LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

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130|||| CHAPTER 2 LIMITS AND DERIVATIVES

55.Prove that f is continuous at a if and only if

lim f !a ! h" ! f !a"

h l0

56.To prove that sine is continuous, we need to show that

limx la sin x ! sin a for every real number a. By Exercise 55 an equivalent statement is that

lim sin!a ! h" ! sin a

h l0

Use (6) to show that this is true.

57.Prove that cosine is a continuous function.

58.(a) Prove Theorem 4, part 3.

(b) Prove Theorem 4, part 5.

59.For what values of x is f continuous?

0	if	x is rational
f !x" ! +1	if	x is irrational
60. For what values of x is t continuous?
0	if	x is rational
t!x" ! +x	if	x is irrational

61. Is there a number that is exactly 1 more than its cube?

62. If a and b are positive numbers, prove that the equation

x3 ! 2x2 " 1 ! x3 ! x " 2 ! 0

has at least one solution in the interval !"1, 1".

63. Show that the function
x4 sin!1#x"	if	x " 0
f !x" ! +0	if	x ! 0

is continuous on !"$, $".

64.(a) Show that the absolute value function F!x" ! & x & is continuous everywhere.

(b)Prove that if f is a continuous function on an interval, then so is & f &.

(c)Is the converse of the statement in part (b) also true? In other words, if & f & is continuous, does it follow that f is continuous? If so, prove it. If not, ﬁnd a counterexample.

65.A Tibetan monk leaves the monastery at 7:00 AM and takes his usual path to the top of the mountain, arriving at 7:00 P M. The following morning, he starts at 7:00 AM at the top and takes the same path back, arriving at the monastery at 7:00 P M. Use the Intermediate Value Theorem to show that there is a point on the path that the monk will cross at exactly the same time of day on both days.

2.6

x	f !x"

0	"1
*1	0
*2	0.600000
*3	0.800000
*4	0.882353
*5	0.923077
*10	0.980198
*50	0.999200
*100	0.999800
*1000	0.999998

FIGURE 1

x l" 3 ! x

because x # 1 l " and 3#x ! 1 l !1 as x l ".

In Sections 2.2 and 2.4 we investigated inﬁnite limits and vertical asymptotes. There we let x approach a number and the result was that the values of y became arbitrarily large (positive or negative). In this section we let x become arbitrarily large (positive or negative) and see what happens to y.

Let’s begin by investigating the behavior of the function f deﬁned by

x2 " 1 f !x" ! x2 ! 1

as x becomes large. The table at the left gives values of this function correct to six decimal places, and the graph of f has been drawn by a computer in Figure 1.

y=1

0 1	y=	≈-1	x
		≈+1

As x grows larger and larger you can see that the values of f !x" get closer and closer to 1. In fact, it seems that we can make the values of f !x" as close as we like to 1 by taking x sufﬁciently large. This situation is expressed symbolically by writing

x2 " 1

x l$

lim x2 ! 1 ! 1

SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES |||| 131

In general, we use the notation

lim f !x" ! L

x l$

to indicate that the values of f !x" become closer and closer to L as x becomes larger and larger.

1 DEFINITION Let f be a function deﬁned on some interval !a, $". Then

lim f !x" ! L

x l$

means that the values of f !x" can be made arbitrarily close to L by taking x sufﬁciently large.

Another notation for limx l$ f !x" ! L is

f !x" l L as x l $

The symbol $ does not represent a number. Nonetheless, the expression lim f !x" ! L is

	often read as	x l$
	often read as	“the limit of f !x", as x approaches inﬁnity, is L”
		“the limit of f !x", as x approaches inﬁnity, is L”
	or	“the limit of f !x", as x becomes inﬁnite, is L”
	or	“the limit of f !x", as x increases without bound, is L”

The meaning of such phrases is given by Deﬁnition 1. A more precise deﬁnition, similar to the ,, - deﬁnition of Section 2.4, is given at the end of this section.

Geometric illustrations of Deﬁnition 1 are shown in Figure 2. Notice that there are many ways for the graph of f to approach the line y ! L (which is called a horizontal asymptote) as we look to the far right of each graph.

y=L

y=Ä

y=L

y=Ä

y=L

FIGURE 2

Examples illustrating lim Ä=L

x `

Referring back to Figure 1, we see that for numerically large negative values of x, the values of f !x" are close to 1. By letting x decrease through negative values without bound, we can make f !x" as close as we like to 1. This is expressed by writing

x2 " 1

x l"$

lim x2 ! 1 ! 1

The general deﬁnition is as follows.

