Calculus III
Exam 3, Summer 2003, Answers
Problems are worth 20 points each. You may use calculators and Tables of Integrals. You must show enough
work to convince me that you know how to do the problems. 1. Find the limits:
a) lim x ln x =
x!0
Answer. L’Hôpital’s rule applies to lnx=(1=x), so
x
x!0 1=(lnx)
lim x ln x = lim
x!0
=
l0 H
(1=x)
= lim (
x!0 ( 1=x2 )
x!0
x) = 0 :
lim
π =2) tanx =
b) lim (x
x!π =2
Answer. If we replace tan x by sin x= cos x, then both numerator and denominator go to infinity, so L’Hôpital’s
rule applies:
lim (x
x!π =2
π =2) sin x
cos x
(x
π =2) tanx = lim
x!π =2
l0 H
=
lim
(x
sin x
x!π =2
π =2) cos x
sin x
=
2. Find the definite integrals:
Z ∞
a)
xe
x2
dx =
0
Answer. First calculate the integral from 0 to A, using the substitution u = x 2 ; du = 2xdx:
Z A
xe
x2
1
2
dx =
0
Now, since e
A2
Z A2
1
(1
2
e u du =
0
e
A2
)
:
! 0 as A ! ∞, when we take the limits we find
Z ∞
xe
x2
dx =
0
1
2
:
Z 2
lnxdx =
b)
0
R
Answer. Since lnxdx = x ln x
x,
Z 2
Z 2
ln xdx = lim
0
ε !0 ε
= 2 ln 2
2
2
lim ε ln ε
= 2 ln2
ε !0
lim ε
ε !0
by Problem 1, part a).
3. Does the series converge or diverge? Give your reasoning.
∞
a)
n+1
3
n=1 n
∑
converges:
lim (ε ln ε
ln xdx = 2 ln2
ε !0
2;
ε)
1:
Answer. Since n + 1 2n, this series converges by comparison with the p-series for p = 2:
2n
n22
n3
n+1
n3
∞
b)
en
∑ ne
diverges:
n=1
Answer. Use the ratio test:
∞
c)
:
2n
∑ (3 + 1 )n
n=1
en+1 ne
(n + 1)e en
n e
!e>1:
n+1
=e
converges:
n
Answer. Since 3 + (1=n) 3,
2n
1
(3 + n )n
2 n
;
3
so the series converges by comparison with the geometric series.
4. Find the radius of convergence of the series:
∞
a)
2n n
∑ x
n=1 n
Answer.
∞
b)
2n+1 n
n + 1 2n
= 2(
n
n+1
)
2n n
∑ x
n=1 n!
Answer.
2n+1 n!
(n + 1)! 2n
=
2
n+1
!2
!0
;
;
so the radius of convergence is 1/2.
so the radius of convergence is ∞.
5. Find the Taylor series centered at the origin for the function
Z x
F (x) =
0
dt
1
t4
:
Answer. We start with the geometric series:
∞
1
1
x
=
∑ xn
Substitute t 4 for x to get
∞
1
1
:
0
t4
=
∑ t 4n
:
0
Now integrate both sides; integrating the right hand side term by term:
Z x
F (x) =
0
∞
dt
1
t4
=
t 4n+1
∑ 4n + 1
0
: