Example 3 illustrates that direct proofs of even rather simple
limits can get complicated. With the limit theorems of §9 we would
just write
lim
4
n
3
+
3
n
n
3
−
6
lim
4
+
3
n
2
1
−
6
n
3
lim 4
+
3
·
lim(
1
n
2
)
lim 1
−
6
·
lim(
1
n
3
)
4
.
Example 4
Show that the sequence
a
n
(
−
1)
n
does not converge.
Discussion
. We will assume that lim(
−
1)
n
a
and obtain a con
tradiction. No matter what
a
is, either 1 or
−
1 will have distance at
least 1 from
a
. Thus the inequality

(
−
1)
n
−
a

<
1 will not hold for
all large
n
.
Formal Proof
Assume that lim(
−
1)
n
a
for some
a
∈
R
. Letting
1 in the
definition of the limit, we see that there exists
N
such that
n > N
implies

(
−
1)
n
−
a

<
1
.
By considering both an even and an odd
n > N
, we see that

1
−
a

<
1
and
 −
1
−
a

<
1
.
2.
Sequences
40
Now by the Triangle Inequality 3.7
2

1
−
(
−
1)


1
−
a
+
a
−
(
−
1)
 ≤ 
1
−
a
+
a
−
(
−
1)

<
1
+
1
2
.
This absurdity shows that our assumption that lim(
−
1)
n
a
must
be wrong, so the sequence (
−
1)
n
does not converge.
Example 5
Let (
s
n
) be a sequence of nonnegative real numbers and suppose
that
s
lim
s
n
. Note that
s
≥
0; see Exercise 8.9(a). Prove that
lim
√
s
n
√
s
.
Discussion
. We must consider
>
0 and show that there exists
N
such that
n > N
implies

√
s
n
−
√
s

<
.
This time we cannot expect to obtain
N
explicitly in terms of
be
cause of the general nature of the problem. But we can hope to show
such
N
exists. The trick here is to violate our training in algebra and
“irrationalize the denominator”:
√
s
n
−
√
s
(
√
s
n
−
√
s
)(
√
s
n
+
√
s
)
√
s
n
+
√
s
s
n
−
s
√
s
n
+
√
s
.
Since
s
n
→
s
we will be able to make the numerator small [for large
n
]. Unfortunately, if
s
0 the denominator will also be small. So we
consider two cases. If
s >
0, the denominator is bounded below by
√
s
and our trick will work:

√
s
n
−
√
s
 ≤

s
n
−
s

√
s
,
so we will select
N
so that

s
n
−
s

<
√
s
for
n > N
. Note that
N
exists, since we can apply the definition of limit to
√
s
just as well
as to
. For
s
0, it can be shown directly that lim
s
n
0 implies
lim
√
s
n
0; the trick of “irrationalizing the denominator” is not
needed in this case.
Formal Proof
Case I:
s >
0. Let
>
0. Since lim
s
n
s
, there exists
N
such that
n > N
implies

s
n
−
s

<
√
s .
§8.
A Discussion about Proofs
41
Now
n > N
implies

√
s
n
−
√
s


s
n
−
s

√
s
n
+
√
s
≤

s
n
−
s

√
s
<
√
s
√
s
.
Case II:
s
0. This case is left to Exercise 8.3.
Example 6
Let (
s
n
) be a convergent sequence of real numbers such that
s
n
0
for all
n
∈
N
and lim
s
n
s
0. Prove that inf
{
s
n

:
n
∈
N
}
>
0.
Discussion
. The idea is that “most” of the terms
s
n
are close to
s
and hence not close to 0. More explicitly, “most” of the terms
s
n
are
within
1
2

s

of
s
, hence most
s
n
satisfy

s
n
 ≥
1
2

s

. This seems clear
from Figure 8.1, but a formal proof will use the triangle inequality.
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 Spring '14
 Limits, Limit of a sequence, Sn, lim sn