2 DEFINITION Let f be a function deﬁned on some interval !"$, a". Then

lim f !x" ! L

x l"$

means that the values of f !x" can be made arbitrarily close to L by taking x sufﬁciently large negative.

132 |||| CHAPTER 2 LIMITS AND DERIVATIVES

	y	y=Ä
	y


	y=L

	0		x
		y
		y
	y=Ä
	y=L

	0		x
FIGURE 3
Examples illustrating lim			Ä=L
		x _`

_ π2

FIGURE 4

y=tanÐ!x
y
	2
0	2	x

FIGURE 5

Again, the symbol !" does not represent a number, but the expression lim f ! x" ! L

is often read as	x l!"
is often read as
	“the limit of f ! x", as x approaches negative inﬁnity, is L”

Deﬁnition 2 is illustrated in Figure 3. Notice that the graph approaches the line y ! L as we look to the far left of each graph.

3 DEFINITION The line y ! L is called a horizontal asymptote of the curve
y ! f ! x" if either
lim f ! x" ! L	or	lim f ! x" ! L
x l"		x l!"

For instance, the curve illustrated in Figure 1 has the line y ! 1 as a horizontal asymptote because

x2 ! 1

x l"

lim x2 # 1 ! 1

An example of a curve with two horizontal asymptotes is y ! tan!1x. (See Figure 4.) In fact,

4	lim	tan!1x ! !	%	lim tan!1x !	%
4	lim	tan!1x ! !	2	lim tan!1x !	2
	x l!"		2	x l"	2

so both of the lines y ! !%#2 and y ! %#2 are horizontal asymptotes. (This follows from the fact that the lines x ! $%#2 are vertical asymptotes of the graph of tan.)

EXAMPLE 1 Find the inﬁnite limits, limits at inﬁnity, and asymptotes for the function f whose graph is shown in Figure 5.

SOLUTION We see that the values of f ! x" become large as x l !1 from both sides, so

lim f ! x" ! "

x l!1

Notice that f ! x" becomes large negative as x approaches 2 from the left, but large positive as x approaches 2 from the right. So

lim f ! x" ! !" and	lim f ! x" ! "
x l2!	x l2#

Thus both of the lines x ! !1 and x ! 2 are vertical asymptotes.

As x becomes large, it appears that f ! x" approaches 4. But as x decreases through negative values, f !x" approaches 2. So

lim f ! x" ! 4	and	lim f ! x" ! 2
x l"		x l!"

This means that both y ! 4 and y ! 2 are horizontal asymptotes.

SECTION 2.6 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES |||| 133

EXAMPLE 2 Find lim

and lim

x l"

x l!"

SOLUTION Observe that when x is large, 1#x is small. For instance,

! 0.01

! 0.0001

! 0.000001

100

10,000

1,000,000

In fact, by taking x large enough, we can make 1#x as close to 0 as we please. Therefore,

according to Deﬁnition 1, we have

lim

! 0

x l"

Similar reasoning shows that when x is large negative, 1#x is small negative, so we also

have

lim

! 0

x l!"

FIGURE 6
lim		1	=0,	lim		1	=0

x	`	x		x	_` x

It follows that the line y ! 0 (the x-axis) is a horizontal asymptote of the curve y ! 1#x. (This is an equilateral hyperbola; see Figure 6.) M

Most of the Limit Laws that were given in Section 2.3 also hold for limits at inﬁnity. It can be proved that the Limit Laws listed in Section 2.3 (with the exception of Laws 9 and 10) are also valid if “x l a” is replaced by “x l "” or “ x l !".” In particular, if we combine Laws 6 and 11 with the results of Example 2, we obtain the following important rule for calculating limits.

5 THEOREM If r & 0 is a rational number, then

lim xr ! 0

x l"

If r & 0 is a rational number such that xr is deﬁned for all x, then

lim xr ! 0

x l!"

V EXAMPLE 3 Evaluate

3x2 ! x ! 2

lim 2

x l" 5x # 4x # 1

and indicate which properties of limits are used at each stage.

SOLUTION As x becomes large, both numerator and denominator become large, so it isn’t obvious what happens to their ratio. We need to do some preliminary algebra.

