Math 1220 (Underdown)
Exam 3
Name:
1. [10 points] Find the limit:
limπ
x→ 2
cos x
x − π2
Solution: This is a 0/0 form, thus L’Hôpital’s Rule applies.
limπ
x→ 2
2. [10 points] Find the limit:
cos x H
sin x
= −1
π = limπ −
x→ 2
x− 2
1
lim
t→∞
t ln t
1 + t2
Solution: This is an ∞/∞ form, thus L’Hôpital’s Rule applies.
t ln t H
ln t + 1 H
1/t
= lim
= lim
= 0
2
t→∞ 1 + t
t→∞
t→∞ 2
2t
lim
April 15, 2015
Math 1220
Exam 3, Page 2 of 8
April 15, 2015
3. [10 points] Find the following limit.
Hint: This is an indeterminate ∞0 form, so let y = (tan x)cos x and find
lim ln y.
x→π/2−
lim − (tan x)cos x
x→π/2
Solution: Let y = (tan x)cos x , then ln y = cos x ln(tan x).
lim ln y = lim − cos x ln(tan x)
x→π/2−
0 · ∞ form
x→π/2
= lim −
x→π/2
ln(tan x)
sec x
∞/∞ form
sec2 x
H
= lim − tan x
x→π/2 sec x tan x
= lim −
sec x
tan2 x
= lim −
cos x
sin2 x
x→π/2
x→π/2
=0
Since we found the limit of ln y we must exponentiate to get the limit of y, e0 = 1 .
Math 1220
Exam 3, Page 3 of 8
4. [10 points] Evaluate the improper integral or show that it diverges.
Hint: u–substitution.
Z∞
ln x
dx
x
e
Solution: Let u = ln x, then du = dx/x,
Z∞
ln x
dx = lim
b→∞
x
Z∞
u du
e
e
b
u2 = lim
b→∞ 2 e
b
1
2
= lim (ln x)
2 b→∞
e
1
= lim (ln b)2 − 1
2 b→∞
= ∞
5. [10 points] Evaluate the improper integral or show that it diverges.
Z2
dx
√
x
0
Solution:
Z2
dx
√ = lim
x t→0+
Z2
0
x−1/2 dx
t
2
1/2
= lim+ 2x
t→0
t
√
√
= 2 lim+ 2 − t
t→0
√
= 2 2
April 15, 2015
Math 1220
Exam 3, Page 4 of 8
April 15, 2015
6. [5 points] Determine the sum of the following geometric series.
∞ n−1
X
4
n=1
5
Solution: Recall the geometric series formula:
∞
X
arn−1 =
n=1
a
1−r
|r| < 1.
Here a = 1, r = 4/5, thus the series sums to:
a
1
1
=
=
= 5
1−r
1 − 4/5
1/5
7. [5 points] Determine whether the following series diverges or converges. Explicitly state
your reasoning. If it converges, find its sum.
∞
X
2n
n=1
3n
Solution: This is another geometric series with ratio, r = 2/3 < 1, thus it converges
and its sum is:
∞ n−1
∞ n
∞
X
X
2
2/3
2
2n X 2
=
=
=
= 2
n
3
3
3
3
1
−
2/3
n=1
n=1
n=1
Math 1220
Exam 3, Page 5 of 8
April 15, 2015
8. [10 points] Use the Limit Comparison Test to determine the convergence or divergence
of the series
∞
X
n2 + 1
22 + 1 32 + 1 42 + 1
=
1
+
+ 3
+ 3
+ ···
3+1
3+1
n
2
3
+
1
4
+
1
n=0
Hint: When n is large adding 1 is insignificant, and each term behaves almost exactly
like a function without the 1s, that is like n2 /n3 . Use this fact to help you choose the
comparison series.
Solution: From the hint, we should compare this series with
X
X
X
bn =
n2 /n3 =
1/n,
that is the Harmonic series, which diverges.
n2 + 1 n
an
n3 + n
= lim 3
=1=L
= lim 3
n→∞ bn
n→∞ n + 1 1
n→∞ n + 1
| {z } |{z}
lim
an
1/bn
Since 0 < L < ∞, according the to the Limit Comparison Test, both series diverge
or converge together. Since the Harmonic series diverges so does the given series.
diverges
Math 1220
Exam 3, Page 6 of 8
April 15, 2015
9. [10 points] Use the Ratio Test to determine whether the following series converges or
diverges. Explicitly state your reasoning.
∞
X
8n
n=1
Solution: Let an =
n!
an+1
8n
, then we must compute lim
.
n→∞ an
n!
an+1
8n+1
n!
= lim
n→∞ an
n→∞ (n + 1)! 8n
| {z } |{z}
lim
an+1
1/an
n
n!
8Z
8 ·Z
n
n→∞ (n + 1)n! Z
8Z
8
= lim
n→∞ n + 1
=0=ρ
= lim
Since ρ < 1 the series converges .
Math 1220
Exam 3, Page 7 of 8
April 15, 2015
10. [10 points] Classify the following series as either: divergent, conditionally convergent or
absolutely convergent. Justify your answer by clearly indicating which test(s) you use
and showing that all hypotheses are met.
Check one:
gent
divergent
conditionally convergent
absolutely conver-
∞
X
(−1)n
3n2 + 5
n=1
Solution: This is an alternating series with terms an = 1/(3n2 + 5). Since
lim an = lim
n→∞
1
= 0,
+5
n→∞ 3n2
we know by the Alternating Series Test that this series converges.
Now to determine whether the convergence is conditional or absolute we must consider the series,
∞
X
1
.
2
3n + 5
n=1
P
We can compare this with the p–series,
1/n2 which converges because 2 = p > 1.
Notice that every term in the series we are examining is smaller than every term in
the p–series with p = 2, that is
1
1
< 2
+5
n
3n2
P
for every n. Therefore by the Ordinary Comparison Test, since the p–series,
1/n2
converges so does our series with all terms made positive. This implies that the given
series converges absoluteley.
Math 1220
Exam 3, Page 8 of 8
Question Points Score
1
10
2
10
3
10
4
10
5
10
6
5
7
5
8
10
9
10
10
10
Total:
90
April 15, 2015