Limits
For Use Only in 2013 – 2014 Pilot Program
2
Limits
2.1 The Idea of Limits
2.2 Definitions of Limits
2.3 Techniques for Computing
Limits
2.4 Infinite Limits
2.5 Limits at Infinity
2.6 Continuity
2.7 Precise Definitions of Limits
Biologists often use mathematical models to describe communities of organisms—
anything from a culture of cells to a herd of zebras. These models result in functions that
approximate the population of the community as it changes in time. An important question
concerns the long-term behavior of these population functions. Does the population approach a steady-state level, which corresponds to a stable community (as in the first figure)? Does the population oscillate continually without settling at some fixed level (as in
the second figure)? Or does the population decrease to zero, indicating that the organism
becomes extinct (as in the third figure)? Such questions about the behavior of functions
are answered using the powerful idea of a limit.
Population increases
and approches a steady state.
300
200
100
Population
Population
400
500
500
400
400
Population
500
300
200
Population oscillates
without approaching
a steady state.
100
0
0
0
20
40
60
Time (months)
80
Population decays
to zero (extinction).
300
200
100
0
0
20
40
60
80
Time (months)
60
Copyright © 2014 Pearson Education, Inc.
0
20
40
60
Time (months)
80
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2.1 The Idea of Limits
61
Chapter Preview
All of calculus is based on the idea of a limit. Not only
are limits important in their own right, but they underlie the two fundamental operations
of calculus: differentiation (calculating derivatives) and integration (evaluating integrals).
Derivatives enable us to talk about the instantaneous rate of change of a function, which,
in turn, leads to concepts such as velocity and acceleration, population growth rates, marginal cost, and flow rates. Integrals enable us to compute areas under curves, surface
areas, and volumes. Because of the incredible reach of this single idea, it is essential to
develop a solid understanding of limits. We first present limits intuitively by showing
how they arise in computing instantaneous velocities and finding slopes of tangent lines.
As the chapter progresses, we build more rigor into the definition of the limit, and we
examine the different ways in which limits exist or fail to exist. The chapter concludes by
introducing the important property called continuity and by giving the formal definition
of a limit.
2.1 The Idea of Limits
This brief opening section illustrates how limits arise in two seemingly unrelated problems: finding the instantaneous velocity of a moving object and finding the slope of a line
tangent to a curve. These two problems provide important insights into limits, and they
reappear in various forms throughout the book.
Average Velocity
Suppose you want to calculate your average velocity as you travel along a straight highway.
If you pass milepost 100 at noon and milepost 130 at 12:30 p.m., you travel 30 miles in a halfhour, so your average velocity over this time interval is 130 mi2>10.5 hr2 = 60 mi>hr.
By contrast, even though your average velocity may be 60 mi >hr, it’s almost certain that
your instantaneous velocity, the speed indicated by the speedometer, varies from one
moment to the next.
EXAMPLE 1 Average velocity A rock is launched vertically upward from the ground
with a speed of 96 ft>s. Neglecting air resistance, a well-known formula from physics
states that the position of the rock after t seconds is given by the function
s1t2 = -16t 2 + 96t.
The position s is measured in feet with s = 0 corresponding to the ground. Find the
average velocity of the rock between each pair of times.
a. t = 1 s and t = 3 s
b. t = 1 s and t = 2 s
SOLUTION Figure 2.1 shows the position of the rock on the time interval 0 … t … 3.
a. The average velocity of the rock over any time interval 3t0, t14 is the change in position divided by the elapsed time:
vav =
s1t12 - s1t02
t1 - t0
Therefore, the average velocity over the interval 31, 34 is
vav =
s132 - s112
144 ft - 80 ft
64 ft
=
=
= 32 ft>s.
3 - 1
3s - 1s
2s
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Chapter 2 r
62
Limits
Height above ground (ft)
s
s(t) ⫽ ⫺16t2 ⫹ 96t
(3, s(3))
s(3) ⫽ 144 ft
144
s(2) ⫽ 128 ft
128
s(1) ⫽ 80 ft
80
(2, s(2))
(1, s(1))
Position
of rock
0
1
2
t
3
Time (s)
FIGURE 2.1
Here is an important observation: As shown in Figure 2.2a, the average velocity is
simply the slope of the line joining the points 11, s1122 and 13, s1322 on the graph of
the position function.
b. The average velocity of the rock over the interval 31, 24 is
vav =
s122 - s112
128 ft - 80 ft
48 ft
=
=
= 48 ft>s.
2 - 1
2s - 1s
1s
Again, the average velocity is the slope of the line joining the points 11, s1122 and
12, s1222 on the graph of the position function (Figure 2.2b).
64 ft
⫽ 32 ft/s
2s
vav ⫽ slope ⫽
s
vav ⫽ slope ⫽
s
48 ft
⫽ 48 ft/s
1s
(3, 144)
(1, 80)
Height above ground (ft)
Change in position
⫽ s(3) ⫺ s(1)
⫽ 64 ft
80
Elapsed time
⫽3s⫺1s⫽2s
0
1
FIGURE 2.2
2
3
t
128
(2, 128)
Change in position
⫽ s(2) ⫺ s(1)
⫽ 48 ft
(1, 80)
80
Elapsed time
⫽2s⫺1s⫽1s
0
1
2
3
t
Time (s)
(b)
Time (s)
(a)
Related Exercises 7–14
➤ See Section 1.1 for a discussion of secant
lines.
In Example 1, what is the average velocity between t = 2 and t = 3?
➤
QUICK CHECK 1
➤
Height above ground (ft)
144
In Example 1, we computed slopes of lines passing through two points on a curve.
Any such line joining two points on a curve is called a secant line. The slope of the secant
line, denoted msec, for the position function in Example 1 on the interval 3t0, t14 is
msec =
s1t12 - s1t02
t1 - t0
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2.1 The Idea of Limits
Height above ground (ft)
Example 1 demonstrates that the average velocity is the slope of a secant line on the graph
of the position function; that is, vav = msec (Figure 2.3).
s(t1) ⫺ s(t0 )
t 1 ⫺ t0
Instantaneous Velocity
Change in position
⫽ s(t1) ⫺ s(t0 )
s
s(t1)
s(t0)
Change in time
⫽ t1 ⫺ t 0
0
t0
t1
Time (s)
t
To compute the average velocity, we use the position of the object at two distinct points
in time. How do we compute the instantaneous velocity at a single point in time? As
illustrated in Example 2, the instantaneous velocity at a point t = t0 is determined by computing average velocities over intervals 3t0, t14 that decrease in length. As t1 approaches
t0, the average velocities typically approach a unique number, which is the instantaneous
velocity. This single number is called a limit.
QUICK CHECK 2
velocity.
FIGURE 2.3
➤
vav ⫽ msec ⫽
63
Explain the difference between average velocity and instantaneous
EXAMPLE 2 Instantaneous velocity Estimate the instantaneous velocity of the rock
in Example 1 at the single point t = 1.
SOLUTION We are interested in the instantaneous velocity at t = 1, so we compute the
average velocity over smaller and smaller time intervals 31, t4 using the formula
Notice that these average velocities are also slopes of secant lines, several of which
are shown in Table 2.1. We see that as t approaches 1, the average velocities appear to
approach 64 ft>s. In fact, we could make the average velocity as close to 64 ft>s as we
like by taking t sufficiently close to 1. Therefore, 64 ft>s is a reasonable estimate of the
instantaneous velocity at t = 1.
Table 2.1
Average
velocity
31, 24
31, 1.54
31, 1.14
31, 1.014
31, 1.0014
31, 1.00014
48 ft>s
56 ft>s
62.4 ft>s
63.84 ft>s
63.984 ft>s
63.998 ft>s
Related Exercises 15–20
In language to be introduced in Section 2.2, we say that the limit of vav as t
approaches 1 equals the instantaneous velocity vinst , which is 64 ft >s. This statement is
written compactly as
vinst = lim vav = lim
t S1
➤ The same instantaneous velocity is
Figure 2.4 gives a graphical illustration of this limit.
obtained as t approaches 1 from the left
(with t 6 1) and as t approaches 1 from
the right (with t 7 1).
0
t
1
t
2
t S1
s1t2 - s112
= 64 ft>s.
t - 1
Position of rock
at various times
t
128 ft
t ⫽ 2 sec
108 ft
t ⫽ 1.5 sec
86.24 ft
t ⫽ 1.1 sec
80 ft
vav ⫽
vav ⫽
...
vav ⫽
s(2) ⫺ s(1)
128 ⫺ 80
⫽ 48 ft/s
⫽
2⫺1
1
s(1.5) ⫺ s(1)
108 ⫺ 80
⫽ 56 ft/s
⫽
0.5
1.5 ⫺ 1
s(1.1) ⫺ s(1)
86.24 ⫺ 80
⫽
⫽ 62.4 ft/s
1.1 ⫺ 1
0.1
t ⫽ 1 sec
As these intervals shrink...
t ⫽ 0 (rock thrown at 96 ft/s)
FIGURE 2.4
Copyright © 2014 Pearson Education, Inc.
...
Time
interval
s1t2 - s112
.
t - 1
➤
vav =
vinst ⫽ 64 ft/s
... the average velocities approach 64 ft/s—
the instantaneous velocity at t ⫽ 1.
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Chapter 2 r
64
Limits
Slope of the Tangent Line
➤ We define tangent lines carefully in
Section 3.1. For the moment, imagine
zooming in on a point P on a smooth
curve. As you zoom in, the curve appears
more and more like a line passing
through P. This line is the tangent line
at P. Because a smooth curve approaches
a line as we zoom in on a point, a smooth
curve is said to be locally linear at any
given point.
Several important conclusions follow from Examples 1 and 2. Each average velocity in
Table 2.1 corresponds to the slope of a secant line on the graph of the position function
(Figure 2.5). Just as the average velocities approach a limit as t approaches 1, the slopes of
the secant lines approach the same limit as t approaches 1. Specifically, as t approaches 1,
two things happen:
1. The secant lines approach a unique line called the tangent line.
2. The slopes of the secant lines msec approach the slope of the tangent line mtan at the
point 11, s1122. Thus, the slope of the tangent line is also expressed as a limit:
mtan = lim msec = lim
tS1
y
tS1
s1t2 - s112
= 64.
t - 1
This limit is the same limit that defines the instantaneous velocity. Therefore, the
instantaneous velocity at t = 1 is the slope of the line tangent to the position curve at t = 1.
P
s
Height above ground (ft)
x
128
Slopes of the secant lines approach
slope of the tangent line.
mtan ⫽ 64
The secant lines approach
the tangent line.
msec ⫽ 62.4
msec ⫽ 56
msec ⫽ 48
(2, 128)
108
(1.5, 108)
(1, 80)
80
(1.1, 86.24)
s(t) ⫽ ⫺16t2 ⫹ 96t
0
0.5
1
1.1
1.5
2.0
t
Time (s)
FIGURE 2.5
QUICK CHECK 3
In Figure 2.5, is mtan at t = 2 greater than or less than mtan at t = 1?
➤
P
The parallels between average and instantaneous velocities, on one hand, and between
slopes of secant lines and tangent lines, on the other, illuminate the power behind the idea
of a limit. As t S 1, slopes of secant lines approach the slope of a tangent line. And as
t S 1, average velocities approach an instantaneous velocity. Figure 2.6 summarizes these
two parallel limit processes. These ideas lie at the foundation of what follows in the coming chapters.
Copyright © 2014 Pearson Education, Inc.
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2.1 The Idea of Limits
AVERAGE VELOCITY
s(t) 16t2 96t
Average velocity is the
change in position divided
by the change in time:
t2s
128
s(t ) s(t0)
.
vav 1
t1 t0
80
t1s
SECANT LINE
s(t) 16t2 96t
s
128
(2, 128)
80
Slope of the secant line is
the change in s divided by
the change in t:
s(t1) s(t0)
.
msec t1 t0
(1, 80)
msec vav 48
0
0.5
1.0
1.5
2.0
t
s
108
80
t 1.5 s
t1s
108
(1.5, 108)
80
(1, 80)
msec vav 56
As the time interval
shrinks, the average
velocity approaches
the instantaneous
velocity at t 1.
0
0.5
1.0
1.5
2.0
t
s
86.24
80
t 1.1 s
t1s
86.24
As the interval on the
t-axis shrinks, the slope of
the secant line approaches
the slope of the tangent
line through (1, 80).
(1.1, 86.24)
(1, 80)
80
msec vav 62.4
0
INSTANTANEOUS VELOCITY
80
៬
t 1
1.0
1.5
2.0
t
TANGENT LINE
s
The instantaneous velocity
at t 1 is the limit of the
average velocities as t
approaches 1.
vinst lim
0.5
t1s
The slope of the tangent
line at (1, 80) is the limit
of the slopes of the secant
lines as t approaches 1.
80
(1, 80)
s(t) s(1)
64 ft/s
t1
mtan lim
0
Instantaneous velocity 64 ft/s
៬
t 1
0.5
1.0
1.5
2.0
Slope of the tangent line 64
FIGURE 2.6
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t
s(t) s(1)
64
t1
65
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Chapter 2 r
66
Limits
SECTION 2.1 EXERCISES
Review Questions
1.
Suppose s1t2 is the position of an object moving along a line at
time t Ú 0. What is the average velocity between the times t = a
and t = b?
2.
Suppose s1t2 is the position of an object moving along a line
at time t Ú 0. Describe a process for finding the instantaneous
velocity at t = a.
3.
