Proof.
Let’s suppose that
f
have more than one limit at
c
, let’s call them
L
1
and
L
2
, by Theorem 20.8 we know that lim
x
→
c
f
(
x
) =
L
iff for every
sequence (
s
n
) in
D
that converges to
c
with
s
n
=
c
for all
n
, the sequence
(
f
(
s
n
)) converges to
L
. Then, if lim
x
→
c
f
(
x
) =
L
1
and lim
x
→
c
f
(
x
) =
L
2
,
all the sequences (
s
n
) that converge to
c
with
s
n
=
c
for all
n
, the sequence
(
f
(
s
n
)) converges to
L
1
and
L
2
, but this contradicts Theorem 16.14.
20.11 Let
f
:
D
→
R
and let
c
be an accumulation point of
D
.
Then the
following are equivalent:
(a)
f
does not have a limit at
c
.
(b) There exist a sequence (
s
n
) in
D
with each
s
n
=
c
such that (
s
n
)
converges to
c
, but (
f
(
s
n
)) is not convergent in
R
.
Proof.
To prove that (a)
=
⇒
(b), suppose that (b) is false.
Let (
s
n
)
be a sequence in
D
with
s
n
→
c
and
s
n
=
c
for all
n
. Since (b) is false,
(
f
(
s
n
)) converges to some value, say
L
. Now, we most show that given any
sequence (
t
n
) in
D
with
t
n
→
c
and
t
n
=
c
for all
n
, we have lim
f
(
t
n
) =
L
.
We only know from the negation of (b) that (
f
(
t
n
)) is convergent.
To
see that lim
f
(
t
n
) =
L
, consider the sequence (
u
n
) = (
s
1
, t
1
, s
2
, t
2
, . . .
),
clearly this sequence converges to
c
and again by the negation of (b),
f
(
u
n
) converges, if
f
(
u
n
) converges to
L
2
=
L
then by Theorem 19.4 all
its subsequences would have to converge to
L
2
, but
f
(
s
n
) is a subsequence
and we assumed that it converges to
L
, then
f
(
u
n
) has to converges to
L
.
Thus, since
f
(
t
n
) is also a subsequence of
f
(
u
n
) by Theorem 19.4,
f
(
t
n
)
also converges to
L
. Now we can rewrite the negation of (b) as follows:
for every sequence
(
s
n
)
in
D
that converges to
c
with
s
n
=
c
for all

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- Fall '08
- Akhmedov,A
- Limits, Limit of a sequence, Sn, subsequence, lim sup sn, Snk