Proof using derivative definition? 
Let $\varphi(h)$ be a function such that $\varphi (h)\neq 0$ for all
$h \neq 0$ and $\lim_{h \rightarrow 0} \varphi(h)=0$. Prove that for
any differentiable function $f(x)$ that
$$ \lim_{h \rightarrow 0}\frac{f(x+\varphi(h))-f(x)}{\varphi(h)} =
f'(x) $$

This seems obviously true to me but I'm having trouble writing the proof. I tried using sequence criteria like so:

$$ Let \space h = \frac{1}{n} $$
$$ f'(x) = \lim_{n \rightarrow
 \infty}\frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}} $$
$$ f'(x) = \lim_{n \rightarrow
 \infty}\frac{f(x+\varphi(\frac{1}{n}))-f(x)}{\varphi(\frac{1}{n})} $$
$$ f'(x) = \lim_{h \rightarrow
 0}\frac{f(x+\varphi(h))-f(x)}{\varphi(h)} $$

But this isn't convincing and I'm not sure if it's accurate. I would appreciate a nudge in the right direction.
 A: We have that by differentiability
$$f(x+h)=f(x_0)+f'(x)h+o(h) \quad \frac{o(h)}h \to 0$$
therefore
$$f(x+\varphi(h))=f(x_0)+f'(x)\varphi(h)+o(\varphi(h))$$
and then
$$\frac{f(x+\varphi(h))-f(x)}{\varphi(h)}=\frac{f'(x)\varphi(h)+o(\varphi(h))}{\varphi(h)} = f'(x)+\frac{o(\varphi(h))}{\varphi(h)} \to f'(x)$$
A: Here is a proof that uses the $(\varepsilon,\delta)$ definition of a limit.
It was assumed that $f$ is differentiable, so $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ exists for every $x\in\text{dom}[f]$. We deduce from the $(\varepsilon,\delta)$ definition of a limit that given $\varepsilon>0$, there exists a $\delta_1>0$ such that
$$\left|\frac{f(x+h)-f(x)}{h}-f'(x)\right|<\varepsilon$$
for all $0<|h-0|<\delta_1$. It was also assumed that $\lim_{h\to0}\varphi(h)=0$, so there exists a $\delta>0$ such that
$$|\varphi(h)-0|<\delta_1$$
for all $0<|h-0|<\delta$.
Notice that we must have $0<|\varphi(h)-0|$ because $\varphi(h)\neq 0$ for every $h\neq 0$. Therefore, if $0<|h-0|<\delta$, then $0<|\varphi(h)-0|<\delta_1$, and since
$$\left|\frac{f(x+h)-f(x)}{h}-f'(x)\right|<\varepsilon$$
is true for every $h$ with $0<|h-0|<\delta_1$, it will also be true for $\varphi(h)$. Thus,
$$\left|\frac{f(x+\varphi(h))-f(x)}{\varphi(h)}-f'(x)\right|<\varepsilon$$
Since $\varepsilon$ was given arbitrarily, the same argument applies to every positive real number. Thus,
$$\lim\limits_{h\to 0}\frac{f(x+\varphi(h))-f(x)}{\varphi(h)}=f'(x)$$
This style of argument can be used to prove a wide variety of limits.
