noted in Section 2.4, “Special Functions and Their Graphs,” many applications aremodeled by piecewise defined functions. To see how we evaluate a limit involving a piece-wise defined function, consider the following example.

! EXAMPLE 7 Limit of a Piecewise Defined Function

Find and if they exist, for

SolutionBecause f(x) is defined by when

Because f(x) is defined by when

And because

does not exist.

Table 9.3 and Figure 9.8 show these results numerically and graphically.

limxS1

f(x)

2 ! limxS1#

f(x) " limxS1$

f(x) ! 3

limxS1$

f(x) ! limxS1$

(x $ 2) ! 3

x ' 1,x $ 2

limxS1#

f(x) ! limxS1#

(x2 $ 1) ! 2

x & 1,x2 $ 1

f(x) ! bx2 $ 1 for x ) 1x $ 2 for x ' 1

limxS1

f(x),limxS1#

f(x), limxS1$

f (x),

Limits of PiecewiseDefined Functions

TABLE 9.3

Left of 1

x f (x) ! x2 # 1

0.1 1.010.9 1.810.99 1.980.999 1.9980.9999 1.9998

Right of 1

x f (x) ! x # 2

1.2 3.21.01 3.011.001 3.0011.0001 3.00011.00001 3.00001

-2 -1 10 2 3 4

1

2

3

4

5

6

x

y

f (x) = x2 + 1 (x ≤ 1)

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

Figure 9.8

We have used graphical, numerical, and algebraic methods to understand and evaluate lim-its. Graphing calculators can be especially effective when we are exploring limits graphi-cally or numerically. "

! EXAMPLE 8 Limits: Graphically, Numerically, and Algebraically

Consider the following limits.

(a) (b)

Investigate each limit by using the following methods.

(i) Graphically: Graph the function with a graphing utility and trace near the limitingx-value.

(ii) Numerically: Use the table feature of a graphing utility to evaluate the function veryclose to the limiting x-value.

(iii) Algebraically: Use properties of limits and algebraic techniques.

limxS#1

2xx $ 1

limxS5

x2 $ 2x # 35x2 # 6x $ 5

Calculator Note

9.1 Limits ! 591

Solution

(a)

(i) Figure 9.9(a) shows the graph of Tracingnear shows y-values getting close to 3.

(ii) Figure 9.9(b) shows a table for withx-values approaching 5 from both sides (note that the function is undefinedat ). Again, the y-values approach 3.

Both (i) and (ii) strongly suggest

(iii) Algebraic evaluation of this limit confirms what the graph and the table suggest.

limxS5

x2 $ 2x # 35x2 # 6x $ 5

! limxS5

(x $ 7)(x # 5)(x # 1)(x # 5)

! limxS5

x $ 7x # 1

!124

! 3

limxS5

x2 $ 2x # 35x2 # 6x $ 5

! 3.

x ! 5

y1 ! (x2 $ 2x # 35)!(x2 # 6x $ 5)x ! 5

y ! (x2 $ 2x # 35)!(x2 # 6x $ 5).

limxS5

x2 $ 2x # 35x2 # 6x $ 5

-4.5 14.5

-6

12

X=4.8 Y=3.1052632

1 Y1X4.94.994.99955.0015.015.1

3.05133.0053.0005ERROR2.99952.9952.9512

X=4.9

Figure 9.9

We could also use the graphing and table features of spreadsheets to explore limits.

-3.4 1.4

-45

45

X=-1.05 Y=42

1 Y1X-1.1-1.01-1.001-1-.999-.99-.9

222022002ERROR-1998-198-18

X=-.9

Figure 9.10

(a) (b)

(a) (b)

(b)

(i) Figure 9.10(a) shows the graph of it indicates a break in thegraph near Evaluation confirms that the break occurs at andalso suggests that the function becomes unbounded near In addition, wecan see that as x approaches from opposite sides, the function is headed indifferent directions. All this suggests that the limit does not exist.

(ii) Figure 9.10(b) shows a graphing calculator table of values for and with x-values approaching The table reinforces our preliminaryconclusions from the graph that the limit does not exist, because the function isunbounded near

(iii) Algebraically we see that this limit has the form Thus limxS#1

2xx $ 1

DNE.#2!0.

x ! #1.

x ! #1.y1 ! 2x!(x $ 1)

#1x ! #1.

x ! #1x ! #1.y ! 2x!(x $ 1);

limxS#1

2xx $ 1

592 ! Chapter 9 Derivatives

1. Yes. For example, Figure 9.2 and Table 9.1 show that this is possible for

Remember that does not depend on f(c).

