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Section 12.2
Techniques for Evaluating Limits
863
12.2 Techniques for Evaluating Limits
What you should learn
• Use the dividing out
technique to evaluate
limits of functions.
• Use the rationalizing
technique to evaluate limits
of functions.
• Approximate limits of
functions graphically and
numerically.
• Evaluate one-sided limits
of functions.
• Evaluate limits of difference
quotients from calculus.
Why you should learn it
Limits can be applied in real-life
situations. For instance, in
Exercise 72 on page 872, you will
determine limits involving the
costs of making photocopies.
Dividing Out Technique
In Section 12.1, you studied several types of functions whose limits can be
evaluated by direct substitution. In this section, you will study several techniques
for evaluating limits of functions for which direct substitution fails.
Suppose you were asked to find the following limit.
x2 x 6
x→3
x3
lim
Direct substitution produces 0 in both the numerator and denominator.
32 3 6 0
Numerator is 0 when x 3.
3 3 0
Denominator is 0 when x 3.
0
0,
The resulting fraction,
has no meaning as a real number. It is called an
indeterminate form because you cannot, from the form alone, determine the
limit. By using a table, however, it appears that the limit of the function as
x → 3 is 5.
x
3.01 3.001 3.0001 3 2.9999 2.999 2.99
x2 x 6
x3
5.01 5.001 5.0001
?
4.9999 4.999 4.99
When you try to evaluate a limit of a rational function by direct substitution
0
and encounter the indeterminate form 0, you can conclude that the numerator and
denominator must have a common factor. After factoring and dividing out,
you should try direct substitution again. Example 1 shows how you can use the
dividing out technique to evaluate limits of these types of functions.
Example 1
Dividing Out Technique
Find the limit: lim
x→3
x2 x 6
.
x3
Solution
From the discussion above, you know that direct substitution fails. So, begin by
factoring the numerator and dividing out any common factors.
Michael Krasowitz/Getty Images
lim
x→3
x2 x 6
x 2x 3
lim
x→3
x3
x3
lim
x→3
x 2x 3
x3
Factor numerator.
Divide out common factor.
lim x 2
Simplify.
3 2 5
Direct substitution and simplify.
x→3
Now try Exercise 7.
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Limits and an Introduction to Calculus
The validity of the dividing out technique stems from the fact that if two
functions agree at all but a single number c, they must have identical limit
behavior at x c. In Example 1, the functions given by
f x x2 x 6
x3
gx x 2
and
agree at all values of x other than x 3. So, you can use gx to find the limit
of f x.
Consider suggesting to your students
that they try making a table of values to
estimate the limit in Example 2 before
finding it algebraically. A range of 0.9
through 1.1 with increment 0.01 is useful.
Example 2
Dividing Out Technique
Find the limit.
lim
x→1 x3
x1
x2 x 1
Solution
Begin by substituting x 1 into the numerator and denominator.
110
Numerator is 0 when x 1.
13 1 2 1 1 0
Denominator is 0 when x 1.
Because both the numerator and denominator are zero when x 1, direct
substitution will not yield the limit. To find the limit, you should factor the
numerator and denominator, divide out any common factors, and then try direct
substitution again.
lim
y
x→1
x1
x 1x 2 1
Divide out common factor.
x−1
x3 − x2 + x − 1
lim
1
x2 1
Simplify.
f is undefined
when x = 1.
1
12 1
Direct substitution
1
2
Simplify.
x→1
f (x ) =
( )
1
1,
2
12.11
x→1
x
1
Factor denominator.
lim
2
FIGURE
x1
x1
lim
x3 x 2 x 1 x→1 x 1x 2 1
2
This result is shown graphically in Figure 12.11.
Now try Exercise 9.