To evaluate the limit at inﬁnity of any rational function, we ﬁrst divide both the numerator and denominator by the highest power of x that occurs in the denominator.

134 |||| CHAPTER 2 LIMITS AND DERIVATIVES

(We may assume that x " case the highest power of

0, since we are interested only in large values of x.) In this x in the denominator is x2, so we have

3x2 ! x ! 2

lim

3x2

! x ! 2

! lim

x l" 5x2 # 4x # 1

x l"

lim

3 !

x l"

(by Limit Law 5)

lim

5 #

x l"

y=0.6

lim 3 ! lim

! 2 lim

x l" x l"

x l" x2

(by 1, 2, and 3)

lim 5 # 4 lim

# lim

x l"

x l" x2

3 ! 0 ! 0

(by 7 and Theorem 5)

5 # 0 # 0

y= 3≈-x-2

A similar calculation shows that the limit as x l !" is also 53 . Figure 7 illustrates the

results of these calculations by showing how the graph of the given rational function

5≈+4x+1

approaches the horizontal asymptote y ! 5 .

EXAMPLE 4 Find the horizontal and vertical asymptotes of the graph of the function

f ! x" !

2x2 # 1

3x ! 5

SOLUTION

Dividing both numerator and denominator by x and using the properties of lim-

its, we have

$ 2 #

lim

2x2 # 1

! lim

(since sx

! x for x & 0)

3x ! 5

3 ! 5

x l"

lim

2 #

lim 2 # lim

x l"

$x l"

x l" x2

s2 # 0

lim

3 !

lim 3 ! 5 lim

3 ! 5 ! 0

x '

x l"

Therefore the line y ! s

#3 is a horizontal asymptote of the graph of f.

In computing the limit as x l !", we must remember that for x ' 0, we have

! % x % ! !x. So when we divide the numerator by x, for x ' 0 we get

s2x2

# 1 !

s2x2 #

1 ! !$2 #

y= Ïã23

y=_ Ïã23

x=53

FIGURE 8

y= Пгггггг2≈+1 3x-5

N We can think of the given function as having a denominator of 1.

y=Пггггг≈+1-x

0			x
	1

FIGURE 9

SECTION 2.6 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES |||| 135

Therefore

2 #

! 2 # lim

lim

s2x2 # 1

! lim

x l!"

! !

x l!"

3x ! 5

x l!"

3 ! 5

lim 1

x l!" x

#3 is also a horizontal asymptote.

Thus the line y ! !

A vertical asymptote is likely to occur when the denominator, 3x ! 5, is 0, that is,

when x ! 35 . If x is close to 35 and x & 35 , then the denominator is close to 0 and 3x ! 5

is positive. The numerator s

2x2 # 1

is always positive, so f ! x" is positive. Therefore

lim

2x2 # 1

! "

x l!5#3"#

3x ! 5

If x is close to 35 but x ' 35 , then 3x ! 5 ' 0 and so f ! x" is large negative. Thus

lim

2x2 # 1

! !"

3x ! 5

x l!5#3"!

The vertical asymptote is x ! 35 . All three asymptotes are shown in Figure 8.

EXAMPLE 5 Compute lim (sx2 # 1 ! x).

x l"

SOLUTION Because both sx2 # 1 and x are large when x is large, it’s difﬁcult to see what happens to their difference, so we use algebra to rewrite the function. We ﬁrst multiply numerator and denominator by the conjugate radical:

! x)

# x

lim (

! x) ! lim (

# 1

x2 # 1

x l" s

x l"

# 1

# x

! lim

! x2 # 1" ! x2

! lim

x l"

sx2 # 1 # x

x l" sx2 # 1 # x

The Squeeze Theorem could be used to show that this limit is 0. But an easier method is to divide numerator and denominator by x. Doing this and using the Limit Laws, we obtain

lim (	s		! x) ! lim
		x2 # 1
x l"			x l"

! lim

x l"

Figure 9 illustrates this result.

sx2 # 1 # x !

$1 # x12 # 1

lim

x l" sx2 # 1 # x

! s1 # 0 # 1 ! 0

The graph of the natural exponential function y ! ex has the line y ! 0 (the x-axis) as a horizontal asymptote. (The same is true of any exponential function with base a & 1.) In

136 |||| CHAPTER 2 LIMITS AND DERIVATIVES

fact, from the graph in Figure 10 and the corresponding table of values, we see that

FIGURE 10

N The problem-solving strategy for Example 6 is introducing something extra (see page 76). Here, the something extra, the auxiliary aid, is the new variable t.

y=þ

FIGURE 11
lim x#=`,		lim x#=_`
x	`	x	_`

6	lim ex ! 0
	x l!"

