I am working on the following exercise:
Prove from the $\epsilon$-$\delta$ definition of a limit that $\lim_{x\to a}f(x)=l$ if and only if $\lim_{h\to 0} f(a+h)=l$.
My Attempt: Assume that $\lim\limits_{x\to a}f(x)=l$. By the definition of a limit, for any $\epsilon >0$ there exists $\delta >0$ such that $|x-a|<\delta$ implies $|f(x)-l|<\epsilon$. Let $h=x-a$. Then, $|x-a|<\delta$ implies $|h|=|h-0|<\delta$, and $|f(x)-l|<\epsilon$ implies $|f(a+h)-l|<\epsilon$. Conversely, assume that $\lim\limits_{h\to 0}f(a+h)=l$. Then, for any $\epsilon>0$ there exists $\delta >0$ such that $|h-0|<\delta$ implies $|f(a+h)-l|<\epsilon$. Let $x=a+h$. Then, $|h-0|<\delta$ implies $|x-a|<\delta$, and $|f(a+h)-l|<\epsilon$ implies $|f(x)-l|<\epsilon$.
I'm wondering if this type of substitution is allowed (i.e. is this a valid proof?).