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Midterm Exam 3
- (20 points) Evaluate lim x→ 0 +^
xx.
We have lim x→ 0 +^
xx^ = lim x→ 0 +^
ex^ ln^ x^ = elimx→^0 +^ x^ ln^ x
if limx→ 0 + x ln x exists. Rewriting this, we have
lim x→ 0 +
ln x 1 /x
which has the indeterminate form −∞/∞ so L’Hˆopital’s rule applies. We obtain
lim x→ 0 +
ln x 1 /x
= lim x→ 0 +
1 /x −x−^2
= lim x→ 0 +
(−x) = 0.
Hence, the original limit is elimx→^0 +^ x^ ln^ x^ = e^0 = 1.
- (20 points) Evaluate (^) ∫ 1 − 1
x^4
dx.
Since (^) x^14 does not exist at x = 0, we break the integral into two pieces and use a limit, obtaining ∫ (^1)
− 1
x^4 dx^ =
− 1
x−^4 dx +
0
x−^4 dx = lim b→ 0 −
∫ (^) b
− 1
x−^4 dx + lim a→ 0 +
a
x−^4 dx
= lim b→ 0 −
[
3 x
− 3
]b
− 1
[
3 x
− 3
] 1
a
= lim b→ 0 −
3 b
3 a
− 3
b−^3 + lim a→ 0 +
a−^3 = ∞.
- (20 points) Prove that an = n+(−1)
n n converges using the^ ε-N^ limit definition. Let ε > 0 be given and choose N = (^1) ε. Then for all n > N we have ∣∣ ∣∣^ n^ + (−1)
n n
∣∣^ (−1)
n n
∣∣ =^1
n
N
1 /ε
= ε.
Hence, an converges to 1.
- (20 points) Solve r + r^3 + r^5 + r^7 + r^9 + · · · =
for r.
If |r| < 1 then we can write the series on the left as the geometric series
r(1 + r^2 + r^4 + · · · ) = r
1 − r^2
Hence, the equation becomes r 1 − r^2
=^5
3 r = 5(1 − r^2 ) 5 r^2 + 3r − 5 = 0
and by the quadratic formula, we have
r = −^3 ±
Since the original geometric series we used only converges for |r| < 1, we must have r =
√ 109 − 3
(20 points) Determine whether ∑∞
n=
n n^2 + 1
converges. Prove your answer.
Since all the terms are positive, we can apply the integral test and consider ∫ (^) ∞
1
x x^2 + 1
dx.
Letting u = x^2 + 1 so du = 2x dx, we obtain ∫ x x^2 + 1 dx^ =
2 u du^ =
2 ln^ |u|^ +^ C^ =
2 ln(x
2 + 1) + C.
Hence, (^) ∫ ∞ 1
x x^2 + 1 dx^ = lim b→∞
∫ (^) b
1
x x^2 + 1 dx^ = lim b→∞
2 ln(b
2 ln 2 =^ ∞.
Therefore, the original series
n=
n n^2 +1 diverges by the integral test.
- (20 points) Determine whether
∑^ ∞
n=
ln
1 + n^1
converges. Prove your answer. (Hint: Use a limit comparison test.)
