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Math 3110 Homework 4 Solutions
Exercise 5.1.4
Claim.
If
a
n
b
n
→
L
,
b
n
6
= 0
for any
n
, and
b
n
→
0
, then
a
n
→
0
.
Proof.
Since lim
n
→∞
a
n
b
n
= 0 and lim
n
→∞
b
n
= 0, then by the product theorem (5.1(2)),
lim
n
→∞
±
a
n
b
n
·
b
n
²
=
L
·
0 = 0
.
But since
a
n
=
a
n
b
n
·
b
n
, then lim
n
→∞
a
n
= lim
n
→∞
±
a
n
b
n
·
b
n
²
, so we have shown that
lim
n
→∞
a
n
= 0
.
Exercise 5.2.1
Claim.
lim
n
→∞
√
n
+ cos
n
√
n
+ 1
= 1
.
Proof.
Since

1
≤
cos
n
≤
1 for all
n
, then
√
n

1
√
n
+ 1
≤
√
n
+ cos
n
√
n
+ 1
≤
√
n
+ 1
√
n
+ 1
for all
n
. Rewriting the left and right hand sides, we have that for all
n
,
r
n
n
+ 1

r
1
n
+ 1
≤
√
n
+ cos
n
√
n
+ 1
≤
r
n
n
+ 1
+
r
1
n
+ 1
.
By the Squeeze Theorem, it suﬃces to show that
lim
n
→∞
³
r
n
n
+ 1

r
1
n
+ 1
!
= 1
and
lim
n
→∞
³
r
n
n
+ 1
+
r
1
n
+ 1
!
= 1
.
1
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View Full Document So by linearity of limits, it suﬃces to show that lim
n
→∞
r
n
n
+ 1
= 1 and lim
n
→∞
r
1
n
+ 1
= 0.
Notice that by the quotient and linearity theorems and the fact that
n
→ ∞
implies
1
n
→
0
(Theorem 5
.
1
∞
), we have
lim
n
→∞
n
n
+ 1
= lim
n
→∞
±
n
n
+ 1
·
1
n
1
n
²
= lim
n
→∞
1
1 +
1
n
=
lim
n
→∞
1
lim
n
→∞
1 + lim
n
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This note was uploaded on 09/29/2011 for the course MATH 3110 at Cornell University (Engineering School).
 '08
 RAMAKRISHNA
 Math

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