Math 135A, Winter 2008
Midterm #2 (Solutions)
(1)
Evaluate
lim
x
→
0
cosh
x

cos
x
sin
x
2
.
Solution:
cosh
x

cos
x
sin
x
2
=
(1 +
x
2
/
2! +
O
(
x
3
))

(1

x
2
/
2! +
O
(
x
3
)
x
2
+
O
(
x
3
)
=
x
2
+
O
(
x
3
)
x
2
+
O
(
x
3
)
→
1 as
x
→
0.
(2)
Is the series
∞
X
k
=1
(

1)
k
ln
k
k
3
+ ln
k
+ 1
absolutely convergent, conditionally convergent, or divergent?
Solution:
The series is absolutely convergent. To justify this, we must show that the series
∑
k
a
k
,
a
k
=
ln
k/
(
k
3
+ ln
k
+ 1), converges.
Let
b
k
= 1
/k
2
, and observe that 0
< a
k
< b
k
for all
k
≥
1. The series
∑
k
b
k
is a
p
series with
p
= 2
>
1;
hence it converges. It follows from the basic comparison test that
∑
k
a
k
also converges.
(3)
Evaluate the integral
Z
+1

1
2
x
1

x
2
dx
or explain why you can’t.
Solution:
This improper integral converges (and therefore can be evaluated) if and only if the improper inte
grals
Z
1
0
2
x
1

x
2
dx
and
Z
0

1
2
x
1

x
2
dx
both converge. But (by definition),
Z
1
0
2
x
1

x
2
dx
=
lim
b
→
1

Z
b
0
2
x
1

x
2
dx
= lim
b
→
1


ln(1

b
2
), which diverges. Hence,
Z
+1

1
2
x
1

x
2
dx
diverges.
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 Fall '08
 Staff
 Math, Bk, improper integral converges

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