Calculus II
Exam 3, Spring 2003, Answers
Remember : you MUST show your work.
1. Find the limits
tan x x π 2
a) lim
x π 2
Answer. First of all, use the identity tan x
of the limits to obtain
x
sin x cos x, and the fact that the limit of a product is the product
π 2 x cos
x
lH
lim sin x lim
x π 2
x π 2
ex 2
∞ e2x
b) lim
Answer. Using the laws of exponents: e x
lim
sin π 2
x π 2
1
sin x
1 e e e , so
lim ee 0 2
2x
2
x
2
x ∞ x
2. Does the integral converge or diverge? Give reasons. If you can, evaluate the integral.
a)
∞
3
x lndxx
2
Answer. Let u
Converges.
dx x. Then integrate from 3 to A:
x lndxx u du u ln x du
∞
ln A
2
3
as A
b)
1 lnA
ln3
2
ln 3
1
ln 3
1
ln A
∞.
x dx1
1
0
2
Diverges.
Answer. We calculate the integral from 0 to c for c slightly less than 1:
x dx1 x 1 1 1 c 1 c
0
as c
1 c
0
2
1.
3. Does the series converge or diverge? Give reasons.
∞
a)
n
e n
e
0 n
∑
Answer. This series converges. Use comparison with the geometric series:
e n
ne
1
e e n
n
∞
1
ln3
and 1 e
∞
b)
1.
3n2
∑ 4n3 5n 25n
n 0
17
1
Answer. This series diverges by comparison with a p-series, p
3n2
4n3
5n 17
25n 1
13
n4
1:
5 n 17 n
25 n 1 n 2
2
1
n
eventually, since the second fraction converges to 3/4.
∞
c)
∑
n 0
n 5n 1
2
2
Answer. This series converges by comparison:
n 5n 1 n5n 2
5
n3
2 2
2
4. What is the interval of convergence of the power series? Show your work.
∑ 5n x ∞
a)
2
n 0
n
Answer. If we write the series as
∑ 5 x
by comparison with the geometric series, this converges if 5 x 2 1, or if x 2 1 5; that is, in the
interval 1 8 x 2 2.
x
b) ∑ 1 n!
Answer. If we rewrite this as
x
x∑
n! we see that this sums to xe everywhere, so the interval of convergence is ∞ ∞ .
∞
n
2
n 0
∞
n
2n 1
n 0
∞
2 n
n 0
x2
5. Find the Maclaurin series for the function. DO a) OR b).
a
1 x
1 4x2
e
x
b
t2
dt
0
Answer. a). Start with the geometric series:
∑ t 1 t
2x 1
∑
1 4x
∞
1
n
n 0
Substitute 2x for t:
∞
2
n 0
n
Multiply by 1
x:
1 x
1 4x2
∑ 2x x ∑ 2x ∑ 2 x ∑ 2 x 1 ∑ 2 2 x ∞
n
n 0
∞
n 0
b) Start with the exponential series:
∞
n
∞
n n
n 0
n n 1
n 0
n 1
∑ xn! t e ∑ 1
n!
∞
et
n
n 0
and substitute t 2 for x:
t2
∞
n
2n 1
n 0
Now integrate, doing the integration on the right term by term:
e
x
0
t2
∑ 1 2nx 2 n! ∞
dt
n 0
n
∞
2n 2
n
n 1
n