2.5 Limits at Infinity
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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.
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For Use Only in 2013 – 2014 Pilot Program
2.5 Limits at Infinity
Limits at Infinity and Horizontal Asymptotes
lim f (x) q
y
x Consider the function f 1x2 = tan-1 x, whose domain is 1- ∞, ∞2 (Figure 2.30). As
x becomes arbitrarily large (denoted x S ∞ ), f 1x2 approaches p>2, and as x becomes
arbitrarily large in magnitude and negative (denoted x S - ∞ ), f 1x2 approaches -p>2.
These limits are expressed as
ᠬ
q
Horizontal
asymptote
f (x) tan1 x
lim tan-1 x =
x
2
97
xS∞
Horizontal
asymptote
p
2
and
p
lim tan-1 x = - .
2
x S -∞
The graph of f approaches the horizontal line y = p>2 as x S ∞ , and it approaches the
horizontal line y = -p>2 as x S - ∞. These lines are called horizontal asymptotes.
q
lim f (x) q
x ᠬ
FIGURE 2.30
DEFINITION Limits at Infinity and Horizontal Asymptotes
y
If f 1x2 becomes arbitrarily close to a finite number L for all sufficiently large and positive x, then we write
lim f (x) L
x ᠬ
L
x
lim f 1x2 = L.
xS∞
y f (x)
f (x)
៬ x
f (x)
៬
x
We say the limit of f 1x2 as x approaches infinity is L. In this case, the line y = L
is a horizontal asymptote of f (Figure 2.31). The limit at negative infinity,
lim f 1x2 = M, is defined analogously. When the limit exists, the horizontal
xS - ∞
asymptote is y = M.
M
lim f (x) M
x QUICK CHECK 1
Evaluate x>1x + 12 for x = 10, 100, and 1000. What is lim
xS∞
x
?
x +1
➤
ᠬ
FIGURE 2.31
EXAMPLE 1 Limits at infinity Evaluate the following limits.
a. lim a2 +
xS - ∞
➤ The limit laws of Theorem 2.3 and the
Squeeze Theorem apply if x S a is
replaced with x S ∞ or x S - ∞.
10
b
x2
b. lim a3 +
xS∞
3 sin x
b
1x
SOLUTION
a. As x becomes large and negative, x 2 becomes large and positive; in turn, 10>x 2
approaches 0. By the limit laws of Theorem 2.3,
y
10
x2
lim a2 +
x S -∞
10
10
b = lim 2 + lim a 2 b = 2 + 0 = 2.
x S -∞
x S -∞ x
x2
(++)++*
c
f (x) 2 equals 2
lim f (x) 2
lim f (x) 2
x x ᠬ
ᠬ
2
FIGURE 2.32
10
b is also equal to 2. Therefore, the graph of y = 2 + 10>x 2
x2
approaches the horizontal asymptote y = 2 as x S ∞ and as x S - ∞ (Figure 2.32).
Notice that lim a2 +
xS∞
y2
2
equals 0
b. The numerator of sin x> 1x is bounded between -1 and 1; therefore, for x 7 0,
x
-
1
sin x
1
…
…
.
1x
1x
1x
As x S ∞, 1x becomes arbitrarily large, which means that
lim
xS∞
-1
1
= lim
= 0.
x S ∞ 1x
1x
It follows by the Squeeze Theorem (Theorem 2.5) that lim
xS∞
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sin x
= 0.
1x
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Chapter 2 r
98
Limits
Using the limit laws of Theorem 2.3,
y
6
lim a3 +
3 sin x
f (x) 3 x
3 sin x
sin x
b = lim 3 + 3 lim a
b = 3.
xS∞
xS∞
1x
(++)+1x
+*
e
5
xS∞
equals 3
4
equals 0
3 sin x
approaches the horizontal asymptote y = 3 as x
1x
becomes large (Figure 2.33). Note that the curve intersects its asymptote infinitely
many times.
Related Exercises 9–14
The graph of y = 3 +
2
lim f (x) 3
x ៬
1
0
0
10
20
30
x
40
FIGURE 2.33
Infinite Limits at Infinity
It is possible for a limit to be both an infinite limit and a limit at infinity. This type of limit
occurs if f 1x2 becomes arbitrarily large in magnitude as x becomes arbitrarily large in
magnitude. Such a limit is called an infinite limit at infinity and is illustrated by the function f 1x2 = x 3 (Figure 2.34).
lim f (x) y
➤
3
x ᠬ
DEFINITION Infinite Limits at Infinity
1
If f 1x2 becomes arbitrarily large as x becomes arbitrarily large, then we write
f (x) x3
lim f 1x2 = ∞.
1
1
1
lim f (x) x ᠬ
FIGURE 2.34
xS∞
x
The limits lim f 1x2 = - ∞, lim f 1x2 = ∞, and lim f 1x2 = - ∞ are
xS∞
xS - ∞
defined similarly.
xS - ∞
Infinite limits at infinity tell us about the behavior of polynomials for large-magnitude
values of x. First, consider power functions f 1x2 = x n, where n is a positive integer.
Figure 2.35 shows that when n is even, lim x n = ∞, and when n is odd, lim x n = ∞
xS { ∞
xS∞
and lim x n = - ∞.
xS - ∞
x
n 0 even:
lim x n n 0 odd:
y
lim x n lim x n y
៬ 60
x
y x6
y x4
៬
x
៬ y x5
yx
7
20
y x3
40
3
2
1
20
y x2
3
2
1
1
2
3
x
FIGURE 2.35
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20
2
3
x
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2.5 Limits at Infinity
99
It follows that reciprocals of power functions f 1x2 = 1>x n = x -n, where n is a positive
integer, behave as follows:
lim
xS∞
1
= lim x -n = 0.
xn
xS - ∞
From here, it is a short step to finding the behavior of any polynomial as x S { ∞. Let
p1x2 = anx n + an - 1x n - 1 + g + a2x 2 + a1x + a0. We now write p in the equivalent
form
a0
an - 2
an - 1
+ 2 + g + n ¢.
x
x
x
"
S0
e
p1x2 = x n °an +
e
S0
S0
Notice that as x becomes large in magnitude, all the terms in p except the first term approach zero. Therefore, as x S {∞, we see that p1x2 ≈ anx n. This means that as
x S { ∞, the behavior of p is determined by the term anx n with the highest power of x.
THEOREM 2.6 Limits at Infinity of Powers and Polynomials
Let n be a positive integer and let p be the polynomial
p1x2 = anx n + an - 1x n - 1 + g + a2x 2 + a1x + a0, where an ≠ 0.
1. lim x n = ∞ when n is even.
xS { ∞
2. lim x n = ∞ and lim x n = - ∞ when n is odd.
xS∞
xS - ∞
1
3. lim n = lim x -n = 0.
S
x {∞ x
xS { ∞
4. lim p1x2 = lim a n x n = ∞ or - ∞, depending on the degree of the
xS { ∞
xS { ∞
polynomial and the sign of the leading coefficient an.
EXAMPLE 2 Limits at infinity Evaluate the limits as x S {∞ of the following
functions.
a. p1x2 = 3x 4 - 6x 2 + x - 10
b. q1x2 = -2x 3 + 3x 2 - 12
SOLUTION
a. We use the fact that the limit is determined by the behavior of the leading term:
lim 13x 4 - 6x 2 + x - 102 = lim 3 x 4 = ∞.
xS∞
b
xS∞
S∞
Similarly,
lim 13x 4 - 6x 2 + x - 102 = lim 3 x 4 = ∞.
xS - ∞
S∞
b
xS - ∞
b. Noting that the leading coefficient is negative, we have
lim 1-2x 3 + 3x 2 - 122 = lim 1-2 x 32 = - ∞
xS∞
S∞
b
xS∞
lim 1-2x 3 + 3x 2 - 122 = lim 1-2 x 32 = ∞.
xS - ∞
xS - ∞
S -∞
b
Describe the behavior
of p1x2 = -3x 3 as x S ∞ and as
x S - ∞.
lim
xS - ∞
Related Exercises 15–24
Copyright © 2014 Pearson Education, Inc.
➤
QUICK CHECK 2
1
= lim x -n = 0 and
xn
xS∞
➤
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Chapter 2 r
100
Limits
End Behavior
The behavior of polynomials as x S {∞ is an example of what is often called end behavior. Having treated polynomials, we now turn to the end behavior of rational, algebraic,
and transcendental functions.
EXAMPLE 3 End behavior of rational functions Determine the end behavior for
the following rational functions.
a. f 1x2 =
3x + 2
x2 - 1
b. g1x2 =
40x 4 + 4x 2 - 1
10x 4 + 8x 2 + 1
c. h1x2 =
x 3 - 2x + 1
2x + 4
SOLUTION
a. An effective approach for evaluating limits of rational functions at infinity is to divide
both the numerator and denominator by x n, where n is the largest power appearing in
the denominator. This strategy forces the terms corresponding to lower powers of x to
approach 0 in the limit. In this case, we divide by x 2:
approaches 0
b
3x + 2
3
2
+ 2
x
3x + 2
0
x2
x
= lim 2
lim 2
= lim
= = 0.
xS∞ x - 1
xS∞ x - 1
xS∞
1
1
1 - 2
2
x
x
approaches 0
➤ Recall that the degree of a polynomial is
xS - ∞
the highest power of x that appears.
y
f (x) 3x 2
x2 1
b. Again we divide both the numerator and denominator by the largest power appearing
in the denominator, which is x 4:
1
1
3x + 2
= 0, and thus the graph of f has the horizontal
x2 - 1
asymptote y = 0. You should confirm that the zeros of the denominator are -1
and 1, which correspond to vertical asymptotes (Figure 2.36). In this example, the
degree of the polynomial in the numerator is less than the degree of the polynomial in
the denominator.
A similar calculation gives lim
x
1
lim f (x) 0
x ᠬ
lim f (x) 0
x ᠬ
40x 4
4x 2
1
+
- 4
4
4
40x + 4x - 1
x
x
x
= lim
lim
x S ∞ 10x 4 + 8x 2 + 1
x S ∞ 10x 4
8x 2
1
+ 4 + 4
4
x
x
x
4
2
Divide the numerator and
denominator by x 4.
approaches 0 approaches 0
y
lim g(x) 4
xS∞
x 4
ᠬ
10 +
b
x ᠬ
= lim
lim g(x) 4
4
x2
8
x2
+
1
x4
1
x4
Simplify.
b
40 +
b
b
FIGURE 2.36
approaches 0 approaches 0
2
4
2
FIGURE 2.37
40x4 4x2 1
g(x) 10x4 8x2 1
2
4
x
=
40 + 0 + 0
= 4.
10 + 0 + 0
Evaluate limits.
Using the same steps (dividing each term by x 4), it can be shown that
40x 4 + 4x 2 - 1
= 4. This function has the horizontal asymptote y = 4
lim
x S - ∞ 10x 4 + 8x 2 + 1
(Figure 2.37). Notice that the degree of the polynomial in the numerator equals the
degree of the polynomial in the denominator.
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2.5 Limits at Infinity
101
c. We divide the numerator and denominator by the largest power of x appearing in the
denominator, which is x, and then take the limit:
x3
2x 1
+
x
x
x
x - 2x + 1
lim
= lim
xS∞
2x + 4
xS∞
2x 4
+
x
x
3
Divide numerator and
denominator by x.
b
b
-
x2
2
+
+
4
x
b
2
constant
= ∞.
1
x
Simplify.
b
= lim
xS∞
approaches 0
b
arbitrarily large constant
approaches 0
Take limits.
As x S ∞, all the terms in this function either approach zero or are constant—except the
x [email protected] in the numerator, which becomes arbitrarily large. Therefore, the limit of the
x 3 - 2x + 1
function does not exist. Using a similar analysis, we find that lim
= ∞.
x S -∞
2x + 4
These limits are not finite, and so the graph of the function has no horizontal asymptote.
In this case, the degree of the polynomial in the numerator is greater than the degree of
the polynomial in the denominator.
➤
Related Exercises 25–34
The conclusions reached in Example 3 can be generalized for all rational functions.
These results are summarized in Theorem 2.7 (Exercise 74).
THEOREM 2.7
End Behavior and Asymptotes of Rational Functions
p1x2
Suppose f 1x2 =
is a rational function, where
q1x2
p1x2 = amx m + am - 1x m - 1 + g + a2x 2 + a1x + a0 and
q1x2 = bnx n + bn - 1x n - 1 + g + b2x 2 + b1x + b0,
with am ≠ 0 and bn ≠ 0.
QUICK CHECK 3 Use Theorem 2.7 to
find the vertical and horizontal
10x
asymptotes of y =
.
3x - 1
a. Degree of numerator less than degree of denominator If m 6 n, then
lim f 1x2 = 0, and y = 0 is a horizontal asymptote of f .
xS { ∞
b. Degree of numerator equals degree of denominator If m = n, then
lim f 1x2 = am >bn, and y = am >bn is a horizontal asymptote of f .
xS { ∞
c. Degree of numerator greater than degree of denominator If m 7 n, then
lim f 1x2 = ∞ or - ∞, and f has no horizontal asymptote.
xS { ∞
d. Assuming that f 1x2 is in reduced form (p and q share no common factors),
vertical asymptotes occur at the zeros of q.
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Chapter 2 r
102
Limits
Although it isn’t stated explicitly, Theorem 2.7 implies that a rational function can
have at most one horizontal asymptote, and whenever there is a horizontal asymptote,
p1x2
p1x2
lim
= lim
. The same cannot be said of other functions, as the next examples
x S ∞ q1x2
x S - ∞ q1x2
show.