2. Not necessarily. Figure 9.4(c) shows the graph of with but does not exist.

3. Not necessarily. Figure 9.3(c) shows the graph of with but

4. Not necessarily. For example, Figure 9.4(c) shows the graph of withbut with so the limit doesn’t exist. Recall that if

then

5. (a)

(b)

(c) Substituting gives so does not exist.

6. Yes, Properties I–IV yield this result.7. Not necessarily. If then this is true. Otherwise, it is not true.8. For both (a) and (b), has the indeterminate form at In this case we

can make no general conclusion about the limit. It is possible for the limit to exist (andbe zero or nonzero) or not to exist. Consider the following indeterminate forms.

(i) (ii)

(iii) which does not exist

9. does not exist and limxSc

h(x)g(x)

! 0limxSc

g(x)h(x)

limxS0

xx2

! limxS0

1x

,

limxS0

x(x $ 1)x

! limxS0

(x $ 1) ! 1limxS0

x2

x! lim

xS0x ! 0

0!0

x ! c.0!0g(x)!h(x)h(c) " 0,

limxS#3!4

4x4x $ 3

#3!0,x ! #3!4

limxS5

x2 # 3x # 3x2 # 8x $ 1

!7

#14! #

12

limxS#3

2x2 $ 5x # 3x2 # 9

! limxS#3

(2x # 1)(x $ 3)(x $ 3)(x # 3)

! limxS#3

2x # 1x # 3

!#7#6

!76

limxSc

f(x) ! L.limxSc#

f(x) ! limxSc$

f(x) ! L,

limxS2$

h(x) ! 2,limxS2#

h(x) ! 0,y ! h(x)

h(2) ! 4.limxS2

h(x) ! 1y ! h(x)

limxS2

h(x)h(2) ! 0,y ! h(x)

limxSc

f(x)f(x) !x2 # x # 6

x $ 2.

! Checkpoint Solutions

Exercises9.1

In Problems 1–6, a graph of y ! f(x) is shown and ac-value is given. For each problem, use the graph to findthe following, whenever they exist.(a) and (b) f(c)

1. 2.

9

y

x

3

–393

y = f (x)

–8

y

x

–44

8

y = f (x)

c ! 6c ! 4

limxSc

f(x)

3. 4.

5. 6.

x

y

2

−2−2 2−6

y = f (x)

x

y

−4−4−8−12

4

−8

y = f (x)

c ! #2c ! #8

x

y

–10 –5–15

15

10

5y = f (x)

x

y

20

10

−1020

y = f (x)

c ! #10c ! 20

9.1 Limits ! 593

In Problems 7–10, use the graph of and thegiven c-value to find the following, whenever they exist.(a) (b) (c) (d) f(c)

7. 8.

9. 10.

In Problems 11–14, complete each table and predict thelimit, if it exists.

11. 12.

x f(x) x f(x)

0.90.990.999

1 ? ?

1.0011.011.1

13.

x f(x)

0.90.990.999

1 ?

1.0011.011.1

cc

TT

limxS1

f(x) ! ?

f(x) ! b5x # 1 for x & 18 # 2x # x2 for x * 1

#0.49#0.499#0.4999

cccc#0.5

TTTT#0.5001#0.501#0.51

f(x) !2x $ 114 # x2

limxS#0.5

f(x) ! ?

f (x) !2 # x # x2

x # 1limxS1

f (x) ! ?

x

y

−2

1

1 2 3

−3

−1

y = f (x)

x

y

−3

3

3–3–6–9

−6

y = f (x)

c ! 2c ! #412

x

y

4

−2−2

2

−4

2

4

y = f (x)-

x

y

5

10

–10–20

y = f (x)

c ! 2c ! #10

limxSc

f(x)lim

xSc#f(x)lim

xSc"f(x)

y ! f(x)14.

x f(x)

?

In Problems 15–36, use properties of limits and algebraicmethods to find the limits, if they exist.15. 16.

17.

18.

19. 20.

21. 22.

23. 24.

25. 26.

27. where

28. where

29. where

30. where

31. 32.

33. 34.

35. 36.

In Problems 37–40, graph each function with a graphingutility and use it to predict the limit. Check your workeither by using the table feature of the graphing utilityor by finding the limit algebraically.

