Formal proof of $\lim \limits_{x \to 0} f(x)=\lim \limits_{x\to a} f(x-a)$ 
$\lim \limits_{x \to 0} f(x)=\lim \limits_{x\to a} f(x-a)$

My strategy (based off an answer to one of my previous questions) has been to set the two limits equal to some number c, so that
$\lim \limits_{x \to 0}$ $f(x)=c$ and $\lim \limits_{x\to a}$ $f(x-a)=c$, and then to show both limits equal each other. So given an $\epsilon >0$, there is a $\delta >0$ such that,
if $0<|x-a|<\delta$, then $|f(x-a)-c|<\epsilon$
and
if $0<|x-0|=|x|<\delta$, then $|f(x)-c|<\epsilon$
How do I proceed from here to show that the two limits are equal?
 A: If you're trying to prove they're equal, then you should not be assuming they're equal by calling them both $c$.
Then you say "So given an $\varepsilon>0$, there is$~\ldots$" etc., but that is something you need to prove, not something you have already shown to be true.
Suppose you let $c=\lim\limits_{x\to0}f(x)$.
Then you need to prove $c=\lim\limits_{x\to a}f(x-a)$.
Given $\varepsilon>0$ you need to prove there exists $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x-a)-c|<\varepsilon$.
You know that there exists $\delta>0$ such that if $0<|w|<\delta$ then $|f(w)-c|<\varepsilon$.
If $0<|x-a|<\delta$ then, letting $w=x-a$, you have $0<|w|<\delta$, so $|f(w)-c|<\varepsilon$.  That says $|f(x-a)-c|<\varepsilon$.
A: It appears you have foobared the definitions a little bit.
As you said put $c = \lim_{x \to a} f(x)$. This translates to saying that $\forall \epsilon > 0$ there exists $\delta > 0$ so that
$$
\lvert x - a \rvert < \delta \implies \lvert f(x) - c \rvert < \epsilon 
$$
(this is not in your question so that's why I believe you foobared the definition).
From here we can but $h:=x-a$ then we have
$$
\lvert h \rvert < \delta \implies \lvert f(h + a) - c \rvert < \epsilon
$$
Since this works for any epsilon this is equivalent to saying that
$$
\lim_{h \to 0} f(h + a) = c
$$
and since we had $\lim_{x \to a} f(x) = c$ we indeed have
$$
\lim_{x \to a} f(x) = \lim_{h \to 0} f(h+a)
$$