In Example 2, the factorization of the denominator can be obtained by
dividing by x 1 or by grouping as follows.
x3 x 2 x 1 x 2x 1 x 1
x 1x 2 1
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Section 12.2
865
Techniques for Evaluating Limits
Rationalizing Technique
Another way to find the limits of some functions is first to rationalize the
numerator of the function. This is called the rationalizing technique. Recall that
rationalizing the numerator means multiplying the numerator and denominator
by the conjugate of the numerator. For instance, the conjugate of x 4 is
x 4.
Example 3
Rationalizing Technique
Find the limit: lim
x 1 1
x
x→0
.
Solution
0
By direct substitution, you obtain the indeterminate form 0.
lim
x 1 1
x
x→0
0 1 1
0
0
0
Indeterminate form
In this case, you can rewrite the fraction by rationalizing the numerator.
x 1 1
x
y
3
2
f (x ) =
1
x
FIGURE
1
12.12
2
x 1 1
xx 1 1
Multiply.
x
xx 1 1
Simplify.
x
x x 1 1
Divide out common factor.
x 1 1
x
1
x 1 1
x 1 1
x 1 1
x0
,
Simplify.
Now you can evaluate the limit by direct substitution.
lim
−1
x+1−1
x
f is undefined
when x = 0.
(0, 12 )
x→0
x 1 1
x
lim
x→0
1
1
1
1
11 2
x 1 1
0 1 1
1
You can reinforce your conclusion that the limit is 2 by constructing a table, as
shown below, or by sketching a graph, as shown in Figure 12.12.
x
f x
0.1
0.01
0.5132
0.5013
0.001
0.5001
0
0.001
0.01
0.1
?
0.4999
0.4988
0.4881
Now try Exercise 17.
The rationalizing technique for evaluating limits is based on multiplication
by a convenient form of 1. In Example 3, the convenient form is
1
x 1 1
x 1 1
.
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Limits and an Introduction to Calculus
Using Technology
The dividing out and rationalizing techniques may not work well for finding
limits of nonalgebraic functions. You often need to use more sophisticated
analytic techniques to find limits of these types of functions.
Example 4
Approximating a Limit
Approximate the limit: lim 1 x1x.
x→0
Numerical Solution
Let f x 1 x1x. Because you are finding
the limit when x 0, use the table feature of a
graphing utility to create a table that shows the
value of f for x starting at x 0.01 and has a
step of 0.001, as shown in Figure 12.13. Because
0 is halfway between 0.001 and 0.001, use the
average of the values of f at these two
x-coordinates to estimate the limit as follows.
lim 1 x1x x→0
2.7196 2.7169
2.71825
2
The actual limit can be found algebraically to be
e 2.71828.
Graphical Solution
To approximate the limit graphically, graph the function
f x 1 x1x, as shown in Figure 12.14. Using the zoom and
trace features of the graphing utility, choose two points on the
graph of f, such as
0.00017, 2.7185
lim 1 x1x x→0
2.7185 2.7181
2.7183.
2
The actual limit can be found algebraically to be e 2.71828.
f(x) = (1 + x)1/x
−2
2
0
Now try Exercise 27.
2.7225
−0.00025
0.00025
2.7150
FIGURE 12.14
Example 5
0.00017, 2.7181
as shown in Figure 12.15. Because the x-coordinates of these two
points are equidistant from 0, you can approximate the limit to be
the average of the y-coordinates. That is,
5
FIGURE 12.13
and
FIGURE 12.15
Approximating a Limit Graphically
Approximate the limit: lim sin x.
x→0
x
Solution
2
−4
sin x
f(x) =
x
4
Direct substitution produces the indeterminate form 00. To approximate the limit,
begin by using a graphing utility to graph f x sin xx, as shown in
Figure 12.16. Then use the zoom and trace features of the graphing utility to
choose a point on each side of 0, such as 0.0012467, 0.9999997 and
0.0012467, 0.9999997. Finally, approximate the limit as the average of the
y-coordinates of these two points, lim sin xx 0.9999997. It can be shown
x→0
−2
FIGURE
12.16
algebraically that this limit is exactly 1.
