proving an iff statement about the definition of a limit using epsilon delta

Question

Let $A$ and $x_0$ be real numbers, and let $f(x)$ be a real-valued function defined in a deleted neighborhood of $x_0$. Use the definition of limit to prove that $\lim \limits_{x \to x_0} f(x) = A $ if and only if $\lim \limits_{x \to 0} f(x_0 + x) = A$.

First we prove $\lim \limits_{x \to x_0} f(x)=A \implies \lim \limits_{x \to 0}f(x_0+x)=A$

$\lvert f(x)-A \rvert <\epsilon$, $\forall x$ with $0<\lvert x-x_0 \rvert < \delta$ or $x_0- \delta <x<x_0 + \delta$

$\lvert f(x_0+x)-A \rvert < \epsilon$, $\forall x $ with $0<\lvert x-0 \rvert < \delta$ or $-\delta < x < \delta$

So I want to choose a $\delta$ so that $x_0$ disappears?

I'm having a hard time understanding this process, can somebody give me a walkthrough?

Answer 1

Suppose $\displaystyle\lim_{x \to x_0}f(x) = A$. Fix $\epsilon>0$. Then there is a $\delta>0$ such that $|x-x_0|<\delta$ implies $|f(x)-A|<\epsilon$.

Therefore, if $\ |(x+x_0)-x_0|<\delta$, then $|f(x+x_0)-A|<\epsilon$. (I suppose this line is the trick)

That is, if $\ |x|<\delta$, then $|f(x+x_0)-A|<\epsilon$.

This means (by definition) that $\displaystyle\lim_{x \to 0}f(x+x_0) = A$