EXAMPLE 4 End behavior of an algebraic function Examine the end behavior of
10x 3 - 3x 2 + 8
f 1x2 =
225x 6 + x 4 + 2
.
SOLUTION The square root in the denominator forces us to revise the strategy used with ra-
tional functions. First, consider the limit as x S ∞. The highest power of the polynomial in
the denominator is 6. However, the polynomial is under a square root, so effectively, the
highest power in the denominator is 2x 6 = x 3. Dividing the numerator and denominator
by x 3, for x 7 0, the limit is evaluated as follows:
10x - 3x + 8
3
lim
xS∞
2
225x 6 + x 4 + 2
10x 3
3x 2
8
- 3 + 3
3
x
x
x
= lim
xS∞
Divide by 2x 6 = x 3.
25x 6
x4
2
+ 6 + 6
6
B x
x
x
approaches 0
approaches 0
= lim
1
x2
25 +
+
8
x3
2
x6
b
A
+
approaches 0
=
➤ Recall that
approaches 0
10
= 2.
125
Evaluate limits.
As x S - ∞, x 3 is negative, so we divide numerator and denominator by 2x 6 = -x 3
(which is positive):
x
if x Ú 0
- x 3 if x 6 0.
Because x is negative as x S - ∞, we have
2x 6 = - x 3.
10x - 3x + 8
3
3
lim
xS - ∞
2
225x + x + 2
6
4
= lim
xS - ∞
10x 3
3x 2
8
+
3
-x
-x 3
-x 3
6
25x
x
2
+ 6 + 6
6
B x
x
x
approaches 0
approaches 0
= lim
xS - ∞
6
10x3 3x2 8
f (x) 兹25x6 x4 2
= -
2
lim f (x) 2
x ᠬ
1
2
lim f (x) 2
ᠬ
FIGURE 2.38
-
1
x2
+
approaches 0
1
x 3
x
b
A
25 +
b
b
-10 +
y
x
Divide by
2x 6 = -x 3 7 0.
4
10
= -2.
125
8
x3
2
x6
Simplify.
b
Therefore,
approaches 0
Evaluate limits.
The limits reveal two asymptotes, y = 2 and y = -2. Observe that the graph crosses
both horizontal asymptotes (Figure 2.38).
Related Exercises 35–38
➤
x if x Ú 0
2x 2 = x = b
- x if x 6 0.
2x 6 = x 3 = b
Simplify.
b
xS∞
b
b
3
x
10 -
EXAMPLE 5 End behavior of transcendental functions Determine the end behavior of the following transcendental functions.
a. f 1x2 = ex and g 1x2 = e-x
b. h1x2 = ln x
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2.5 Limits at Infinity
103
SOLUTION
y
lim ex x ᠬ
f (x) ex
a. The graph of f 1x2 = ex (Figure 2.39) makes it clear that as x S ∞, ex increases without bound. All exponential functions bx with b 7 1 behave this way, because raising
a number greater than 1 to ever-larger powers produces numbers that increase without
bound. The figure also suggests that as x S - ∞, the graph of ex approaches the horizontal asymptote y = 0. This claim is confirmed analytically by recognizing that
lim ex 0
1
x = 0.
xS∞ e
lim ex = lim e-x = lim
x ᠬ
xS - ∞
x
0
x
xS∞
xS - ∞
lim e-x = 0 and lim e-x = ∞.
FIGURE 2.39
xS∞
f (x) ex
yx
lim ln x x
៬
h(x) f 1(x) ln x
1
x
1
Reflection of y e x
across line y x
x
lim ln x ៬0
xS - ∞
b. The domain of ln x is 5 x: x 7 06 , so we evaluate lim+ ln x and lim ln x to determine
xS0
xS0
It is not obvious whether the graph of ln x approaches a horizontal asymptote or
whether the function grows without bound as x S ∞. Furthermore, the numerical
evidence (Table 2.9) is inconclusive because ln x increases very slowly. The inverse
relation between ex and ln x is again useful. The fact that the domain of ex is 1 - ∞,∞2
implies that the range of ln x is also 1- ∞,∞2. Therefore, the values of ln x lie in the
interval 1- ∞, ∞2, and it follows that lim ln x = ∞.
xS∞
c. The cosine function oscillates between -1 and 1 as x approaches infinity (Figure 2.41).
Therefore, lim cos x does not exist. For the same reason, lim cos x does not exist.
xS∞
xS - ∞
Table 2.9
x
10
105
1010
1050
1099
T
∞
xS∞
end behavior. For the first limit, recall that ln x is the inverse of ex (Figure 2.40), and
the graph of ln x is a reflection of the graph of e x across the line y = x. The horizontal
asymptote 1y = 02 of ex is also reflected across y = x, becoming a vertical asymptote
1x = 02 for ln x. These observations imply that lim+ ln x = - ∞.
y
ln x
2.302
11.513
23.026
115.129
227.956
T
???
f (x) cos x
1
x
1
lim cos x does not exist.
x FIGURE 2.41
៬
lim cos x does not exist.
x ៬
Related Exercises 39–44
➤
y
FIGURE 2.40
xS∞
Therefore, lim e = ∞ and lim e = 0. Because e-x = 1>ex, it follows that
x
The end behavior of exponential and logarithmic functions are important in upcoming
work. We summarize these results in the following theorem.
End Behavior of ex, e-x, and ln x
The end behavior for ex and e-x on 1- ∞, ∞2 and ln x on 10, ∞2 is given by the
following limits:
THEOREM 2.8
QUICK CHECK 4
How do the functions
e10x and e-10x behave as x S ∞ and as
x S - ∞?
lim e x = ∞
and
lim e -x = 0
and
lim ln x = - ∞
and
xS∞
xS∞
xS0 +
Copyright © 2014 Pearson Education, Inc.
lim e x = 0,
xS - ∞
lim e -x = ∞,
xS - ∞
lim ln x = ∞.
xS∞
➤
For Use Only in 2013 – 2014 Pilot Program
Chapter 2 r
104
Limits
SECTION 2.5 EXERCISES
Review Questions
35–38. Algebraic functions Evaluate lim f 1x2 and lim f 1x2 for the
xS∞
1.
Explain the meaning of lim f 1x2 = 10.
2.
What is a horizontal asymptote?
f 1x2
Determine lim
if f 1x2 S 100,000 and g1x2 S ∞ as x S ∞.
x S ∞ g1x2
35. f 1x2 =
4.
Describe the end behavior of g1x2 = e-2x.
36. f 1x2 =
5.
Describe the end behavior of f 1x2 = - 2x 3.
6.
The text describes three cases that arise when examining the end
behavior of a rational function f 1x2 = p1x2>q1x2. Describe the
end behavior associated with each case.
3.
xS - ∞
following functions. Then give the horizontal asymptote(s) of f (if any).
xS - ∞
7.
Evaluate lim ex, lim ex, and lim e-x.
8.
Use a sketch to find the end behavior of f 1x2 = ln x.
37. f 1x2 =
4x 3 + 1
2x + 216x 6 + 1
3
2x 2 + 1
2x + 1
3
2x 6 + 8
4x 2 + 23x 4 + 1
38. f 1x2 = 4x 1 3x - 29x 2 + 1 2
Basic Skills
39–44. Transcendental functions Determine the end behavior of the
following transcendental functions by evaluating appropriate limits.
Then provide a simple sketch of the associated graph, showing asymptotes if they exist.
9–14. Limits at infinity Evaluate the following limits.
39. f 1x2 = - 3e-x
40. f 1x2 = 2x
41. f 1x2 = 1 - ln x
42. f 1x2 = ln x 43. f 1x2 = sin x
44. f 1x2 =
9.
xS - ∞
xS∞
10
lim a3 + 2 b
xS∞
x
11. lim
uS ∞
xS∞
1
10
10. lim a5 + + 2 b
x
xS∞
x
cos u
u2
12. lim
xS∞
cos x 5
x S ∞ 1x
14.
13. lim
3 + 2x + 4x 2
x2
100
sin4 x 3
b
+
x
x2
lim a5 +
xS - ∞
15–24. Infinite limits at infinity Determine the following limits.
15. lim x 12
16.
xS∞
17. lim x
-6
18.
xS∞
19. lim 13x
xS∞
21.
12
- 9x 2
7
lim 1- 3x 16 + 22
xS - ∞
23. lim 1-12x -52
xS∞
20.
22.
24.
45. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
a. The graph of a function can never cross one of its horizontal
asymptotes.
b. A rational function f can have both lim f 1x2 = L and
c. The graph of any function can have at most two horizontal
asymptotes.
lim x - 11
x S -∞
lim 13x 7 + x 22
xS - ∞
lim 2x -8
xS - ∞
lim 12x -8 + 4x 32
xS - ∞
25–34. Rational functions Evaluate lim f 1x2 and lim f 1x2 for the
xS - ∞
following rational functions. Then give the horizontal asymptote of f
(if any).
3x 2 - 7
x 2 + 5x
46–55. Horizontal and vertical asymptotes
a. Evaluate lim f 1x2 and lim f 1x2, and then identify any
xS - ∞
xSa
2
xSa
x - 4x + 3
46. f 1x2 =
x - 1
48. f 1x2 =
216x 4 + 64x 2 + x 2
2x 2 - 4
49. f 1x2 =
3x 4 + 3x 3 - 36x 2
x 4 - 25x 2 + 144
26. f 1x2 =
27. f 1x2 =
6x 2 - 9x + 8
3x 2 + 2
28. f 1x2 =
4x 2 - 7
8x + 5x + 2
29. f 1x2 =
3x 3 - 7
x 4 + 5x 2
30. f 1x2 =
x4 + 7
x5 + x2 - x
51. f 1x2 =
31. f 1x2 =
2x + 1
3x 4 - 2
32. f 1x2 =
12x 8 - 3
3x 8 - 2x 7
52. f 1x2 =
33. f 1x2 =
40x 5 + x 2
16x 4 - 2x
34. f 1x2 =
-x 3 + 1
2x + 8
53. f 1x2 =
2
xS∞
horizontal asymptotes.
b. Find the vertical asymptotes. For each vertical asymptote x = a,
evaluate lim- f 1x2 and lim+ f 1x2.
4x
20x + 1
25. f 1x2 =
xS∞
xS - ∞
x S -∞
xS∞
Further Explorations
lim f 1x2 = ∞ .
lim 3 x 11
50
e2x
47. f 1x2 =
50. f 1x2 = 16x 2 1 4x 2 - 216x 4 + 1 2
x2 - 9
x1x - 32
x - 1
x 2>3 - 1
2x 2 + 2x + 6 - 3
x - 1
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2x 3 + 10x 2 + 12x
x 3 + 2x 2
For Use Only in 2013 – 2014 Pilot Program
2.5 Limits at Infinity
54. f 1x2 =
105
61. lim+ f 1x2 = ∞ , lim- f 1x2 = - ∞ , lim f 1x2 = 1,
1 - x2 xS0
x1x + 12
xS0
lim f 1x2 = -2
xS - ∞
55. f 1x2 = 2 x - 2 x - 1 62. Asymptotes Find the vertical and horizontal asymptotes of
f 1x2 = e1>x.
56–59. End behavior for transcendental functions
56. The central branch of f 1x2 = tan x is shown in the figure.
a. Evaluate lim -tan x and
x S p>2
lim tan x. Are these
x S -p>2 +
infinite limits or limits at infinity?
b. Sketch a graph of g1x2 = tan-1x by reflecting the graph of
f over the line y = x, and use it to evaluate lim tan-1x and
xS∞
lim tan-1 x.
xS - ∞
y
xS∞
63. Asymptotes Find the vertical and horizontal asymptotes of
cos x + 21x
.
f 1x2 =
1x
Applications
64–69. Steady states If a function f represents a system that varies in
time, the existence of lim f 1t2 means that the system reaches a steady
tS ∞
state (or equilibrium). For the following systems, determine if a steady
state exists and give the steady-state value.
f (x) tan x
64. The population of a bacteria culture is given by p1t2 =
2500
.
t + 1
65. The population of a culture of tumor cells is given by p1t2 =
3500t
.
t + 1
1
q
q
1
66. The amount of drug (in milligrams) in the blood after an IV tube
is inserted is m1t2 = 20011 - 2-t2.
x
67. The value of an investment in dollars is given by v1t2 = 1000e 0.065t.
68. The population of a colony of squirrels is given by
1500
p1t2 =
.
3 + 2e-0.1t
T
57. Graph y = sec -1 x and evaluate the following limits using the
graph. Assume the domain is 5 x: x Ú 16.
a. lim sec -1 x
xS∞
b.
lim sec -1 x
xS - ∞
58. The hyperbolic cosine function, denoted cosh x, is used to model
the shape of a hanging cable (a telephone wire, for example). It is
ex + e-x
.
defined as cosh x =
2
a. Determine its end behavior by evaluating lim cosh x and
xS∞
lim cosh x.
xS - ∞
b. Evaluate cosh 0. Use symmetry and part (a) to sketch a plausible graph for y = cosh x.
59. The hyperbolic sine function is defined as sinh x =
ex - e-x
.
2
a. Determine its end behavior by evaluating lim sinh x and
xS∞
lim sinh x.
xS - ∞
b. Evaluate sinh 0. Use symmetry and part (a) to sketch a plausible graph for y = sinh x.
60–61. Sketching graphs Sketch a possible graph of a function f that
satisfies all the given conditions. Be sure to identify all vertical and
horizontal asymptotes.