37. 38. limxS#3

x4 $ 3x3

2x4 # 18x2lim

xS10

x2 # 19x $ 903x2 # 30x

limhS0

2(x $ h)2 # 2x2

hlimhS0

(x $ h)3 # x3

h

limxS3

x2 $ 2x # 3x # 3

limxS#1

x2 $ 5x $ 6x $ 1

limxS5

x2 # 6x $ 8x # 5

limxS2

x2 $ 6x $ 9x # 2

f(x) ! d x3 # 4x # 3

for x ) 2

3 # x2

xfor x ' 2

limxS2

f(x),

f (x) !cx2 $

4x

for x ) #1

3x3 # x # 1 for x ' #1lim

xS#1f(x),

f(x) ! b7x # 10 for x & 525 for x * 5

limxS5

f(x),

f(x) ! b10 # 2x for x & 3x2 # x for x * 3

limxS3

f(x),

limxS10

x2 # 8x # 20x2 # 11x $ 10

limxS#2

x2 $ 4x $ 4x2 $ 3x $ 2

limxS#5

x2 $ 8x $ 15x2 $ 5x

limxS7

x2 # 8x $ 7x2 # 6x # 7

limxS#4

x2 # 16x $ 4

limxS3

x2 # 9x # 3

limxS#1!3

1 # 3x9x2 $ 1

limxS#1!2

4x # 24x2 $ 1

limxS3

(2x3 # 12x2 $ 5x $ 3)

limxS#1

(4x3 # 2x2 $ 2)

limxS80

(82 # x)limxS#35

(34 $ x)

#1.99#1.999

cc#2

TT#2.001#2.01#2.1

limxS#2

f(x) ! ?

f (x) ! b4 # x2 for x ) #2x2 $ 2x for x ' #2

594 ! Chapter 9 Derivatives

39. 40.

In Problems 41–44, use the table feature of a graphingutility to predict each limit. Check your work by usingeither a graphical or an algebraic approach.

41. 42.

43. where

44. where

45. Use values 0.1, 0.01, 0.001, 0.0001, and 0.00001 withyour calculator to approximate

to three decimal places. This limit equals the specialnumber e that is discussed in Section 5.1, “ExponentialFunctions,” and Section 6.2, “Compound Interest; Geo-metric Sequences.”

46. If and find

(a)

(b)

(c)

47. If and find

(a) (b)

(c) (d)

48. (a) If and find

if it exists. Explain your conclusions.

(b) If and find

if it exists. Explain your conclusions.

APPLICATIONS

49. Revenue The total revenue for a product is given by

where x is the number of units sold. What is

50. Profit If the profit function for a product is given by

find

51. Sales and training The average monthly sales volume(in thousands of dollars) for a firm depends on the num-ber of hours x of training of its sales staff, according to

S(x) !4x

$ 30 $x4

, 4 ) x ) 100

limxS40

P(x).

P(x) ! 92x # x2 # 1760

limxS100

R(x)?

R(x) ! 1600x # x2

limxS0

f(x),

f (0) ! 0,limxS0$

f(x) ! 3, limxS0#

f(x) ! 0,

limxS2

f(x),

f (2) ! 0,limxS2$

f(x) ! 5, limxS2#

f(x) ! 5,

limxS3

g(x)f(x)

limxS33 f(x) $ g(x) 4 lim

xS33 f(x) # g(x) 4lim

xS33 f(x) $ g(x) 4 limxS3

g(x) ! #2,limxS3

f(x) ! 4

limxS2

3g(x)f(x) # g(x)

limxS25 3 f(x) 42 # 3g(x) 426lim

xS2f (x)

limxS2

g(x) ! 11,limxS23 f(x) $ g(x) 4 ! 5

limaS0

(1 $ a)1!a

f(x) ! b2 $ x # x2 for x ) 713 # 9x for x ' 7

limxS7

f(x),

f(x) !c12 #

34

x for x ) 4

x2 # 7 for x ' 4limxS4

f(x),

limxS#4

x3 $ 4x2

2x2 $ 7x # 4lim

xS#2

x4 # 4x2

x2 $ 8x $ 12

limxS5

x2 # 7x $ 10x2 # 10x $ 25

limxS#1

x3 # xx2 $ 2x $ 1

(a) Find (b) Find

52. Sales and training During the first 4 months ofemployment, the monthly sales S (in thousands of dol-lars) for a new salesperson depend on the number ofhours x of training, as follows:

(a) Find (b) Find

53. Advertising and sales Suppose that the daily sales S(in dollars) t days after the end of an advertising cam-paign are

(a) Find S(0). (b) Find (c) Find

54. Advertising and sales Sales y (in thousands of dol-lars) are related to advertising expenses x (in thousandsof dollars) according to

(a) Find (b) Find

55. Productivity During an 8-hour shift, the rate ofchange of productivity (in units per hour) of children’sphonographs assembled after t hours on the job is

(a) Find (b) Find

(c) Is the rate of productivity higher near the lunchbreak (at ) or near quitting time (at )?

