I cant understand this prove explanation on limit series

[transgalactic]

there is a continues function f(x) and bounded on (x_0,+infinity)
proove that for every T there is a sequence
X_n=+infinity
so
lim [f(x_n +T) - f(x_n)]=0
n->+infinity

i was told:
uppose that \lim_{x\to\infty}f(x)=a. So we know that given any \varepsilon>0 there exists a \eta>0 such that \eta< x\implies |f(x)-a|<\varepsilon~(1), and since x_n\to \infty we may find a \delta>0 such that \delta<n\implies \eta<x_n~(2). Now suppose that T>0 (the proof for the other cases is analgous), then choose \delta such that (2)\implies (1) then \left[f\left(x+T\right)-f(x)\right|\leqslant \left|f\left(x+T\right)-a\right|+|f(x)-a|<2\varepsilon, this implies the result.


but whrn i read this article on limit proves
http://www.mathhelpforum.com/math-h...e-never-learnt-well-epsilon-delta-proofs.html

i learned from the delta proofes article that when you define the delta

<br /> \delta&gt;0<br />
it needs to come with
<br /> |x-x_3|&lt;\delta<br />
in our case x_3 goes to infinity
so the inqueality that i presented not logical

but on the other hand
it how its done on the article limit proove

??

 

[Mark44]

Limits in which x approaches infinity are proved differently than those for which x approaches some finite number a.

\lim_{x \rightarrow \infty} f(x) = L

means that for any \epsilon &gt; 0 there exists a number M > 0, such that for any x > M, then |f(x) - L| < \epsilon

If I want to prove the limit above to you, you tell me how close f(x) has to be to L (you give me \epsilon), and I tell you a number M.

If you're satisfied, we shake hands and go about our business.

If you're not satisfied, you tell me another \epsilon that's even smaller, and I have to find another M (even larger).

And so on, until you're convinced that I can force f(x) as close to L as you like, by specifiying how big x has to be.

Got it?

 

[transgalactic]

in your explanation M is delta>0
??

 

[Mark44]

in your explanation M is delta>0
??

No. M is generally a pretty large number, while \delta is usually very small. A big difference is the definition I showed doesn't try to get x within \delta of infinity.

 

[transgalactic]

but he does use delta

for what purpose
??

 

[Mark44]

If the limit is as x approaches a finite number a, you use \delta, since you want to make x very close to a. I.e., you want to make |x - a| < \delta.
If the limit is as x approaches infinity, that's a different matter, since as I explained earlier, you can't get x within \delta of infinity. You can, however, make x larger than some (presumably large) number M.

I don't think I can make it any clearer than that.