60. f 1- 12 = - 2, f 112 = 2, f 102 = 0, lim f 1x2 = 1,
lim f 1x2 = - 1
69. The amplitude of an oscillator is given by a1t2 = 2 a
t + sin t
b.
t
70–73. Looking ahead to sequences A sequence is an infinite, ordered
list of numbers that is often defined by a function. For example, the
sequence 5 2, 4, 6, 8, c6 is specified by the function f 1n2 = 2n,
where n = 1, 2, 3, c . The limit of such a sequence is lim f 1n2,
nS ∞
provided the limit exists. All the limit laws for limits at infinity may
be applied to limits of sequences. Find the limit of the following
sequences, or state that the limit does not exist.
4
4 2
4
70. e 4, 2, , 1, , , cf, which is defined by f 1n2 = , for
n
3
5 3
n = 1, 2, 3, c
1 2 3
n - 1
71. e 0, , , , c f, which is defined by f 1n2 =
,
n
2 3 4
for n = 1, 2, 3, c
n2
1 4 9 16
,
72. e , , , , cf, which is defined by f 1n2 =
2 3 4 5
n + 1
for n = 1, 2, 3, c
n + 1
3 4 5
73. e 2, , , , c f, which is defined by f 1n2 =
,
4 9 16
n2
for n = 1, 2, 3, c
xS∞
xS - ∞
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Limits
Additional Exercises
T
p1x2
74. End behavior of a rational function Suppose f 1x2 =
q1x2
is a rational function, where
p1x2 = a mx m + a m - 1x m - 1 + g + a 2x 2 + a 1x + a 0,
q1x2 = b nx n + b n - 1x n - 1 + g + b 2x 2 + b 1x + b 0, a m ≠ 0,
and b n ≠ 0.
am
a. Prove that if m = n, then lim f 1x2 =
.
xS { ∞
bn
QUICK CHECK ANSWERS
1. 10>11, 100>101, 1000>1001, 1 2. p1x2 S - ∞ as
x S ∞ and p1x2 S ∞ as x S - ∞ 3. Horizontal
1
asymptote is y = 10
3 ; vertical asymptote is x = 3 .
4. lim e 10x = ∞, lim e 10x = 0, lim e -10x = 0,
b. Prove that if m 6 n, then lim f 1x2 = 0.
xS { ∞
75–76. Limits of exponentials Evaluate lim f 1x2 and lim f 1x2.
xS∞
xS∞
x S -∞
lim e -10x = ∞
Then state the horizontal asymptote(s) of f . Confirm your findings by
plotting f .
75. f 1x2 =
2e x + 3e 2x
e 2x + e 3x
76. f 1x2 =
xS - ∞
xS∞
xS - ∞
3e x + e - x
ex + e - x
2.6 Continuity
The graphs of many functions encountered in this text contain no holes, jumps, or breaks.
For example, if L = f 1t2 represents the length of a fish t years after it is hatched, then the
length of the fish changes gradually as t increases. Consequently, the graph of L = f 1t2
contains no breaks (Figure 2.42a). Some functions, however, do contain abrupt changes
in their values. Consider a parking meter that accepts only quarters and each quarter buys
15 minutes of parking. Letting c1t2 be the cost (in dollars) of parking for t minutes, the
graph of c has breaks at integer multiples of 15 minutes (Figure 2.42b).
25
1.25
Cost (dollars)
y
Length (in)
L
L ⫽ f (t)
5
1.00
y ⫽ c(t)
0.75
0.50
0.25
0
1
2
3
4
t
0
15
30
45
60
t
Time (min)
(b)
Time (yr)
(a)
FIGURE 2.42
QUICK CHECK 1
For what values of t in
10, 602 does the graph of y = c1t2 in
Figure 2.42b have a discontinuity?
➤
T
77. Subtle asymptotes Use analytical methods to identify all the
ln 19 - x 22
. Then confirm your results by
asymptotes of f 1x2 =
2e x - e - x
locating the asymptotes using a graphing calculator.
➤
Chapter 2 r
106
Informally, we say that a function f is continuous at a if the graph of f contains no
holes or breaks at a (that is, if the graph near a can be drawn without lifting the pencil). If
a function is not continuous at a, then a is a point of discontinuity.
Continuity at a Point
This informal description of continuity is sufficient for determining the continuity of simple functions, but it is not precise enough to deal with more complicated functions such as
h1x2 = W
x sin
1
x
0
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if x = 0.
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2.6 Continuity
107
It is difficult to determine whether the graph of h has a break at 0 because it oscillates rapidly as x approaches 0 (Figure 2.43). We need a better definition.
y
0.2
Is h continuous
at x 0?
0.1
0.2
x
0.2
0.1
h (x) 0.2
x sin
0
1
x
if x 0
if x 0
FIGURE 2.43
DEFINITION Continuity at a Point
A function f is continuous at a if lim f 1x2 = f 1a2. If f is not continuous at a, then a
xSa
is a point of discontinuity.
There is more to this definition than first appears. If lim f 1x2 = f 1a2, then f 1a2 and
xSa
lim f 1x2 must both exist, and they must be equal. The following checklist is helpful in
xSa
determining whether a function is continuous at a.
Continuity Checklist
In order for f to be continuous at a, the following three conditions must hold.
1. f 1a2 is defined 1a is in the domain of f 2.
2. lim f 1x2 exists.
xSa
y
3. lim f 1x2 = f 1a2 1the value of f equals the limit of f at a2.
xSa
5
y f (x)
If any item in the continuity checklist fails to hold, the function fails to be continuous at a.
From this definition, we see that continuity has an important practical consequence:
3
If f is continuous at a, then lim f 1x2 = f 1a2, and direct substitution may be used to
1
0
1
3
5
7
FIGURE 2.44
➤ In Example 1, the discontinuities at
x = 1 and x = 2 are called removable
discontinuities because they can be
removed by redefining the function
at these points (in this case f 112 = 3
and f 122 = 1). The discontinuity at
x = 3 is called a jump discontinuity.
The discontinuity at x = 5 is called an
infinite discontinuity. These terms are
discussed in Exercises 95–101.
x
evaluate lim f 1x2.
xSa
xSa
EXAMPLE 1 Points of discontinuity Use the graph of f in Figure 2.44 to identify
values of x on the interval 10, 72 at which f has a discontinuity.
SOLUTION The function f has discontinuities at x = 1, 2, 3, and 5 because the graph
contains holes or breaks at each of these locations. These claims are verified using the
continuity checklist.
r f 112 is not defined.
r f 122 = 3 and lim f 1x2 = 1. Therefore, f 122 and lim f 1x2 exist but are not equal.
xS2
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Limits
r lim f 1x2 does not exist because the left-sided limit lim- f 1x2 = 2 differs from the
xS3
xS3
right-sided limit lim+ f 1x2 = 1.
xS3
r Neither lim f 1x2 nor f 152 exists.
Related Exercises 9–12
xS5
EXAMPLE 2 Identifying discontinuities Determine whether the following functions
are continuous at a. Justify each answer using the continuity checklist.
3x 2 + 2x + 1
;
x - 1
3x 2 + 2x + 1
b. g1x2 =
;
x - 1
1
if x
x sin
x
c. h1x2 = W
0
if x
a. f 1x2 =
a = 1
a = 2
≠ 0
;a = 0
= 0
SOLUTION
a. The function f is not continuous at 1 because f 112 is undefined.
b. Because g is a rational function and the denominator is nonzero at 2, it follows by
Theorem 2.3 that lim g1x2 = g122 = 17. Therefore, g is continuous at 2.
xS2
c. By definition, h102 = 0. In Exercise 55 of Section 2.3, we used the Squeeze Theorem
1
to show that lim x sin = 0. Therefore, lim h1x2 = h102, which implies that h is
S
x
x 0
xS0
continuous at 0.
Related Exercises 13–20
➤
Chapter 2 r
➤
108
The following theorems make it easier to test various combinations of functions for
continuity at a point.
THEOREM 2.9 Continuity Rules
If f and g are continuous at a, then the following functions are also continuous at a.
Assume c is a constant and n 7 0 is an integer.
a. f + g
c. cf
e. f >g, provided g1a2 ≠ 0
b. f - g
d. fg
f. 1 f 1x22n
To prove the first result, note that if f and g are continuous at a, then lim f 1x2 = f 1a2
xSa
and lim g1x2 = g1a2. From the limit laws of Theorem 2.3, it follows that
xSa
lim 1 f 1x2 + g1x22 = f 1a2 + g1a2.
xSa
Therefore, f + g is continuous at a. Similar arguments lead to the continuity of differences, products, quotients, and powers of continuous functions. The next theorem is a
direct consequence of Theorem 2.9.
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2.6 Continuity
THEOREM 2.10
109
Polynomial and Rational Functions
a. A polynomial function is continuous for all x.
p
b. A rational function (a function of the form , where p and q are polynomials)
q
is continuous for all x for which q1x2 ≠ 0.
EXAMPLE 3 Applying the continuity theorems For what values of x is the function
20
f 1x2 =
2
6
x
x
continuous?
x 2 - 7x + 12
SOLUTION
a. Because f is rational, Theorem 2.10b implies it is continuous for all x at which the
denominator is nonzero. The denominator factors as 1x - 321x - 42, so it is zero at
x = 3 and x = 4. Therefore, f is continuous for all x except x = 3 and x = 4
(Figure 2.45).
Related Exercises 21–26
20
Continuous everywhere
except x 3 and x 4
FIGURE 2.45
➤
y
x
f (x) 2
x 7x 12
The following theorem allows us to determine when a composition of two functions is
continuous at a point. Its proof is informative and is outlined in Exercise 102.
THEOREM 2.11
Continuity of Composite Functions at a Point
If g is continuous at a and f is continuous at g1a2, then the composite function
f ∘ g is continuous at a.
Evaluate
lim 2x 2 + 9 and 2lim 1x 2 + 92.
xS4
xS4
How do these results illustrate that the
order of a function evaluation and a
limit may be switched for continuous
functions?
Theorem 2.11 is useful because it allows us to conclude that the composition of
two continuous functions is continuous at a point. For example, the composite function
3
x
a
b is continuous for all x ≠ 1. The theorem also says that under the stated condix - 1
tions on f and g, the limit of their composition is evaluated by direct substitution; that is,
lim f 1g1x22 = f 1g1a22.
➤
xSa
EXAMPLE 4 Limit of a composition Evaluate lim a
xS0
x 4 - 2x + 2 10
b .
x 6 + 2x 4 + 1
x 4 - 2x + 2
is continuous for all x because its
x 6 + 2x 4 + 1
x 4 - 2x + 2 10
denominator is always positive (Theorem 2.10b). Therefore, a 6
b , which
x + 2x 4 + 1
is the composition of the continuous function f 1x2 = x 10 and a continuous rational
function, is continuous for all x by Theorem 2.11. By direct substitution,
SOLUTION The rational function
lim a
xS0
x 4 - 2x + 2 10
04 - 2 # 0 + 2 10
b = a 6
b = 210 = 1024.
6
4
x + 2x + 1
0 + 2 # 04 + 1
Related Exercises 27–30
➤
QUICK CHECK 2
Closely related to Theorem 2.11 are two results dealing with limits of composite functions; they are used frequently in upcoming chapters. We present these two results—one a
more general version of the other—in a single theorem.
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Limits
THEOREM 2.12
Limits of Composite Functions
1. If g is continuous at a and f is continuous at g1a2, then
1
2
lim f 1g1x22 = f lim g1x2 .
xSa
xSa
2. If lim g1x2 = L and f is continuous at L, then
xSa
1
2
lim f 1g1x22 = f lim g1x2 .
xSa
xSa
Proof: The first statement follows directly from Theorem 2.11, which states that
lim f 1g1x22 = f 1g1a22. If g is continuous at a, then lim g1x2 = g1a2, and it follows that
xSa
xSa
1
2
lim f 1g1x22 = f 1g1a22 = f lim g1x2 .
xSa
b
xSa
lim g1x2
xSa
The proof of the second statement relies on the formal definition of a limit, which is discussed in Section 2.7.
Both statements of Theorem 2.12 justify interchanging the order of a limit and a function evaluation. By the second statement, the inner function of the composition needn’t be
continuous at the point of interest, but it must have a limit at that point.
EXAMPLE 5 Limits of composite functions Evaluate the following limits.
a. lim 22x 2 - 1
b. lim cos a
x S -1
xS2
x2 - 4
b
x - 2
SOLUTION
a. We show later in this section that 1x is continuous for x Ú 0. The inner function of
the composite function 22x 2 - 1 is 2x 2 - 1 and it is continuous and positive at -1.
By the first statement of Theorem 2.12,
lim 22x 2 - 1 =
lim 2x 2 - 1 = 11 = 1.
A x S -1
e
x S -1
1
b. We show later in this section that cos x is continuous at all points of its domain. The
x2 - 4
x2 - 4
inner function of the composite function cos a
b is
, which is not
x - 2
x - 2
continuous at 2. However,
lim a
xS2
1x - 221x + 22
x2 - 4
b = lim
= lim 1x + 22 = 4.
x - 2
xS2
x - 2
xS2
Therefore, by the second statement of Theorem 2.12,
lim cos a
g
xS2
x2 - 4
x2 - 4
b = cos a lim a
b b = cos 4 ≈ -0.654.
x - 2
xS2 x - 2
4
Related Exercises 31–34
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➤
Chapter 2 r
➤
110
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2.6 Continuity
y
111
Continuity on an Interval
Continuous
on [a, b)
A function is continuous on an interval if it is continuous at every point in that interval.