What is the slope of the secant line between the points 1a, f 1a22
and 1b, f 1b22 on the graph of f ?
4.
Describe a process for finding the slope of the line tangent to the
graph of f at 1a, f 1a22.
5.
6.
Describe the parallels between finding the instantaneous velocity
of an object at a point in time and finding the slope of the line tangent to the graph of a function at a point on the graph.
7.
Average velocity The function s1t2 represents the position of an
object at time t moving along a line. Suppose s122 = 136 and
s132 = 156. Find the average velocity of the object over the interval of time 32, 34 .
8.
Average velocity The function s1t2 represents the position of
an object at time t moving along a line. Suppose s112 = 84 and
s142 = 144. Find the average velocity of the object over the interval of time 31, 44 .
9.
Average velocity The position of an object moving along a line
is given by the function s1t2 = -16t 2 + 128t. Find the average
velocity of the object over the following intervals.
10. Average velocity The position of an object moving along a line
is given by the function s1t2 = - 4.9t 2 + 30t + 20. Find the
average velocity of the object over the following intervals.
30, 24
30, 1.54
30, 14
30, 0.54
0
0.5
1
1.5
2
s1t2
0
30
52
66
72
s(t)
46
0.5
1
1.5
2
2.5
t
T
13. Average velocity Consider the position function
s1t2 = -16t 2 + 100t representing the position of an object
moving along a line. Sketch a graph of s with the secant line passing through 10.5, s10.522 and 12, s1222. Determine the slope of the
secant line and explain its relationship to the moving object.
T
14. Average velocity Consider the position function s1t2 = sin pt
representing the position of an object moving along a line on the
end of a spring. Sketch a graph of s together with a secant line
passing through 10, s1022 and 10.5, s10.522. Determine the slope of
the secant line and explain its relationship to the moving object.
T
15. Instantaneous velocity Consider the position function
s1t2 = - 16t 2 + 128t (Exercise 9). Complete the following table
with the appropriate average velocities. Then make a conjecture
about the value of the instantaneous velocity at t = 1.
Time
interval
31, 24 31, 1.54 31, 1.14 31, 1.014 31, 1.0014
Average
velocity
T
16. Instantaneous velocity Consider the position function
s1t2 = -4.9t 2 + 30t + 20 (Exercise 10). Complete the following
table with the appropriate average velocities. Then make a conjecture about the value of the instantaneous velocity at t = 2.
Time
interval
32, 34 32, 2.54 32, 2.14 32, 2.014 32, 2.0014
Average
velocity
T
t
d. 30.5, 14
84
11. Average velocity The table gives the position s1t2 of an object
moving along a line at time t, over a two-second interval. Find the
average velocity of the object over the following intervals.
a.
b.
c.
d.
c. 30.5, 1.54
s
b. 30, 24
d. 30, h4, where h 7 0 is a real number
a. 30, 34
c. 30, 14
b. 30.5, 24
150
136
114
b. 31, 34
d. 31, 1 + h4, where h 7 0 is a real number
a. 31, 44
c. 31, 24
T
a. 30.5, 2.54
Graph the parabola f 1x2 = x 2. Explain why the secant lines
between the points 1- a, f 1- a22 and 1a, f 1a22 have zero slope.
What is the slope of the tangent line at x = 0?
Basic Skills
T
12. Average velocity The graph gives the position s1t2 of an object
moving along a line at time t, over a 2.5-second interval. Find the
average velocity of the object over the following intervals.
17. Instantaneous velocity The following table gives the position
s1t2 of an object moving along a line at time t. Determine the
average velocities over the time intervals 31, 1.014 , 31, 1.0014 ,
and 31, 1.00014 . Then make a conjecture about the value of the
instantaneous velocity at t = 1.
t
1
1.0001
1.001
1.01
s1t2
64
64.00479984
64.047984
64.4784
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For Use Only in 2013 – 2014 Pilot Program
2.1 The Idea of Limits
T
18. Instantaneous velocity The following table gives the position
s1t2 of an object moving along a line at time t. Determine the
average velocities over the time intervals 32, 2.014, 32, 2.0014,
and 32, 2.00014. Then make a conjecture about the value of the
instantaneous velocity at t = 2.
t
2
2.0001
2.001
2.01
s1t2
56
55.99959984
55.995984
55.9584
T
67
29. Tangent lines with zero slope
a. Graph the function f 1x2 = x 2 - 4x + 3.
b. Identify the point 1a, f 1a22 at which the function has a tangent
line with zero slope.
c. Confirm your answer to part (b) by making a table of slopes
of secant lines to approximate the slope of the tangent line at
this point.
30. Tangent lines with zero slope
T
a. Graph the function f 1x2 = 4 - x 2.
b. Identify the point 1a, f 1a22 at which the function has a tangent
line with zero slope.
c. Consider the point 1a, f 1a22 found in part (b). Is it true that the
secant line between 1a - h, f 1a - h22 and 1a + h, f 1a + h22
has slope zero for any value of h ≠ 0?
19. Instantaneous velocity Consider the position function
s1t2 = -16t 2 + 100t. Complete the following table with the
appropriate average velocities. Then make a conjecture about
the value of the instantaneous velocity at t = 3.
Time interval
Average velocity
T
32, 34
32.9, 34
a. Graph the position function, for 0 … t … 9.
b. From the graph of the position function, identify the time at
which the projectile has an instantaneous velocity of zero; call
this time t = a.
c. Confirm your answer to part (b) by making a table of average
velocities to approximate the instantaneous velocity at t = a.
d. For what values of t on the interval 30, 94 is the instantaneous
velocity positive (the projectile moves upward)?
e. For what values of t on the interval 30, 94 is the instantaneous
velocity negative (the projectile moves downward)?
32.99, 34
32.999, 34
32.9999, 34
T
20. Instantaneous velocity Consider the position function s1t2 = 3 sin t
that describes a block bouncing vertically on a spring. Complete
the following table with the appropriate average velocities. Then
make a conjecture about the value of the instantaneous velocity at
t = p>2.
Time interval
T
Average velocity
3p>2, p4
3p>2, p>2 + 0.14
3p>2, p>2 + 0.0014
T
3p>2, p>2 + 0.00014
Further Explorations
21–24. Instantaneous velocity For the following position functions,
make a table of average velocities similar to those in Exercises 19 and
20 and make a conjecture about the instantaneous velocity at the
indicated time.
21. s1t2 = - 16t 2 + 80t + 60
22. s1t2 = 20 cos t
at t = p>2
23. s1t2 = 40 sin 2t
at t = 0
24. s1t2 = 20>1t + 12
at t = 3
33. Slope of tangent line Given the function f 1x2 = 1 - cos x and
the points A1p>2, f 1p>222, B1p>2 + 0.05, f 1p>2 + 0.0522,
C1p>2 + 0.5, f 1p>2 + 0.522, and D1p, f 1p22 (see figure), find
the slopes of the secant lines through A and D, A and C, and A
and B. Then use your calculations to make a conjecture about the
slope of the line tangent to the graph of f at x = p>2.
y
D
2
y ⫽ 1 ⫺ cos x
C
A
1
B
at t = 0
25–28. Slopes of tangent lines For the following functions, make a
table of slopes of secant lines and make a conjecture about the slope
of the tangent line at the indicated point.
25. f 1x2 = 2x 2 at x = 2
26. f 1x2 = 3 cos x at x = p>2
27. f 1x2 = e x at x = 0
28. f 1x2 = x 3 - x at x = 1
0
q
q ⫹ 0.5
␲
x
q ⫹ 0.05
QUICK CHECK ANSWERS
1. 16 ft>s 2. Average velocity is the velocity over an interval
of time. Instantaneous velocity is the velocity at one point of
time. 3. Less than
➤
T
32. Impact speed A rock is dropped off the edge of a cliff, and its
distance s (in feet) from the top of the cliff after t seconds is
s1t2 = 16t 2. Assume the distance from the top of the cliff to the
ground is 96 ft.
a. When will the rock strike the ground?
b. Make a table of average velocities and approximate the velocity at which the rock strikes the ground.
3p>2, p>2 + 0.014
T
31. Zero velocity A projectile is fired vertically upward and has a
position given by s1t2 = -16t 2 + 128t + 192, for 0 … t … 9.
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Chapter 2 r
68
Limits
2.2 Definitions of Limits
Computing tangent lines and instantaneous velocities (Section 2.1) are just two of many
important calculus problems that rely on limits. We now put these two problems aside
until Chapter 3 and begin with a preliminary definition of the limit of a function.
DEFINITION Limit of a Function (Preliminary)
Suppose the function f is defined for all x near a except possibly at a. If f 1x2 is
arbitrarily close to L (as close to L as we like) for all x sufficiently close (but not equal)
to a, we write
➤ The terms arbitrarily close and
sufficiently close will be made precise
when rigorous definitions of limits are
given in Section 2.7.
lim f 1x2 = L
xSa
and say the limit of f 1x2 as x approaches a equals L.
Informally, we say that lim f 1x2 = L if f 1x2 gets closer and closer to L as x gets
xSa
closer and closer to a from both sides of a. The value of lim f 1x2 (if it exists) depends
xSa
upon the values of f near a, but it does not depend on the value of f 1a2. In some cases, the
limit lim f 1x2 equals f 1a2. In other instances, lim f 1x2 and f 1a2 differ, or f 1a2 may not
xSa
xSa
even be defined.
EXAMPLE 1 Finding limits from a graph Use the graph of f (Figure 2.7) to determine the following values, if possible.
y
6
a. f 112 and lim f 1x2
y ⫽ f (x)
5
xS1
b. f 122 and lim f 1x2
xS2
c. f 132 and lim f 1x2
xS3
4
SOLUTION
3
a. We see that f 112 = 2. As x approaches 1 from either side, the values of f 1x2
approach 2 (Figure 2.8). Therefore, lim f 1x2 = 2.
xS1
2
1
0
1
2
3
4
5
6
x
b. We see that f 122 = 5. However, as x approaches 2 from either side, f 1x2 approaches
3 because the points on the graph of f approach the open circle at 12, 32 (Figure 2.9).
Therefore, lim f 1x2 = 3 even though f 122 = 5.
xS2
c. In this case, f 132 is undefined. We see that f 1x2 approaches 4 as x approaches 3 from
either side (Figure 2.10). Therefore, lim f 1x2 = 4 even though f 132 does not exist.
FIGURE 2.7
xS3
y
y
y
f (1) ⫽ 2
f (2) ⫽ 5
6
f (3) undefined
6
6
y ⫽ f (x)
y ⫽ f (x)
y ⫽ f (x)
4
... f (x) approaches 2.
1
As x approaches 1...
FIGURE 2.8
6
x
0
2
6
x
As x approaches 2...
FIGURE 2.9
0
3
6
x
As x approaches 3...
FIGURE 2.10
Related Exercises 7–10
Copyright © 2014 Pearson Education, Inc.
➤
2
0
... f (x) approaches 4.
... f (x) approaches 3.
3
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2.2 Definitions of Limits
69
QUICK CHECK 1
In Example 1, suppose we redefine the function at one point so that
f 112 = 1. Does this change the value of lim f 1x2?
➤
xS1
In Example 1, we worked with the graph of a function. Let’s now work with tabulated
values of a function.
1x - 1
x - 1
corresponding to values of x near 1. Then make a conjecture about the value of lim f 1x2.
EXAMPLE 2 Finding limits from a table Create a table of values of f 1x2 =
➤ In Example 2, we have not stated with
certainty that lim f 1x2 = 0.5. But this is
xS1
our best guess based upon the numerical
evidence. Methods for calculating limits
precisely are introduced in Section 2.3.
xS1
SOLUTION Table 2.2 lists values of f corresponding to values of x approaching 1 from
both sides. The numerical evidence suggests that f 1x2 approaches 0.5 as x approaches 1.
Therefore, we make the conjecture that lim f 1x2 = 0.5.
xS1
x
f 1x2 =
1x − 1
x − 1
S
1
d
0.9
0.99
0.999
0.9999
1.0001
1.001
1.01
1.1
0.5131670
0.5012563
0.5001251
0.5000125
0.4999875
0.4998751
0.4987562
0.4880885
Related Exercises 11–14
➤
Table 2.2
One-Sided Limits
The limit lim f 1x2 = L is referred to as a two-sided limit because f 1x2 approaches L as x
xSa
approaches a for values of x less than a and for values of x greater than a. For some functions, it makes sense to examine one-sided limits called left-sided and right-sided limits.
DEFINITION One-Sided Limits
➤ As with two-sided limits, the value of a
one-sided limit (if it exists) depends on
the values of f 1x2 near a but not on the
value of f 1a2.
1. Right-sided limit Suppose f is defined for all x near a with x 7 a. If f 1x2 is
arbitrarily close to L for all x sufficiently close to a with x 7 a, we write
lim f 1x2 = L
x S a+
and say the limit of f 1x2 as x approaches a from the right equals L.
2. Left-sided limit Suppose f is defined for all x near a with x 6 a. If f 1x2 is
arbitrarily close to L for all x sufficiently close to a with x 6 a, we write
lim f 1x2 = L
x S a-
and say the limit of f 1x2 as x approaches a from the left equals L.
x3 - 8
.
41x - 22
Use tables and graphs to make a conjecture about the values of lim+ f 1x2, lim- f 1x2, and
EXAMPLE 3 Examining limits graphically and numerically Let f 1x2 =
➤ Computer-generated graphs and tables
help us understand the idea of a limit.