Notice that the values of ex approach 0 very rapidly.

y=«

V EXAMPLE 6 Evaluate lim e1#x.

x l0!

x	ex
0	1.00000
!1	0.36788
!2	0.13534
!3	0.04979
!5	0.00674
!8	0.00034
!10	0.00005

SOLUTION	If we let t ! 1#x, we know that t l !" as x l 0!. Therefore, by (6),
	lim e1#x ! lim et ! 0
	x l0!	t l!"
(See Exercise 71.)			M
EXAMPLE 7 Evaluate lim sin x.
	x l"
SOLUTION	As x increases, the values of sin x oscillate between 1 and !1 inﬁnitely often
and so they don’t approach any deﬁnite number. Thus limx l" sin x does not exist.			M

INFINITE LIMITS AT INFINITY

The notation

lim f ! x" ! "

x l"

is used to indicate that the values of f ! x" become large as x becomes large. Similar meanings are attached to the following symbols:

lim f ! x" ! "	lim f !x" ! !"	lim f !x" ! !"
x l!"	x l"	x l!"
EXAMPLE 8 Find lim x3 and	lim x3.
x l"	x l!"
SOLUTION When x becomes large, x3 also becomes large. For instance,
103 ! 1000	1003 ! 1,000,000	10003 ! 1,000,000,000
In fact, we can make x3 as big as we like by taking x large enough. Therefore we can
write
	lim x3 ! "
	x l"
Similarly, when x is large negative, so is x3. Thus
	lim x3 ! !"
	x l!"
These limit statements can also be seen from the graph of y ! x3 in Figure 11.			M

y=«

100												y=þ
100

0 1												x

FIGURE 12

« is much larger than þ when x is large.

2 x

_16

FIGURE 13 y=(x-2)$(x +1)#(x-1)

SECTION 2.6 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES |||| 137

Looking at Figure 10 we see that

lim ex ! "

x l"

but, as Figure 12 demonstrates, y ! ex becomes large as x l " at a much faster rate than y ! x3.

EXAMPLE 9 Find lim ! x2 ! x".

x l"

| SOLUTION It would be wrong to write

lim ! x2 ! x" ! lim x2 ! lim x ! " ! "

x l"

The Limit Laws can’t be applied to inﬁnite limits because " is not a number (" ! " can’t be deﬁned). However, we can write

lim ! x2 ! x" ! lim x! x ! 1" ! "

		x l"	x l"
because both x and x ! 1 become arbitrarily large and so their product does too.				M
EXAMPLE 10 Find lim	x2	# x	.
EXAMPLE 10 Find lim	3	! x	.
x l"	3	! x

SOLUTION As in Example 3, we divide the numerator and denominator by the highest power of x in the denominator, which is just x:

The next example shows that by using inﬁnite limits at inﬁnity, together with intercepts, we can get a rough idea of the graph of a polynomial without having to plot a large number of points.

V EXAMPLE 11 Sketch the graph of y ! !x ! 2"4! x # 1"3! x ! 1" by ﬁnding its intercepts and its limits as x l " and as x l !".

SOLUTION The y-intercept is f !0" ! !!2"4!1"3!!1" ! !16 and the x-intercepts are

found by setting y ! 0: x ! 2, !1, 1. Notice that since ! x ! 2"4 is positive, the function doesn’t change sign at 2; thus the graph doesn’t cross the x-axis at 2. The graph crosses the axis at !1 and 1.

When x is large positive, all three factors are large, so

lim ! x ! 2"4!x # 1"3!x ! 1" ! "

x l"

When x is large negative, the ﬁrst factor is large positive and the second and third factors are both large negative, so

lim ! x ! 2"4! x # 1"3!x ! 1" ! "

x l!"
Combining this information, we give a rough sketch of the graph in Figure 13.	M

138 |||| CHAPTER 2 LIMITS AND DERIVATIVES

PRECISE DEFINITIONS

Deﬁnition 1 can be stated precisely as follows.