Consider the functions f and g whose graphs are shown in Figure 2.46. Both these functions are continuous for all x in 1a, b2, but what about the endpoints? To answer this question, we introduce the ideas of left-continuity and right-continuity.
y f (x)
DEFINITION Continuity at Endpoints
O
a
A function f is continuous from the left (or left-continuous) at a if lim- f 1x2 = f 1a2
x
b
xSa
(a)
and f is continuous from the right (or right-continuous) at a if lim+ f 1x2 = f 1a2.
xSa
y
Continuous
on (a, b]
Combining the definitions of left-continuous and right-continuous with the definition
of continuity at a point, we define what it means for a function to be continuous on an
interval.
y g(x)
DEFINITION Continuity on an Interval
O
a
A function f is continuous on an interval I if it is continuous at all points of I. If I
contains its endpoints, continuity on I means continuous from the right or left at the
endpoints.
x
b
(b)
FIGURE 2.46
To illustrate these definitions, consider again the functions in Figure 2.46. In Figure 2.46a,
f is continuous from the right at a because lim+ f 1x2 = f 1a2; but it is not continuous from
xSa
the left at b because f 1b2 is not defined. Therefore, f is continuous on the interval 3a, b2. The
behavior of the function g in Figure 2.46b is the opposite: It is continuous from the left at b,
but it is not continuous from the right at a. Therefore, g is continuous on 1a, b4 .
QUICK CHECK 3 Modify the graphs of the functions f and g in Figure 2.46 to obtain
functions that are continuous on 3a, b4 .
➤
EXAMPLE 6 Intervals of continuity Determine the intervals of continuity for
f 1x2 = e
y
SOLUTION This piecewise function consists of two polynomials that describe a parabola
10
y f (x)
x 2 + 1 if x … 0
3x + 5 if x 7 0.
and a line (Figure 2.47). By Theorem 2.10, f is continuous for all x ≠ 0. From its graph,
it appears that f is left-continuous at 0. This observation is verified by noting that
Continuous
on (0, )
lim f 1x2 = lim- 1x 2 + 12 = 1,
x S 0-
2
FIGURE 2.47
2
which means that lim- f 1x2 = f 102. However, because
Left-continuous
at x 0
2
xS0
x
lim f 1x2 = lim+ 13x + 52 = 5 ≠ f 102,
x S 0+
xS0
we see that f is not right-continuous at 0. Therefore, we can also say that f is continuous
on 1 - ∞, 04 and on 10, ∞2.
Related Exercises 35–40
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➤
Continuous
on (, 0]
xS0
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112
Chapter 2 r
Limits
Functions Involving Roots
Recall that Limit Law 7 of Theorem 2.3 states
3
4
lim 3 f 1x24 n>m = lim f 1x2 n>m,
xSa
xSa
provided f 1x2 Ú 0, for x near a, if m is even and n>m is reduced. Therefore, if m is odd
and f is continuous at a, then 3 f 1x24 n>m is continuous at a, because
3
4
lim 3 f 1x24 n>m = lim f 1x2
xSa
xSa
n>m
= 3 f 1a24 n>m.
When m is even, the continuity of 3f 1x24 n>m must be handled more carefully because
this function is defined only when f 1x2 Ú 0. Exercise 59 of Section 2.7 establishes an
important fact:
If f is continuous at a and f 1a2 7 0, then f is positive for all values of x in the
domain sufficiently close to a.
Combining this fact with Theorem 2.11 (the continuity of composite functions), it follows that 3 f 1x24 n>m is continuous at a provided f 1a2 7 0. At points where f 1a2 = 0, the
behavior of 3 f 1x24 n>m varies. Often we find that 3 f 1x24 n>m is left- or right-continuous at
that point, or it may be continuous from both sides.
THEOREM 2.13
Continuity of Functions with Roots
Assume that m and n are positive integers with no common factors. If m is an odd
integer, then 3 f 1x24 n>m is continuous at all points at which f is continuous.
If m is even, then 3 f 1x24 n>m is continuous at all points a at which f is continuous
and f 1a2 7 0.
EXAMPLE 6 Continuity with roots For what values of x are the following functions
Continuous on [3, 3]
continuous?
y
a. g1x2 = 29 - x 2
4
SOLUTION
g(x) 兹9 x2
2
3
Right-continuous
at x 3
x
Left-continuous
at x 3
xS3
that g is left-continuous at 3. Similarly, g is right-continuous at -3 because
lim + 29 - x 2 = 0 = g1-32. Therefore, g is continuous on 3 -3, 34 .
FIGURE 2.48
xS - 3
b. The polynomial x 2 - 2x + 4 is continuous for all x by Theorem 2.10a. Because f
involves an odd root (m = 3, n = 2 in Theorem 2.13), f is continuous for all x.
QUICK CHECK 4
On what interval is
continuous?
On what
f 1x2 = x
interval is f 1x2 = x 2>5 continuous?
a. The graph of g is the upper half of the circle x 2 + y 2 = 9 (which can be verified
by solving x 2 + y 2 = 9 for y). From Figure 2.48, it appears that g is continuous
on 3 -3, 34 . To verify this fact, note that g involves an even root 1m = 2, n = 1
in Theorem 2.13). If -3 6 x 6 3, then 9 - x 2 7 0 and by Theorem 2.13, g is
continuous for all x on 1-3, 32.
At the right endpoint, lim- 29 - x 2 = 0 = g132 by Limit Law 7, which implies
1>4
Related Exercises 41–50
➤
3
b. f 1x2 = 1x 2 - 2x + 422>3
➤
Continuity of Transcendental Functions
The understanding of continuity that we have developed with algebraic functions may
now be applied to transcendental functions.
Trigonometric Functions In Example 8 of Section 2.3, we used the Squeeze Theorem
to show that lim sin x = 0 and lim cos x = 1. Because sin 0 = 0 and cos 0 = 1, these
xS0
xS0
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113
limits imply that sin x and cos x are continuous at 0. The graph of y = sin x (Figure 2.49)
suggests that lim sin x = sin a for any value of a, which means that sin x is continuous
xSa
everywhere. The graph of y = cos x also indicates that cos x is continuous for all x.
Exercise 105 outlines a proof of these results.
With these facts in hand, we appeal to Theorem 2.9e to discover that the remaining trigonometric functions are continuous on their domains. For example, because
sec x = 1>cos x, the secant function is continuous for all x for which cos x ≠ 0 (for all
x except odd multiples of p>2) (Figure 2.50). Likewise, the tangent, cotangent, and cosecant functions are continuous at all points of their domains.
y ⫽ sec x
y
y
3
y ⫽ sin x
1
sin a
(a, sin a)
... sin x
⫺w
៬ sin a
⫺q
q
w
x
⫺3
As x
FIGURE 2.49
៬
f (x) 4x
43 = 4 # 4 # 4 = 64; 4-2 =
(0, 1)
x
FIGURE 2.51
sec x is continuous at all
points of its domain.
FIGURE 2.50
Exponential Functions The continuity of exponential functions of the form
f 1x2 = b x, with 0 6 b 6 1 or b 7 1, raises an important question. Consider the function
f 1x2 = 4x (Figure 2.51). Evaluating f is routine if x is rational:
y
Exponential functions are defined
for all real numbers and are
continuous on (, ), as shown
in Chapter 6
x
a
a...
1
1
1
=
; 43>2 = 243 = 8; and 4-1>3 = 3 .
16
42
24
But what is meant by 4x when x is an irrational number, such as 12? In order for
f 1x2 = 4x to be continuous for all real numbers, it must also be defined when x is an
irrational number. Providing a working definition for an expression such as 412 requires
mathematical results that don’t appear until Chapter 6. Until then, we assume without
proof that the domain of f 1x2 = b x is the set of all real numbers and that f is continuous
at all points of its domain.
Inverse Functions Suppose a function f is continuous and one-to-one on an interval I. Reflecting the graph of f through the line y = x generates the graph of f -1. The
reflection process introduces no discontinuities in the graph of f -1, so it is plausible (and
indeed, true) that f -1 is continuous on the interval corresponding to I. We state this fact
without a formal proof.
THEOREM 2.14
Continuity of Inverse Functions
If a continuous function f has an inverse on an interval I, then its inverse f -1 is
also continuous (on the interval consisting of the points f 1x2, where x is in I).
Because all the trigonometric functions are continuous on their domains, they are also
continuous when their domains are restricted for the purpose of defining inverse functions. Therefore, by Theorem 2.14, the inverse trigonometric functions are continuous at
all points of their domains.
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Limits
Logarithmic functions of the form f 1x2 = logb x are continuous at all points of
their domains for the same reason: They are inverses of exponential functions, which are
one-to-one and continuous. Collecting all these facts together, we have the following theorem.
THEOREM 2.15
Continuity of Transcendental Functions
The following functions are continuous at all points of their domains.
Trigonometric
cos x
sin x
tan x
cot x
sec x
csc x
Inverse Trigonometric
cos-1 x
sin-1 x
cot-1 x
tan-1 x
-1
csc -1 x
sec x
Exponential
ex
bx
Logarithmic
logb x
ln x
For each function listed in Theorem 2.15, we have lim f 1x2 = f 1a2, provided a is in
xSa
the domain of the function. This means that limits involving these functions may be evaluated by direct substitution at points in the domain.
EXAMPLE 7 Limits involving transcendental functions Evaluate the following
limits after determining the continuity of the functions involved.
cos2 x - 1
4
a. lim
b. lim 1 2
ln x + tan-1 x 2
x S 0 cos x - 1
xS1
SOLUTION
➤ Limits like the one in Example 7a
are denoted 0/0 and are known as
indeterminate forms, to be studied
further in Section 4.7.
a. Both cos2 x - 1 and cos x - 1 are continuous for all x by Theorems 2.9 and 2.15.
However, the ratio of these functions is continuous only when cos x - 1 ≠ 0, which
occurs when x is not an integer multiple of 2p. Note that both the numerator and denominator
cos2 x - 1
of
approach 0 as x S 0. To evaluate the limit, we factor and simplify:
cos x - 1
lim
xS0
1cos x - 121cos x + 12
cos2 x - 1
= lim
= lim 1cos x + 12
cos x - 1
xS0
cos x - 1
xS0
(where cos x - 1 may be canceled because it is nonzero as x approaches 0). The limit
on the right is now evaluated using direct substitution:
lim 1cos x + 12 = cos 0 + 1 = 2.
xS0
4
Show that
f 1x2 = 2ln x is right-continuous
at x = 1.
xS1
xS1
does not exist.
Related Exercises 51–56
➤
QUICK CHECK 5
b. By Theorem 2.15, ln x is continuous on its domain 10, ∞2. However, ln x 7 0 only
4
when x 7 1, so Theorem 2.13 implies 2
ln x is continuous on 11, ∞2. At x = 1,
4
2ln x is right-continuous (Quick Check 5). The domain of tan-1 x is all real
4
numbers, and it is continuous on 1- ∞, ∞2. Therefore, f 1x2 = 2
ln x + tan-1 x is
continuous on 31, ∞2. Because the domain of f does not include points with x 6 1,
4
4
lim- 1 2
ln x + tan-1 x 2 does not exist, which implies that lim 1 2
ln x + tan-1 x 2
We close this section with an important theorem that has both practical and theoretical uses.
The Intermediate Value Theorem
A common problem in mathematics is finding solutions to equations of the form f 1x2 = L.
Before attempting to find values of x satisfying this equation, it is worthwhile to determine whether a solution exists.
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2.6 Continuity
115
The existence of solutions is often established using a result known as the Intermediate
Value Theorem. Given a function f and a constant L, we assume L lies between f 1a2 and
f 1b2. The Intermediate Value Theorem says that if f is continuous on 3a, b4 , then the graph
of f must cross the horizontal line y = L at least once (Figure 2.52). Although this theorem
is easily illustrated, its proof goes beyond the scope of this text.
Intermediate Value Theorem
y
y
f (b)
y f (x)
f (a)
y f (x)
L
L
f (b)
f (a)
O
a
c
b
x
O
a c1
c2
c3 b
x
In (a, b), there is at least one number c such that f (c) L,
where L is between f (a) and f (b).
FIGURE 2.52
THEOREM 2.16 The Intermediate Value Theorem
f is not continuous on [a, b]...
f (b)
Suppose f is continuous on the interval 3a, b4 and L is a number strictly between
f 1a2 and f 1b2. Then there exists at least one number c in 1a, b2 satisfying f 1c2 = L.
y ⫽ f (x)
L
The importance of continuity in Theorem 2.16 is illustrated in Figure 2.53, where we
see a function f that is not continuous on 3a, b4 . For the value of L shown in the figure,
there is no value of c in 1a, b2 satisfying f 1c2 = L. The next example illustrates a practical application of the Intermediate Value Theorem.
f (a)
O
a
x
b
EXAMPLE 8 Finding an interest rate Suppose you invest $1000 in a special 5-year
... and there is no number c
in (a, b) such that f (c) ⫽ L.
savings account with a fixed annual interest rate r, with monthly compounding. The
r 60
amount of money A in the account after 5 years (60 months) is A1r2 = 1000a1 +
b .
12
Your goal is to have $1400 in the account after 5 years.
FIGURE 2.53
QUICK CHECK 6
Does the equation
f 1x2 = x 3 + x + 1 = 0 have a solution on the interval 3 -1, 14 ? Explain.
a. Use the Intermediate Value Theorem to show there is a value of r in (0, 0.08)—that is,
an interest rate between 0% and 8%—for which A1r2 = 1400.
➤
b. Use a graphing utility to illustrate your explanation in part (a), and then estimate the
interest rate required to reach your goal.
Amount of money
(dollars) after 5 years
y
SOLUTION
y ⫽ A(r)
2000
y ⫽ 1400
Interest rate that yields
$1400 after 5 years
500
r 60
b is continuous
a. As a polynomial in r (of degree 60), A1r2 = 1000a1 +
12
for all r.
