I wanted to verify that for definition of the derivative it is true that:
$$ \lim_{x \to a}\frac{f(x)-f(a)}{x-a}= \lim_{h\to 0}\frac{f(a+h) -f(a)}{h}$$
If I denote $h=x-a$, we can let $x\to a$, this means that $h \to 0$. Also notice that $f(x)=f(a+h)$ $$ \lim_{x \to a}\frac{f(x)-f(a)}{x-a}=\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}$$ As desired.
Is the main idea here really just to use a variable substitution?
I suppose there is some formal theorem I also use here that when $f(x) \to y$ we can let $g(f(x)) \to g(y)$ if the function $g$ is continuous.
If I understand the problem correctly, the question is if it is possible to prove the following equality:
$$\lim\limits_{x\to a}g(x)=\lim\limits_{h\to 0}g(a+h)$$
Because in this case you can use: $$g(x) = \frac{f(x)-f(a)}{x-a}$$
If you want to use the formal definition of the limit:
$$\lim\limits_{x\to a}g(x)=y$$ means that for every $\epsilon>0$ given we can find a value $\delta>0$ so that the following inequation applies for all $w\in[-\delta,\delta]$: $$|g(\underbrace{a+w}_{"x\to a"})-y|<\epsilon$$
However, because $a+w=a+(0+w)$ this also means that: $$|g(a+\underbrace{(0+w)}_{"h\to 0"})-y|=|g(\underbrace{a+w}_{"x\to a"})-y|<\epsilon$$
This however means that: $$\lim\limits_{h\to 0}g(a+h)=y$$ ... because we can also find a value $\delta$ for each given value $\epsilon$ ...