Help with a bit more rigorous proof of the chain rule. We need to show that $D_x(f\circ g)(a)=D_xf(g(a)) D_xg(a)$
$D_xf(g(a)) D_xg(a)=(\lim_{y\to g(a)} \frac{f(y)-f(g(a))}{y-g(a)}) (\lim_{x\to a} \frac{g(x)-(g(a)}{x-a}) $
Substituting $y=g(x)$
$D_xf(g(a)) D_xg(a)=(\lim_{g(x)\to g(a)} \frac{f(g(x))-f(g(a))}{g(x)-g(a)}) (\lim_{x\to a} \frac{g(x)-(g(a)}{x-a}) $
We know that $\forall \epsilon \exists \epsilon' $ such that 
$0<|g(x)-g(a)|< \epsilon' \implies |\frac{f(g(x))-f(g(a))}{g(x)-g(a)} - D_xf(g(a))| < \epsilon$
We also know that $\forall \epsilon' \exists \delta $ such that 
$0<|x-a|< \delta \implies |\frac{g(x)-g(a)}{x-a}| < \epsilon'$
Thus, we can conclude $\forall \epsilon \exists \delta $ such that 
$0<|x-a|< \delta \implies |\frac{f(g(x))-f(g(a))}{g(x)-g(a)}| < \epsilon$
Thus, 
$D_xf(g(a)) D_xg(a)=(\lim_{x\to a} \frac{f(g(x))-f(g(a))}{g(x)-g(a)}) (\lim_{x\to a} \frac{g(x)-(g(a)}{x-a}) $
Ignoring $g(x)=g(a)$ (For now) and applying the rule of product of limits, we get 
$D_xf(g(a)) D_xg(a)=D_x(f \circ g)(a)$
Here's where I need help. I can't find a way to get around the case where $g(x)=g(a)$. It would be really helpful if someone were to verify my proof so far as well.
 A: $\newcommand{\R}{\mathbb{R}}\newcommand\dx{\frac{d}{dx}}$
$\newcommand\vara{a}$
Recall
For a function $f$ differentiable at $b$, we may write a linear approximation for $f$ in a neighborhood of $b$:
\begin{equation}
\newcommand{\linapprox}[2]{f'({#1})({#2}-{#1}) + f({#1}) + o({#2})}
f(x)= \linapprox{\vara}{x} \qquad(*)
\end{equation}
Where $o(x)$ is the difference between $f'(\vara)(x-\vara) + f(\vara)$ and $f(x)$. The differentiability of $f$ implies that $\frac{o(x)-o(\vara)}{x-\vara}$ converges to $0$, and we will use this fact in a moment.

Let $f,g$ be continuous functions, $g$ differentiable at a point $a$, and $f$ differentiable at $g(a)$. Then, $f\circ g$ is also differentiable at $a$ and $(f\circ g)'(a) = (f'(g(a))g'(a)$
Proof
We write the definition of $(f\circ g)':$
\begin{equation}
\lim\limits_{x\rightarrow a}{\frac{f(g(x))-f(g(a))}{x-a}}\qquad(1)
\end{equation}

Now, we will apply $(*)$ to get a linear approximation of $f$ in a neighborhood of $g(a)$ so that $(1)$ becomes the following:
\begin{align}
&\lim\limits_{x\rightarrow a}
\frac{\linapprox{g(a)}{g(x)}
-(\linapprox{g(a)}{g(a)})}{x-a}\\
&=
\lim\limits_{x\rightarrow a}\frac{
f'(g(a))(g(x)-g(a))+o(g(x)) +o(g(a))}{x-a}\qquad\text{after cancelling like terms}
\end{align}
In the above, $\frac{o(x)-o(a)}{x-a}$ converges to $0$ as $x$ converges to $a$. Furthermore, $f'(g(a))$ is a constant, so we can move it outside of the limit.
\begin{equation}
(f\circ g)'(a)= f'(g(a))
\left[\lim\limits_{x\rightarrow a}
\frac{g(x)-g(a)}{x-a}
\right]+
\lim\limits_{x\rightarrow a}\frac{o(x)-o(a)}{x-a}\\\\
=f'(g(a))g'(a)
\end{equation}
Thus $(f\circ g)'(a)$ exists and is equal to $f'(g(a))g'(a)$
