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S E C T I O N 2.7 Limits at Infinity 141
x c 0:01 c 0:001 c C 0:001 c C 0:
sin x sin c
x c
Here c D 6 and cos^ c^ D
p 3 2 ^ 0:866025.
2.7 Limits at Infinity
Preliminary Questions
1. Assume that
lim x!
f .x/ D L and lim x!L
g.x/ D 1
Which of the following statements are correct?
(a) x D L is a vertical asymptote of g.x/.
(b) y D L is a horizontal asymptote of g.x/.
(c) x D L is a vertical asymptote of f .x/.
(d) y D L is a horizontal asymptote of f .x/.
SOLUTION
(a) Because lim x!L
g.x/ D 1, x D L is a vertical asymptote of g.x/. This statement is correct.
(b) This statement is not correct.
(c) This statement is not correct.
(d) Because lim x!
f .x/ D L, y D L is a horizontal asymptote of f .x/. This statement is correct.
2. What are the following limits? (a) lim x!
x^3 (b) lim x! 1
x^3 (c) lim x! 1
x^4
SOLUTION
(a) limx!1 x^3 D 1
(b) limx! 1 x^3 D 1
(c) limx! 1 x^4 D 1
3. Sketch the graph of a function that approaches a limit as x! 1 but does not approach a limit (either finite or infinite) as
x! 1.
SOLUTION
y
x
4. What is the sign of a if f .x/ D ax^3 C x C 1 satisfies lim x! 1
f .x/ D 1?
SOLUTION Because lim x! 1
x^3 D 1 , a must be negative to have lim x! 1
f .x/ D 1.
5. What is the sign of the leading coefficient a 7 if f .x/ is a polynomial of degree 7 such that lim x! 1
f .x/ D 1?
SOLUTION The behavior of f .x/ as x! 1 is controlled by the leading term; that is, limx! 1 f .x/ D limx! 1 a 7 x^7.
Because x^7! 1 as x! 1 , a 7 must be negative to have limx! 1 f .x/ D 1.
6. Explain why lim x!
sin 1 x exists but lim x! 0
sin 1 x does not exist. What is lim x!
sin 1 x
SOLUTION As x! 1, (^) x^1! 0 , so
lim x!
sin
x
D sin 0 D 0:
On the other hand, 1 x! ˙1^ as^ x^!^0 , and as^
1 x! ˙1, sin^
1 x oscillates infinitely often. Thus
lim x! 0
sin
x
does not exist.
142 C H A P T E R 2 LIMITS
Exercises
1. What are the horizontal asymptotes of the function in Figure 1?
− 20 20 40 60 80
x
1
2
y
y =^ f ( x )
FIGURE 1
SOLUTION Because
lim x! 1
f .x/ D 1 and lim x!
f .x/ D 2;
the function f .x/ has horizontal asymptotes of y D 1 and y D 2.
2. Sketch the graph of a function f .x/ that has both y D 1 and y D 5 as horizontal asymptotes.
SOLUTION
− 1 − 10 − 5 5 10
1
2
3
4
5
y
x
3. Sketch the graph of a function f .x/ with a single horizontal asymptote y D 3.
SOLUTION
− 13
− 9
− 5
− 4 − 2 −^12
3
y
x
4. Sketch the graphs of two functions f .x/ and g.x/ that have both y D 2 and y D 4 as horizontal asymptotes but lim x!
f .x/ ¤ lim x!
g.x/.
SOLUTION
− 1 − 2
− 10 − 5 5 10
1
2
3
4
y
x
y = f ( x )
− 1 − 2
− 10 − 5 5 10
1
2
3
y
x
y = g ( x )
5. Investigate the asymptotic behavior of f .x/ D
x^3
x^3 C x
numerically and graphically:
(a) Make a table of values of f .x/ for x D ˙ 50 , ˙ 100 , ˙ 500 , ˙ 1000.
(b) Plot the graph of f .x/.
(c) What are the horizontal asymptotes of f .x/?
SOLUTION
(a) From the table below, it appears that
lim x!˙
x 3
x^3 C x
D 1:
144 C H A P T E R 2 LIMITS
9. lim x!
3x^2 C 20x
2x^4 C 3x^3
SOLUTION
lim x!
3x^2 C 20x
2x^4 C 3x^3
D lim x!
x 4 .3x^2 C 20x/
x 4 .2x^4 C 3x^3 29/
D lim x!
3 x^2
C 20
x^3 2 C 3 x
29 x^4
D
D 0:
10. lim x!
x C 5
SOLUTION
lim x!
x C 5
D lim x!
x 1 .4/
x 1 .x C 5/
D lim x!
4 x 1 C 5 x
D
D 0:
11. lim x!
7x 9
4x C 3
SOLUTION
lim x!