7 DEFINITION Let f be a function deﬁned on some interval !a, "". Then

lim f !x" ! L

x l"

means that for every ( & 0 there is a corresponding number N such that if x & N then % f !x" ! L % ' (

FIGURE 14

lim Ä=L

x `

FIGURE 15

lim Ä=L

x `

In words, this says that the values of f !x" can be made arbitrarily close to L (within a distance (, where ( is any positive number) by taking x sufﬁciently large (larger than N, where N depends on (). Graphically it says that by choosing x large enough (larger than some number N ) we can make the graph of f lie between the given horizontal lines y ! L ! ( and y ! L # ( as in Figure 14. This must be true no matter how small we choose (. Figure 15 shows that if a smaller value of ( is chosen, then a larger value of N may be required.

y
	y=L+∑	y=Ä
	y=L+∑			Ä is
L	∑			Ä is
L	∑			in here
	y=L-∑

0		N	x
			when x is in here

y=Ä

y=L+∑

y=L-∑

0	N	x
0

Similarly, a precise version of Deﬁnition 2 is given by Deﬁnition 8, which is illustrated in Figure 16.

8 DEFINITION Let f be a function deﬁned on some interval !!", a". Then

lim f ! x" ! L

x l!"

means that for every ( & 0 there is a corresponding number N such that if x ' N then % f ! x" ! L % ' (

SECTION 2.6 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES |||| 139

y=Ä

y=L+∑

y=L-∑

FIGURE 16

lim Ä=L

_ `

In Example 3 we calculated that

lim

3x2

! x ! 2

5x2

# 4x # 1

x l"

In the next example we use a graphing device to relate this statement to Deﬁnition 7 with

L ! 53 and ( ! 0.1.

EXAMPLE 12 Use a graph to ﬁnd a number N such that

3x2 ! x ! 2

x & N

then

(

! 0.6 ( ' 0.1

5x2 # 4x # 1

SOLUTION We rewrite the given inequality as

0.5 '

3x2 ! x ! 2

' 0.7

5x2 # 4x # 1

We need to determine the values of x for which the given curve lies between the horizon-

tal lines y ! 0.5 and y ! 0.7. So we graph the curve and these lines in Figure 17. Then

y=0.7

we use the cursor to estimate that the curve crosses the line y ! 0.5 when x ) 6.7. To

y=0.5

the right of this number the curve stays between the lines y ! 0.5 and y ! 0.7. Round-

3≈-x-2

ing to be safe, we can say that

5≈+4x+1

! x ! 2

x & 7

then

(

! 0.6 ( ' 0.1

5x2 # 4x # 1

FIGURE 17

In other words, for ( ! 0.1 we can choose N ! 7 (or any larger number) in Deﬁnition 7.

EXAMPLE 13 Use Deﬁnition 7 to prove that lim

! 0.

x l"

SOLUTION Given ( & 0, we want to ﬁnd N such that

x & N

then

(

! 0 ( ' (

In computing the limit we may assume that x & choose N ! 1#(. So

if x & N !	1	then
	(

0. Then 1#x ' ( &? x & 1#(. Let’s

( 1x ! 0 ( ! 1x ' (

140 |||| CHAPTER 2 LIMITS AND DERIVATIVES

Therefore, by Deﬁnition 7,

lim 1 ! 0

x l" x

Figure 18 illustrates the proof by showing some values of ( and the corresponding values of N.

y y y

∑=1		∑=0.2		∑=0.1
0 N=1			N=5 x		N=10 x
	x	0		0

FIGURE 18 M

Finally we note that an inﬁnite limit at inﬁnity can be deﬁned as follows. The geomet-

ric illustration is given in Figure 19.

y=M

9 DEFINITION Let f be a function deﬁned on some interval !a, "". Then

lim f ! x" ! "

x l"

means that for every positive number M there is a corresponding positive number

N such that

f ! x" & M

FIGURE 19

if x & N

then

lim Ä=`

Similar deﬁnitions apply when the symbol " is replaced by !". (See Exercise 70.)

2.6EXERCISES

1.Explain in your own words the meaning of each of the following.

(a) lim f ! x" ! 5	(b) lim f ! x" ! 3
x l"	x l!"

2.(a) Can the graph of y ! f ! x" intersect a vertical asymptote? Can it intersect a horizontal asymptote? Illustrate by sketching graphs.

(b)How many horizontal asymptotes can the graph of y ! f ! x" have? Sketch graphs to illustrate the possibilities.

3.For the function f whose graph is given, state the following.

(a) lim f ! x"	(b) lim f ! x"
x l2	x l!1!

