Proof
Let
M >
0. Select a real number
m
so that 0
< m <
lim
t
n
. Whether
lim
t
n
+∞
or not, it is clear that there exists
N
1
such that
n > N
1
implies
t
n
> m
;
see Exercise 8.10. Since lim
s
n
+∞
, there exists
N
2
so that
n > N
2
implies
s
n
>
M
m
.
Put
N
max
{
N
1
, N
2
}
. Then
n > N
implies
s
n
t
n
>
M
m
·
m
M
.
Example 7
Use Theorem 9.9 to prove that lim
n
2
+
3
n
+
1
+∞
; see Example 6.
Solution
We observe that
n
2
+
3
n
+
1
n
+
3
n
1
+
1
n
s
n
t
n
where
s
n
n
+
3
n
and
t
n
1
1
+
1
n
. It
is easy to show that lim
s
n
+∞
and lim
t
n
1. So by Theorem 9.9,
we have lim
s
n
t
n
+∞
.
Here is another useful theorem.
9.10 Theorem.
For a sequence
(
s
n
)
of positive real numbers, we have
lim
s
n
+∞
if
and only if
lim(
1
s
n
)
0
.
Proof
Let (
s
n
) be a sequence of positive real numbers. We have to show
lim
s
n
+∞
implies
lim
1
s
n
0
(1)
and
lim
1
s
n
0
implies
lim
s
n
+∞
.
(2)
In this case the proofs will appear very similar, but the thought
processes will be quite different.
2.
Sequences
52
To prove (1), suppose that lim
s
n
+∞
. Let
>
0 and let
M
1
.
Since lim
s
n
+∞
, there exists
N
such that
n > N
implies
s
n
> M
1
. Therefore
n > N
implies
>
1
s
n
>
0, so
n > N
implies
1
s
n
−
0
<
.
That is, lim(
1
s
n
)
0. This proves (1).
To prove (2), we abandon the notation of the last paragraph and
begin anew. Suppose that lim(
1
s
n
)
0. Let
M >
0 and let
1
M
. Then
>
0, so there exists
N
such that
n > N
implies

1
s
n
−
0

<
1
M
.
Since
s
n
>
0, we can write
n > N
implies
0
<
1
s
n
<
1
M
and hence
n > N
implies
M < s
n
.
That is, lim
s
n
+∞
and (2) holds.
Exercises
9.1.
Using the limit theorems 9.2–9.6 and 9.7, prove the following.
Justify all steps.
(a)
lim
n
+
1
n
1
(b)
lim
3
n
+
7
6
n
−
5
1
2
(c)
lim
17
n
5
+
73
n
4
−
18
n
2
+
3
23
n
5
+
13
n
3
17
23
9.2.
Suppose that lim
x
n
3, lim
y
n
7 and that all
y
n
are nonzero.
Determine the following limits:
(a)
lim(
x
n
+
y
n
)
(b)
lim
3
y
n
−
x
n
y
2
n
9.3.
Suppose that lim
a
n
a
, lim
b
n
b
, and that
s
n
a
3
n
+
4
a
n
b
2
n
+
1
. Prove
lim
s
n
a
3
+
4
a
b
2
+
1
carefully, using the limit theorems.
9.4.
Let
s
1
1 and for
n
≥
1 let
s
n
+
1
√
s
n
+
1.
(a)
List the first four terms of (
s
n
).
(b)
It turns out that (
s
n
) converges. Assume this fact and prove
that the limit is
1
2
(1
+
√
5).
Exercises
53
9.5.
Let
t
1
1 and
t
n
+
1
t
2
n
+
2
2
t
n
for
n
≥
1. Assume that (
t
n
) converges
and find the limit.
9.6.
Let
x
1
1 and
x
n
+
1
3
x
2
n
for
n
≥
1.
(a)
Show that if
a
lim
x
n
, then
a
1
3
or
a
0.
(b)
Does lim
x
n
exist? Explain.
(c)
Discuss the apparent contradiction between parts (a) and (b).
9.7.
Complete the proof of 9.7(c), i.e., give the standard argument
needed to show that lim
s
n
0.
9.8.
Give the following when they exist. Otherwise assert “NOT EXIST.”
(a)
lim
n
3
(b)
lim(
−
n
3
)
(c)
lim(
−
n
)
n
(d)
lim(1
.
01)
n
(e)
lim
n
n
9.9.
Suppose that there exists
N
0
such that
s
n
≤
t
n
for all
n > N
0
.
(a)
Prove that if lim
s
n
+∞
, then lim
t
n
+∞
.
(b)
Prove that if lim
t
n
−∞
, then lim
s
n
−∞
.
(c)
Prove that if lim
s
n
and lim
t
n
exist, then lim
s
n
≤
lim
t
n
.
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 Spring '14
 Limits, Limit of a sequence, Sn, lim sn