Now try Exercise 31.
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Section 12.2
Techniques for Evaluating Limits
867
Te c h n o l o g y
The graphs shown in Figures 12.14 and 12.16 appear to be continuous at x 0.
However, when you try to use the trace or the value feature of a graphing utility
to determine the value of y when x 0, there is no value given. Some graphing
utilities can show breaks or holes in a graph when an appropriate viewing window
is used. Because the holes in the graphs in Figures 12.14 and 12.16 occur on the
y-axis, the holes are not visible.
One-Sided Limits
In Section 12.1, you saw that one way in which a limit can fail to exist is when
a function approaches a different value from the left side of c than it approaches
from the right side of c. This type of behavior can be described more concisely
with the concept of a one-sided limit.
lim f x L1 or f x → L1 as x → c
Limit from the left
lim f x L2 or f x → L2 as x → c
Limit from the right
x→c x→c You might wish to illustrate the concept
of one-sided limits (and why they are
necessary) with tables or graphs.
Example 6
Find the limit as x → 0 from the left and the limit as x → 0 from the right for
f x y
f(x) = 2
−2
f (x ) =
1
2x
x
−1
2
lim
x→0
2x 2.
x
Limit from the left: f x → 2 as x → 0
Because f x 2 for all x > 0, the limit from the right is
f(x) = − 2
lim
x→0
FIGURE
x
From the graph of f , shown in Figure 12.17, you can see that f x 2 for all
x < 0. Therefore, the limit from the left is
x
−1
2x.
Solution
2
1
Evaluating One-Sided Limits
2x 2.
x
Limit from the right: f x → 2 as x → 0
Now try Exercise 43.
12.17
In Example 6, note that the function approaches different limits from the left
and from the right. In such cases, the limit of f x as x → c does not exist. For
the limit of a function to exist as x → c, it must be true that both one-sided
limits exist and are equal.
Existence of a Limit
If f is a function and c and L are real numbers, then
lim f x L
x→c
if and only if both the left and right limits exist and are equal to L.
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Limits and an Introduction to Calculus
Example 7
Finding One-Sided Limits
Find the limit of f x as x approaches 1.
f x 44xx,x ,
2
x < 1
x > 1
Solution
Remember that you are concerned about the value of f near x 1 rather than at
x 1. So, for x < 1, f x is given by 4 x, and you can use direct substitution
to obtain
y
7
lim f x lim 4 x
x→1
f(x) = 4 − x, x < 1
6
x→1
41
f(x) = 4x − x 2, x > 1
5
3.
4
For x > 1, f x is given by 4x x 2, and you can use direct substitution to obtain
3
lim f x lim 4x x2
2
x→1
1
x
−2 −1
−1
FIGURE
x→1
41 12
1
2
3
5
6
3.
Because the one-sided limits both exist and are equal to 3, it follows that
12.18
lim f x 3.
x→1
The graph in Figure 12.18 confirms this conclusion.
Now try Exercise 47.
Example 8
Comparing Limits from the Left and Right
To ship a package overnight, a delivery service charges $8 for the first pound and
$2 for each additional pound or portion of a pound. Let x represent the weight of
a package and let f x represent the shipping cost. Show that the limit of f x as
x → 2 does not exist.
8, 0 < x ≤ 1
f x 10, 1 < x ≤ 2
12, 2 < x ≤ 3
Overnight Delivery
Shipping cost (in dollars)
y
Solution
13
12
11
10
9
8
7
The graph of f is shown in Figure 12.19. The limit of f x as x approaches 2 from
the left is
For 2 < x ≤ 3, f(x) = 12
lim f x 10
For 1 < x ≤ 2, f (x) = 10
x→2
whereas the limit of f x as x approaches 2 from the right is
For 0 < x ≤ 1, f(x) = 8
lim f x 12.
x
1
2
3
Weight (in pounds)
FIGURE
12.19
x→2
Because these one-sided limits are not equal, the limit of f x as x → 2 does not
exist.