(c)	lim	f ! x"	(d)	lim f ! x"
	x l!1#			x l"
(e)	lim f ! x"		(f)	The equations of the asymptotes
	x l!"
			y
			1
				1	x

SECTION 2.6 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES |||| 141

4. For the function t whose graph is given, state the following.

13–14 Evaluate the limit and justify each step by indicating the

(a) lim t! x"

(b)

lim t! x"

appropriate properties of limits.

x l"

x l!"

3x2 ! x # 4

12x 3 ! 5x # 2

(d)

lim t! x"

13.

lim

14.

lim

# 5x ! 8

# 3x

x l3

x l0

x l"

x l"$ 1 # 4x

(e)

lim# t! x"

(f)

The equations of the asymptotes

x l!2

15–36 Find the limit.

3x # 5

15.

lim

16.

lim

x l"

2x # 3

x l"

x ! 4

17.

lim

1 ! x ! x2

18.

lim

2 ! 3y2

5y2 # 4y

x l!" 2x2 ! 7

y l"

19.

lim

x3 # 5x

20.

lim

t2 # 2

! x

# 4

# t

! 1

x l"

t l!"

21.

lim

4u4 # 5

22.

lim

x # 2

!u2 ! 2"!2u2 ! 1"

s9x2 # 1

u l"

x l"

5 –10 Sketch the graph of an example of a function f that satisﬁes all of the given conditions.

5.	f !0" ! 0,	f !1" ! 1,		lim f ! x" ! 0,		f	is odd
				x l"
6.	lim f ! x" ! ",		lim f ! x" ! !",		lim f ! x" ! 1,
	x l0#		x l0!		x l"
	lim f ! x" ! 1
	x l!"
7.	lim f ! x" ! !",		lim f ! x" ! ",		lim f ! x" ! 0,
	x l2		x l"		x l!"
	lim f ! x" ! ",		lim f ! x" ! !"
	x l0#		x l0!
8.	lim f ! x" ! ",		lim f ! x" ! 3,		lim f ! x" ! !3
	x l!2		x l!"		x l"
9.	f !0" ! 3,	lim	f ! x" ! 4, lim f ! x" ! 2,
		x l0!		x l0#
	lim f ! x" ! !", lim			f ! x" ! !",		lim	f ! x" ! ",
	x l!"		x l4!			x l4#
	lim f ! x" ! 3
	x l"
10.	lim f ! x" ! !",		lim f ! x" ! 2,		f !0" ! 0, f is even
	x l3		x l"

23.

lim

9x6

! x

24.

lim

9x6 ! x

x3 # 1

x l"

x l!"

lim (

! 3x)

26.

lim

(x #

)

25.

9x2

# x

# 2x

x l"

x l!"

27.lim (sx2 # ax ! sx2 # bx )

xl"

28.lim cos x

xl"

		x # x3 # x5
29.	lim	x # x3 # x5	30. lim			x2 # 1
29.	lim	1 ! x2 # x4	30. lim			x2 # 1
	x l"	1 ! x2 # x4		x l" s
31.	lim ! x4 # x5 "		32.	lim	x3 ! 2x # 3
31.	lim ! x4 # x5 "		32.	lim		5 ! 2x2
	x l!"			x l"		5 ! 2x2
33.	lim	1 ! ex	34.	lim tan!1! x2 ! x4 "
33.	lim	1 # 2ex	34.	lim tan!1! x2 ! x4 "
	x l"	1 # 2ex		x l"
35.	lim !e!2x cos x"		36.	lim		etan x
	x l"			x l! # 2"		#
				%		#

; 11.	Guess the value of the limit
	lim	x2
	lim	2x
	x l"	2x
	by evaluating the function f ! x" ! x2#2x for x ! 0, 1, 2, 3,
	4, 5, 6, 7, 8, 9, 10, 20, 50, and 100. Then use a graph of f
	to support your guess.
;12.	(a) Use a graph of
	f ! x" ! &1 !		2	'x
	f ! x" ! &1 !		x	'x

to estimate the value of limx l" f ! x" correct to two decimal places.

(b)Use a table of values of f ! x" to estimate the limit to four decimal places.

;37. (a) Estimate the value of

lim (sx2 # x # 1 # x)

x l!"

by graphing the function f ! x" ! sx2 # x # 1 # x.

(b)Use a table of values of f ! x" to guess the value of the limit.