Evaluating A1r2 at the endpoints of the interval 30, 0.084 , we have
A102 = 1000 and A10.082 ≈ 1489.85. Therefore,
A102 6 1400 6 A10.082,
0
0.02
0.0675
Interest rate
FIGURE 2.54
0.10
r
and it follows, by the Intermediate Value Theorem, that there is a value of r
in 10, 0.082 for which A1r2 = 1400.
b. The graphs of y = A1r2 and the horizontal line y = 1400 are shown in Figure 2.54; it
is evident that they intersect between r = 0 and r = 0.08. Solving A1r2 = 1400 algebraically or using a root finder reveals that the curve and line intersect at r ≈ 0.0675.
Therefore, an interest rate of approximately 6.75% is required for the investment to be
worth $1400 after 5 years.
Related Exercises 57–64
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116
Limits
SECTION 2.6 EXERCISES
13–20. Continuity at a point Determine whether the following
functions are continuous at a. Use the continuity checklist to
justify your answer.
Review Questions
1.
Which of the following functions are continuous for all values in
their domain? Justify your answers.
a. a1t2 = altitude of a skydiver t seconds after jumping from a
plane
b. n1t2 = number of quarters needed to park in a metered parking space for t minutes
c. T1t2 = temperature t minutes after midnight in Chicago on
January 1
d. p1t2 = number of points scored by a basketball player after
t minutes of a basketball game
2.
What does it mean for a function to be continuous on an interval?
4.
We informally describe a function f to be continuous at a if its
graph contains no holes or breaks at a. Explain why this is not an
adequate definition of continuity.
5.
2x 2 + 3x + 1
; a = 5
x 2 + 5x
14. f 1x2 =
2x 2 + 3x + 1
; a = -5
x 2 + 5x
15. f 1x2 = 1x - 2; a = 1
Give the three conditions that must be satisfied by a function to be
continuous at a point.
3.
13. f 1x2 =
16. g1x2 =
1
; a = 3
x - 3
x2 - 1
17. f 1x2 = c x - 1
3
if x ≠ 1
; a = 1
if x = 1
Complete the following sentences.
x 2 - 4x + 3
x - 3
18. f 1x2 = c
2
a. A function is continuous from the left at a if __________.
b. A function is continuous from the right at a if __________.
19. f 1x2 =
if x ≠ 3
; a = 3
if x = 3
5x - 2
; a = 4
x - 9x + 20
2
6.
Describe the points (if any) at which a rational function fails to be
continuous.
7.
What is the domain of f 1x2 = e x >x and where is f continuous?
x2 + x
20. f 1x2 = c x + 1
2
8.
Explain in words and pictures what the Intermediate Value Theorem says.
21–26. Continuity on intervals Use Theorem 2.10 to determine the
intervals on which the following functions are continuous.
Basic Skills
9–12. Discontinuities from a graph Determine the points at which
the following functions f have discontinuities. For each point, state the
conditions in the continuity checklist that are violated.
9.
10.
y
4
3
3
2
2
1
1
0
11.
4
y ⫽ f (x)
1
2
3
4
5
0
x
y
12.
5
4
3
3
2
2
1
1
1
2
3
4
1
2
3
5
x
4
5
0
22. g1x2 =
3x 2 - 6x + 7
x2 + x + 1
23. f 1x2 =
x 5 + 6x + 17
x2 - 9
24. s1x2 =
x 2 - 4x + 3
x2 - 1
25. f 1x2 =
1
x2 - 4
26. f 1t2 =
t + 2
t2 - 4
27–30. Limits of compositions Evaluate the following limits and
justify your answer.
4
3
27. lim 1x 8 - 3x 6 - 1240
b
28. lim a 5
xS0
x S 2 2x - 4x 2 - 50
x
y
4
0
y ⫽ f (x)
29. lim a
xS1
x + 5 4
b
x + 2
30. lim a
xS∞
2x + 1 3
b
x
31–34. Limits of composite functions Evaluate the following limits
and justify your answer.
5
y ⫽ f (x)
21. p1x2 = 4x 5 - 3x 2 + 1
y
5
5
if x ≠ -1
; a = -1
if x = - 1
y ⫽ f (x)
1
2
3
4
5
x 3 - 2x 2 - 8x
xS4 B
x - 4
x
31. lim
32. lim tan
sin x
33. lim ln a2
b
x
xS0
34. lim a
tS4
xS0
t - 4
2t - 2
x
216x + 1 - 1
b
1>3
35–38. Intervals of continuity Determine the intervals of continuity
for the following functions.
35. The graph of Exercise 9
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36. The graph of Exercise 10
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2.6 Continuity
37. The graph of Exercise 11
38. The graph of Exercise 12
b. Use a graph to illustrate your explanation in part (a); then
approximate the interest rate required to reach your goal.
39. Intervals of continuity Let
T
x 2 + 3x if x Ú 1
f 1x2 = e
2x
if x 6 1.
a. Use the continuity checklist to show that f is not continuous at 0.
b. Is f continuous from the left or right at 0?
c. State the interval(s) of continuity.
42. g1x2 = 2x 4 - 1
3 2
43. f 1x2 = 2
x - 2x - 3
44. f 1t2 = 1t 2 - 123>2
45. f 1x2 = 12x - 322>3
46. f 1z2 = 1z - 123>4
49. lim
xS3
1 2x 2
48.
+ 72
lim
xS - 1
50. lim
tS2
1x
2
52. f 1x2 = e 1x;
lim f 1x2;
xS4
53. f 1x2 =
1 + sin x
;
cos x
54. f 1x2 =
ln x
;
sin-1 x
x S 1-
ex
55. f 1x2 =
;
1 - ex
x S 0-
56. f 1x2 =
T
lim f 1x2;
x S p>4
- 4 + 2x - 9 2
3
2
e 2x - 1
;
ex - 1
61. x 3 - 5x 2 + 2x = - 1; 1-1, 52
62. -x 5 - 4x 2 + 21x + 5 = 0; 10, 32
1 + 2t 2 + 5
64. x ln x - 1 = 0; 11, e2
lim f 1x2
lim f 1x2
x S 4p>3
lim f 1x2
lim f 1x2;
60. 2x 4 + 25x 3 + 10 = 5; 10, 12
63. x + e x = 0; 1-1, 02
x S 2p-
f 1x2;
a. Use the Intermediate Value Theorem to show that the following
equations have a solution on the given interval.
t2 + 5
lim f 1x2
lim
59–64. Applying the Intermediate Value Theorem
59. 2x 3 + x - 2 = 0; 1-1, 12
x S 0+
x S p>2-
T
c. Illustrate your answers with an appropriate graph.
51–56. Continuity and limits with transcendental functions
Determine the interval(s) on which the following functions are
continuous; then evaluate the given limits.
51. f 1x2 = csc x;
,
b. Use a graphing utility to find all the solutions to the equation on
the given interval.
47–50. Limits with roots Determine the following limits and justify
your answers.
4x + 10
47. lim
x S 2 A 2x - 2
1 - 11 + r>122-360
a. Use the Intermediate Value Theorem to show there is a value
of r in (0.06, 0.08)—an interest rate between 6% and 8%—that
allows you to make monthly payments of $1000 per month.
b. Use a graph to illustrate your explanation to part (a). Then
determine the interest rate you need for monthly payments of
$1000.
x 3 + 4x + 1 if x … 0
if x 7 0.
2x 3
41. f 1x2 = 22x 2 - 16
150,0001r>122
where r is the annual interest rate. Suppose banks are currently
offering interest rates between 6% and 8%.
40. Intervals of continuity Let
41–46. Functions with roots Determine the interval(s) on which the
following functions are continuous. Be sure to consider right- and
left-continuity at the endpoints.
58. Intermediate Value Theorem and mortgage payments You
are shopping for a $150,000, 30-year (360-month) loan to buy a
house. The monthly payment is
m1r2 =
a. Use the continuity checklist to show that f is not continuous at 1.
b. Is f continuous from the left or right at 1?
c. State the interval(s) of continuity.
f 1x2 = e
117
Further Explorations
65. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
a. If a function is left-continuous and right-continuous at a, then
it is continuous at a.
b. If a function is continuous at a, then it is left-continuous and
right-continuous at a.
c. If a 6 b and f 1a2 … L … f 1b2, then there is some value of c
in 1a, b2 for which f 1c2 = L.
d. Suppose f is continuous on 3a, b4. Then there is a point c in
1a, b2 such that f 1c2 = 1 f 1a2 + f 1b22>2.
66. Continuity of the absolute value function Prove that the
absolute value function x is continuous for all values of x.
(Hint: Using the definition of the absolute value function,
compute lim- x and lim+ x .)
lim f 1x2
x S 0+
xS0
lim f 1x2
xS0
57. Intermediate Value Theorem and interest rates Suppose $5000
is invested in a savings account for 10 years (120 months), with
an annual interest rate of r, compounded monthly. The amount of
money in the account after 10 years is A1r2 = 500011 + r>122120.
a. Use the Intermediate Value Theorem to show there is a value
of r in (0, 0.08)—an interest rate between 0% and 8%—that
allows you to reach your savings goal of $7000 in 10 years.
xS0
67–70. Continuity of functions with absolute values Use the
continuity of the absolute value function (Exercise 66) to determine
the interval(s) on which the following functions are continuous.
67. f 1x2 = x 2 + 3x - 18 69. h1x2 = `
1
`
1x - 4
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68. g1x2 = `
x + 4
`
x2 - 4
70. h1x2 = x 2 + 2x + 5 + 1x
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Limits
71–80. Miscellaneous limits Evaluate the following limits.
cos2 x + 3 cos x + 2
cos x + 1
71. lim
xSp
sin x - 1
73. lim
x S p>2 1sin x - 1
75. lim
xS0
cos x - 1
sin2 x
72.
lim
x S 3p>2
sin2 x + 6 sin x + 5
sin2 x - 1
1
1
2 + sin u
2
74. lim
uS0
sin u
76. lim+
xS0
-1
tan x
x
78. lim
cos t
e 3t
79. lim-
x
ln x
80. lim+
x
ln x
xS1
tS ∞
xS0
T
1 - cos2 x
sin x
77. lim
xS∞
a. Determine the value of a for which g is continuous from the
left at 1.
b. Determine the value of a for which g is continuous from the
right at 1.
c. Is there a value of a for which g is continuous at 1? Explain.
86. Asymptotes of a function containing exponentials Let
2e x + 5e 3x
f 1x2 = 2x
. Evaluate lim- f 1x2, lim+ f 1x2, lim f 1x2,
xS0
xS0
x S -∞
e - e 3x
and lim f 1x2. Then give the horizontal and vertical asymptotes
xS∞
of f. Plot f to verify your results.
T
81. Pitfalls using technology The graph of the sawtooth function
y = x - : x ; , where : x ; is the greatest integer function or floor
function (Exercise 37, Section 2.2), was obtained using a graphing utility (see figure). Identify any inaccuracies appearing in the
graph and then plot an accurate graph by hand.
y ⫽ x ⫺ :x;
87. Asymptotes of a function containing exponentials Let
2e x + 10e - x
. Evaluate lim f 1x2, lim f 1x2, and
f 1x2 =
ex + e - x
xS0
x S -∞
lim f 1x2. Then give the horizontal and vertical asymptotes of f.
xS∞
Plot f to verify your results.
T
88–89. Applying the Intermediate Value Theorem Use the
Intermediate Value Theorem to verify that the following equations have
three solutions on the given interval. Use a graphing utility to find the
approximate roots.
88. x 3 + 10x 2 - 100x + 50 = 0; 1-20, 102
1.5
89. 70x 3 - 87x 2 + 32x - 3 = 0; 10, 12
Applications
⫺2
90. Parking costs Determine the intervals of continuity for the
parking cost function c introduced at the outset of this section
(see figure). Consider 0 … t … 60.
2
y
⫺0.5
82. Pitfalls using technology Graph the function f 1x2 =
a graphing window of 3- p, p4 * 30, 24 .
sin x
using
x
a. Sketch a copy of the graph obtained with your graphing device
and describe any inaccuracies appearing in the graph.
b. Sketch an accurate graph of the function. Is f continuous at 0?
sin x
c. What is the value of lim
.
S
x 0 x
1.25
Cost (dollars)
T
1.00
0.50
0.25
0
a. Sketch the graph of a function that is not continuous at 1, but is
defined at 1.
b. Sketch the graph of a function that is not continuous at 1, but
has a limit at 1.
84. An unknown constant Determine the value of the constant a for
which the function
if x ≠ - 1
if x = - 1
is continuous at -1.
85. An unknown constant Let
x2 + x
g1x2 = W a
3x + 5
15
30
45
60
t
Time (min)
83. Sketching functions
x 2 + 3x + 2
x + 1
f 1x2 = c
a
y ⫽ c(t)
0.75
if x 6 1
if x = 1
if x 7 1.
91. Investment problem Assume you invest $250 at the end of each
year for 10 years at an annual interest rate of r. The amount of money
250111 + r210 - 12
in your account after 10 years is A =
.
r
Assume your goal is to have $3500 in your account after 10 years.
a. Use the Intermediate Value Theorem to show that there is an
interest rate r in the interval 10.01, 0.102—between 1% and
10%—that allows you to reach your financial goal.
b. Use a calculator to estimate the interest rate required to reach
your financial goal.
