Homework 9 Solutions
§4.4 # 11 Find the limit. Use Hospital’s Rule where appropriate. If there is more
elementary method,consider using it.If l’ Hospital’s Rule doesn’t apply, explain
why.
This limit has the form 00 .
limx→(π/2)+
cos x
1−sin x
=LH limx→(π/2)+
− sin x
− cos x
= limx→(π/2)+ tan x = −∞.
§4.4 # 49 Find the limit. Use Hospital’s Rule where appropriate. If there is more
elementary method, consider using it. If l’ Hospital’s Rule doesn’t apply, explain
why.
x
1
x−1 − ln x =
x( x1 )+ln x−1
=LH limx→1 (x−1)(
1
x )+ln x
limx→1
ln x−(x−1)
limx→1 x (x−1)
ln x
ln x
= limx→1 1−(+1 )+ln
=LH limx→1
x
x
1
x
1
+ x1
x2
·
x2
x2
x
= limx→1 1+x
=
1/2.
§4.4 # 58 Find the limit.Use Hospital’s Rule where appropriate.If there is more
elementary method,consider using it.If l’ Hospital’s Rule doesn’t apply,explain why.
bx
limx→+∞ 1 + xa
bx
Set y = 1 + xa
and note ln y = bx ln(1 + xa ).
Thus limx→∞ ln y =
limx→∞
1
a ) (−a)
(1+ x
−b−1
limx→∞ bx ln(1+ xa )
=
ln(1+ a )
limx→∞ b−1 x−1x
LH
=
= ab.
Thus limx→+∞ 1 +
a bx
x
= limx→+∞ y = limx→+∞ eln y = eab .
1
limx→∞
1
−2
)
a ) (−ax
(1+ x
−b−1 x−2
=
§4.5 # 51 Use the guidelines of this section to sketch the curve. y = xe−1/x
Let f (x) = xe−1/x .
S
A. D = (−∞, 0) (0, ∞).
B. No intercept.
C. No symmetry.
D. limx→0−
e−1/x
1/x
=LH limx→0−
e−1/x (1/x2 )
−1/x2
= − limx→0− e−1/x = −∞, so x = 0 is a
V.A. Also, limx→0+ xe−1/x = 0, so the graph approaches the origin as x → 0+ .
E. f 0 (x) = xe−1/x ( x12 ) + e−1/x (1) = e−1/x ( x1 + 1) =
x+1
xe1/x
> 0 ⇔ x < −1 or x > 0,
so f is increasing on (−∞, −1) and (0, ∞), and f is decreasing on (−1, 0).
F. Local maximum value f (−1) = −e, no local minimum value.
G. f 0 (x) = e−1/x ( x1 + 1) ⇒ f 00 (x) = e−1/x (− x12 ) + ( x1 + 1)e−1/x ( x12 ) =
( x1 + 1)] =
1
x3 e1/x
1 −1/x
[−1 +
x2 e
> 0 ⇔ x > 0, so f is CU on (0, ∞) and CD on (−∞, 0). No IP.
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Figure 1: Graph of y = xe−1/x .
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§4.6 # 12 (a) Graph the function. (b) Use Hospital’s Rule to explain the behavior
as x → 0. (c) Estimate the minimum value and intervals of concavity. Then use
calculus to find the exact values.
Let f (x) = xe1/x .
a)
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2
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2
4
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Figure 2: Graph of y = xe1/x .
1/x
b) limx→0+ xe1/x = limx→0+ e1/x =LH limx→0+ e
(−x−2 )
−x−2
1/x
= limx→0+ e1/x = +∞
limx→0− xe1/x = 0 ∗ 0 = 0.
1/x
c) f 0 (x) = e1/x + xe1/x ( −1
(1 − x1 ) shows that f 0 > 0 (and hence f is
x2 ) = e
increasing) for x < 0 and for x > 1, while f 0 < 0 (and hence f is decreasing) for
0 < x < 1. Hence f has a local min at x = 1 with f (1) = e.
f 00 (x) =
e1/x
x3
shows that f 00 > 0 (and thus f is concave up) for x > 0, while
f 00 < 0 (and thus f is concave down) for x < 0. Note that there is NOT a POI at
x = 0 since the function f is not defined there.
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