(c)Prove that your guess is correct.

;38. (a) Use a graph of

f ! x" ! s3x2 # 8x # 6 ! s3x2 # 3x # 1

to estimate the value of limx l" f ! x" to one decimal place.

(b)Use a table of values of f ! x" to estimate the limit to four decimal places.

(c)Find the exact value of the limit.

142 |||| CHAPTER 2 LIMITS AND DERIVATIVES

39– 44 Find the horizontal and vertical asymptotes of each curve. If you have a graphing device, check your work by graphing the curve and estimating the asymptotes.

2x # 1

39. y ! x " 2

2x2 # x " 1

41. y ! x2 # x " 2

x3 " x

43. y ! x2 " 6x # 5

x2 # 1

40. y ! 2x2 " 3x " 2

1# x4

42.y ! x2 " x4

2ex

44. y ! ex " 5

;45. Estimate the horizontal asymptote of the function

3x3 # 500x2

f "x# ! x3 # 500x2 # 100x # 2000

by graphing f for "10 ' x ' 10. Then calculate the equation of the asymptote by evaluating the limit. How do you explain the discrepancy?

;46. (a) Graph the function

f "x# ! s2x2 # 1 3x " 5

How many horizontal and vertical asymptotes do you observe? Use the graph to estimate the values of the limits

lim	s	2x2	# 1	and lim	s	2x2	# 1
lim	3x " 5			and lim	3x " 5
x l!	3x " 5			x l"!	3x " 5

(b)By calculating values of f "x#, give numerical estimates of the limits in part (a).

(c)Calculate the exact values of the limits in part (a). Did you get the same value or different values for these two limits? [In view of your answer to part (a), you might have to check your calculation for the second limit.]

47.Find a formula for a function f that satisﬁes the following conditions:

lim f "x# ! 0,	lim f "x# ! "!, f "2# ! 0,
x l &!	x l0
lim f "x# ! !,	lim f "x# ! "!
x l3"	x l3#

48.Find a formula for a function that has vertical asymptotes x ! 1 and x ! 3 and horizontal asymptote y ! 1.

49–52 Find the limits as x l ! and as x l"!. Use this information, together with intercepts, to give a rough sketch of the graph as in Example 11.

49. y ! x4 " x6

50. y ! x3"x # 2#2"x " 1#

51.y ! "3 " x#"1 # x#2"1 " x#4

52.y ! x2"x2 " 1#2"x # 2#

53. (a) Use the Squeeze Theorem to evaluate lim	sin x .
x l!	x

;(b) Graph f "x# ! "sin x#!x. How many times does the graph cross the asymptote?

;54. By the end behavior of a function we mean the behavior of its values as x l ! and as x l "!.

(a)Describe and compare the end behavior of the functions

P"x# ! 3x5 " 5x3 # 2x

Q"x# ! 3x5

by graphing both functions in the viewing rectangles $"2, 2% by $"2, 2% and $"10, 10% by $"10,000, 10,000%.

(b)Two functions are said to have the same end behavior if their ratio approaches 1 as x l !. Show that P and Q have the same end behavior.

55.Let P and Q be polynomials. Find

lim P"x# x l! Q"x#

if the degree of P is (a) less than the degree of Q and

(b) greater than the degree of Q.

56.Make a rough sketch of the curve y ! xn (n an integer) for the following ﬁve cases:

(i)	n ! 0	(ii)	n % 0, n odd
(iii)	n % 0, n even	(iv)	n $ 0, n odd
(v)	n $ 0, n even

Then use these sketches to ﬁnd the following limits.

(a) lim xn	(b)	lim xn
x l0#		x l0"
(c) lim xn	(d)	lim xn
x l!		x l"!

57. Find limx l! f "x# if, for all x % 1,

10ex " 21		5s
	$ f "x# $		x

2ex		sx " 1

58.(a) A tank contains 5000 L of pure water. Brine that contains 30 g of salt per liter of water is pumped into the tank at a rate of 25 L!min. Show that the concentration of salt after t minutes (in grams per liter) is

C"t# ! 30t

200 # t

(b) What happens to the concentration as t l !?

59.In Chapter 9 we will be able to show, under certain assumptions, that the velocity v"t# of a falling raindrop at time t is

v"t# ! v*"1 " e"tt!v*#

where t is the acceleration due to gravity and v* is the terminal velocity of the raindrop.

(a) Find limt l! v"t#.

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