Why is this famous proof of the chain rule called "technically incorrect" in this pdf? So I was looking through various proofs of the chain rule...and I came across this paper. The first proof given is complete and quite well-explained. But another simplistic proof is given in the end...which is mentioned as "technically incorrect". Can anyone tell me why?
Here is the incorrect proof in question:
$$\begin{aligned}
(f \circ g)'(x) &= \lim_{h \to 0}\frac{f(g(x+h)) - f(g(x))}{h} \\
\implies (f \circ g)'(x) \cdot \left(\frac{1}{g'(x)}\right) &= \lim_{h \to 0}\left(\frac{f(g(x+h)) - f(g(x))}{h}\right)\cdot\left(\frac{h}{g(x+h)-g(x)}\right)\\
&= \lim_{h \to 0}\left(\frac{f(g(x+h)) - f(g(x))}{g(x+h)-g(x)}\right) \\
&= f'(g(x)) \\
\end{aligned}$$
 A: What if $g$ is a constant function?
A: If $g'(x) = 0$, the proof fails, but as mentioned in the comments, this can be handled as a separate case.
A more serious issue is the calculation
$$(f \circ g)'(x) \cdot \left(\frac{1}{g'(x)}\right) = \lim_{h \to 0}\left(\frac{f(g(x+h)) - f(g(x))}{h}\right)\cdot\left(\frac{h}{g(x+h)-g(x)}\right)$$
If we write 
$$a(h) = \frac{f(g(x+h))-f(g(x))}{h}$$
and
$$b(h) = \frac{h}{g(x+h)-g(x)}$$
then the above calculation is the assertion that
$$\lim_{h \to 0}\ a(h) \cdot \lim_{h \to 0}\ b(h) = \lim_{h \to 0}\ a(h)b(h),$$
in other words, that the limit of a product is the product of the limits.
This is true, provided that both limits on the left-hand side exist. But the existence of $\displaystyle \lim_{h \to 0}\ a(h)$ is exactly what we are trying to prove, so the argument is circular.

Edit to add:
Another issue worth noting is the final assertion, namely
$$f'(g(x)) = \lim_{h \to 0}\frac{f(g(x+h)) - f(g(x))}{g(x+h)-g(x)}$$
In fact, the definition of the derivative of $f$, evaluated at $g(x)$, gives us
$$f'(g(x)) = \lim_{k \to 0}\frac{f(g(x)+k) - f(g(x))}{k}$$
The only reason we are able to conclude that these two expressions are equal (even after handling the $g(x+h) = g(x)$ case properly) is because $g$ is continuous at $x$. This of course follows from the differentiability of $g$ at $x$, but a careful proof would point this out.