Keep in mind, however, that computers
are not infallible and they may produce
incorrect results, even for simple
functions (see Example 5).
xS2
xS2
lim f 1x2, if they exist.
xS2
SOLUTION Figure 2.11a shows the graph of f obtained with a graphing utility. The graph
is misleading because f 122 is undefined, which means there should be a hole in the graph
at 12, 32 (Figure 2.11b).
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For Use Only in 2013 – 2014 Pilot Program
Chapter 2 r
70
Limits
y
f (x) ⫽
y
x3 ⫺ 8
4(x ⫺ 2)
x3 ⫺ 8
4(x ⫺ 2)
f (x) ⫽
3
3
2
2
The hole in the graph
at x ⫽ 2 indicates that
the function is
undefined at this point.
This computergenerated graph is
inaccurate because f
is undefined at x ⫽ 2.
0
1
x
2
FIGURE 2.11
0
1
x
2
(a)
(b)
The graph in Figure 2.12a and the function values in Table 2.3 suggest that f 1x2
approaches 3 as x approaches 2 from the right. Therefore, we write
lim f 1x2 = 3,
x S 2+
which says the limit of f 1x2 as x approaches 2 from the right equals 3.
y
f (x) ⫽
y
x3 ⫺ 8
4(x ⫺ 2)
f (x) ⫽
x3 ⫺ 8
4(x ⫺ 2)
... f (x) approaches 3.
3
3
2
2
... f (x) approaches 3.
0
1
2
x
x
0
As x approaches
2 from the right...
FIGURE 2.12
➤ Remember that the value of the limit does
not depend upon the value of f 122. In this
case, lim f 1x2 = 3 despite the fact that
1
x
x
2
As x approaches
2 from the left...
(a)
(b)
Similarly, Figure 2.12b and Table 2.3 suggest that as x approaches 2 from the left,
f 1x2 approaches 3. So, we write
lim f 1x2 = 3,
xS2
f 122 is undefined.
x S 2-
which says the limit of f 1x2 as x approaches 2 from the left equals 3. Because f 1x2
approaches 3 as x approaches 2 from either side, we write lim f 1x2 = 3.
xS2
Table 2.3
S
x
x − 8
41x − 22
d
1.99
1.999
1.9999
2.0001
2.001
2.01
2.1
2.8525
2.985025
2.99850025
2.99985000
3.00015000
3.00150025
3.015025
3.1525
3
Related Exercises 15–18
Copyright © 2014 Pearson Education, Inc.
➤
f 1x2 =
2
1.9
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2.2 Definitions of Limits
71
Based upon the previous example, you might wonder whether the limits
lim- f 1x2, lim+ f 1x2, and lim f 1x2 always exist and are equal. The remaining examples
xSa
xSa
xSa
demonstrate that these limits may have different values, and in other cases, some or all
of these limits may not exist. The following theorem is useful when comparing one-sided
and two-sided limits.
➤ If P and Q are statements, we write P if
THEOREM 2.1 Relationship Between One-Sided and Two-Sided Limits
Assume f is defined for all x near a except possibly at a. Then lim f 1x2 = L if
xSa
and only if lim+ f 1x2 = L and lim- f 1x2 = L.
and only if Q when P implies Q and Q
implies P.
xSa
xSa
A proof of Theorem 2.1 is outlined in Exercise 44 of Section 2.7. Using this theorem,
it follows that lim f 1x2 ≠ L if either lim+ f 1x2 ≠ L or lim- f 1x2 ≠ L (or both).
xSa
xSa
xSa
Furthermore, if either lim+ f 1x2 or lim- f 1x2 does not exist, then lim f 1x2 does not exist.
xSa
xSa
xSa
We put these ideas to work in the next two examples.
EXAMPLE 4 A function with a jump Given the graph of g in Figure 2.13, find the
y
following limits, if they exist.
6
a. lim- g1x2
xS2
b. lim+ g1x2
c. lim g1x2
xS2
xS2
SOLUTION
4
a. As x approaches 2 from the left, g1x2 approaches 4. Therefore, lim- g1x2 = 4.
y g(x)
xS2
b. Because g1x2 = 1, for all x Ú 2, lim+ g1x2 = 1.
2
xS2
c. By Theorem 2.1, lim g1x2 does not exist because lim- g1x2 ≠ lim+ g1x2.
0
2
4
xS2
xS2
Related Exercises 19–24
x
➤
xS2
EXAMPLE 5 Some strange behavior Examine lim cos 11>x2.
FIGURE 2.13
xS0
SOLUTION From the first three values of cos 11>x2 in Table 2.4, it is tempting to conclude
that lim+ cos 11>x2 = -1. But this conclusion is not confirmed when we evaluate cos 11>x2
xS0
for values of x closer to 0.
Table 2.4
cos 11 , x2
x
0.001
0.0001
0.00001
0.000001
0.0000001
0.00000001
0.56238
-0.95216 ¶
-0.99936
0.93675
-0.90727
-0.36338
We might incorrectly
conclude that cos 11>x2
approaches - 1 as x
approaches 0 from the
right.
The behavior of cos 11>x2 near 0 is better understood by letting x = 1>1np2, where n
is a positive integer. By making this substitution, we can sample the function at discrete
points that approach zero. In this case
cos
QUICK CHECK 2
➤
Why is the graph of
y = cos 11>x2 difficult to plot near
x = 0, as suggested by Figure 2.14?
1
1
if n is even
= cos np = b
x
-1 if n is odd.
As n increases, the values of x = 1>1np2 approach zero, while the values of cos 11>x2
oscillate between -1 and 1 (Figure 2.14). Therefore, cos 11>x2 does not approach a
single number as x approaches 0 from the right. We conclude that lim+ cos 11>x2 does
xS0
not exist, which implies that lim cos 11>x2 does not exist.
xS0
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For Use Only in 2013 – 2014 Pilot Program
Chapter 2 r
72
Limits
The values of cos (1/x) oscillate between 1
and 1, over shorter and shorter intervals, as
x approaches 0 from the right.
y
1
y cos(1/x)
0
1
6␲
1
5␲
1
4␲
1
3␲
x
1
2␲
1
Related Exercises 25–26
➤
FIGURE 2.14
Using tables and graphs to make conjectures for the values of limits worked well until
Example 5. The limitation of technology in this example is not an isolated incident. For
this reason, analytical techniques (paper-and-pencil methods) for finding limits are developed in the next section.
SECTION 2.2 EXERCISES
Review Questions
8.
1.
Explain the meaning of lim f 1x2 = L.
2.
True or false: When lim f 1x2 exists, it always equals f 1a2. Explain.
xSa
Finding limits from a graph Use the graph of g in the figure to
find the following values, if they exist.
a. g102
b. lim g1x2
c. g112
d. lim g1x2
xS0
xSa
xS1
y
3.
Explain the meaning of lim+ f 1x2 = L.
4.
Explain the meaning of lim- f 1x2 = L.
5.
If lim- f 1x2 = L and lim+ f 1x2 = M, where L and M are finite
xSa
xSa
xSa
xSa
real numbers, then how are L and M related if lim f 1x2 exists?
xSa
6.
What are the potential problems of using a graphing utility to
estimate lim f 1x2?
2
y g(x)
xSa
Basic Skills
7.
Finding limits from a graph Use the graph of h in the figure to
find the following values, if they exist.
a. h122
b. lim h1x2
xS2
c. h142
d. lim h1x2
e. lim h1x2
xS4
xS5
1
9.
Finding limits from a graph Use the graph of f in the figure to
find the following values, if they exist.
a. f 112
b. lim f 1x2
c. f 102
d. lim f 1x2
xS1
y
x
1
xS0
y
5
y f (x)
y h(x)
3
1
1
1
1
0
1
2
4
x
Copyright © 2014 Pearson Education, Inc.
x
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2.2 Definitions of Limits
10. Finding limits from a graph Use the graph of f in the figure to
find the following values, if they exist.
a. f 122
b. lim f 1x2
c. lim f 1x2
xS2
T
xS5
y
6
xS0
c. What mathematical constant does lim 11 + x21>x appear to
xS0
equal?
y f (x)
4
T
2
xS2
b. Evaluate f 1x2 for values of x near 2 to support your conjecture
in part (a).
0
1
2
4
6
11. Estimating a limit from tables Let f 1x2 =
x
T
x2 - 4
.
x - 2
a. Calculate f 1x2 for each value of x in the following table.
x2 - 4
b. Make a conjecture about the value of lim
.
xS2 x - 2
x
f 1x2 =
T
1.9
1.99
1.999
1.9999
2.1
2.01
2.001
2.0001
xS0
T
x − 4
x − 2
b. Evaluate f 1x2 for values of x near 1 to support your conjecture
in part (a).
T
12. Estimating a limit from tables Let f 1x2 =
x3 - 1
.
x - 1
xS0
T
0.99
0.999
0.9999
x3 − 1
x − 1
x
f 1x2 =
b. Evaluate g1x2 for values of x near 0 to support your conjecture
in part (a).
T
0.9
18. Estimating a limit graphically and numerically
3 sin x - 2 cos x + 2
Let g1x2 =
.
x
a. Plot a graph of g to estimate lim g1x2.
a. Calculate f 1x2 for each value of x in the following table.
x3 - 1
b. Make a conjecture about the value of lim
.
xS1 x - 1
f 1x2 =
17. Estimating a limit graphically and numerically
1 - cos12x - 22
Let f 1x2 =
.
1x - 122
a. Plot a graph of f to estimate lim f 1x2.
xS1
x2 − 4
x − 2
x
16. Estimating a limit graphically and numerically
e 2x - 2x - 1
Let g1x2 =
.
x2
a. Plot a graph of g to estimate lim g1x2.
b. Evaluate g1x2 for values of x near 0 to support your conjecture
in part (a).
2
x
f 1x2 =
15. Estimating a limit graphically and numerically
x - 2
Let f 1x2 =
.
ln x - 2 a. Plot a graph of f to estimate lim f 1x2.
1
T
14. Estimating a limit of a function Let f 1x2 = 11 + x21>x.
a. Make two tables, one showing the values of f for x = 0.01,
0.001, 0.0001, and 0.00001 and one showing values of f for
x = - 0.01, -0.001, -0.0001, and -0.00001. Round your
answers to five digits.
b. Estimate the value of lim 11 + x21>x.
d. lim f 1x2
xS4
73
x 2 - 25
. Use tables
x - 5
and graphs to make a conjecture about the values of lim+ f 1x2,
xS5
lim- f 1x2, and lim f 1x2, if they exist.
19. One-sided and two-sided limits Let f 1x2 =
xS5
1.1
1.01
1.001
1.0001
x3 − 1
x − 1
T
xS5
x - 100
. Use
1x - 10
tables and graphs to make a conjecture about the values of
lim + g1x2, lim - g1x2, and lim g1x2, if they exist.
20. One-sided and two-sided limits Let g1x2 =
x S 100
t - 9
.
1t - 3
a. Make two tables, one showing the values of g for
t = 8.9, 8.99, and 8.999 and one showing values of g for
t = 9.1, 9.01, and 9.001.
t - 9
.
b. Make a conjecture about the value of lim
S
t 9 1t - 3
13. Estimating a limit of a function Let g1t2 =
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x S 100
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Chapter 2 r
74
Limits
21. One-sided and two-sided limits Use the graph of f in the figure
to find the following values, if they exist. If a limit does not exist,
explain why.
a. f 112
b. lim- f 1x2
c. lim+ f 1x2
xS1
y
5
y f (x)
d. lim f 1x2
xS1
xS1
3
y
2
1
1
y f (x)
2
3
x
5
24. Finding limits from a graph Use the graph of g in the figure to
find the following values, if they exist. If a limit does not exist,
explain why.
1
0
x
1
a. g1- 12
b.
d. lim g1x2
e. g112
f. lim g1x2
g. lim g1x2
h. g152
i.
xS - 1
22. One-sided and two-sided limits Use the graph of g in the figure
to find the following values, if they exist. If a limit does not exist,
explain why.
a. g122
b. lim- g1x2
c. lim+ g1x2
d. lim g1x2
e. g132
f. lim- g1x2
g. lim+ g1x2
h. g142
i. lim g1x2
xS2
xS2
xS3
xS3
lim g1x2
x S - 1-
c.
lim g1x2
x S - 1+
xS1
lim g1x2
x S 5-
y
6
xS2
5
xS3
4
xS4
y g(x)
2
y
1
5
3
y g(x)
T
25. Strange behavior near x=0
x
2 2 2
,
,
,
p 3p 5p
2
2
2 2
,
, and
. Describe the pattern of values you observe.
7p 9p
11p
b. Why does a graphing utility have difficulty plotting the graph
of y = sin 11>x2 near x = 0 (see figure)?
c. What do you conclude about lim sin 11>x2?
1
2
3
x
5
xS0
y
23. Finding limits from a graph Use the graph of f in the figure to
find the following values, if they exist. If a limit does not exist,
explain why.
a. f 112
b. lim- f 1x2
c. lim+ f 1x2
d. lim f 1x2
e. f 132
f. lim- f 1x2
g. lim+ f 1x2
h. lim f 1x2
i. f 122
j. lim- f 1x2
k. lim+ f 1x2
l. lim f 1x2
xS2
5
a. Create a table of values of sin 1 1>x 2 , for x =
0
xS3
1
3
1
xS1
1
xS1
xS3
xS2
1
xS1
0
xS3
xS2
2
9␲
2
7␲
2
5␲
x
1
T
26. Strange behavior near x=0
a. Create a table of values of tan 13>x2 for x = 12>p, 12>13p2,
12>15p2, c, 12>111p2. Describe the general pattern in the
values you observe.
b. Use a graphing utility to graph y = tan 13>x2. Why does a
graphing utility have difficulty plotting the graph near x = 0?
c. What do you conclude about lim tan 13>x2?
xS0
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2.2 Definitions of Limits
75
e. For what values of a does lim : x ; exist? Explain.
