Prove that $\lim_{x \to 0} f(x)=0$ if and only if $\lim_{x \to c} f(x-c) = 0$
Consider the statement in the forward direction (1. $\Rightarrow$ 2.). If $\lim_{x \to 0} f(x)=0$ (if $f: A \subseteq \mathbb{R} \rightarrow \mathbb{R}$), then we will have that $\forall \epsilon>0$, $\exists \delta_{1}>0$ such that $\forall x \in A, x \neq 0$ then $|x| < \delta_{1}$ implies $|f(x)|<\epsilon$ by definition of the limit.
I was thinking of letting $x-c = t$ and then we would have $|x-c|=|t|<\delta$ (if we choose $\delta$ to be $\delta < \delta_{1}$ ). Now, $|x-c| = |t| \neq 0$ as we require $x \neq c$. By definition, this would imply that $|f(t)| < \epsilon \Rightarrow |f(x-c)|<\epsilon$, thus completing the proof in one direction. However, I don't really understand if we can transfer this argument for $x$ to $t$ (what if $t=x-c \notin A$?).
What would be a more intuitive way to solve this problem? Or am I missing something simple? Thank you for your time.