7x 9 4x C 3
D lim x!
x 1 .7x 9/
x 1 .4x C 3/
D lim x!
9 x 4 C 3 x
D
12. lim x!
9x 2 2 6 29x
SOLUTION
lim x!
9x 2 2 6 29x
D lim x!
x 1 .9x 2 2/ x 1 .6 29x/
D lim x!
9x 2 x 6 x
D
D 1 :
13. lim x! 1
7x 2 9 4x C 3
SOLUTION
lim x! 1
7x^2
4x C 3
D lim x! 1
x 1 .7x^2 9/
x 1 .4x C 3/
D lim x! 1
7x 9 x 4 C 3 x
D 1 :
14. lim x! 1
5x 9
4x^3 C 2x C 7
SOLUTION
lim x! 1
5x 9
4x^3 C 2x C 7
D lim x! 1
x 3 .5x 9/
x 3 .4x^3 C 2x C 7/
D lim x! 1
5 x^2
9 x^3 4 C 2 x^2
C
7 x^3
D
D 0:
15. lim x! 1
3x^3
x C 4
SOLUTION
lim x! 1
3x^3
x C 4
D lim x! 1
x 1 .3x^3 10/
x 1 .x C 4/
D lim x! 1
3x 2 10 x 1 C 4 x
D
D 1:
16. lim x! 1
2x^5 C 3x^4 31x
8x^4 31x^2 C 12
SOLUTION
lim x! 1
2x^5 C 3x^4 31x
8x^4 31x^2 C 12
D lim x! 1
x 4 .2x^5 C 3x^4 31x/
x 4 .8x^4 31x^2 C 12/
D lim x! 1
2x C 3 31 x^3 8 31 x^2
C
12 x^4
D
D 1 :
In Exercises 17–22, find the horizontal asymptotes.
17. f .x/ D
2x 2 3x 8x^2 C 8
S E C T I O N 2.7 Limits at Infinity 145
SOLUTION First calculate the limits as x! ˙1. For x! 1,
lim x!
2x^2 3x
8x^2 C 8
D lim x!
3 x 8 C 8 x^2
D
D
Similarly,
lim x! 1
2x^2 3x
8x^2 C 8
D lim x! 1
3 x 8 C 8 x^2
D
D
Thus, the horizontal asymptote of f .x/ is y D 1
18. f .x/ D
8x^3 x^2
7 C 11x 4x^4
SOLUTION First calculate the limits as x! ˙1. For x! 1,
lim x!
8x 3 x 2
7 C 11x 4x^4
D lim x!
8 x
1 x^2 7 x^4
C 11
x^3
D 0:
Similarly,
lim x! 1
8x^3 x^2
7 C 11x 4x^4
D lim x! 1
8 x
1 x^2 7 x^4
C
11 x^3
D 0:
Thus, the horizontal asymptote of f .x/ is y D 0.
19. f .x/ D
p 36x^2 C 7 9x C 4
SOLUTION For x > 0, x 1 D jx 1 j D
p x 2 , so
lim x!
p 36x^2 C 7 9x C 4
D lim x!
q 36 C 7 x^2
9 C 4 x
D
p 36 9
D
On the other hand, for x < 0, x 1 D jx 1 j D
p x 2 , so
lim x! 1
p 36x^2 C 7 9x C 4
D lim x! 1
q 36 C 7 x^2 9 C 4 x
D
p 36 9
D
Thus, the horizontal asymptotes of f .x/ are y D 2 3 and^ y^ D^
2
20. f .x/ D
p 36x^4 C 7
9x^2 C 4
SOLUTION For all x ¤ 0 , x 2 D jx 2 j D
p x 4 , so
lim x!
p 36x^4 C 7 9x^2 C 4
D lim x!
q 36 C 7 x^4 9 C 4 x^2
D
p 36 9
D
Similarly,
lim x! 1
p 36x^4 C 7 9x^2 C 4
D lim x! 1
q 36 C 7 x^4
9 C 4 x^2
D
p 36 9
D
Thus, the horizontal asymptote of f .x/ is y D 23.
21. f .t/ D
et
1 C e t
SOLUTION With
lim t!
et 1 C e t^
D
D 1
and
lim t! 1
e t
1 C e t^
D 0;
the function f .t/ has one horizontal asymptote, y D 0.
S E C T I O N 2.7 Limits at Infinity 147
30. lim t! 1
4 C 6e2t
5 9e3t
SOLUTION Because
lim t! 1
e 2t D lim t! 1
e 3t D 0;
it follows that
lim t! 1
4 C 6e2t
5 9e3t^
D
4 C 0
D
31. Determine lim x!
tan 1 x. Explain geometrically.