Now try Exercise 69.
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Section 12.2
Techniques for Evaluating Limits
869
A Limit from Calculus
In the next section, you will study an important type of limit from calculus—the
limit of a difference quotient.
Example 9
Evaluating a Limit from Calculus
For the function given by f x x 2 1, find
lim
h→0
f 3 h f 3
.
h
Solution
Example 9 previews the derivative that
is introduced in Section 12.3.
Direct substitution produces an indeterminate form.
lim
h→0
f 3 h f 3
3 h2 1 32 1
lim
h→0
h
h
2
lim 9 6h h 1 9 1
h→0
h
Group Activity
Write a limit problem (be sure the limit
exists) and exchange it with that of a
partner. Use a numerical approach to
estimate the limit, and use an algebraic
approach to verify your estimate.
Discuss your results with your partner.
2
lim 6h h
h→0
h
0
0
By factoring and dividing out, you obtain the following.
lim
h→0
f 3 h f 3
6h h2
h6 h
lim
lim
h→0
h→0
h
h
h
lim 6 h
h→0
60
6
So, the limit is 6.
Now try Exercise 63.
Note that for any x-value, the limit of a difference quotient is an expression
of the form
lim
h→ 0
f x h f x .
h
Direct substitution into the difference quotient always produces the indeterminate
form 00 . For instance,
For a review of evaluating
difference quotients, refer to
Section 1.4.
lim
h→0
f x h f x f x 0 f x
h
0
f x f x
0
0
.
0
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Limits and an Introduction to Calculus
12.2 Exercises
VOCABULARY CHECK: Fill in the blanks.
1. To evaluate the limit of a rational function that has common factors in its numerator and denominator,
use the _______ _______ _______ .
0
2. The fraction 0 has no meaning as a real number and therefore is called an _______ _______ .
3. The limit lim f x L1 is an example of a _______ _______ .
x→c
4. The limit of a _______ _______ is an expression of the form lim
h→0
f x h f x
.
h
PREREQUISITE SKILLS REVIEW: Practice and review algebra skills needed for this section at www.Eduspace.com.
In Exercises 1–4, use the graph to determine each limit
visually (if it exists). Then identify another function that
agrees with the given function at all but one point.
2x 2 x
1. gx x
x 2 3x
2. hx x
4
−2
−2
−2
4
x
−2
(b) lim gx
(b) lim hx
(c) lim gx
(c) lim hx
x→0
11. lim
x5 32
x2
5 y 5
y
y→0
15. lim
x 3 3
x
x→0
4
2
2
2
4
17. lim
2
−4
(a) lim gx
(a) lim f x
(b) lim gx
(b) lim f x
(c) lim gx
(c) lim f x
x→1
x→2
x→1
2x 1 1
x
x→0
x
−2
x
x→0
t3 27
t3
13. lim
y
4
x→1
10. lim
x→2
x2 1
4. f x x1
6
x→1
2x2 3x 2
x2
x→3
y
−2
8. lim
t→3
x→2
x3 x
3. gx x1
−2
1 2x 3x 2
1x
x→1
t3 8
t→2 t 2
(a) lim hx
x→2
7. lim
9. lim
(a) lim gx
x→1
5x
x2 25
x→2
−6
x→0
6. lim
4
x
2
x6
x2 36
x→5
2
6
5. lim
x→6
y
y
In Exercises 5–26, find the limit (if it exists). Use a graphing
utility to verify your result graphically.
4
19. lim
x 7 2
x→3
x3
x4 1
x→1 x 1
12. lim
14. lim
7 z 7
z
z→0
16. lim
x 4 2
x
x→0
18. lim
3 x
x9
20. lim
4 18 x
x2
x→9
x→2
1
1
x1
21. lim
x→0
x
1
1
4x 4
22. lim
x→0
x
1
1
x4 4
23. lim
x→0
x
1
1
2x 2
24. lim
x→0
x
25. lim
x→0
sec x
tan x
26. lim
x→2
1 sin x
cos x
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Section 12.2
In Exercises 27–38, use a graphing utility to graph the
function and approximate the limit accurate to three
decimal places.
