Math 414
Professor Lieberman
March 10, 2003
HOMEWORK #7 SOLUTIONS
Section 2.5
18. (a) a1 = 1, a2 = 3/2, a3 = 7/5, so a1 < a2 and a2 < a3 which means that (an ) is not
monotone.
(b) First, we observe that an ≥ 1 implies that an+1 ≥ 1. Since a1 = 1, we see by induction
that an ≥ 1 for all n. Therefore
1
1 |an+1 − an |
1
|an+2 − an+1 | = −
=
≤ |an+1 − an |.
1 + an+1 1 + an
(1 + an )(1 + an+1 )
4
Therefore (an ) is contractive with constant k = 1/4 < 1.
(c) By taking the limit as n goes to infinity in the equation for an+1 in terms of an , we see
that the limit A satisfies the equation
A = 1 + 1/(1 + A), so A2 = 2. Since all an ’s are
√
positive, we conclude that A = 2.
(d) There is no answer for this one. You have to convince yourself.
Section 2.6
8. By problem 7, there is a subsequence (bn ) which converges to B, and then Theorem 2.6.5
implies that A = B.
Section 2.7
18. True. Just use the limit theorems:
√
√
√
n
lim nk = lim ( n n)k = ( lim n n)k = 1k = 1.
n→∞
n→∞
n→∞
21. False. If an = 1/n and bn = (−1)n , then an bn → 0 but bn /an = (−1)n n diverges and so
does bn .
29. False. If an = 0 and bn = 1/n, then limn→∞ an = 0 = limn→∞ bn , but an 6= bn for all n.
36. False. If an = (−1)n , then
bn =
(
0
−1
n
if n is eve, n
if n is odd,
so (an ) diverges, but bn converges to zero.
47. False. If an = bn = (−1)n , then (an ) and (bn ) both diverge but (an bn ) converges because
an bn = 1 for all n.
1
2
Section 2.8
7. A little calculation shows that αn = 0 if n is even and αn = (n + 1)/(2n + 1) if n is odd.
It follows that the subsequence (α2n ) converges to 0, but (α2n+1 ) converges to 1/2. By
Theorem 2.6.5, this implies that (αn ) does not converge, so (an ) is not (C, 1) summable.
8. By algebra,
n−1
an
a1 + · · · + an−1 an
+
= αn−1
+ ,
αn =
n
n
n
n
so
an
1
= αn − αn−1 1 −
.
n
n
Now we write A for limn→∞ αn , and observe that
1
1
lim αn−1 1 −
= lim αn−1
lim 1 −
= A · 1 = A.
n→∞
n→∞
n→∞
n
n
It follows that
an
1
lim
= lim αn − αn−1 1 −
= A − A = 0.
n→∞ n
n→∞
n
To use this result to show that (an ) is not (C, 1) summable, we note that an /n = −1/2
if n is even, so limn→∞ an /n can’t be zero.
Section 3.1
5. (a) We use the limit theorems to see that
−x2
−1
1
= lim
=− .
x→∞ 2x2 − 3
x→∞ 2 − 3x−2
2
(i) For x < 3, we have |x − 3| = (3 − x), so (x − 3)/|x − 3| = −1 and therefore the limit is
−1.
lim
6 (a) Remember that the function has to be defined on all of R, so one possibility is
(
1
if x 6= 0,
f (x) = x
0 if x = 0.
(b) There are two cases (with subcases) to consider: (1) limx→∞ f (x) exists either (a) as
a finite number, (b) as +∞ or (c) as −∞ (2) limt→−∞ f (−t) exists either (a) as a finite
number, (b) as +∞ or (c) as −∞. There’s no point in working out all six cases because
they all are basically the same. Here’s a proof for (1)(a).
Because limx→∞ f (x) exists as a finite number, say L, we know that, for any ε > 0, there
is a positive number M such that, if x > M , then |f (x) − L|ε. If t < −M , then −t > M ,
so |f (−t) − L| < ε, so limt→−∞ f (x) = L.
(c) Because 2 < e, we have
x
2
x −x
lim 2 e = lim
= ∞.
x→−∞
x→−∞ e
8. (c) There’s no asymptote because the remainder term is zero.
3
Section 3.2
1. (b) By plugging in, the limit is 2. To prove this, let ε > 0 be given and set δ =
min{1/4, 2ε/9}. If 0 < |x − 1| < δ, then
2
x − x − 2
x−3 9
2x − 3 − 2 = 2x − 3 |x − 1| ≤ 2 |x − 1| < ε.
(f) The limit is 2. Given ε > 0, choose δ = min{1, ε}. If 0 < |x − 1| < δ, then
1−x
√
|x − 1|
1 − √x − 2 = | x − 1| = √x + 1 ≤ |x − 1| < ε.
5. (a) The difficulty is that
lim sin
x→0
1
x
does not exist.
Section 3.3
4. (a) For x > 2, we have |x − 2|/(x − 2) = 1, so
lim
x→2+
|x − 2|
= 1.
x−2
9. (b) We have
lim f (x) = 1,
x→∞
lim f (x) = 1, lim f (x) = ∞, lim f (x) = 0,
x→−∞
x→0+
x→0−
so the asymptotes are y = 1 (horizontal) and x = 0 (vertical). There are no roots.