Further Explorations
xSa
27. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
y
x2 - 9
does not exist.
a. The value of lim
xS3 x - 3
b. The value of lim f 1x2 is always found by computing f 1a2.
3
xSa
c. The value of lim f 1x2 does not exist if f 1a2 is undefined.
d. lim 1x = 0
1
xSa
xS0
e.
3
lim cot x = 0
1
x S p>2
3
x
y :x;
28–29. Sketching graphs of functions Sketch the graph of a function
with the given properties. You do not need to find a formula for the
function.
3
28. f 112 = 0, f 122 = 4, f 132 = 6, lim- f 1x2 = - 3, lim + f 1x2 = 5
xS2
xS2
29. g112 = 0, g122 = 1, g132 = - 2, lim g1x2 = 0,
38. The ceiling function For any real number x, the ceiling function
< x = is the least integer greater than or equal to x.
xS2
lim g1x2 = -1, lim+ g1x2 = - 2
x S 3-
a. Graph the ceiling function y = < x = , for -2 … x … 3.
xS3
30. h1- 12 = 2, lim - h1x2 = 0, lim + h1x2 = 3,
x S -1
b. Evaluate lim- < x = , lim+ < x = , and lim < x = .
x S -1
h112 = lim- h1x2 = 1, lim+ h1x2 = 4
xS1
31. p102 = 2, lim p1x2 = 0, lim p1x2 does not exist,
xS0
xSa
Applications
xS2
32–35. Calculator limits Estimate the value of the following limits by
creating a table of function values for h = 0.01, 0.001, and 0.0001,
and h = - 0.01, - 0.001, and - 0.0001.
32. lim 11 + 2h21>h
hS0
2 - 1
h
h
hS0
35. lim
ln 11 + h2
hS0
w S 3.3
h
c. Interpret the limits lim+ f 1w2 and lim- f 1w2.
wS1
x
, for x ≠ 0.
x
36. A step function Let f 1x2 =
39. Postage rates Assume that postage for sending a first-class letter
in the United States is $0.44 for the first ounce (up to and including 1 oz) plus $0.17 for each additional ounce (up to and including each additional ounce).
a. Graph the function p = f 1w2 that gives the postage p for
sending a letter that weighs w ounces, for 0 6 w … 5.
b. Evaluate lim f 1w2.
33. lim 11 + 3h22>h
hS0
34. lim
wS4
xS0
40. The Heaviside function The Heaviside function is used in engineering applications to model flipping a switch. It is defined as
examining lim- f 1x2 and lim+ f 1x2.
xS0
H1x2 = b
37. The floor function For any real number x, the floor function (or
greatest integer function) : x ; is the greatest integer less than or
equal to x (see figure).
xS - 1
x S 2.3
x S 2.3
xS2
b. Compute lim - : x ; , lim + : x ; , and lim : x ; .
xS0
x S 2.3
xSa
xSa
d. In general, if a is not an integer, state the values of lim- : x ;
xSa
and lim+ : x ; .
xSa
examining lim- H1x2 and lim+ H1x2.
xS2
c. For a given integer a, state the values of lim- : x ; and lim+ : x ; .
0 if x 6 0
1 if x Ú 0.
a. Sketch a graph of H on the interval 3-1, 24.
b. Does lim H1x2 exist? Explain your reasoning after first
a. Compute lim - : x ; , lim + : x ; , lim- : x ; , and lim+ : x ; .
xS - 1
wS1
d. Does lim f 1w2 exist? Explain.
a. Sketch a graph of f on the interval 3-2, 24.
b. Does lim f 1x2 exist? Explain your reasoning after first
xS0
x S 1.5
xS2
p122 = lim+ p1x2 = 1
T
xS1
xS2
c. For what values of a does lim < x = exist? Explain.
xS1
xS0
xS0
Additional Exercises
41. Limits of even functions A function f is even if f 1-x2 = f 1x2,
for all x in the domain of f. If f is even, with lim+ f 1x2 = 5 and
xS2
lim- f 1x2 = 8, find the following limits.
xS2
a.
lim
x S - 2+
f 1x2
b.
lim
x S - 2-
f 1x2
42. Limits of odd functions A function g is odd if g1- x2 = -g1x2,
for all x in the domain of g. If g is odd, with lim+ g1x2 = 5 and
xS2
lim- g1x2 = 8, find the following limits.
xS2
a.
lim g1x2
x S - 2+
Copyright © 2014 Pearson Education, Inc.
b.
lim g1x2
x S - 2-
For Use Only in 2013 – 2014 Pilot Program
Limits
Technology Excercises
T
43. lim x sin
xS0
45. lim
xS1
T
T
43–46. Limit by graphing Use the zoom and trace features of a
graphing utility to approximate the following limits.
1
x
44. lim
xS1
3
9 1 22x - x 4 - 2
x2
1 - x 3>4
3
1812x - 12
x3 - 1
6x - 3x
x S 0 x ln 2
46. lim
47. Limits by graphs
tan 2x
,
a. Use a graphing utility to estimate lim
x S 0 sin x
tan 3x
tan 4x
lim
, and lim
.
x S 0 sin x
x S 0 sin x
tan px
, for any real
b. Make a conjecture about the value of lim
S
x 0 sin x
constant p.
T
sin px
for
sin qx
at least three different pairs of nonzero constants p and q of your
sin px
choice. Estimate lim
in each case. Then use your work to
x S 0 sin qx
sin px
make a conjecture about the value of lim
for any nonzero
x S 0 sin qx
values of p and q.
49. Limits by graphs Use a graphing utility to plot y =
sin nx
, for n = 1, 2, 3, and 4
x
(four graphs). Use the window 3-1, 14 * 30, 54.
sin x
sin 2x
sin 3x
sin 4x
a. Estimate lim
, lim
, lim
, and lim
.
x xS0 x
x
xS0 x
xS0
xS0
sin px
b. Make a conjecture about the value of lim
, for any real
x
xS0
constant p.
48. Limits by graphs Graph f 1x2 =
QUICK CHECK ANSWERS
1. The value of lim f 1x2 depends on the value of f only
xS1
near 1, not at 1. Therefore, changing the value of f 112 will
not change the value of lim f 1x2. 2. A graphing device
xS1
has difficulty plotting y = cos 11>x2 near 0 because values
of the function vary between -1 and 1 over shorter and
shorter intervals as x approaches 0.
➤
Chapter 2 r
76
2.3 Techniques for Computing Limits
Graphical and numerical techniques for estimating limits, like those presented in the previous section, provide intuition about limits. These techniques, however, occasionally lead
to incorrect results. Therefore, we turn our attention to analytical methods for evaluating
limits precisely.
Limits of Linear Functions
The graph of f 1x2 = mx + b is a line with slope m and y-intercept b. From Figure 2.15,
we see that f 1x2 approaches f 1a2 as x approaches a. Therefore, if f is a linear function
we have lim f 1x2 = f 1a2. It follows that for linear functions, lim f 1x2 is found by direct
xSa
xSa
substitution of x = a into f 1x2. This observation leads to the following theorem, which is
proved in Exercise 28 of Section 2.7.
y
y
y f (x)
y f (x)
f (x)
f (a)
(a, f (a))
... f (x) approaches f (a)
f (a)
(a, f (a))
... f (x) approaches f (a)
f (x)
O
x
a
As x approaches a
from the left...
x
lim f (x) f (a) because f (x)
ᠬ
x a
FIGURE 2.15
O
a
x
As x approaches a
from the right...
៬ f (a) as x ៬ a from both sides of a.
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x
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2.3 Techniques for Computing Limits
77
THEOREM 2.2 Limits of Linear Functions
Let a, b, and m be real numbers. For linear functions f 1x2 = mx + b,
lim f 1x2 = f 1a2 = ma + b.
xSa
EXAMPLE 1 Limits of linear functions Evaluate the following limits.
a. lim f 1x2, where f 1x2 = 12 x - 7
b. lim g1x2, where g1x2 = 6
xS3
xS2
SOLUTION
xS3
xS3
1 12 x
- 7 2 = f 132 = - 11
2.
b. lim g1x2 = lim 6 = g122 = 6.
xS2
xS2
Related Exercises 11–16
➤
a. lim f 1x2 = lim
Limit Laws
The following limit laws greatly simplify the evaluation of many limits.
THEOREM 2.3 Limit Laws
Assume lim f 1x2 and lim g1x2 exist. The following properties hold, where c is a
xSa
xSa
real number, and m 7 0 and n 7 0 are integers.
1. Sum lim 3 f 1x2 + g1x24 = lim f 1x2 + lim g1x2
xSa
xSa
xSa
2. Difference lim 3 f 1x2 - g1x24 = lim f 1x2 - lim g1x2
xSa
xSa
xSa
3. Constant multiple lim 3cf 1x24 = c lim f 1x2
xSa
xSa
3
43
4
4. Product lim 3 f 1x2g1x24 = lim f 1x2 lim g1x2
xSa
5. Quotient lim c
xSa
➤ Law 6 is a special case of Law 7. Letting
m = 1 in Law 7 gives Law 6.
xSa
xSa
lim f 1x2
f 1x2
xSa
d =
, provided lim g1x2 ≠ 0
g1x2
lim g1x2
xSa
xSa
3
4
6. Power lim 3 f 1x24 = lim f 1x2
n
xSa
xSa
n
3
4
7. Fractional power lim 3 f 1x24 n>m = lim f 1x2
xSa
xSa
n>m
, provided f 1x2 Ú 0, for
x near a, if m is even and n>m is reduced to lowest terms
A proof of Law 1 is outlined in Section 2.7. Laws 2–5 are proved in Appendix B.
Law 6 is proved from Law 4 as follows. For a positive integer n, if lim f 1x2 exists, we
xSa
have
lim 3 f 1x24 n = lim 3 f 1x2 f 1x2 g f 1x24
x S a (+++)+++*
xSa
3
n factors of f 1x2
43
4
3
4
= lim f 1x2 lim f 1x2 g lim f 1x2
xSa
xSa
xSa
(++
++++++)++++++
++*
n factors of lim f 1x2
3
4
n
= lim f 1x2 .
xSa
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xSa
Repeated use of Law 4
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Chapter 2 r
Limits
➤ Recall that to take even roots of a number
(for example, square roots or fourth
roots), the number must be nonnegative if
the result is to be real.
In Law 7, the limit of 3 f 1x24 n>m involves the mth root of f 1x2 when x is near a. If the fraction n>m is in lowest terms and m is even, this root is undefined unless f 1x2 is nonnegative for all x near a, which explains the restrictions shown.
EXAMPLE 2 Evaluating limits Suppose lim f 1x2 = 4, lim g1x2 = 5, and
xS2
xS2
lim h1x2 = 8. Use the limit laws in Theorem 2.3 to compute each limit.
xS2
a. lim
xS2
f 1x2 - g1x2
h1x2
b. lim 36f 1x2g1x2 + h1x24
c. lim 3g1x24 3
xS2
xS2
SOLUTION
a. lim
xS2
lim 3 f 1x2 - g1x24
f 1x2 - g1x2
xS2
=
h1x2
lim h1x2
Law 5
xS2
lim f 1x2 - lim g1x2
xS2
=
xS2
Law 2
lim h1x2
xS2
4 - 5
1
= - .
8
8
=
b. lim 36 f 1x2g1x2 + h1x24 = lim 36 f 1x2g1x24 + lim h1x2
xS2
xS2
= 6 # lim 3 f 1x2g1x24 + lim h1x2
Law 3
= 6 # lim f 1x2
Law 4
xS2
3
xS2
4 # 3 lim g1x24 + lim h1x2
xS2
xS2
xS2
= 6 # 4 # 5 + 8 = 128.
3
4
c. lim 3g1x24 3 = lim g1x2
xS2
xS2
Law 1
xS2
= 53 = 125.
3
Law 6
Related Exercises 17–24
➤
78
Limits of Polynomial and Rational Functions
The limit laws are now used to find the limits of polynomial and rational functions. For
example, to evaluate the limit of the polynomial p1x2 = 7x 3 + 3x 2 + 4x + 2 at an
arbitrary point a, we proceed as follows:
lim p1x2 = lim 17x 3 + 3x 2 + 4x + 22
xSa
xSa
= lim 17x 32 + lim 13x 22 + lim 14x + 22
Law 1
= 7 lim 1x 32 + 3 lim 1x 22 + lim 14x + 22
Law 3
xSa
xSa
xSa
xSa
xSa
3
1
2
+ 3 lim x
xSa
s
a
2
+ lim 14x + 22
xSa
u
2
= 7 lim x
xSa
s
1
xSa
a
4a + 2
= 7a + 3a + 4a + 2 = p1a2.
3
2
Law 6
Theorem 2.2
As in the case of linear functions, the limit of a polynomial is found by direct substitution;
that is, lim p1x2 = p1a2 (Exercise 91).
xSa
It is now a short step to evaluating limits of rational functions of the form
f 1x2 = p1x2>q1x2, where p and q are polynomials. Applying Law 5, we have
lim
xSa
lim p1x2
p1x2
p1a2
xSa
=
=
, provided q1a2 ≠ 0,
q1x2
lim q1x2
q1a2
xSa
which shows that limits of rational functions are also evaluated by direct substitution.
