) = 0.
8, page 70.
(
⇒
) First assume that lim
x
→∞
f
(
x
) =
L
and let
x
n
∈
(
a,
∞
) such that
x
n
→ ∞
as
n
→ ∞
. We want to show that
f
(
x
n
)
→
L
.
Let
>
0. Since lim
x
→∞
f
(
x
) =
L
, there exists
M
∈
R
such that for any
x > M
,

f
(
x
)

L

<
.
Because
x
n
→ ∞
, there exists a rank
N
0
∈
N
such that
∀
n
≥
N
0
,
x
n
> M
.
Combining these, it follows that
∀
n
≥
N
0
,

f
(
x
n
)

L

<
. Thus
f
(
x
n
)
→
L
.
(
⇐
) Conversely, now assume that
f
(
x
n
)
→
L
for any sequence
x
n
∈
(
a,
∞
)
such that
x
n
→ ∞
, and we’d like to prove that lim
x
→∞
f
(
x
) =
L
. We do
this using a proof by contradiction.
Suppose that
f
(
x
) does not converge
to
L
as
x
approaches
a
.
Then there exists an
0
>
0 such that for any
M
∈
R
, there exists a real number
x
M
(which depends on
M
) such that
x
M
> M
and

f
(
x
M
)

L
 ≥
0
. In particular, for each natural number
n > a
(taken as
M
in the above), we’ll find a real number
x
n
, such that
x
n
> n
and

f
(
x
n
)

L
 ≥
0
.
The inequality
x
n
> n
for any
n > a
implies that
the sequence
x
n
→ ∞
, thus by assumption
f
(
x
n
)
→
L
as
n
→ ∞
.
Thus

f
(
x
n
)

L

<
0
, for
n
sufficiently large. But this is in contradiction with

f
(
x
n
)

L
 ≥
0
, for any
n
∈
N
,
n > a
.
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 Fall '12
 Draghic
 Natural number, Limit of a sequence, xm

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