Proving limit of functions involving neighborhoods 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 $\lim \limits_{x \to 0} f(x_0 + x) = A$.
so assume that $\forall \epsilon>0$ $\exists \delta=\delta(\epsilon,x_0) $ such that $|f(x)-A|<\epsilon$ for all $x$ where $0<|x-x_0|<\delta$
Therefore $x_0-\delta<x<x_0+\delta$ and $A-\epsilon<f(x)<A+\epsilon$ 
Can we set $\epsilon $ equal to $x_0 $, since $x_0$ is a some real number?
I'm having a lot of trouble understanding this process, can anyone walk me through it?

If $\lim_{x \to x_{0}}f(x) = A$ and if $\varepsilon > 0$, then there is some $\delta > 0$ such that $0 < |x-x_{0}| < \delta$ only if $|f(x) - A| < \varepsilon$; note that $0 < |x| < \min \{ |2x_{0}-\delta|, |2x_{0}+\delta| \} =: M$ only if $2x_{0} - \delta - x_{0} < x < 2x_{0} + \delta - x_{0}$ and only if  $x_{0} - \delta < x < x_{0} + \delta$; hence $0 < |x| < M$ only if $|f(x+x_{0}) - A| < \varepsilon$.
Can you complete the other way around?