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2.3 Techniques for Computing Limits
xSa
2
used to evaluate a limit become clear in
Section 2.6, when the important property
of continuity is discussed.
THEOREM 2.4 Limits of Polynomial and Rational Functions
Assume p and q are polynomials and a is a constant.
a. Polynomial functions:
b. Rational functions:
lim p1x2 = p1a2
xSa
p1a2
p1x2
=
, provided q1a2 ≠ 0
x S a q1x2
q1a2
lim
3x 2 - 4x
.
5x 3 - 36
QUICK CHECK 1
EXAMPLE 3 Limit of a rational function Evaluate lim
xS2
SOLUTION Notice that the denominator of this function is nonzero at x = 2. Using
x S -1
x - 1
.
x
➤
lim
xS2
Theorem 2.4b,
lim
xS2
31222 - 4122
3x 2 - 4x
=
= 1.
5x 3 - 36
51232 - 36
Related Exercises 25–27
QUICK CHECK 2
Use Theorem 2.4b to compute lim
xS1
5x 4 - 3x 2 + 8x - 6
.
x + 1
EXAMPLE 4 An algebraic function Evaluate lim
xS2
SOLUTION Using Theorems 2.3 and 2.4, we have
22x 3 + 9 + 3x - 1
lim
=
xS2
4x + 1
=
➤
Evaluate
lim 12x 4 - 8x - 162 and
22x 3 + 9 + 3x - 1
.
4x + 1
lim 1 22x 3 + 9 + 3x - 1 2
xS2
Law 5
lim 14x + 12
xS2
2lim 12x 3 + 92 + lim 13x - 12
xS2
xS2
Laws 1 and 7
lim 14x + 12
xS2
212122 + 92 + 13122 - 12
14122 + 12
125 + 5
10
=
=
.
9
9
3
=
Theorem 2.4
Notice that the limit at x = 2 equals the value of the function at x = 2.
Related Exercises 28–32
➤
1
substitution lim f 1x2 = f 1a2 can be
➤
➤ The conditions under which direct
79
One-Sided Limits
Theorem 2.2, Limit Laws 1–6, and Theorem 2.4 also hold for left-sided and right-sided
limits. In other words, these laws remain valid if we replace lim with lim+ or lim-. Law 7
xSa
xSa
must be modified slightly for one-sided limits, as shown in the next theorem.
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Chapter 2 r
80
Limits
THEOREM 2.3 (CONTINUED) Limit Laws for One-Sided Limits
Laws 1–6 hold with lim replaced by lim+ or lim-. Law 7 is modified as follows.
xSa
xSa
xSa
Assume m 7 0 and n 7 0 are integers.
7. Fractional power
4
a. lim+ 3 f 1x24 n>m =
3 lim
f 1x2 n>m, provided f 1x2 Ú 0, for x near a with x 7 a, if
m is even and n>m is reduced to lowest terms.
b. lim- 3 f 1x24 n>m =
3 lim
f 1x2 n>m,
xSa
x S a+
x S a-
xSa
4
provided f 1x2 Ú 0, for x near a with x 6 a, if
m is even and n>m is reduced to lowest terms.
EXAMPLE 5 Calculating left- and right-sided limits Let
y
2x 4 if x 1
x 1 if x 1
f (x) 4
f 1x2 = b
Find the values of lim- f 1x2, lim+ f 1x2, and lim f 1x2, or state that they do not exist.
xS1
lim f (x) 2
xS1
xS1
xS1
We verify this observation analytically by applying the limit laws. For x … 1,
f 1x2 = -2x + 4; therefore,
lim f (x) 0
x 1
៬
lim f 1x2 = lim- 1-2x + 42 = 2.
x S 1-
x
xS1
Theorem 2.2
For x 7 1, note that x - 1 7 0; it follows that
lim f 1x2 = lim+ 1x - 1 = 0.
x S 1+
xS1
Law 7
Because lim- f 1x2 = 2 and lim+ f 1x2 = 0, lim f 1x2 does not exist by Theorem 2.1.
xS1
xS1
xS1
Related Exercises 33–38
➤
1
FIGURE 2.16
xS1
SOLUTION The graph of f (Figure 2.16) suggests that lim- f 1x2 = 2 and lim+ f 1x2 = 0.
៬
x 1
2
-2x + 4 if x … 1
1x - 1 if x 7 1.
Other Techniques
So far, we have evaluated limits by direct substitution. A more challenging problem
is finding lim f 1x2 when the limit exists, but lim f 1x2 ≠ f 1a2. Two typical cases are
xSa
xSa
shown in Figure 2.17. In the first case, f 1a2 is defined, but it is not equal to lim f 1x2; in
xSa
the second case, f 1a2 is not defined at all.
y
y
lim f (x) f (a)
ᠬ
lim f (x) exists, but f (a) is undefined.
ᠬ
x a
x a
y f (x)
f (a)
O
a
x
FIGURE 2.17
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y f (x)
O
a
x
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2.3 Techniques for Computing Limits
81
EXAMPLE 6 Other techniques Evaluate the following limits.
a. lim
xS2
x 2 - 6x + 8
x2 - 4
b. lim
xS1
1x - 1
x - 1
SOLUTION
a. Factor and Cancel This limit cannot be found by direct substitution because the
denominator is zero when x = 2. Instead, the numerator and denominator are
factored; then, assuming x ≠ 2, we cancel like factors:
1x - 221x - 42
x 2 - 6x + 8
x - 4
=
=
.
2
1x - 221x + 22
x + 2
x - 4
➤ The argument used in this example
is common. In the limit process, x
approaches 2, but x ≠ 2. Therefore, we
may cancel like factors.
x - 4
x 2 - 6x + 8
=
whenever x ≠ 2, the two functions have the same
2
x + 2
x - 4
limit as x approaches 2 (Figure 2.18). Therefore,
Because
x 2 - 6x + 8
x - 4
2 - 4
1
= lim
=
= - .
2
xS2
xS2 x + 2
2 + 2
2
x - 4
lim
y
y
1
1
1
1
4
y
x
x2 6x 8
x2 4
2
1
1
4
y
x
x4
x2
2
2
lim x 6x 8 lim x 4 q
x 2
x 2 x2
x2 4
ᠬ
ᠬ
FIGURE 2.18
b. Use Conjugates This limit was approximated numerically in Example 2 of Section 2.2;
we conjectured that the value of the limit is 12. Direct substitution fails in this case
because the denominator is zero at x = 1. Instead, we first simplify the function
by multiplying the numerator and denominator by the algebraic conjugate of the
numerator. The conjugate of 1x - 1 is 1x + 1; therefore,
➤ We multiply the given function by
1 =
1x + 1
1x + 1
.
11x - 1211x + 12
1x - 1
=
x - 1
1x - 1211x + 12
=
x + 1x - 1x - 1
1x - 1211x + 12
x - 1
1x - 1211x + 12
1
.
=
1x + 1
=
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Rationalize the numerator; multiply by 1.
Expand the numerator.
Simplify.
Cancel like factors when x ≠ 1.
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Chapter 2 r
82
Limits
The limit can now be evaluated:
Related Exercises 39–52
QUICK CHECK 3
➤
1
1x - 1
1
1
=
= lim
= .
x - 1
x S 1 1x + 1
1 + 1
2
lim
xS1
An Important Limit
Evaluate
x 2 - 7x + 10
lim
.
xS5
x - 5
Despite our success in evaluating limits using direct substitution, algebraic manipulation,
and the limit laws, there are important limits for which these techniques do not work. One
such limit arises when investigating the slope of a line tangent to the graph of an exponential function.
➤
EXAMPLE 7 Slope of a line tangent to f 1 x2 = 2x Estimate the slope of the line
tangent to the graph of f 1x2 = 2x at the point P10, 12.
SOLUTION In Section 2.1, the slope of a tangent line was obtained by finding the limit
of slopes of secant lines; the same strategy is employed here. We begin by selecting a
point Q near P on the graph of f with coordinates 1x, 2x2. The secant line joining the
points P10, 12 and Q1x, 2x2 is an approximation to the tangent line. To compute the slope
of the tangent line (denoted by mtan) at x = 0, we look at the slope of the secant line
msec = 12x - 12>x and take the limit as x approaches 0.
y
y
f (x) ⫽ 2x
f (x) ⫽ 2x
Q(x, 2x )
As x 0⫹,
Q approaches P, ...
៬
... the secant lines approach the
tangent line, and msec mtan.
tangent
line
៬
tangent line
P(0, 1)
0
As x 0⫺,
Q approaches P, ...
Approaching P
from the right
x
... the secant lines approach the
tangent line, and msec mtan.
៬
P(0, 1)
Q(x, 2x )
x
x
x
0
(a)
៬
Approaching P
from the left
(b)
FIGURE 2.19
2x - 1
exists only if it has the same value as x S 0 + (Figure 2.19a)
x
xS0
and as x S 0- (Figure 2.19b). Because it is not an elementary limit, it cannot be evaluated using the limit laws of this section. Instead, we investigate the limit using numerical
evidence. Choosing positive values of x near 0 results in Table 2.5.
The limit lim
Table 2.5
x
msec
2x − 1
=
x
1.0
0.1
0.01
0.001
0.0001
0.00001
1.000000
0.7177
0.6956
0.6934
0.6932
0.6931
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2.3 Techniques for Computing Limits
2x - 1
lim
≈ 0.693, which is
xS0
x
approximately ln 2. The connection
between the natural logarithm and slopes
of lines tangent to exponential curves is
made clear in Chapters 3 and 6.
We see that as x approaches 0 from the right, the slopes of the secant lines approach the slope of the tangent line, which is approximately 0.693. A similar calculation
(Exercise 53) gives the same approximation for the limit as x approaches 0 from the left.
Because the left-sided and right-sided limits are the same, we conclude that
lim 12x - 12>x ≈ 0.693 (Theorem 2.1). Therefore, the slope of the line tangent to
xS0
f 1x2 = 2x at x = 0 is approximately 0.693.
Related Exercises 53–54
➤ The Squeeze Theorem is also called
The Squeeze Theorem
the Pinching Theorem or the Sandwich
Theorem.
y
f (x) ⱕ g(x) ⱕ h(x)
➤
➤ Example 7 shows that
83
y ⫽ h(x)
The Squeeze Theorem provides another useful method for calculating limits. Suppose the
functions f and h have the same limit L at a and assume the function g is trapped between
f and h (Figure 2.20). The Squeeze Theorem says that g must also have the limit L at a. A
proof of this theorem is outlined in Exercise 54 of Section 2.7.
y ⫽ g(x)
L
y ⫽ f (x)
O
៬
FIGURE 2.20
៬
៬
xSa
xSa
xSa
x
a
Squeeze Theorem:
As x a, h(x) L and f (x)
Therefore, g(x) L.
THEOREM 2.5 The Squeeze Theorem
Assume the functions f, g, and h satisfy f 1x2 … g1x2 … h1x2 for all values of x
near a, except possibly at a. If lim f 1x2 = lim h1x2 = L, then lim g1x2 = L.
EXAMPLE 8 Sine and cosine limits A geometric argument (Exercise 90) may be
៬
used to show that for -p>2 6 x 6 p>2,
L.
- x … sin x … x and 0 … 1 - cos x … x .
Use the Squeeze Theorem to confirm the following limits.
a. lim sin x = 0
b. lim cos x = 1
xS0
xS0
SOLUTION
a. Letting f 1x2 = - x , g1x2 = sin x, and h1x2 = x , we see that g is trapped between
f and h on -p>2 6 x 6 p>2 (Figure 2.21a). Because lim f 1x2 = lim h1x2 = 0
xS0
xS0
(Exercise 37), the Squeeze Theorem implies that lim g1x2 = lim sin x = 0.
xS0
y
兩x兩 sin x 兩x兩
y 兩x兩
y
y sin x
1
xS0
y 兩x兩
1
on q x q
y 1 cos x
➤ The two limits in Example 8 play a
q
crucial role in establishing fundamental
properties of the trigonometric functions.
The limits reappear in Section 2.6.
q
1
x
y 兩x兩
(a)
FIGURE 2.21
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y0
q
q
x
0 1 cos x 兩x兩
1
on q x q
(b)
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Chapter 2 r
84
Limits
b. In this case, we let f 1x2 = 0, g1x2 = 1 - cos x, and h1x2 = x (Figure 2.21b).
Because lim f 1x2 = lim h1x2 = 0, the Squeeze Theorem implies that
xS0
xS0
lim g1x2 = lim 11 - cos x2 = 0. By the limit laws, it follows that
lim 1 - lim cos x = 0, or lim cos x = 1.
xS0
xS0
y
EXAMPLE 9 Squeeze Theorem for an important limit
1
y cos x
1
y
Related Exercises 55–58
xS0
➤
xS0
xS0
a. Use a graphing utility to confirm that
sin x
x
cos x …
sin x
1
, for 0 6 x … 1.
…
x
cos x
b. Use part (a) and the Squeeze Theorem to prove that
y cos x
lim
0.5
xS0
sin x
= 1.
x
SOLUTION
a. Figure 2.22 shows the graphs of y = cos x (lower curve), y =
1
0
1
x
sin x
(middle curve),
x
1
(upper curve) on the interval -1 … x … 1. These graphs confirm the
cos x
given inequalities.
and y =
FIGURE 2.22
b. By part (a), cos x …
QUICK CHECK 4
Suppose f satisfies
x2
1 … f 1x2 … 1 +
for all values of x
6
near zero. Find lim f 1x2, if possible.
