Calculus II
Practice Exam 3, Answers
In problems 1-4, find the limits.
1. lim
x 0
cos x 1
x2
lH
Answer.
limx
0
sinx
2x
1
2
x π
2. lim
sin x x π
3 x π
lim
Answer. cos x 1
3
x π
2
lH
lH
x π
3. lim x5 e
x ∞
Answer.
6x π
lim
x π
sin x
lH
lim
x π
6
cos x
6
x
0
x5
∞ ex
lim
x
which converges to zero since the exponential grows faster than any polynomial.
1 x 4. lim
2
x ∞
x
Answer.
lim
x ∞
x
1
x2
1 1 0 since x 2 0 as x ∞. We arrived at the second formulation from the first by dividing both numerator and
denominator by x. Observe that, although l’Hôpital’s rule applies, it doesn’t get us anywhere.
In problems 5-7: Does the integral converge or diverge? If you can, find the value of the integral.
e
using the substitution u x du 5.
∞
x2 dx
0 xe
1 ∞
2 0
u du
2
6. Answer.
∞
0
x2
x3
1
1
2
2xdx and a known computation (see example 8.16).
dx diverges since
1
1
1 x 2x
for x sufficiently large, and our knowledge that dx x diverges.
x2
1
x2
x3
∞
0
7.
1
0
dx
x9 10
Answer.
lim
1
dx
lim
a 0 a x9 10
a 0
lim 10x1
a 0
10 1
a
10
8. Does the sequence converge or diverge?
a) an
n2
n!
Answer. an
n n 1n n 2 ! 1 1 n 1 2 ! n2
n!
2
0
2
n
since the first factor converges to 1, while the second converges to 0.
n n!1
n!
Answer. b n n!1 n 1
b) bn
2
n
2
4
∞
because the expression under the square root sign goes to infinity (which we can show by an argument similar
to that in part a).
n3 50n 1
n4 123n3 1
1 50
n3 50n 1
n2
Answer. cn
n4 123n3 1 n 1 123
n
since every factor converges to 1 except that n
c) cn
∞.
9. Does the series converge or diverge?
∞ 2
n
a) ∑
1 n!
1
n3
1
n4
0
n 1 n! 1 1
n 1 ! n 1 n n 1 2
Answer. This converges by the ratio test:
2
2
which is less than 1.
∞
b) ∑
1
n n!1
2
Answer. This diverges by 9b: the general term does not go to 0.
∞
c) ∑
20
n3 50n 1
n4 123n3 1
Answer. This diverges because
1
n 1 n3 50n 1
n4 123n3 1
50
n2
123
n
eventually. By comparison with ∑ 1 n the series diverges.
1
n3
1
n4
1
2n
0
10. Does the series converge or diverge?
∞
3n 1
converges
5 2
1 n
by comparison with the series ∑ 1 n3
a) ∑
:
2
3
4
n n
3n 1
n5 2
∞
b) ∑
1
n
3 2
3 2
n 31 !5n! n
converges
1
by comparison with the geometric series:
1
n
n 31 !5n! 1 n 1 1 5 3 35 n
∞
c) ∑
1
2n ! n 1
2n 1 !
n
n
n
1
n 1!
n
diverges
since the general term does not converge to 0:
2n ! n 1 n 1
2n 1 ! 2n 1 ∞
1
converges
1 2 3n 1
n
1
by comparison with the series ∑ 1 n3
d) ∑
:
1
2
2
n1
1
3n 1 2
1
3n3
2
11. Find the radius of convergence of the series:
∑n n
∞
2 x Answer. . We observe that this is the thrice differentiated geometric series, so R the ratio test for the coefficients:
n 1 n n 1 n 1
n 2 1
n n 1 n 2
b) ∑ 2 1 x
a)
n 3
∞
n
1 n
n 3
1. However we can use
n
0
Answer. Write down the ratio of successive coefficients and divide numerator and denominator by 2 n :
2
1
2n 1 1
2n 1
1
2n
1
2n
2
so the radius of convergence is 1/2.
3n 1 x 1
n 1
Answer. The coefficient looks like 3 n and so the series converges if x 1 interval. Thus R 1.
∞
c) ∑
1
2
n
3
1, and diverges outside this
x 12. Find the Maclaurin series for 1
3.
Answer. Starting with the geometric series, substitute
x for x:
1 x ∑ 1 x
1 x ∑ 1 nx ∑ 1 n n 1 x 2 1 x
1 1 x 12 ∑ 1 n 2 n 1 x
∞
1
n n
n 0
Now, differentiate twice:
∞
2
n
n 1
n 1
∞
3
n
n 2
n 2
so
∞
3
13. Find the Maclaurin series for
Answer. We start by substituting
n
x
0 arctantdt.
x 2 for x in the geometric series:
1 x
∑
1 x
x arctanx ∑ 1 2n 1
arctantdt ∑ 1 2n x2 2n 1 ∞
1
2
Now integrate twice:
n
n 0
n 2n
n 0
∞
2n 1
n
n 0
∞
x
0
2n 2
n
n 0
1 .
14. Find the Maclaurin series for x ln x
Answer. Once again start with the geometric series, with
1 x ∞
∑
1
Integrate and multiply by x:
x for x
1 x !
n n
x ∑ 1 nx 1 ∑ 1 nx 1 !
∞
ln x
n 0
n
∞
n 1
n 0
n
n 2
n 0
15. Find the terms up to fourth order for the Maclaurin series for
ex
1
x
x , explicitly, that is, term by term, up to
#""$"
Answer. We write down the Maclaurin series for each of e x 1 1
the fourth order:
x2 x3 x4
ex 1 x
2
6 24
%"$""
1
1 x x2 x3 x4
1 x
Now, we multiply these together as if they were polynomials, relegating all terms of order greater than 4 to
the
:
ex
x2 x3 x4
1 x
1 x x 2 x3 x4
1 x
2
6 24
""$"
% ""$" %""$" 1 x x x x & x x x x & x2 x2 x2 & x6 x6 '
3
4
2
3
4
2
3
4
#"""
3
4
#""$"
x4
24
where we have done the multiplication by successively multiplying the second series by the terms of the first.
Now we collect terms;
1 2 1 3 9 4
ex
1
x
x
x
1 x
2
3
24
(Why have all the terms in the first two parentheses , except 1,cancelled?)
2