92. Applying the Intermediate Value Theorem Suppose you park
your car at a trailhead in a national park and begin a 2-hr hike to a
lake at 7 a.m. on a Friday morning. On Sunday morning, you leave
the lake at 7 a.m. and start the 2-hr hike back to your car. Assume
the lake is 3 mi from your car. Let f 1t2 be your distance from the
car t hours after 7 a.m. on Friday morning and let g1t2 be your distance from the car t hours after 7 a.m. on Sunday morning.
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2.6 Continuity
a. Evaluate f 102, f 122, g102, and g122.
b. Let h1t2 = f 1t2 - g1t2. Find h102 and h122.
c. Use the Intermediate Value Theorem to show that there is
some point along the trail that you will pass at exactly the
same time of morning on both days.
93. The monk and the mountain A monk set out from a monastery in
the valley at dawn. He walked all day up a winding path, stopping
for lunch and taking a nap along the way. At dusk, he arrived at a
temple on the mountaintop. The next day, the monk made the return
walk to the valley, leaving the temple at dawn, walking the same
path for the entire day, and arriving at the monastery in the evening.
Must there be one point along the path that the monk occupied at
the same time of day on both the ascent and descent? (Hint: The
question can be answered without the Intermediate Value Theorem.)
(Source: Arthur Koestler, The Act of Creation.)
94. Does continuity of f imply continuity of f? Let
1
if x Ú 0
- 1 if x 6 0.
Write a formula for g1x2 .
Is g continuous at x = 0? Explain.
Is g continuous at x = 0? Explain.
For any function f, if f is continuous at a, does it necessarily follow that f is continuous at a? Explain.
95–96. Classifying discontinuities The discontinuities in graphs
(a) and (b) are removable discontinuities because they disappear if we
define or redefine f at a so that f 1a2 = lim f 1x2. The function in
xSa
graph (c) has a jump discontinuity because left and right limits exist at
a but are unequal. The discontinuity in graph (d) is an infinite discontinuity because the function has a vertical asymptote at a.
y
y
y ⫽ f (x)
y
a
(a)
T
100–101. Classifying discontinuities Classify the discontinuities in
the following functions at the given points. See Exercises 95–96.
100. f 1x2 =
O
x - 2
x 3 - 4x 2 + 4x
; x = 0 and x = 1
x1x - 12
102. Continuity of composite functions Prove Theorem 2.11: If g is
continuous at a and f is continuous at g1a2, then the composition
f ∘ g is continuous at a. (Hint: Write the definition of continuity
for f and g separately; then combine them to form the definition
of continuity for f ∘ g.)
103. Continuity of compositions
a. Find functions f and g such that each function is continuous at
0, but f ∘ g is not continuous at 0.
b. Explain why examples satisfying part (a) do not contradict
Theorem 2.11.
104. Violation of the Intermediate Value Theorem? Let
x
f 1x2 =
. Then f 1-22 = - 1 and f 122 = 1. Therefore,
x
f 1-22 6 0 6 f 122, but there is no value of c between - 2 and 2
for which f 1c2 = 0. Does this fact violate the Intermediate Value
Theorem? Explain.
xS0
x
a
(b)
xS0
y ⫽ f (x)
y ⫽ f (x)
a
(c)
x
Infinite
discontinuity
O
a
(d)
x
95. Is the discontinuity at a in graph (c) removable? Explain.
96. Is the discontinuity at a in graph (d) removable? Explain.
97–98. Removable discontinuities Show that the following functions
have a removable discontinuity at the given point. See Exercises 95–96.
97. f 1x2 =
xSa
QUICK CHECK ANSWERS
1. t = 15, 30, 45 2. Both expressions have a value of 5,
showing that lim f 1g1x22 = f 1lim g1x22. 3. Fill
xSa
in the endpoints.
O
xSa
thereby establishing that sin x is continuous for all x. (Hint: Let
h = x - a so that x = a + h and note that h S 0 as x S a.)
b. Use the identity cos 1a + h2 = cos a cos h - sin a sin h with
the fact that lim cos x = 1 to prove that lim cos x = cos a.
y
Jump
discontinuity
; x = 2
a. Use the identity sin 1a + h2 = sin a cos h + cos a sin h with
the fact that lim sin x = 0 to prove that lim sin x = sin a,
Removable
discontinuity
x
x - 2
105. Continuity of sin x and cos x
y ⫽ f (x)
Removable
discontinuity
O
a. Does the function f 1x2 = x sin 11>x2 have a removable
discontinuity at x = 0?
b. Does the function g1x2 = sin 11>x2 have a removable
discontinuity at x = 0?
4
xSa
4. 30, ∞2; 1- ∞, ∞2
5. Note that
4
lim 2ln x = 2 lim ln x = 0 and f 112 = 2
ln 1 = 0.
xS1 +
4
x S 1+
Because the limit from the right and the value of the
function at x = 1 are equal, the function is right-continuous
at x = 1. 6. The equation has a solution on the interval
3 -1, 14 because f is continuous on 3 -1, 14 and
f 1-12 6 0 6 f 112.
x 2 - 7x + 10
; x = 2
x - 2
Copyright © 2014 Pearson Education, Inc.
➤
g1x2 = e
if x ≠ 1
; x = 1
if x = 1
99. Do removable discontinuities exist? Refer to Exercises 95–96.
101. h1x2 =
Additional Exercises
a.
b.
c.
d.
x2 - 1
98. g1x2 = c 1 - x
3
119
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Chapter 2 r
120
Limits
2.7 Precise Definitions of Limits
The limit definitions already encountered in this chapter are adequate for most elementary
limits. However, some of the terminology used, such as sufficiently close and arbitrarily
large, needs clarification. The goal of this section is to give limits a solid mathematical foundation by transforming the previous limit definitions into precise mathematical
statements.
Moving Toward a Precise Definition
Assume the function f is defined for all x near a, except possibly at a. Recall that
lim f 1x2 = L means that f 1x2 is arbitrarily close to L for all x sufficiently close (but not
xSa
equal) to a. This limit definition is made precise by observing that the distance between
f 1x2 and L is f 1x2 - L and that the distance between x and a is x - a . Therefore, we
write lim f 1x2 = L if we can make f 1x2 - L arbitrarily small for any x, distinct from a,
xSa
with x - a sufficiently small. For instance, if we want f 1x2 - L to be less than 0.1,
then we must find a number d 7 0 such that
➤ The phrase for all x near a means for all
f 1x2 - L 6 0.1 whenever
x in an open interval containing a.
x - a 6 d and x ≠ a.
If, instead, we want f 1x2 - L to be less than 0.001, then we must find another number
d 7 0 such that
➤ The Greek letters d (delta) and e
(epsilon) represent small positive
numbers when discussing limits.
f 1x2 - L 6 0.001 whenever 0 6 x - a 6 d.
For the limit to exist, it must be true that for any e 7 0, we can always find a d 7 0
such that
➤ The two conditions x - a 6 d
and x ≠ a are written concisely as
0 6 x - a 6 d.
f 1x2 - L 6 e whenever 0 6 x - a 6 d.
y
EXAMPLE 1 Determining values of D from a graph Figure 2.55 shows the graph of
7
a linear function f with lim f 1x2 = 5. For each value of e 7 0, determine a value of
xS3
d 7 0 satisfying the statement
y f (x)
5
f 1x2 - 5 6 e whenever 0 6 x - 3 6 d.
lim f (x) 5
ᠬ
x 3
a. e = 1
3
b. e =
1
0
1
2
SOLUTION
1
3
5
7
FIGURE 2.55
➤ The founders of calculus, Isaac Newton
(1642–1727) and Gottfried Leibniz
(1646–1716), developed the core ideas
of calculus without using a precise
definition of a limit. It was not until the
19th century that a rigorous definition
was introduced by Louis Cauchy
(1789–1857) and later refined by Karl
Weierstrass (1815–1897).
x
a. With e = 1, we want f 1x2 to be less than 1 unit from 5, which means f 1x2 is between
4 and 6. To determine a corresponding value of d, draw the horizontal lines y = 4 and
y = 6 (Figure 2.56a). Then sketch vertical lines passing through the points where the
horizontal lines and the graph of f intersect (Figure 2.56b). We see that the vertical
lines intersect the x-axis at x = 1 and x = 5. Note that f 1x2 is less than 1 unit from
5 on the y-axis if x is within 2 units of 3 on the x-axis. So for e = 1, we let d = 2 or
any smaller positive value.
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2.7 Precise Definitions of Limits
f (x) 5 1
y
y
6
6
5
5
4
4
0
1
3
5
7
x
121
Values of x such that
f (x) 5 1
f (x) 5 1
0
1
3
5
7
x
0 x 3 2
(b)
(a)
FIGURE 2.56
➤ Once an acceptable value of d is found
satisfying the statement
f 1x2 - L 6 e whenever
0 6 x - a 6 d,
b. With e = 12, we want f 1x2 to lie within a half-unit of 5 or, equivalently, f 1x2 must lie
between 4.5 and 5.5. Proceeding as in part (a), we see that f 1x2 is within a half-unit
of 5 on the y-axis (Figure 2.57a) if x is less than 1 unit from 3 (Figure 2.57b). So for
e = 12, we let d = 1 or any smaller positive number.
any smaller positive value of d also works.
y
5.5
5
4.5
5.5
5
4.5
0
2
3
x
4
FIGURE 2.57
5Ω
3
x
4
The idea of a limit, as illustrated in Example 1, may be described in terms of a contest
between two people named Epp and Del. First, Epp picks a particular number e 7 0; then
he challenges Del to find a corresponding value of d 7 0 such that
f 1x2 - 5 6 e whenever 0 6 x - 3 6 d.
5
f (x) 5 Ω
4
0
2
Related Exercises 9–12
Values of x such that
f (x) 5 Ω
6
5Ω
f (x) 5 q
0 x 3 1
(b)
(a)
y
0
Values of x such that
f (x) 5 q
➤
f (x) 5 q
y
2
3~
3
3~
0 x 3 ~
FIGURE 2.58
4
x
(1)
To illustrate, suppose Epp chooses e = 1. From Example 1, we know that Del will satisfy (1) by choosing 0 6 d … 2. If Epp chooses e = 12, then (by Example 1) Del responds by letting 0 6 d … 1. If Epp lets e = 18, then Del chooses 0 6 d … 14
(Figure 2.58). In fact, there is a pattern: For any e 7 0 that Epp chooses, no matter how small, Del will satisfy (1) by choosing a positive value of d satisfying
0 6 d … 2e. Del has discovered a mathematical relationship: If 0 6 d … 2e and
0 6 x - 3 6 d, then f 1x2 - 5 6 e, for any e 7 0. This conversation illustrates the
general procedure for proving that lim f 1x2 = L.
xSa
Copyright © 2014 Pearson Education, Inc.
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Limits
QUICK CHECK 1
In Example 1, find a positive number d satisfying the statement
f 1x2 - 5 6
1
100
whenever 0 6 x - 3 6 d.
➤
Chapter 2 r
122
A Precise Definition
Example 1 dealt with a linear function, but it points the way to a precise definition of a
limit for any function. As shown in Figure 2.59, lim f 1x2 = L means that for any positive
xSa
number e, there is another positive number d such that
f 1x2 - L 6 e whenever 0 6 x - a 6 d.
In all limit proofs, the goal is to find a relationship between e and d that gives an admissible value of d, in terms of e only. This relationship must work for any positive value of e.
lim f (x) L
ᠬ
x a
y
L⑀
f (x)
L
... then f (x) L ⑀.
L⑀
y f (x)
a␦ x a
O
a␦
x
If 0 x a ␦...
FIGURE 2.59
DEFINITION Limit of a Function
➤ The value of d in the precise definition of
a limit depends only on e.
Assume that f 1x2 exists for all x in some open interval containing a, except possibly at
a. We say that the limit of f 1x2 as x approaches a is L, written
lim f 1x2 = L,
xSa
➤ Definitions of the one-sided limits
lim f 1x2 = L and lim- f 1x2 = L are
x S a+
xSa
if for any number e 7 0 there is a corresponding number d 7 0 such that
f 1x2 - L 6 e whenever 0 6 x - a 6 d.
discussed in Exercises 39–43.
EXAMPLE 2 Finding D for a given E using a graphing utility Let
f 1x2 = x 3 - 6x 2 + 12x - 5 and demonstrate that lim f 1x2 = 3 as follows.
xS2
For the given values of e, use a graphing utility to find a value of d 7 0 such that
f 1x2 - 3 6 e whenever 0 6 x - 2 6 d.
a. e = 1
b. e =
1
2
SOLUTION
a. The condition f 1x2 - 3 6 e = 1 implies that f 1x2 lies between 2 and 4. Using
a graphing utility, we graph f and the lines y = 2 and y = 4 (Figure 2.60). These
lines intersect the graph of f at x = 1 and at x = 3. We now sketch the vertical lines
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2.7 Precise Definitions of Limits
123
x = 1 and x = 3 and observe that f 1x2 is within 1 unit of 3 whenever x is within
1 unit of 2 on the x-axis (Figure 2.60). Therefore, with e = 1, we can choose any d
with 0 6 d … 1.
b. The condition f 1x2 - 3 6 e = 12 implies that f 1x2 lies between 2.5 and 3.5 on the
y-axis. We now find that the lines y = 2.5 and y = 3.5 intersect the graph of f at
x ≈ 1.21 and x ≈ 2.79 (Figure 2.61). Observe that if x is less than 0.79 units from
2 on the x-axis, then f 1x2 is less than a half-unit from 3 on the y-axis. Therefore, with
e = 12 we can choose any d with 0 6 d … 0.79.
y
y
5
y f (x)
y f (x)
4
3.5
f (x) 3 1
3
f (x) 3 q
3
2.5
2
1
0
1
2
3
x
0
0 x 2 1
FIGURE 2.60
1.21
2
2.79
4
0 x 2 0.79
FIGURE 2.61
Related Exercises 13–14
For the function f
given in Example 2, estimate a value of
d 7 0 satisfying f 1x2 - 3 6 0.25
whenever 0 6 x - 2 6 d.
