Math 3110
Spring ’09
Homework 4:
Selected Solutions
[5.11 (b)]:
Claim
:
lim
n
→∞
n

3
2
n

1
¶
(
2

n
)
= 0
Proof
Use linearity, then the quotient rule and finally the product rule for limits:
lim
n
→∞
n

3
2
n

1
¶
(
2

n
)
= lim
n
→∞
1
n
¶
1
n
¶
n

3
2
n

1
¶
(
2

n
)
= lim
n
→∞
1

3
n
2

1
n
(
2

n
)
= 0
[5.14]:
Claim
:
If
a
n
b
n
→
L
,
b
n
6
= 0 for all
n
and
b
n
→
0, the
a
n
→
0.
Proof
There were two main ways that students tried to prove this  the first, using the product
rule, is very easy whereas the second, using the quotient rule, is a little more involved. I’ll show
both versions below.
Product Rule Version:
Let
c
n
=
a
n
b
n
. We are given that both
c
n
and
b
n
converge to finite limits,
so by the product rule
lim
n
→∞
c
n
b
n
= lim
c
n
lim
n
→∞
b
n
=
L
·
0 = 0
.
And, of course,
c
n
b
n
=
a
n
b
n
·
b
n
=
a
n
,
so
lim
n
→∞
a
n
= 0
Quotient Rule Version:
Using the quotient rule directly (without employing the product rule as
above) means that we can no longer assume that
a
n
converges at all. The natural way to proceed
here is to rule out all possibilities other than
lim
n
→∞
a
n
= 0.
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 Fall '08
 RAMAKRISHNA
 Calculus, Derivative, Product Rule, Quotient Rule, Limits, lim, Reciprocal rule

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