27. lim
x→0
e2x 1
x
29. lim x ln x
x→0
x→0
35. lim
x→1
x→0
tan x
x
34. lim
3 x
1
1x
36. lim
x→0
x→0
37. lim 1 x
2x
x→0
41. lim x→16
4 x
x 16
x→6
x 6
x6
1
45. lim 2
x→1 x 1
x→2
x→1
58. (a) lim
x→0
x
cos x
5x
40. lim
x→5 25 x2
x 2 2
x
x→0
44. lim
x2
46. lim
1
x2 1
x→1
In Exercises 59– 66, find lim
h→ 0
x ≤ 1
x > 1
2
x ≤ 0
x > 0
x→0
f x h f x
.
h
59. f x 3x 1
60. f x 5 6x
61. f x x
62. f x x 2
63. f x x 3x
64. f x 4 2x x 2
2
65. f x 1
x2
66. f x 1
x1
Free-Falling Object In Exercises 67 and 68, use the position function st 16t 2 128, which gives the height
(in feet) of a free-falling object. The velocity at time t a
seconds is given by lim [sa st] /a t.
t→a
68. Find the velocity when t 2 seconds.
69. Salary Contract A union contract guarantees a 10%
salary increase yearly for 3 years. For a current salary of
$28,000, the salary f t (in thousands of dollars) for the
next 3 years is given by
where t represents the time in years. Show that the limit of
f as t → 2 does not exist.
x < 1
x ≥ 1
2
1 cos x
x
28.00, 0 < t ≤ 1
f t 30.80, 1 < t ≤ 2
33.88, 2 < t ≤ 3
x 1,
x ≤ 2
2x 3, x > 2
2
(b) lim
67. Find the velocity when t 1 second.
x 2
x→2
sin x 2
x2
x→0
x→0
42. lim
1
x
(b) lim
x→0
x1
70. Consumer Awareness The cost of sending a package
overnight is $10.75 for the first pound and $3.95 for each
additional pound or portion of a pound. A plastic mailing
bag can hold up to 3 pounds. The cost f x of sending a
package in a plastic mailing bag is given by
In Exercises 51–56, use a graphing utility to graph the function and the equations y x and y x in the same
viewing window. Use the graph to find lim f x.
51. f x x cos x
x→0
3 x x
38. lim 1 2x1x
x→1
x→0
56. f x x cos
1 cos 2x
x
x→ 1
2x 1,
48. lim f x where f x 4x,
4x ,
49. lim f x where f x 3 x,
4x,
50. lim f x where f x x 4,
47. lim f x where f x 1
55. f x x sin
x
57. (a) lim x 2 sin x 2
In Exercises 43–50, graph the function. Determine the limit
(if it exists) by evaluating the corresponding one-sided
limits.
43. lim
54. f x x cos x
sin 3x
x
Graphical, Numerical, and Algebraic Analysis In Exercises
39– 42, (a) graphically approximate the limit (if it exists)
by using a graphing utility to graph the function,
(b) numerically approximate the limit (if it exists) by using
the table feature of a graphing utility to create a table, and
(c) algebraically evaluate the limit (if it exists) by the
appropriate technique(s).
x1
39. lim 2
x→1 x 1
53. f x x sin x
In Exercises 57 and 58, state which limit can be evaluated
by using direct substitution. Then evaluate or approximate
each limit.
30. lim x2 ln x
32. lim
x→0
33. lim
x→0
sin 2x
x
31. lim
1 ex
x
28. lim
871
Techniques for Evaluating Limits
52. f x x sin x
10.75, 0 < x ≤ 1
f x 14.70, 1 < x ≤ 2
18.65, 2 < x ≤ 3
where x represents the weight of the package (in pounds).