➤
This procedure could be repeated for smaller and smaller values of e 7 0. For each
value of e, there exists a corresponding value of d, proving that the limit exists.
QUICK CHECK 2
x
The inequality 0 6 x - a 6 d means that x lies between a - d and a + d with
x ≠ a. We say that the interval 1a - d, a + d2 is symmetric about a because a is the
midpoint of the interval. Symmetric intervals are convenient, but Example 3 demonstrates
that we don’t always get symmetric intervals without a bit of extra work.
➤
EXAMPLE 3 Finding a symmetric interval Figure 2.62 shows the graph of g with
y
lim g1x2 = 3. For each value of e, find the corresponding values of d 7 0 that satisfy
the condition
xS2
6
g1x2 - 3 6 e whenever 0 6 x - 2 6 d.
y g(x)
a. e = 2
b. e = 1
c. For any given value of e, make a conjecture about the corresponding values of d that
satisfy the limit condition.
3
2
1
SOLUTION
0
1
FIGURE 2.62
2
6
x
a. With e = 2, we need a value of d 7 0 such that g1x2 is within 2 units of 3, which means
between 1 and 5, whenever x is less than d units from 2. The horizontal lines y = 1 and
y = 5 intersect the graph of g at x = 1 and x = 6. Therefore, g1x2 - 3 6 2 if x lies
in the interval 11, 62 with x ≠ 2 (Figure 2.63a). However, we want x to lie in an interval
that is symmetric about 2. We can guarantee that g1x2 - 3 6 2 only if x is less than
1 unit away from 2, on either side of 2 (Figure 2.63b). Therefore, with e = 2, we take
d = 1 or any smaller positive number.
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124
Chapter 2 r
Limits
g(x) 3 2
y
y
5
5
y g(x)
y g(x)
3
3
1
1
0
1
2
3
6
x
0
1
2
3
6
x
Symmetric interval 0 x 2 1
that guarantees g(x) 3 2
Values of x such that
g(x) 3 2
(a)
(b)
FIGURE 2.63
b. With e = 1, g1x2 must lie between 2 and 4 (Figure 2.64a). This implies that x must
be within a half-unit to the left of 2 and within 2 units to the right of 2. Therefore,
g1x2 - 3 6 1 provided x lies in the interval 11.5, 42. To obtain a symmetric interval about 2, we take d = 12 or any smaller positive number. Then we are guaranteed
that g1x2 - 3 6 1 when 0 6 x - 2 6 12 (Figure 2.64b).
y
y
4
g(x) 3 1
4
y g(x)
3
3
2
2
0
1.5
4
x
0
y g(x)
1.5
2.5
4
x
Symmetric interval 0 x 2 q
that guarantees g(x) 3 1
Values of x such that
g(x) 3 1
(b)
(a)
FIGURE 2.64
c. From parts (a) and (b), it appears that if we choose d … e>2, the limit condition is satisfied for any e 7 0.
Limit Proofs
We use the following two-step process to prove that lim f 1x2 = L.
xSa
Steps for proving that lim f 1x2 = L
xSa
➤ The first step of the limit-proving process
is the preliminary work of finding a
candidate for d. The second step verifies
that the d found in the first step actually
works.
1. Find D. Let e be an arbitrary positive number. Use the inequality f 1x2 - L 6 e
to find a condition of the form x - a 6 d, where d depends only on the
value of e.
2. Write a proof. For any e 7 0, assume 0 6 x - a 6 d and use the relationship between e and d found in Step 1 to prove that f 1x2 - L 6 e.
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➤
Related Exercises 15–18
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2.7 Precise Definitions of Limits
125
EXAMPLE 4 Limit of a linear function Prove that lim 14x - 152 = 1 using the
xS4
precise definition of a limit.
SOLUTION
Step 1: Find d. In this case, a = 4 and L = 1. Assuming e 7 0 is given, we use
14x - 152 - 1 6 e to find an inequality of the form x - 4 6 d. If
14x - 152 - 1 6 e, then
4x - 16 6 e
4 x - 4 6 e Factor 4x - 16.
e
x - 4 6 . Divide by 4 and identify d = e>4.
4
We have shown that 14x - 152 - 1 6 e implies x - 4 6 e>4. Therefore,
a plausible relationship between d and e is d = e>4. We now write the actual
proof.
Step 2: Write a proof. Let e 7 0 be given and assume 0 6 x - 4 6 d where
d = e>4. The aim is to show that 14x - 152 - 1 6 e for all x such that
0 6 x - 4 6 d. We simplify 14x - 152 - 1 and isolate the x - 4 term:
14x - 152 - 1 = 4x - 16 = 4 x - 4
(+)+*
less than d = e>4
e
6 4a b = e.
4
We have shown that for any e 7 0,
f 1x2 - L = 14x - 152 - 1 6 e whenever 0 6 x - 4 6 d,
xS4
Related Exercises 19–24
➤
provided 0 6 d … e>4. Therefore, lim 14x - 152 = 1.
Justifying Limit Laws
The precise definition of a limit is used to prove the limit laws in Theorem 2.3. Essential
in several of these proofs is the triangle inequality, which states that
x + y … x + y , for all real numbers x and y.
EXAMPLE 5 Proof of Limit Law 1 Prove that if lim f 1x2 and lim g1x2 exist, then
xSa
xSa
lim 3 f 1x2 + g1x24 = lim f 1x2 + lim g1x2.
xSa
➤ Because lim f 1x2 exists, if there exists a
xSa
d 7 0 for any given e 7 0, then there
also exists a d 7 0 for any given 2e .
xSa
xSa
SOLUTION Assume that e 7 0 is given. Let lim f 1x2 = L, which implies that there
xSa
exists a d1 7 0 such that
f 1x2 - L 6
e
2
whenever 0 6 x - a 6 d1.
Similarly, let lim g1x2 = M, which implies there exists a d2 7 0 such that
xSa
g1x2 - M 6
e
2
whenever 0 6 x - a 6 d2.
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Limits
➤ The minimum value of a and b is denoted
min 5a, b6. If x = min 5a, b6, then x is
the smaller of a and b. If a = b, then
x equals the common value of a and b.
In either case, x … a and x … b.
Let d = min 5d1, d26 and suppose 0 6 x - a 6 d. Because d … d1, it follows that
0 6 x - a 6 d1 and f 1x2 - L 6 e>2. Similarly, because d … d2, it follows that
0 6 x - a 6 d2 and g1x2 - M 6 e>2. Therefore,
0 1 f 1x2 + g1x22 - 1L + M2 0 = 0 1 f 1x2 - L2 + 1g1x2 - M2 0 Rearrange terms.
… 0 f 1x2 - L 0 + 0 g1x2 - M 0
Triangle inequality.
6
➤ Proofs of other limit laws are outlined in
Exercises 25 and 26.
We have shown that given any e 7 0, if 0 6 x - a 6 d, then
0 1 f 1x2 + g1x22 - 1L + M2 0 6 e, which implies that lim 3 f 1x2 + g1x24 =
xSa
L + M = lim f 1x2 + lim g1x2.
Related Exercises 25–28
xSa
➤ Notice that for infinite limits, N plays
the role that e plays for regular limits. It
sets a tolerance or bound for the function
values f 1x2.
e
e
+ = e.
2
2
xSa
➤
Chapter 2 r
126
Infinite Limits
In Section 2.4, we stated that lim f 1x2 = ∞ if f 1x2 grows arbitrarily large as x approaches
xSa
a. More precisely, this means that for any positive number N (no matter how large), f 1x2
is larger than N if x is sufficiently close to a but not equal to a.
DEFINITION Two-Sided Infinite Limit
The infinite limit lim f 1x2 = ∞ means that for any positive number N, there exists a
xSa
corresponding d 7 0 such that
f 1x2 7 N whenever 0 6 x - a 6 d.
As shown in Figure 2.65, to prove that lim f 1x2 = ∞, we let N represent any positive
xSa
number. Then we find a value of d 7 0, depending only on N, such that
f 1x2 7 N whenever 0 6 x - a 6 d.
This process is similar to the two-step process for finite limits.
➤ Precise definitions for lim f 1x2 = - ∞,
y
xSa
lim+ f 1x2 = - ∞, lim+ f 1x2 = ∞,
xSa
xSa
lim f 1x2 = - ∞, and lim- f 1x2 = ∞
x S a-
xSa
are given in Exercises 45–49.
f (x)
f (x) N
N
O
a␦
x
a
a␦
0 x a Values of x such that f (x) N
FIGURE 2.65
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x
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2.7 Precise Definitions of Limits
127
Steps for proving that lim f 1x2 = H
xSa
1. Find D. Let N be an arbitrary positive number. Use the statement f 1x2 7 N
to find an inequality of the form x - a 6 d, where d depends only on N.
2. Write a proof. For any N 7 0, assume 0 6 x - a 6 d and use the relationship between N and d found in Step 1 to prove that f 1x2 7 N.
EXAMPLE 6 An Infinite Limit Proof Let f 1x2 =
lim f 1x2 = ∞.
1
. Prove that
1x - 222
xS2
SOLUTION
1
7 N to find d,
1x - 222
where d depends only on N. Taking reciprocals of this inequality, it follows that
Step 1: Find d 7 0. Assuming N 7 0, we use the inequality
1
N
1
x - 2 6
.
1N
1x - 222 6
➤ Recall that 2x 2 = x .
Take the square root of both sides.
1
1
has the form x - 2 6 d if we let d =
.
1N
1N
We now write a proof based on this relationship between d and N.
The inequality x - 2 6
1
Step 2: Write a proof. Suppose N 7 0 is given. Let d =
and assume
1N
1
0 6 x - 2 6 d =
. Squaring both sides of the inequality
1N
1
x - 2 6
and taking reciprocals, we have
1N
1
N
Square both sides.
1
7 N. Take reciprocals of both sides.
1x - 222
1
We see that for any positive N, if 0 6 x - 2 6 d =
, then
1N
1
1
f 1x2 =
7 N. It follows that lim
= ∞. Note that
2
S
x
2
1x - 22
1x - 222
1
because d =
, d decreases as N increases.
1N
QUICK CHECK 3
In Example 6, if N is
increased by a factor of 100, how must
d change?
➤
Related Exercises 29–32
➤
1x - 222 6
Limits at Infinity
Precise definitions can also be written for the limits at infinity lim f 1x2 = L and
S
lim f 1x2 = L. For discussion and examples, see Exercises 50 and 51.x ∞
x S -∞
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128
Chapter 2 r
Limits
SECTION 2.7 EXERCISES
11. Determining values of D from a graph The function f in the figure satisfies lim f 1x2 = 6. Determine the largest value of d 7 0
Review Questions
1.
Suppose x lies in the interval 11, 32 with x ≠ 2. Find the smallest
positive value of d such that the inequality 0 6 x - 2 6 d
is true.
2.
Suppose f 1x2 lies in the interval 12, 62. What is the smallest value
of e such that f 1x2 - 4 6 e?
3.
Which one of the following intervals is not symmetric about x = 5?
a. 11, 92
b. 14, 62
c. 13, 82
xS3
satisfying each statement.
a. If 0 6 x - 3 6 d, then f 1x2 - 6 6 3.
b. If 0 6 x - 3 6 d, then f 1x2 - 6 6 1.
y
9
d. 14.5, 5.52
4.
Does the set 5x: 0 6 x - a 6 d6 include the point x = a?
Explain.
5.
State the precise definition of lim f 1x2 = L.
6.
Interpret f 1x2 - L 6 e in words.
7.
Suppose f 1x2 - 5 6 0.1 whenever 0 6 x 6 5. Find
all values of d 7 0 such that f 1x2 - 5 6 0.1 whenever
0 6 x - 2 6 d.
8.
Give the definition of lim f 1x2 = ∞ and interpret it using
xSa
pictures.
y ⫽ f (x)
6
xSa
1
0
Determining values of D from a graph The function f in the
figure satisfies lim f 1x2 = 5. Determine the largest value
xS4
of d 7 0 satisfying each statement.
6 d,
6 2.
6 d,
6 1.
x
6
d 7 0 satisfying each statement.
xS2
a. If 0 6 x - 2 then f 1x2 - 5 b. If 0 6 x - 2 then f 1x2 - 5 3
12. Determining values of D from a graph The function f in the
figure satisfies lim f 1x2 = 5. Determine the largest value of
Basic Skills
9.
1
a. If 0 6 x - 4 6 d, then f 1x2 - 5 6 1.
b. If 0 6 x - 4 6 d, then f 1x2 - 5 6 0.5.
y
y
8
8
y ⫽ f (x)
5
5
y ⫽ f (x)
1
0
1
1
x
2
0
10. Determining values of D from a graph The function f in the
figure satisfies lim f 1x2 = 4. Determine the largest value
xS2
T
of d 7 0 satisfying each statement.
a. If 0 6 x - 2 then f 1x2 - 4 b. If 0 6 x - 2 then f 1x2 - 4 6 d,
6 1.
6 d,
6 1>2.
1
4
8
x
13. Finding D for a given E using a graph Let f 1x2 = x 3 + 3 and
note that lim f 1x2 = 3. For each value of e, use a graphing utility
xS0
y
to find a value of d 7 0 such that f 1x2 - 3 6 e whenever
0 6 x - 0 6 d. Sketch graphs illustrating your work.
