A basic calculus question. $\Big\{ \lim\limits_{x \to a} f(x-a) = L\Big\} \Longleftrightarrow \Big\{ \lim\limits_{h \to 0} f(h) = L\Big\}$
True or false? 

Edit: Many thanks to the people who provided answers for this question.
I was confused about the domain of $f(x-a)$. If $D$ is the domain of $f(x)$, then the domain of $f(x-a)$ is taken to be $a + D$. For some reason I had it in my head that in some situations the domain of $f(x-a)$ might be taken to be some proper subset of $a + D$, for example $(a + D) \cap \mathbb{Q}$. I thought that then maybe the existence of $\lim\limits_{x \to a} f(x-a)$ might have been calculated on $(a + D) \cap \mathbb{Q}$, say, instead of $a + D$.
I should have asked that question (the default domain of $f(x - a)$) instead of this question...
 A: Let $g(x)=f(x-a)$ for each $x$ such that $x-a$ is in the domain of $f$. Note that this implies that $x$ is in the domain of $f$ if and only if $x+a$ is in the domain of $g$, and $f(x)=g(x+a)$.
Suppose that $\lim\limits_{h\to 0}f(h)=L$. This means that for each $\epsilon>0$ there is a $\delta_\epsilon>0$ such that $|f(h)-L|<\epsilon$ whenever $0<|h|<\delta_\epsilon$ and $x$ is in the domain of $f$. Let $x=h+a$, and suppose that $0<|x-a|<\delta_\epsilon$ and $x$ is in the domain of $g$. Then $h=x-a$ is in the domain of $f$, and $0<|h|<\delta_\epsilon$, so $|g(x)-L|=|f(x-a)-L|=|f(h)-L|<\epsilon$, and it follows that $\lim\limits_{x\to a}f(x-a)=L$.
The argument that $\lim\limits_{x\to a}f(x-a)=L$ implies that $\lim\limits_{h\to 0}f(h)=L$ is very similar; you might want to see whether you can work it out on your own.
A: Let $h=x-a$. Then as $x\rightarrow a$, $h\rightarrow 0$.
This means that
\begin{align*}
L & = \lim_{x\rightarrow a}f\left(x-a\right)\\
 & = \lim_{h\rightarrow0}f\left(h\right).
\end{align*}