SOLUTION As an angle increases from 0 to 2 , its tangent x D tan approaches 1. Therefore,
lim x!
tan 1 x D
Geometrically, this means that the graph of y D tan 1 x has a horizontal asymptote at y D 2
32. Show that lim x!
p x^2 C 1 x/ D 0. Hint: Observe that
p x^2 C 1 x D
p x^2 C 1 C x
SOLUTION Rationalizing the ”numerator,” we find
p x^2 C 1 x D.
p x^2 C 1 x/
p x^2 C 1 C x p x^2 C 1 C x
D
.x 2 C 1/ x 2 p x^2 C 1 C x
D
p x^2 C 1 C x
Thus,
lim x!
p x^2 C 1 x/ D lim x!
p x^2 C 1 C x
D 0:
33. According to the Michaelis–Menten equation (Figure 7), when an enzyme is combined with a substrate of concentration s (in
millimolars), the reaction rate (in micromolars/min) is
R.s/ D
As
K C s
.A, K constants)
(a) Show, by computing lim s!
R.s/, that A is the limiting reaction rate as the concentration s approaches 1.
(b) Show that the reaction rate R.s/ attains one-half of the limiting value A when s D K.
(c) For a certain reaction, K D 1:25 mM and A D 0:1. For which concentration s is R.s/ equal to 75 % of its limiting value?
Leonor Michaelis 1875−
Maud Menten 1879−
FIGURE 2 Canadian-born biochemist Maud Menten is best known for her fundamental work on enzyme kinetics with German
scientist Leonor Michaelis. She was also an accomplished painter, clarinetist, mountain climber, and master of numerous
languages.
SOLUTION
(a) lim s!
R.s/ D lim s!
As K C s
D lim s!
A
1 C
K s
D A.
148 C H A P T E R 2 LIMITS
(b) Observe that
R.K/ D
AK
K C K
D
AK
2K
D
A
have of the limiting value.
(c) By part (a), the limiting value is 0.1, so we need to determine the value of s that satisfies
R.s/ D
0:1s
1:25 C s
D 0:075:
Solving this equation for s yields
s D
D 3:75 mM:
34. Suppose that the average temperature of the earth is T .t/ D 283 C 3.1 e 0:03t^ / kelvins, where t is the number of years
since 2000.
(a) Calculate the long-term average L D lim t!
T .t/.
(b) At what time is T .t/ within one-half a degree of its limiting value?
SOLUTION
(a) L D lim t!
T .t/ D lim t!
.283 C 3.1 e 0:03t // D 286 kelvins.
(b) We need to solve the equation
T .t/ D 283 C 3.1 e 0:03t / D 285:5:
This yields
t D
ln 6 59:73:
The average temperature of the earth will be within one-half a degree of its limiting value in roughly 2060.
In Exercises 35–42, calculate the limit.
35. lim x!
p 4x^4 C 9x 2x
SOLUTION Write
p 4x^4 C 9x 2x 2 D
p 4x^4 C 9x 2x 2
p 4x^4 C 9x C 2x 2 p 4x^4 C 9x C 2x^2
D
.4x^4 C 9x/ 4x^4 p 4x^4 C 9x C 2x^2
D
9x p 4x^4 C 9x C 2x^2
Thus,
lim x!
p 4x^4 C 9x 2x 2 / D lim x!
9x p 4x^4 C 9x C 2x^2
D 0:
36. lim x!
p 9x^3 C x x3=2/
SOLUTION Write
p 9x^3 C x x 3= D
p 9x^3 C x x 3=
p 9x^3 C x C x 3= p 9x^3 C x C x3=
D
.9x 3 C x/ x 3 p 9x^3 C x C x3=^
D
8x 3 C x p 9x^3 C x C x3=^
Thus,
lim x!
p 9x^3 C x x 3= / D lim t!
8x 3 C x p 9x^3 C x C x3=^
D 1:
37. lim x!
p x
p x C 2
150 C H A P T E R 2 LIMITS
43. Let P .n/ be the perimeter of an n-gon inscribed in a unit circle (Figure 3).
(a) Explain, intuitively, why P .n/ approaches 2 as n! 1.
(b) Show that P .n/ D 2n sin n
(c) Combine (a) and (b) to conclude that lim n!
n sin^
n
D 1.
(d) Use this to give another argument that lim ! 0
sin
D 1.
n =^6 n =^9 n =^12
FIGURE 3
SOLUTION
(a) As n! 1, the n-gon approaches a circle of radius 1. Therefore, the perimeter of the n-gon approaches the circumference of
the unit circle as n! 1. That is, P .n/! 2 as n! 1.