8
a. e = 1
T
y ⫽ f (x)
14. Finding D for a given E using a graph Let
g1x2 = 2x 3 - 12x 2 + 26x + 4 and note that lim g1x2 = 24.
xS2
For each value of e, use a graphing utility to find a value of d 7 0
such that g1x2 - 24 6 e whenever 0 6 x - 2 6 d. Sketch
graphs illustrating your work.
4
a. e = 1
1
0
b. e = 0.5
1
2
4
x
b. e = 0.5
15. Finding a symmetric interval The function f in the figure
satisfies lim f 1x2 = 3. For each value of e, find a value of d 7 0
xS2
such that
f 1x2 - 3 6 e whenever 0 6 x - 2 6 d.
a. e = 1
Copyright © 2014 Pearson Education, Inc.
b. e =
1
2
(2)
For Use Only in 2013 – 2014 Pilot Program
2.7 Precise Definitions of Limits
c. For any e 7 0, make a conjecture about the corresponding
value of d satisfying (2).
129
x 2 - 7x + 12
= -1
xS3
x - 3
22. lim
y
23. lim x 2 = 0 (Hint: Use the identity 2x 2 = x .)
6
24. lim 1x - 322 = 0 (Hint: Use the identity 2x 2 = x .)
xS0
xS3
25. Proof of Limit Law 2 Suppose lim f 1x2 = L and lim g1x2 = M.
y ⫽ f (x)
xSa
xSa
Prove that lim 3f 1x2 - g1x24 = L - M.
xSa
3
26. Proof of Limit Law 3 Suppose lim f 1x2 = L. Prove that
xSa
lim 3cf 1x24 = cL, where c is a constant.
1
xSa
0
1
2
27. Limit of a constant function and f 1x2 = x Give proofs of the
following theorems.
x
6
16. Finding a symmetric interval The function f in the figure satisfies lim f 1x2 = 5. For each value of e, find a value of d 7 0
xS4
such that
f 1x2 - 5 6 e whenever 0 6 x - 4 6 d.
(3)
a. e = 2
b. e = 1
c. For any e 7 0, make a conjecture about the corresponding
value of d satisfying (3).
y
a. lim c = c
for any constant c
b. lim x = a
for any constant a
xSa
xSa
28. Continuity of linear functions Prove Theorem 2.2: If f 1x2 =
mx + b, then lim f 1x2 = ma + b for constants m and b. (Hint:
xSa
For a given e 7 0, let d = e> m .) Explain why this result implies
that linear functions are continuous.
29–32. Limit proofs for infinite limits Use the precise definition of
infinite limits to prove the following limits.
7
29. lim
6
xS4
1
= ∞
1x - 422
5
31. lim a
4
xS0
1
+ 1b = ∞
x2
30. lim
1
= ∞
1x + 124
32. lim a
1
- sin xb = ∞
x4
x S -1
xS0
3
2
Further Explorations
1
33. 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 and assume lim f 1x2 = L.
0
T
1
2
3
4
5
6
7
x
2x 2 - 2
17. Finding a symmetric interval Let f 1x2 =
and note
x - 1
that lim f 1x2 = 4. For each value of e, use a graphing utility
xS1
to find a value of d 7 0 such that f 1x2 - 4 6 e whenever
0 6 x - 1 6 d.
a. e = 2
b. e = 1
c. For any e 7 0, make a conjecture about the value of d that
satisfies the preceding inequality.
x + 1 if x … 3
1
if x 7 3
2x + 2
and note that lim f 1x2 = 2. For each value of e, use a graphing
1
T
18. Finding a symmetric interval Let f 1x2 = b 31
xS3
utility to find a value of d 7 0 such that f 1x2 - 2 6 e
whenever 0 6 x - 3 6 d.
b. e = 14
a. e = 12
c. For any e 7 0, make a conjecture about the value of d that
satisfies the preceding inequality.
19–24. Limit proofs Use the precise definition of a limit to prove the
following limits.
19. lim 18x + 52 = 13
xS1
20. lim 1- 2x + 82 = 2
xS3
21. lim
xSa
we can always find an e 7 0 such that f 1x2 - L 6 e whenever 0 6 x - a 6 d.
c. The limit lim f 1x2 = L means that for any arbitrary e 7 0,
xSa
we can always find a d 7 0 such that f 1x2 - L 6 e whenever 0 6 x - a 6 d.
d. If x - a 6 d, then a - d 6 x 6 a + d.
34. Finding D algebraically Let f 1x2 = x 2 - 2x + 3.
a. For e = 0.25, find a corresponding value of d 7 0 satisfying
the statement
f 1x2 - 2 6 e whenever 0 6 x - 1 6 d.
b. Verify that lim f 1x2 = 2 as follows. For any e 7 0, find a
xS1
corresponding value of d 7 0 satisfying the statement
f 1x2 - 2 6 e whenever 0 6 x - 1 6 d.
35–38. Challenging limit proofs Use the definition of a limit to prove
the following results.
1
1
= (Hint: As x S 3, eventually the distance between x
x
3
and 3 will be less than 1. Start by assuming x - 3 6 1 and
1
1
6 .2
show
2
x
Copyright © 2014 Pearson Education, Inc.
x - 16
= 8 (Hint: Factor and simplify.)
xS4 x - 4
2
xSa
a. For a given e 7 0, there is one value of d 7 0 for which
f 1x2 - L 6 e whenever 0 6 x - a 6 d.
b. The limit lim f 1x2 = L means that given an arbitrary d 7 0,
35. lim
xS3
For Use Only in 2013 – 2014 Pilot Program
Chapter 2 r
130
Limits
x - 4
= 4 (Hint: Multiply the numerator and
1x - 2
denominator by 1x + 2.)
36. lim
xS4
37.
lim
x S 1>10
1
= 10 (Hint: To find d, you will need to bound x away
x
Additional Exercises
44. The relationship between one-sided and two-sided limits
Prove the following statements to establish the fact that
lim f 1x2 = L if and only if lim- f 1x2 = L and lim+ f 1x2 = L.
xSa
xSa
xSa
xSa
Assume f exists for all values of x near a with x 6 a. We say that
the limit of f 1 x2 as x approaches a from the left of a is L and write
lim- f 1x2 = L, if for any e 7 0 there exists d 7 0 such that
xSa
f 1x2 6 N whenever a 6 x 6 a + d.
a. Write an analogous formal definition for lim+ f 1x2 = ∞ .
xSa
b. Write an analogous formal definition for lim- f 1x2 = - ∞ .
xSa
c. Write an analogous formal definition for lim- f 1x2 = ∞ .
xSa
46–47. One-sided infinite limits Use the definitions given in Exercise 45
to prove the following infinite limits.
46.
f 1x2 - L 6 e whenever 0 6 a - x 6 d.
39. Comparing definitions Why is the last inequality in the definition of lim f 1x2 = L, namely, 0 6 x - a 6 d, replaced
xSa
lim
x S 1+
xSa
41. One-sided limit proofs Prove the following limits for
3x - 4 if x 6 0
f 1x2 = b
2x - 4 if x Ú 0.
a. lim+ f 1x2 = - 4
xS0
xS0
c. lim f 1x2 = -4
xS0
42. Determining values of D from a graph The function f in the
figure satisfies lim+ f 1x2 = 0 and lim- f 1x2 = 1. Determine a
xS2
xS2
value of d 7 0 satisfying each statement.
6
6
6
6
2
1
2
1
whenever
whenever
whenever
whenever
0
0
0
0
6
6
6
6
x
x
2
2
-
2
2
x
x
6
6
6
6
d
d
d
d
Use this definition to prove the following statements.
48. lim
xS1
-2
= -∞
1x - 122
49. lim
x S -2
-10
= -∞
1x + 224
50–51. Definition of a limit at infinity The limit at infinity
lim f 1x2 = L means that for any e 7 0, there exists N 7 0 such that
f 1x2 - L 6 e whenever x 7 N.
Use this definition to prove the following statements.
50. lim
10
= 0
x
51. lim
2x + 1
= 2
x
xS∞
xS∞
52–53. Definition of infinite limits at infinity We say that
lim f 1x2 = ∞ if for any positive number M, there is a
xS∞
corresponding N 7 0 such that
f 1x2 7 M whenever x 7 N.
y
Use this definition to prove the following statements.
6
52. lim
x
= ∞
100
53. lim
x2 + x
= ∞
x
xS∞
y ⫽ f (x)
xS∞
1
0
1
= ∞
1 - x
f 1x2 6 M whenever 0 6 x - a 6 d.
xS∞
b. lim- f 1x2 = - 4
lim
x S 1-
xSa
xSa
0 6 a - x 6 d in the definition of lim- f 1x2 = L?
47.
any negative number M there exists a d 7 0 such that
xSa
40. Comparing definitions Why is the last inequality in the definition of lim f 1x2 = L, namely, 0 6 x - a 6 d, replaced with
1
= -∞
1 - x
48–49. Definition of an infinite limit We write lim f 1x2 = - ∞ if for
with 0 6 x - a 6 d in the definition of lim+ f 1x2 = L?
0
0
1
1
xSa
d 7 0 such that
f 1x2 - L 6 e whenever 0 6 x - a 6 d.
-
xSa
45. Definition of one-sided infinite limits We say that
lim+ f 1x2 = - ∞ if for any negative number N, there exists
39–43. Precise definitions for left- and right-sided limits
Use the following definitions.
Assume f exists for all x near a with x 7 a. We say that the
limit of f 1 x2 as x approaches a from the right of a is L and write
lim f 1x2 = L, if for any e 7 0 there exists d 7 0 such that
f 1x2
f 1x2
f 1x2
f 1x2
xSa
xSa
1
1
38. lim 2 =
S
x 5 x
25
a.
b.
c.
d.
xSa
b. If lim f 1x2 = L, then lim- f 1x2 = L and lim+ f 1x2 = L.
1
1
` 6
.)
from 0. So let ` x 10
20
x S a+
xSa
a. If lim- f 1x2 = L and lim+ f 1x2 = L, then lim f 1x2 = L.
1
2
6
43. One-sided limit proof Prove that lim+ 1x = 0.
xS0
x
54. Proof of the Squeeze Theorem Assume the functions f, g, and h
satisfy the inequality f 1x2 … g1x2 … h1x2 for all values of x near a,
except possibly at a. Prove that if lim f 1x2 = lim h1x2 = L, then
xSa
xSa
lim g1x2 = L.
xSa
Copyright © 2014 Pearson Education, Inc.
For Use Only in 2013 – 2014 Pilot Program
Review Exercises
55. Limit proof Suppose f is defined for all values of x near a,
except possibly at a. Assume for any integer N 7 0, there is
another integer M 7 0 such that f 1x2 - L 6 1>N whenever
x - a 6 1>M. Prove that lim f 1x2 = L using the precise
xSa
definition of a limit.
57. Prove that lim
xS0
f 1x2 = e
56. For the following function, note that lim f 1x2 ≠ 3. Find a value
xS2
of e 7 0 for which the preceding condition for nonexistence is
satisfied.
0 if x is rational
1 if x is irrational.
Prove that lim f 1x2 does not exist for any value of a. (Hint:
xSa
Assume lim f 1x2 = L for some values of a and L and let e = 12.)
xSa
f 1x2 - L 6 e whenever 0 6 x - a 6 d.
x
does not exist.
x
58. Let
56–58. Proving that lim f 1x2 3 L Use the following definition
xSa
for the nonexistence of a limit. Assume f is defined for all values of x
near a, except possibly at a. We say that lim f 1x2 ≠ L if for some
e 7 0 there is no value of d 7 0 satisfying the condition
xSa
59. A continuity proof Suppose f is continuous at a and assume
f 1a2 7 0. Show that there is a positive number d 7 0 for which
f 1x2 7 0 for all values of x in 1a - d, a + d2. (In other words, f
is positive for all values of x in the domain sufficiently close to a.)
QUICK CHECK ANSWERS
1
1. d = 50
or smaller 2. d = 0.62 or smaller
decrease by a factor of 1100 = 10 (at least).
y
131
3. d must
➤
6
5
4
y ⫽ f (x)
3
2
1
0
CHAPTER 2
1.
1
2
3
4
x
REVIEW EXERCISES
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
x - 1
has vertical asymptotes at
a. The rational function 2
x - 1
x = -1 and x = 1.
b. Numerical or graphical methods always produce good
estimates of lim f 1x2.
2.
Estimating limits graphically Use the graph of f in the figure to
find the following values, if possible.
a. f 1-12
b. lim - f 1x2
c.
e. f 112
f. lim f 1x2
g. lim f 1x2
i.
lim+ f 1x2
xS3
x S -1
xS1
y
d. If lim f 1x2 = ∞ or lim f 1x2 = - ∞ , then lim f 1x2 does
6
xSa
xSa
not exist.
e. If lim f 1x2 does not exist, then either lim f 1x2 = ∞ or
xSa
lim f 1x2 = - ∞ .
x S -1
h. lim- f 1x2
xS2
xS3
xS3
c. The value of lim f 1x2, if it exists, is found by calculating f 1a2.
xSa
d. lim f 1x2
j. lim f 1x2
xSa
xSa
lim f 1x2
x S -1+
y ⫽ f (x)
5
xSa
4
xSa
f. If a function is continuous on the intervals 1a, b2 and 1b, c2, where
a 6 b 6 c, then the function is also continuous on 1a, c2.
g. If lim f 1x2 can be calculated by direct substitution, then f is
3
2
xSa
continuous at x = a.
1
⫺1
Copyright © 2014 Pearson Education, Inc.
1
2
3
4
x
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