56. Revenue If the revenue for a product is and the average revenue per unit is

find (a) and (b)

57. Cost-benefit Suppose that the cost C of obtainingwater that contains p percent impurities is given by

(a) Find if it exists. Interpret this result.

(b) Find if it exists.

(c) Is complete purity possible? Explain.

limpS0$

C(p),

limpS100#

C(p),

C(p) !120,000

p# 1200

limxS0$

R(x)x

.limxS100

R(x)x

R(x) !R(x)

x, x ' 0

100x # 0.1x2,R(x) !

t ! 8t ! 4

limtS8#

r(t).limtS4

r(t).

r(t) !128(t2 $ 6t)

(t2 $ 6t $ 18)2, 0 ) t ) 8

limxS0$

y(x).limxS10

y(x).

y ! y(x) !200x

x $ 10, x * 0

limtS14

S(t).limtS7

S(t).

S ! S(t) ! 400 $2400t $ 1

limxS10

S(x).limxS4$

S(x).

S ! S(x) !9x

$ 10 $x4

, x * 4

limxS100#

S(x).limxS4$

S(x).

9.1 Limits ! 595

58. Cost-benefit Suppose that the cost C of removingp percent of the particulate pollution from the smoke-stacks of an industrial plant is given by

(a) Find

(b) Find if it exists.

(c) Can 100% of the particulate pollution be removed?Explain.

59. Federal income tax Use the following tax rate sched-ule for single taxpayers, and create a table of values thatcould be used to find the following limits, if they exist.Let x represent the amount of taxable income, and letT(x) represent the tax due.(a)

(b)

(c)

Source: Internal Revenue Service, 2004, Form 1040 Instructions

60. Parking costs The Ace Parking Garage charges $5.00for parking for 2 hours or less, and $1.50 for each extrahour or part of an hour after the 2-hour minimum. Theparking charges for the first 5 hours could be written asa function of the time as follows:

(a) Find if it exists.

(b) Find if it exists.limtS2

f(t),

limtS1

f(t),

f(t) ! d$5.00 if 0 & t ) 2$6.50 if 2 & t ) 3$8.00 if 3 & t ) 4$9.50 if 4 & t ) 5

If your taxableincome is:

The tax is:

Over––But notover––

$07,150

29,05070,350

146,750319,100

$7,15029,05070,350

146,750319,100

of theamountover––

$07,150

29,05070,350

146,750319,100

10%$715.00 + 15%

4,000.00 + 25%14,325.00 + 28%35,717.00 + 33%92,592.50 + 35%

Schedule X––Use if your filing status is Single

limxS29,050

T(x)

limxS29,050$

T(x)

limxS29,050#

T(x)

limpS100#

C(p),

limpS80

C(p).

C(p) !730,000100 # p

# 7300

61. Municipal water rates The Corner Water Corp. ofShippenville, Pennsylvania has the following rates per1000 gallons of water used.

Cost per 1000 Gallons Usage (x) (C(x))

First 10,000 gallons $7.98Next 110,000 gallons 6.78Over 120,000 gallons 5.43

If Corner Water has a monthly service fee of $3.59,write a function that models the charges(where x is thousands of gallons) and find (that is, as usage approaches 10,000 gallons).

62. Phone card charges A certain calling card costs3.7 cents per minute to make a call. However, when thecard is used at a pay phone there is a 10-minute chargefor the first minute of the call, and then the regularcharge thereafter. If is the charge from a payphone for a call lasting t minutes, create a table ofcharges for calls lasting close to 1 minute and use it tofind the following limits, if they exist.(a) (b) (c)

Dow Jones Industrial Average The graph in the follow-ing figure shows the Dow Jones Industrial Average(DJIA) at 1-minute intervals for Monday, October 5,2004. Use the graph for Problems 63 and 64, with t asthe time of day and D(t) as the DJIA at time t.63. Estimate if it exists. Explain what this

limit corresponds to.64. Estimate if it exists. Explain what this

limit corresponds to.

Source: Bloomberg Financial Markets, The New York Times, October 6, 2004. Copyright © 2004. The New York Times Co. Reprinted by permission.

limtS4:00PM#

D(t),

limtS9:30AM$

D(t),

limtS1

C(t)limtS1$

C(t)limtS1#

C(t)

C ! C(t)

limxS10

C(x)C ! C(x)