- #1

transgalactic

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

[itex]

\delta>0

[/itex]

it needs to come with

[itex]

|x-x_3|<\delta

[/itex]

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

??