Problem Set 7
Calculus 1
Abdullah Khalid, Hassan Bukhari
January 3, 2010
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any time.
1. Apply the Mean Value Theorem to the function
f
on the interval (

b, b
) to find a
point
c
such that
f
0
(
c
) =
f
(
b
)

f
(

b
)
2
b
Since
f
is odd,
f
(
b
) =

f
(

b
). Hence
f
0
(
c
) =
f
(
b
)
/b
.
2.
(a) Applying L’Hospital’s rule we get
lim
x
→∞
ln ln
x
x
= lim
x
→∞
1
x
ln
x
= 0
(b) The limit does not exist. A very intuitive reason why the limit does not exist
is because sin oscillates between 1 and 1 where as ln
x
approaches
∞
as
x
approaches
∞
.
(c) We know that
lim
x
→
0
+
x
ln
x
= lim
x
→
0
+
ln
x
x

1
= lim
x
→
0
+

x
2
x
= lim
x
→
0
+

x
= 0
So we can rewrite
lim
x
→
0
+
sin
x
ln
x
= lim
x
→
0
+
sin
x
x
x
ln
x
= lim
x
→
0
+
sin
x
x
lim
x
→∞
x
ln
x
= 0
1
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(d) We want to find lim
x
→∞
f
(
x
) where
f
(
x
) =
x
1
/x
.
Let us first compute the
limit of ln
f
(
x
).
lim
x
→∞
ln
f
(
x
) = lim
x
→∞
ln
x
x
= 0
where the last equality follows by a simple application of L’Hospital’s rule.
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 Spring '10
 RazaSuleman
 Calculus, Derivative, lim, lim x ln, lim lim, lim =− lim

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