➤
xS0
lim cos x = lim
xS0
xS0
sin x
1
, for x near 0, except at 0. Furthermore,
…
x
cos x
1
= 1. The conditions of the Squeeze Theorem are satisfied and
cos x
we conclude that
lim
xS0
sin x
= 1.
x
This important limit is used in Chapter 3 to discover derivative rules for trigonometric
functions.
Related Exercises 55–58
➤
SECTION 2.3 EXERCISES
10. Suppose
Review Questions
1.
How is lim f 1x2 calculated if f is a polynomial function?
2.
How are lim- f 1x2 and lim+ f 1x2 calculated if f is a polynomial
xSa
xSa
f 1x2 = b
xSa
function?
4
if x … 3
x + 2 if x 7 3.
Compute lim- f 1x2 and lim+ f 1x2.
3.
For what values of a does lim r1x2 = r1a2 if r is a rational
xSa
function?
4.
Assume lim g1x2 = 4 and f 1x2 = g1x2 whenever x ≠ 3.
11–16. Limits of linear functions Evaluate the following limits.
Evaluate lim f 1x2, if possible.
11. lim 13x - 72
12. lim 1-2x + 52
13.
14. lim 1-3x2
15. lim 4
16.
xS3
x 2 - 7x + 12
= lim 1x - 42.
xS3
x - 3
xS3
5.
Explain why lim
6.
If lim f 1x2 = -8, find lim 3 f 1x24 2>3.
7.
8.
xS0
xS2
Evaluate lim 2x 2 - 9.
lim 5x
x S -9
lim p
x S -5
17–24. Applying limit laws Assume lim f 1x2 = 8, lim g1x2 = 3,
xS1
xS1
and lim h1x2 = 2. Compute the following limits and state the limit
p1x2
q1x2
= 10 and
Suppose lim f 1x2 = lim h1x2 = 5. Find lim g1x2, where
xS5
xS6
laws used to justify your computations.
xS2
f 1x2
17. lim 34f 1x24
18. lim c
19. lim 3 f 1x2 - g1x24
20. lim 3 f 1x2h1x24
xS1
f 1x2 … g1x2 … h1x2, for all x.
9.
xS1
xS1
Suppose p and q are polynomials. If lim
xS2
Basic Skills
xS2
xS2
q102 = 2, find p102.
xS3
xS4
xS3
xS2
xS3
xS1
xS1
21. lim c
xS1
f 1x2g1x2
h1x2
d
23. lim 3h1x24 5
xS1
Copyright © 2014 Pearson
Education, Inc.
h1x2
d
xS1
22. lim c
xS1
f 1x2
g1x2 - h1x2
d
3
24. lim 2
f 1x2g1x2 + 3
xS1
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2.3 Techniques for Computing Limits
25–32. Evaluating limits Evaluate the following limits.
25. lim 12x 3 - 3x 2 + 4x + 52
xS1
5x 2 + 6x + 1
xS1
8x - 4
45.
26. lim 1t 2 + 5t + 72
lim
x S -1
t S -2
27. lim
3 2
28. lim 2
t - 10
47. lim
3b
29. lim
b S 2 14b + 1 - 1
30. lim 1x - x2
49. lim
31. lim
xS3
xS9
tS3
2
5
xS2
- 5x
14x - 3
32. lim
hS0
3
116 + 3h + 4
lim f 1x2
b. lim + f 1x2
x S -1
34. One-sided limits Let
lim f 1x2
b. lim + f 1x2
c. lim f 1x2
e. lim+ f 1x2
f. lim f 1x2
x S -5
d. lim- f 1x2
xS5
x S -5
xS5
x
T
xSa
xS2
xS2
36. One-sided limits
x - 3
.
A2 - x
x - 3
b. Explain why lim+
does not exist.
xS3 A 2 - x
a. Evaluate lim-
x
xS3
-1
-0.1
-0.01 -0.001
-0.1
-0.01 - 0.001 -0.0001 0.0001 0.001
T
55. Applying the Squeeze Theorem
1
… x , for x ≠ 0.
x
b. Illustrate the inequalities in part (a) with a graph.
a. Show that - x … x sin
c. Use the Squeeze Theorem to show that lim x sin
38. Absolute value limit Show that lim x = a , for any real
xS0
xSa
T
39–52. Other techniques Evaluate the following limits, where a and b
are fixed real numbers.
x2 - 1
xS1 x - 1
x 2 - 16
41. lim
S
x 4 4 - x
43. lim
xSb
1x - b250 - x + b
x - b
x 2 - 2x - 3
xS3
x - 3
3t 2 - 7t + 2
42. lim
tS2
2 - t
40. lim
44.
lim
x S -b
1x + b27 + 1x + b210
41x + b2
0.01
3 − 1
x
x
if x Ú 0
x = b
- x if x 6 0.
39. lim
- 0.0001 - 0.00001
x
xS0
number. (Hint: Consider the cases a 6 0 and a Ú 0.)
x3 - a
x - a
a. Sketch a graph of y = 3x and carefully draw four secant lines
connecting the points P10, 12 and Q1x, 3x2, for x = - 2, - 1, 1,
and 2.
b. Find the slope of the line that joins P10, 12 and Q1x, 3x2, for
x ≠ 0.
c. Complete the table and make a conjecture about the value
3x - 1
of lim
.
S
x
x 0
b. Explain why lim- 1x - 2 does not exist.
xS0
x 2 - a2
,a 7 0
x S a 1x - 1a
54. Slope of a tangent line
a. Evaluate lim+ 1x - 2.
xS0
2
b16 + t - t 22
t - 3
2 − 1
x
35. One-sided limits
lim- x and lim+ x . Recall that
tS3
x
xS5
37. Absolute value limit Show that lim x = 0 by first evaluating
1
5
h
48. lim a4t -
52. lim
Compute the following limits, or state that they do not exist.
x S -5-
1x - 3
x - 9
116 + h - 4
h
0
if x … - 5
f 1x2 = c 225 - x 2 if - 5 6 x 6 5
3x
if x Ú 5.
a.
hS0
-
a. Sketch a graph of y = 2x and carefully draw three secant lines
connecting the points P10, 12 and Q1x, 2x2, for x = - 3, - 2,
and -1.
b. Find the slope of the line that joins P10, 12 and Q1x, 2x2, for
x ≠ 0.
c. Complete the table and make a conjecture about the value of
2x - 1
lim.
x
xS0
c. lim f 1x2
x S -1
1
5 + h
53. Slope of a tangent line
Compute the following limits or state that they do not exist.
x S -1-
x + 1
50. lim
hS0
x 2 + 1 if x 6 - 1
f 1x2 = b
1x + 1 if x Ú - 1.
a.
46. lim
x - a
,a 7 0
x S a 1x - 1a
51. lim
T
33. One-sided limits Let
12x - 122 - 9
85
1
= 0.
x
56. A cosine limit by the Squeeze Theorem It can be shown that
x2
… cos x … 1, for x near 0.
1 2
a. Illustrate these inequalities with a graph.
b. Use these inequalities to find lim cos x.
xS0
T
57. A sine limit by the Squeeze Theorem It can be shown that
sin x
x2
…
1 … 1, for x near 0.
x
6
a. Illustrate these inequalities with a graph.
sin x
.
b. Use these inequalities to find lim
xS0 x
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86
T
Limits
58. A logarithm limit by the Squeeze Theorem
a. Draw a graph to verify that - x … x ln x … x , for
- 1 … x … 1, where x ≠ 0.
b. Use the Squeeze Theorem to determine lim x 2 ln x 2.
2
2
xS0
x5
xSa x
xn
74. lim
S
x a x
73. lim
Further Explorations
75. lim
59. Explain why or why not Determine whether the following
statements are true and give an explanation or counterexample.
Assume a and L are finite numbers.
76.
a. If lim f 1x2 = L, then f 1a2 = L.
xSa
xSa
c. If lim f 1x2 = L and lim g1x2 = L, then f 1a2 = g1a2.
xSa
xSa
f 1x2
77. lim
xS1
does not exist if g1a2 = 0.
79. lim
e. If lim+ 2f 1x2 = 4 lim+ f 1x2, it follows that
xS1
xS1
f 1x2.
lim 2f 1x2 = 4xlim
S1
xS1
80. lim
d. The limit lim
xSa
g1x2
xS4
xS0
60–67. Evaluating limits Evaluate the following limits, where c and k
are constants.
100
h S 0 110h - 1211 + 2
61. lim 15x - 623>2
60. lim
xS2
1
1
15
x 2 + 2x
62. lim
xS3
x - 3
64. lim a
xS2
63. lim
xS1
1
2
b
- 2
x - 2
x - 2x
65. lim
110x - 9 - 1
x - 1
15 + h22 - 25
hS0
- 11232
4
2
x - 2
x - 16
x - 1
1x - 1
78. lim
xS1
x - 1
14x + 5 - 3
31x - 421x + 5
3 - 1x + 5
x
, where c is a nonzero constant
1cx + 1 - 1
81. Creating functions satisfying given limit conditions Find functions f and g such that lim f 1x2 = 0 and lim 1 f 1x2 g1x22 = 5.
xS1
xS1
82. Creating functions satisfying given limit conditions Find a
f 1x2
function f satisfying lim a
b = 2.
xS1 x - 1
83. Finding constants Find constants b and c in the polynomial
p1x2
p1x2 = x 2 + bx + c such that lim
= 6. Are the
xS2 x - 2
constants unique?
h
Applications
x 2 - 2cx + c 2
x - c
xSc
84. A problem from relativity theory Suppose a spaceship of
length L 0 is traveling at a high speed v relative to an observer. To
the observer, the ship appears to have a smaller length given by
the Lorentz contraction formula
66. lim
67.
x S 16
3
12
x23
77–80. Limits involving conjugates Evaluate the following limits.
b. If lim- f 1x2 = L, then lim+ f 1x2 = L.
xSa
a5
a
an
, for any positive integer n
a
3
2
x - 1
(Hint: x - 1 =
x - 1
xS1
lim
-
w 2 + 5kw + 4k 2
, for k ≠ 0
wS - k
w 2 + kw
lim
68. Finding a constant Suppose
f 1x2 = b
L = L0
3x + b if x … 2
x - 2 if x 7 2.
B
1 -
v2
,
c2
where c is the speed of light.
Determine a value of the constant a for which lim g1x2 exists
xS - 1
and state the value of the limit, if possible.
a. What is the observed length L of the ship if it is traveling at
50% of the speed of light?
b. What is the observed length L of the ship if it is traveling at
75% of the speed of light?
c. In parts (a) and (b), what happens to L as the speed of the ship
increases?
v2
d. Find lim- L 0 1 - 2 and explain the significance of this
vSc
B
c
limit.
70–76. Useful factorization formula Calculate the following limits
using the factorization formula
85. Limit of the radius of a cylinder A right circular cylinder with a
height of 10 cm and a surface area of S cm2 has a radius given by
Determine a value of the constant b for which lim f 1x2 exists and
xS2
state the value of the limit, if possible.
69. Finding a constant Suppose
g1x2 = b
x 2 - 5x if x … -1
ax 3 - 7 if x 7 - 1.
x n - a n = 1x - a21x n - 1 + x n - 2a + x n - 3a 2 + g + xa n - 2 + a n - 12,
r1S2 =
where n is a positive integer and a is a real number.
x 5 - 32
S
x 2 x - 2
70. lim
72.
2S
1
a 100 +
- 10b.
p
2 A
Find lim+ r1S2 and interpret your result.
x6 - 1
S
x 1 x - 1
SS0
71. lim
x7 + 1
(Hint: Use the formula for x 7 - a 7 with a = -1.)
xS - 1 x + 1
lim
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2.4 Infinite Limits
b. Show that sin u 6 u , for - p>2 6 u 6 p>2. 1Hint:
The length of arc AB is u, if 0 … u 6 p>2, and -u, if
- p>2 6 u 6 0.2
c. Conclude that - u … sin u … u , for - p>2 6 u 6 p>2.
d. Show that 0 … 1 - cos u … u , for -p>2 6 u 6 p>2.
86. Torricelli’s Law A cylindrical tank is filled with water to
a depth of 9 meters. At t = 0, a drain in the bottom of the
tank is opened and water flows out of the tank. The depth
of water in the tank (measured from the bottom of the tank)
t seconds after the drain is opened is approximated by
d1t2 = 13 - 0.015t22, for 0 … t … 200.
Evaluate and interpret lim - d1t2.
t S 200
2 ␪ 0
A
87. Electric field The magnitude of the electric field at a point x
meters from the midpoint of a 0.1m line of charge is given by
4.35
E1x2 =
(in units of newtons per coulomb, N>C).
x2x 2 + 0.01
Evaluate lim E1x2.
