1. series limit proove question..

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 tried to solve it like this:
the bound is L
if f(x) is bounded than there is e>0 |f(x)-L|<e
but i dont know what to do next
??

2. Originally Posted by 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 tried to solve it like this:
the bound is L
if f(x) is bounded than there is e>0 |f(x)-L|<e
but i dont know what to do next
??
I am interpreting this that if $f(x)$ and $x_n\to \infty$ prove that for any $T$, $\lim_{n\to\infty}\left[ f\left(x_n+T\right)-f(x_n)\right]=0$

Suppose 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. 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.

Note: If this was a wrong interpretation of your question or this is unclear feel completely free to ask

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

$
\delta>0
$

it needs to come with[/COLOR]
$
|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

??