(b) Each side of the n-gon is the third side of an isosceles triangle with equal length sides of length 1 and angle D 2 n between the equal length sides. The length of each side of the n-gon is therefore
r
12 C 12 2 cos
n
D
r
2.1 cos
n
/ D
r
4 sin
n
D 2 sin
n
Finally,
P .n/ D 2n sin
n
(c) Combining parts (a) and (b),
lim n!
P .n/ D lim n!
2n sin
n
D 2:
Dividing both sides of this last expression by 2 yields
lim n!
n
sin
n
D 1:
(d) Let D n. Then^ ^!^0 as^ n^! 1,
n
sin
n
D
sin D
sin
and
lim n!
n
sin
n
D lim ! 0
sin
D 1:
44. Physicists have observed that Einstein’s theory of special relativity reduces to Newtonian mechanics in the limit as c! 1,
where c is the speed of light. This is illustrated by a stone tossed up vertically from ground level so that it returns to earth one
second later. Using Newton’s Laws, we find that the stone’s maximum height is h D g=8 meters (g D 9:8 m/s^2 ). According to
special relativity, the stone’s mass depends on its velocity divided by c, and the maximum height is
h.c/ D c
q c^2 =g^2 C 1=4 c^2 =g
Prove that lim c!
h.c/ D g=8.
SOLUTION Write
h.c/ D c
q c^2 =g^2 C 1=4 c 2 =g D
c
q c^2 =g^2 C 1=4 c 2 =g
(^) c
p c^2 =g^2 C 1=4 C c^2 =g
c
p c^2 =g^2 C 1=4 C c^2 =g
D
c^2 .c^2 =g^2 C 1=4/ c^4 =g^2
c
p c^2 =g^2 C 1=4 C c^2 =g
D
c^2 =
c
p c^2 =g^2 C 1=4 C c^2 =g
Thus,
lim c!
h.c/ D lim c!
c^2 =
c
p c^2 =g^2 C 1=4 C c^2 =g
D
c^2 =
2c^2 =g
D
g 8
S E C T I O N 2.7 Limits at Infinity 151
Further Insights and Challenges
45. Every limit as x! 1 can be rewritten as a one-sided limit as t! 0 C, where t D x 1 . Setting g.t/ D f .t 1 /, we have
lim x!
f .x/ D lim t! 0 C
g.t/
Show that lim x!
3x^2 x
2x^2 C 5
D lim t! 0 C
3 t
2 C 5t^2
, and evaluate using the Quotient Law.
SOLUTION Let t D x 1. Then x D t 1 , t! 0 C as x! 1, and
3x^2 x
2x^2 C 5
D
3t 2 t 1
2t 2 C 5
D
3 t
2 C 5t^2
Thus,
lim x!
3x^2 x
2x^2 C 5
D lim t! 0 C
3 t
2 C 5t^2
D
46. Rewrite the following as one-sided limits as in Exercise 45 and evaluate.
(a) lim x!
3 12x^3
4x^3 C 3x C 1
(b) lim x!
e1=x
(c) lim x!
x sin
x
(d) lim x!
ln
x C 1
x 1
SOLUTION
(a) Let t D x 1
. Then x D t 1 , t! 0 C as x! 1, and
3 12x 3
4x^3 C 3x C 1
D
3 12t 3
4t 3 C 3t 1 C 1
D
3t 3 12
4 C 3t^2 C t^3
Thus,
lim x!
3 12x^3
4x^3 C 3x C 1
D lim t! 0 C
3t^3
4 C 3t^2 C t^3
D
D 3:
(b) Let t D x 1
. Then x D t 1 , t! 0 C as x! 1, and e 1=x D e t . Thus,
lim x!
e 1=x D lim t! 0 C
e t D e 0 D 1:
(c) Let t D x 1. Then x D t 1 , t! 0 C as x! 1, and
x sin
x
D
t
sin t D
sin t
t
Thus,
lim x!
x sin
x
D lim t! 0 C
sin t t
D 1:
(d) Let t D x 1. Then x D t 1 , t! 0 C as x! 1, and
x C 1
x 1
D
t 1 C 1
t 1 1
D
1 C t
1 t
Thus,
lim x!
ln
x C 1 x 1
D lim t! 0 C
ln
1 C t 1 t
D ln 1 D 0:
47. Let G.b/ D lim x!
.1 C bx^ /1=x^ for b 0. Investigate G.b/ numerically and graphically for b D 0:2, 0:8, 2 , 3 , 5 (and additional
values if necessary). Then make a conjecture for the value of G.b/ as a function of b. Draw a graph of y D G.b/. Does G.b/ appear
to be continuous? We will evaluate G.b/ using L’Hˆopital’s Rule in Section 4.5 (see Exercise 69 in Section 4.5).
SOLUTION
(^) b D 0:2:
x 5 10 50 100
f .x/ 1.000064 1.000000 1.000000 1.