C
1
␪
x S 10
O
1
B
C
88–89. Limits of composite functions
p1x2 = b n x n + b n - 1x n - 1 + g + b 1x + b 0,
xS - 1
89. Suppose g1x2 = f 11 - x2, for all x, lim+ f 1x2 = 4, and
prove that lim p1x2 = p1a2 for any value of a.
xS1
xSa
lim- f 1x2 = 6. Find lim+ g1x2 and lim- g1x2.
xS0
A
91. Theorem 2.4a Given the polynomial
88. If lim f 1x2 = 4, find lim f 1x 22.
xS1
B
␪
0␪ 2
Additional Exercises
xS1
O
xS0
90. Two trigonometric inequalities Consider the angle u in standard
position in a unit circle, where 0 … u 6 p>2 or - p>2 6 u 6 0
(use both figures).
a. Show that AC = sin u , for -p>2 6 u 6 p>2. 1Hint:
Consider the cases 0 … u 6 p>2 and - p>2 6 u 6 0
separately.2
QUICK CHECK ANSWERS
1. 0, 2
2. 2
3. 3
4. 1
➤
T
87
2.4 Infinite Limits
Two more limit scenarios are frequently encountered in calculus and are discussed in this
and the following section. An infinite limit occurs when function values increase or decrease without bound near a point. The other type of limit, known as a limit at infinity,
occurs when the independent variable x increases or decreases without bound. The ideas
behind infinite limits and limits at infinity are quite different. Therefore, it is important to
distinguish these limits and the methods used to calculate them.
An Overview
To illustrate the differences between infinite limits and limits at infinity, consider the values of f 1x2 = 1>x 2 in Table 2.6. As x approaches 0 from either side, f 1x2 grows larger
and larger. Because f 1x2 does not approach a finite number as x approaches 0, lim f 1x2
xS0
does not exist. Nevertheless, we use limit notation and write lim f 1x2 = ∞. The infinity
xS0
symbol indicates that f 1x2 grows arbitrarily large as x approaches 0. This is an example of
an infinite limit; in general, the dependent variable becomes arbitrarily large in magnitude
as the independent variable approaches a finite number. y
Table 2.6
y
x
f 1x2 = 1>x2
{0.1
{0.01
{0.001
T
0
100
10,000
1,000,000
T
∞
0
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ᠬ
1
x2
x
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88
Limits
With limits at infinity, the opposite occurs: The dependent variable approaches a
finite number as the independent variable becomes arbitrarily large in magnitude. In
Table 2.7, we see that f 1x2 = 1>x 2 approaches 0 as x increases. In this case, we write
lim f 1x2 = 0.
xS∞
y
Table 2.7
x
10
100
1000
T
∞
y
f 1x2 = 1 , x2
0.01
0.0001
0.000001
T
0
1
0
x2
O
lim
lim
x ៬
x
៬
1
x2
1
0
x2
x
A general picture of these two limit scenarios is shown in Figure 2.23.
y
Infinite limit
y as x a
៬
Limit at infinity
y L as x ៬
៬
៬
L
O
x
a
M
y
Limit at infinity
y M as x f (x) large
and positive
៬
៬
FIGURE 2.23
y f (x)
O
Infinite Limits
x
a
The following definition of infinite limits is informal, but it is adequate for most functions
encountered in this book. A precise definition is given in Section 2.7.
x approaches a
DEFINITION Infinite Limits
(a)
y
O
Suppose f is defined for all x near a. If f 1x2 grows arbitrarily large for all x sufficiently close (but not equal) to a (Figure 2.24a), we write
x approaches a
x
a
y f (x)
lim f 1x2 = ∞.
xSa
We say the limit of f 1x2 as x approaches a is infinity.
If f 1x2 is negative and grows arbitrarily large in magnitude for all x sufficiently
close (but not equal) to a (Figure 2.24b), we write
lim f 1x2 = - ∞.
f (x) large
and negative
(b)
xSa
In this case, we say the limit of f 1x2 as x approaches a is negative infinity. In both
cases, the limit does not exist.
FIGURE 2.24
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2.4 Infinite Limits
EXAMPLE 1 Infinite limits Evaluate lim
xS1
89
x
x
and lim
using the
x S - 1 1x 2 - 122
1x 2 - 122
graph of the function.
x
(Figure 2.25) shows that as x approaches 1
1x 2 - 122
(from either side), the values of f grow arbitrarily large. Therefore, the limit does not
exist and we write
SOLUTION The graph of f 1x2 =
lim
lim f (x) 5
xS1
ᠬ
x 1
As x approaches -1, the values of f are negative and grow arbitrarily large in
magnitude; therefore,
x
f (x) 2
(x 1)2
lim
1
1
x
= ∞.
1x - 122
2
xS - 1
x
x
= - ∞.
1x 2 - 122
Related Exercises 7–8
➤
y
Example 1 illustrates two-sided infinite limits. As with finite limits, we also need to
work with right-sided and left-sided infinite limits.
lim f (x) x 1
ᠬ
FIGURE 2.25
5
DEFINITION One-Sided Infinite Limits
Suppose f is defined for all x near a with x 7 a. If f 1x2 becomes arbitrarily large
for all x sufficiently close to a with x 7 a, we write lim+ f 1x2 = ∞ (Figure 2.26a).
xSa
The one-sided infinite limits lim+ f 1x2 = - ∞ (Figure 2.26b), lim- f 1x2 = ∞
xSa
xSa
(Figure 2.26c), and lim- f 1x2 = - ∞ (Figure 2.26d) are defined analogously.
xSa
y
y
y f (x)
O
O
a
y f (x)
x
lim f (x) lim f (x) x a
x a
ᠬ
ᠬ
(a)
y
(b)
y
y f (x)
O
O
x
a
a
lim f (x) x a
ᠬ
(c)
FIGURE 2.26
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a
y f (x)
lim f (x) x a
ᠬ
(d)
x
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90
Limits
In all the infinite limits illustrated in Figure 2.26, the line x = a is called a vertical
asymptote; it is a vertical line that is approached by the graph of f as x approaches a.
QUICK CHECK 1 Sketch the graph
of a function and its vertical
asymptote that satisfies the conditions lim+ f 1x2 = - ∞ and
lim- f 1x2 = ∞.
xS2
DEFINITION Vertical Asymptote
If lim f 1x2 = { ∞, lim+ f 1x2 = { ∞, or lim- f 1x2 = { ∞, the line x = a is
xSa
➤
xS2
xSa
EXAMPLE 2 Determining limits graphically The vertical lines x = 1 and
lim g(x) ᠬ
x 1
y
xSa
called a vertical asymptote of f .
x = 3 are vertical asymptotes of the function g1x2 =
lim g(x) ᠬ
x 3
x - 2
. Use
1x - 122 1x - 32
Figure 2.27 to analyze the following limits.
g(x) x2
(x 1)2(x 3)
a. lim g1x2
b. lim- g1x2
xS1
c. lim g1x2
xS3
xS3
SOLUTION
1
a. The values of g grow arbitrarily large as x approaches 1 from either side.
Therefore, lim g1x2 = ∞.
x
3
xS1
b. The values of g are negative and grow arbitrarily large in magnitude as x
approaches 3 from the left, so lim- g1x2 = - ∞.
lim g(x) xS3
x 3
ᠬ
c. Note that lim+ g1x2 = ∞ and lim- g1x2 = - ∞. Therefore, lim g1x2 does not
FIGURE 2.27
xS3
xS3
xS3
exist. Because g behaves differently as x S 3- and as x S 3+ , we do not write
lim g1x2 = ∞ . We simply say that the limit does not exist.
Related Exercises 9–16
➤
xS3
Finding Infinite Limits Analytically
Table 2.8
x
0.01
0.001
0.0001
T
0+
5 + x
x
5.01
= 501
0.01
5.001
= 5001
0.001
5.0001
= 50,001
0.0001
T
∞
Many infinite limits are analyzed using a simple arithmetic property: The fraction a>b
grows arbitrarily large in magnitude if b approaches 0 while a remains nonzero and relatively constant. For example, consider the fraction 15 + x2>x for values of x approaching
0 from the right (Table 2.8).
5 + x
S ∞ as x S 0 + because the numerator 5 + x approaches 5 while
We see that
x
5 + x
the denominator is positive and approaches 0. Therefore, we write lim+
= ∞.
x
xS0
5 + x
Similarly, lim= - ∞ because the numerator approaches 5 while the denominator
x
xS0
approaches 0 through negative values.
EXAMPLE 3 Evaluating limits analytically Evaluate the following limits.
a. lim+
xS3
2 - 5x
x - 3
b. limxS3
2 - 5x
x - 3
SOLUTION
QUICK CHECK 2
Evaluate lim+
xS0
x - 5
x
approaches - 13
c
x - 5
by determining the sign
x
xS0
of the numerator and denominator.
and lim-
a. As x S 3 + , the numerator 2 - 5x approaches 2 - 5132 = -13 while the denominator x - 3 is positive and approaches 0. Therefore,
lim+
xS3
2 - 5x
= - ∞.
x - 3
s
positive and
approaches 0
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2.4 Infinite Limits
91
b. As x S 3 -, 2 - 5x approaches 2 - 5132 = -13 while x - 3 is negative and
approaches 0. Therefore,
approaches - 13
c
lim-
xS3
s
2 - 5x
= ∞.
x - 3
negative and
approaches 0
Related Exercises 17–28
EXAMPLE 4 Evaluating limits analytically Evaluate lim +
xS - 4
➤ We can assume that x ≠ 0 because we
➤
These limits imply that the given function has a vertical asymptote at x = 3.
-x 3 + 5x 2 - 6x
.
-x 3 - 4x 2
SOLUTION First we factor and simplify, assuming x ≠ 0:
are considering function values near
x = - 4.
1x - 221x - 32
-x1x - 221x - 32
-x 3 + 5x 2 - 6x
=
=
.
3
2
2
x1x + 42
-x - 4x
-x 1x + 42
As x S -4 + , we find that
approaches 42
f
lim +
xS - 4
1x - 221x - 32
-x 3 + 5x 2 - 6x
= lim +
= - ∞.
3
2
xS - 4
x1x + 42
-x - 4x
u
negative and
approaches 0
Verify that
x1x + 42 S 0 through negative
values as x S -4 + .
This limit implies that the given function has a vertical asymptote at x = -4.
➤
Related Exercises 17–28
➤ Example 5 illustrates that f 1x2>g1x2
might not grow arbitrarily large in
magnitude if both f 1x2 and g1x2
approach 0. Such limits are called
indeterminate forms and are examined in
detail in Section 4.7.
x 2 - 4x + 3
. Evaluate
x2 - 1
the following limits and find the vertical asymptotes of f . Verify your work with a graphing utility.
EXAMPLE 5 Location of vertical asymptotes Let f 1x2 =
a. lim f 1x2
b.
xS1
lim
x S - 1-
f 1x2
c.
lim
x S - 1+
f 1x2
SOLUTION
a. Notice that as x S 1, both the numerator and denominator of f approach 0, and the
function is undefined at x = 1. To compute lim f 1x2, we first factor:
xS1
➤ It is permissible to cancel the x - 1
1x - 121x - 32
because
1x - 121x + 12
x approaches 1 but is not equal to 1.
Therefore, x - 1 ≠ 0.
factors in lim
xS1
➤
QUICK CHECK 3
x 2 - 4x + 3
lim f 1x2 = lim
xS1
xS1
x2 - 1
1x - 121x - 32
= lim
x S 1 1x - 121x + 12
1x - 32
= lim
S
x 1 1x + 12
=
1 - 3
= -1.
1 + 1
Factor.
Cancel like factors, x ≠ 1.
Substitute x = 1.
Therefore, lim f 1x2 = -1 (even though f 112 is undefined). The line x = 1 is not a
xS1
vertical asymptote of f .
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92
Chapter 2 r
Limits
b. In part (a), we showed that
f 1x2 =
x - 3
x 2 - 4x + 3
=
, provided x ≠ 1.
x + 1
x2 - 1
We use this fact again. As x approaches -1 from the left, the one-sided limit is
approaches - 4
c
lim - f 1x2 = lim x S -1
x S -1
x-3
= ∞.
x + 1
s
negative and
approaches 0
c. As x approaches -1 from the right, the one-sided limit is
approaches - 4
c
lim f 1x2 = lim +
x S -1 +
x S -1
x-3
= - ∞.
x + 1
s
positive and
approaches 0
The infinite limits lim + f 1x2 = - ∞ and lim - f 1x2 = ∞ each imply that the line
xS - 1
xS - 1
x = -1 is a vertical asymptote of f . The graph of f generated by a graphing utility may
appear as shown in Figure 2.28a. If so, two corrections must be made. A hole should
appear in the graph at 11, -12 because lim f 1x2 = -1, but f 112 is undefined. It is also a
xS1
good idea to replace the solid vertical line with a dashed line to emphasize that the vertical asymptote is not a part of the graph of f (Figure 2.28b).
➤ Graphing utilities vary in how they
Two versions of the graph of y display vertical asymptotes. The errors
shown in Figure 2.28a do not occur on all
graphing utilities.
x2 4x 3
x2 1
y
y
1
1
x
2
x
2
Calculator graph
Correct graph
(a)
(b)
FIGURE 2.28
QUICK CHECK 4
➤
Why not?
The line x = 2 is not a vertical asymptote of y =
Copyright © 2014 Pearson Education, Inc.
1x - 121x - 22
.
x - 2
➤
Related Exercises 29–34
For Use Only in 2013 – 2014 Pilot Program
2.4 Infinite Limits
y
93
EXAMPLE 6 Limits of trigonometric functions Evaluate the following limits.
lim cot ␪ ␪ 0
ᠬ
a. lim+ cot u
b. lim- cot u
uS0
uS0
SOLUTION
y cot ␪
a. Recall that cot u = cos u>sin u. Furthermore (Example 8, Section 2.3), lim+ cos u = 1
1
uS0
d
1
and sin u is positive and approaches 0 as u S 0+. Therefore, as u S 0+, cot u
becomes arbitrarily large and positive, which means lim+ cot u = ∞. This limit is con-
␪
q
uS0
firmed by the graph of cot u (Figure 2.29), which has a vertical asymptote at u = 0.
b. In this case, lim- cos u = 1 and as u S 0 -, sin u S 0 with sin u 6 0. Therefore, as
uS0
lim cot ␪ u S 0 -, cot u is negative and becomes arbitrarily large in magnitude. It follows that
lim- cot u = - ∞, as confirmed by the graph of cot u.
