# I cant understand this prove explanation on limit series

1. Jan 21, 2009

### 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.

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

$\delta>0$
it needs to come with
$|x-x_3|<\delta$
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

??

2. Jan 22, 2009

### Staff: Mentor

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 $\epsilon > 0$ 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 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?

3. Jan 22, 2009

### transgalactic

in your explanation M is delta>0
??

4. Jan 22, 2009

### Staff: Mentor

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.

5. Jan 22, 2009

### transgalactic

but he does use delta

for what purpose
??

6. Jan 22, 2009

### Staff: Mentor

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.