I've been working on understanding limits thoroughly, so I'm rewriting how I understand the chain rule. Please help me fill in my gaps in understanding.
$f$ is some function. Then
$f'(x) = \lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}$
Now I might want to evaluate something like
$$\left(f(g(x))\right)' = \lim\limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}$$
Evaluating this is tricky, so we need a way to do it.
$\left(f(g(x))\right)' = (\lim\limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}\cdot\frac{g(x+h)-g(x)}{h}$)
$\left(f(g(x))\right)' = (\lim\limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)})\cdot\lim\limits_{h \to 0}(\frac{g(x+h)-g(x)}{h})$
if $k=g(x+h)-g(x)$, then
$g(x+h)=k+g(x)$, so
$\left(f(g(x))\right)' = (\lim\limits_{h \to 0}\frac{f(g(x)+k)-f(g(x))}{k})\cdot(\lim\limits_{h \to 0}\frac{g(x+h)-g(x)}{h})$
and, assuming I can go ahead and just change the limit variable on the left term, then
$=(\lim\limits_{k \to 0}\frac{f(g(x)+k)-f(g(x))}{k}) \cdot (\lim\limits_{h \to 0}\frac{g(x+h)-g(x)}{h})$
$=f'(g(x))\cdot g'(x)$
Which is easier to figure out, and is also the chain rule.
Is that correct?