Show that the limit of f as x → 1 does not exist.
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Limits and an Introduction to Calculus
71. Communications The cost of a telephone call between
two cities is $0.75 for the first minute and $0.50 for each
additional minute or portion of a minute. A model for the
cost C is given by Ct 0.75 0.50 t 1 where t
is the time in minutes. (Recall from Section 1.6 that
f x x the greatest integer less than or equal to x.)
(a) Sketch the graph of C for 0 < t ≤ 5.
(b) Complete the table and observe the behavior of C as t
approaches 3.5. Use the graph from part (a) and the
table to find lim Ct.
t→3.5
3
t
3.3
3.4
C
3.5
3.6
3.7
4
?
74. If f c L, then lim f x L.
x→c
75. Think About It
(a) Sketch the graph of a function for which f 2 is defined
but for which the limit of f x as x approaches 2 does
not exist.
(b) Sketch the graph of a function for which the limit of
f x as x approaches 1 is 4 but for which f 1 4.
76. Writing Consider the limit of the rational function given
by pxqx. What conclusion can you make if direct
substitution produces each expression? Write a short
paragraph explaining your reasoning.
(a) lim
px 0
qx 1
(b) lim
px 1
qx 1
(c) lim
px 1
qx 0
(d) lim
px 0
qx 0
x→c
(c) Complete the table and observe the behavior of C as t
approaches 3. Does the limit of Ct as t approaches 3
exist? Explain.
x→c
x→c
x→c
Skills Review
t
2
2.5
2.9
C
3
3.1
3.5
4
?
78. Write an equation of the line that passes through 1, 1
and is parallel to the line that passes through 3, 3 and
5, 2.
Model It
72. Consumer Awareness The cost C (in dollars) of
making x photocopies at a copy shop is given by the
function
0.15x, 0 < x ≤ 25
0.10x, 25 < x ≤ 100
Cx 0.07x, 100 < x ≤ 500
0.05x, x > 500
x→15
x→99
(iii) lim Cx
x→305
(c) Create a table of values to show numerically that
each limit does not exist.
(i) lim Cx
x→25
(ii) lim Cx
x→100
80. Arc Length Find the length of the arc on a circle with a
radius of 26 millimeters intercepted by a central angle of
101.
82. Circular Sector Find the area of the sector of a circle
with a radius of 3.4 feet and central angle 81.
(b) Find each limit and interpret your result in the
context of the situation.
(ii) lim Cx
79. Arc Length Find the length of the arc on a circle with a
radius of 8.5 inches intercepted by a central angle of 45.
81. Circular Sector Find the area of the sector of a circle
with a radius of 9 centimeters and central angle 135.
(a) Sketch a graph of the function.
(i) lim Cx
77. Write an equation of the line that passes through 6, 10
and is perpendicular to the line that passes through 4, 6
and 3, 4.
(iii) lim Cx
x→500
(d) Explain how you can use the graph in part (a) to
verify that the limits in part (c) do not exist.
In Exercises 83– 88, identify the type of conic algebraically.
Then use a graphing utility to graph the conic.
83. r 3
1 cos 84. r 12
3 2 sin 85. r 9
2 3 cos 86. r 4
4 cos 87. r 5
1 sin 88. r 6
3 4 sin Synthesis
In Exercises 89–92, determine whether the vectors are
orthogonal, parallel, or neither.
True or False? In Exercises 73 and 74, determine whether
the statement is true or false. Justify your answer.
89. 7, 2, 3, 1, 4, 5
73. When your attempt to find the limit of a rational function
yields the indeterminate form 00, the rational function’s
numerator and denominator have a common factor.
91. 4, 3, 6, 12, 9, 18
90. 5, 5, 0, 0, 5, 1
92. 2, 3, 1, 2, 2, 2