␪ 0
ᠬ
FIGURE 2.29
uS0
Related Exercises 35–40
SECTION 2.4 EXERCISES
Review Questions
8.
Use a graph to explain the meaning of lim+ f 1x2 = - ∞ .
2.
Use a graph to explain the meaning of lim f 1x2 = ∞ .
y
3.
What is a vertical asymptote?
1
4.
Consider the function F 1x2 = f 1x2>g1x2 with g1a2 = 0. Does
F necessarily have a vertical asymptote at x = a? Explain your
reasoning.
xSa
xSa
5.
Suppose f 1x2 S 100 and g1x2 S 0, with g1x2 6 0, as x S 2.
f 1x2
Determine lim
.
x S 2 g1x2
6.
Evaluate limxS3
1
1
and lim+
.
S
x - 3
x 3 x - 3
7.
Analyzing infinite limits numerically Compute the values of
x + 1
f 1x2 =
in the following table and use them to discuss
1x - 122
lim f 1x2.
x + 1
1x − 1 2 2
3
a. lim- f 1x2
b. lim+ f 1x2
c. lim f 1x2
d. lim- f 1x2
e. lim+ f 1x2
f. lim f 1x2
xS1
xS2
xS1
xS1
xS2
xS2
y
x + 1
x
1.1
0.9
1.01
0.99
1.001
0.999
1.0001
0.9999
x
Analyzing infinite limits graphically The graph of f in the
figure has vertical asymptotes at x = 1 and x = 2. Analyze
the following limits.
xS1
x
y f (x)
1
9.
Basic Skills
T
Analyzing infinite limits graphically Use the graph of
x
to discuss lim f 1x2 and lim f 1x2.
f 1x2 = 2
xS - 1
xS3
1x - 2x - 322
1.
y f (x)
1x − 1 2 2
1
Copyright © 2014 Pearson Education, Inc.
2
x
➤
q
For Use Only in 2013 – 2014 Pilot Program
Chapter 2 r
94
Limits
10. Analyzing infinite limits graphically The graph of g in the
figure has vertical asymptotes at x = 2 and x = 4. Analyze the
following limits.
a. lim- g1x2
b. lim+ g1x2
c. lim g1x2
d. lim- g1x2
e. lim+ g1x2
f. lim g1x2
xS2
xS2
xS4
T
xS2
xS4
xS4
y
x
4
a.
lim - h1x2
xS - 2
d. lim- h1x2
xS3
b.
a.
lim + h1x2
c. lim h1x2
xS - 2
xS - 2
e. lim+ h1x2
f. lim h1x2
xS3
xS3
y
b. lim+ f 1x2
x S 0-
lim
x S - 2+
f 1x2
b. lim f 1x2
f 112 = 0,
lim f 1x2 = - ∞ ,
d. lim- p1x2
xS3
b.
lim p1x2
f. lim p1x2
xS3
xS3
y
x
y p(x)
lim f 1x2 = ∞
x S 4-
xS2
g152 = -1,
lim g1x2 = - ∞ ,
xS4
lim g1x2 = - ∞
x S 7+
b. lim-
1
x - 2
c. lim
1
x - 2
18. a. lim+
2
1x - 323
b. lim-
2
1x - 323
c. lim
2
1x - 323
19. a. lim+
x - 5
1x - 422
b. lim-
x - 5
1x - 422
c. lim
x - 5
- 422
20. a. lim+
x - 2
1x - 123
b. lim-
x - 2
1x - 123
c. lim
x - 2
- 123
xS3
xS4
xS1
21. a. lim+
3
lim f 1x2 = 1,
xS3
1
x - 2
c. lim
lim +
xS - 2
xS0
25. lim+
xS1
xS4
xS1
xS3
x S 4 1x
x S 1 1x
b. lim-
1x - 32
1x - 121x - 22
xS3
1x - 32
1x - 32
1x - 42
x1x + 22
x - 5x
x2
3
23. lim
xS3
xS2
1x - 121x - 22
xS3
22. a.
xS2
1x - 121x - 22
xS3
2
xS0
17. a. lim+
xS - 2
e. lim+ p1x2
d. lim+ f 1x2
16. Sketching graphs Sketch a possible graph of a function g,
together with vertical asymptotes, satisfying all the following
conditions.
c. lim p1x2
x S - 2+
lim f 1x2
lim f 1x2 = ∞ ,
x S 0+
xS2
12. Analyzing infinite limits graphically The graph of p in the
figure has vertical asymptotes at x = - 2 and x = 3. Investigate
the following limits.
lim p1x2
xS1
17–28. Evaluating limits analytically Evaluate the following limits or
state that they do not exist.
x
y h(x)
x S - 2-
d. lim+ f 1x2
x S 0-
f 132 is undefined,
lim g1x2 = ∞ ,
a.
c.
xS - 2
x S 7-
3
lim f 1x2
x S 1-
15. Sketching graphs Sketch a possible graph of a function f ,
together with vertical asymptotes, satisfying all the following
conditions on 30, 44 .
g122 = 1,
2
c.
xS0
14. Analyzing infinite limits graphically Graph the function
e -x
f 1x2 =
using a graphing utility. (Experiment with
x1x + 222
your choice of a graphing window.) Use your graph to discuss
the following limits.
y g(x)
11. Analyzing infinite limits graphically The graph of h in the figure has vertical asymptotes at x = - 2 and x = 3. Investigate the
following limits.
lim f 1x2
a.
T
2
13. Analyzing infinite limits graphically Graph the function
1
using a graphing utility with the window
f 1x2 = 2
x - x
3-1, 24 * 3-10, 104. Use your graph to discuss the following
limits.
b. lim xS - 2
1x - 42
x1x + 22
Copyright © 2014 Pearson Education, Inc.
xS - 2
1x - 42
x1x + 22
24. lim
4t - 100
t - 5
26. lim
z - 5
1z - 10z + 2422
2
x 2 - 5x + 6
x - 1
c. lim
2
tS5
zS4
2
For Use Only in 2013 – 2014 Pilot Program
2.4 Infinite Limits
x 2 - 4x + 3
xS2
1x - 222
2
x - 4x + 3
c. lim
xS2
1x - 222
b. lim-
27. a. lim+
xS2
x - 5x + 6x
x 4 - 4x 2
x 3 - 5x 2 + 6x
c. lim
xS - 2
x 4 - 4x 2
3
28. a.
2
lim
x S - 2+
x 2 - 4x + 3
1x - 222
x - 5x + 6x
x 4 - 4x 2
x - 5x 2 + 6x
d. lim
xS2
x 4 - 4x 2
3
b.
2
lim
x S - 23
29. Location of vertical asymptotes Analyze the following limits
x - 5
.
and find the vertical asymptotes of f 1x2 = 2
x - 25
b. lim - f 1x2
c. lim + f 1x2
a. lim f 1x2
xS5
x S -5
x S -5
30. Location of vertical asymptotes Analyze the following limits
x + 7
and find the vertical asymptotes of f 1x2 = 4
.
x - 49x 2
c. lim f 1x2 d. lim f 1x2
a. lim- f 1x2 b. lim+ f 1x2
xS7
xS7
xS - 7
42. Finding a function with vertical asymptotes Find polynomials
p and q such that f = p>q is undefined at 1 and 2, but p>q has a
vertical asymptote only at 2. Sketch a graph of your function.
43. Finding a function with infinite limits Give a formula for a
function f that satisfies lim+ f 1x2 = ∞ and lim- f 1x2 = - ∞ .
xS6
xSa
x
x2 + 1
1
c. f 1x2 = 2
x - 1
1
e. f 1x2 =
1x - 122
x
x2 - 1
x
d. f 1x2 =
1x - 122
x
f. f 1x2 =
x + 1
a. f 1x2 =
A.
b. f 1x2 =
B.
y
xS0
y
1
1
1
1
x
1
1
x
1
x
1
x
xSa
x 2 - 9x + 14
31. f 1x2 = 2
x - 5x + 6
33. f 1x2 =
xS6
44. Matching Match functions a–f with graphs A–F in the figure
without using a graphing utility.
31–34. Finding vertical asymptotes Find all vertical asymptotes
x = a of the following functions. For each value of a, discuss
lim+ f 1x2, lim- f 1x2, and lim f 1x2.
xSa
95
x + 1
x 3 - 4x 2 + 4x
32. f 1x2 =
cos x
x + 2x
34. f 1x2 =
x 3 - 10x 2 + 16x
x 2 - 8x
2
C.
D.
y
y
35–38. Trigonometric limits Investigate the following limits.
35.
lim csc u
xS0
37. lim+ 1- 10 cot x2
xS0
T
c.
T
38.
lim
u S p>2 +
lim + tan x
x S p>2
lim
x S - p>2
+
tan x
b.
d.
c.
lim sec x tan x
x S p>2+
lim
x S - p>2+
sec x tan x
b.
d.
E.
1
x
F.
y
y
lim - tan x
x S p>2
lim
x S - p>2-
tan x
1
40. Analyzing infinite limits graphically Graph the function
y = sec x tan x with the window 3-p, p4 * 3- 10, 104. Use the
graph to analyze the following limits.
a.
1
1
tan u
3
39. Analyzing infinite limits graphically Graph the function
y = tan x with the window 3- p, p4 * 3- 10, 104. Use the graph
to analyze the following limits.
a.
1
36. lim- csc x
u S 0+
1
1
1
x
1
lim sec x tan x
x S p>2-
lim
x S - p>2-
sec x tan x
Further Explorations
Additional Exercises
41. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
45. Limits with a parameter Let f 1x2 =
a. The line x = 1 is a vertical asymptote of the function
x 2 - 7x + 6
.
f 1x2 =
x2 - 1
b. The line x = -1 is a vertical asymptote of the function
x 2 - 7x + 6
f 1x2 =
.
x2 - 1
c. If g has a vertical asymptote at x = 1 and lim+ g1x2 = ∞ ,
xS1
then lim- g1x2 = ∞ .
x 2 - 7x + 12
.
x - a
a. For what values of a, if any, does lim+ f 1x2 equal a finite
xSa
number?
b. For what values of a, if any, does lim+ f 1x2 = ∞ ?
xSa
c. For what values of a, if any, does lim+ f 1x2 = - ∞ ?
xS1
Copyright © 2014 Pearson Education, Inc.
xSa
For Use Only in 2013 – 2014 Pilot Program
96
Chapter 2 r
Limits
b. Create a graph that gives a more complete representation of f.
46–47. Steep secant lines
a. Given the graph of f in the following figures, find the slope of the
secant line that passes through 10, 02 and 1h, f 1h22 in terms of h,
for h 7 0 and h 6 0.
b. Evaluate the limit of the slope of the secant line found in part (a)
as h S 0+ and h S 0-. What does this tell you about the line tangent
to the curve at 10, 02?
46. f 1x2 = x
y
20
15
1>3
10
y
y⫽
2000
50 ⫹ 100 x2
2
4
5
(h,
h1/3)
(0, 0)
⫺4
x
h
T
47. f 1x2 = x 2>3
x2/3
(h, h2/3)
49. f 1x2 =
x 2 - 3x + 2
x 10 - x 9
50. g1x2 = 2 - ln x 2
51. h1x2 =
ex
1x + 123
px
52. p1x2 = sec a b , for x 6 2
2
53. g1u2 = tan a
(0, 0)
x
h
55. f 1x2 =
T
48. Care with graphing The figure shows the graph of the function
2000
graphed in the window 3 -4, 44 * 30, 204.
f 1x2 =
50 + 100x 2
xS0
pu
b
10
1
1x sec x
54. q1s2 =
p
s - sin s
56. g1x2 = e1>x
57. Can a graph intersect a vertical asymptote? A common misconception is that the graph of a function never intersects its vertical asymptotes. Let
4
f 1x2 = W x - 1
x2
a. Evaluate lim+ f 1x2, lim- f 1x2, and lim f 1x2.
xS0
if x 6 1
if x Ú 1 .
Explain why x = 1 is a vertical asymptote of the graph of f and
show that the graph of f intersects the line x = 1.
QUICK CHECK ANSWERS
1. Answers will vary, but all graphs should have a vertical
asymptote at x = 2. 2. - ∞; ∞ 3. As x S -4 + , x 6 0 and
1x + 42 7 0, so x1x + 42 S 0 through negative values.
1x - 121x - 22
4. lim
= lim 1x - 12 = 1, which is not
xS2
x - 2
xS2
an infinite limit, so x = 2 is not a vertical asymptote.
➤
xS0
x
49–56. Asymptotes Use analytical methods and/or a graphing utility
to identify the vertical asymptotes (if any) of the following functions.
y
T
0
Technology Exercises
y ⫽ x1/3
y⫽
⫺2
2.5 Limits at Infinity
Limits at infinity—as opposed to infinite limits—occur when the independent variable
becomes large in magnitude. For this reason, limits at infinity determine what is called the
end behavior of a function. An application of these limits is to determine whether a system (such as an ecosystem or a large oscillating structure) reaches a steady state as time
increases.
Copyright © 2014 Pearson Education, Inc.
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