Why does the result follow? How does this theorem follow?

Theorem. If $g$ is differentiable at $a$ and $g(a) \neq 0$, then $\phi = 1/g$ is also differentiable at $a$, and $$\phi'(a) = (1/g)'(a) = -\frac{g'(a)}{[g(a)]^2}.$$
Proof. The result follows from $$\frac{\phi(a+h)-\phi(a)}{h} = \frac{g(a)-g(a+h)}{hg(a)g(a+h)}.$$


$$
  \lim_{h \to 0} \frac{g(a)-g(a+h)}{hg(a)g(a+h)}=  \lim_{h \to 0} \left( \frac{g(a)-g(a+h)}{h} \cdot \frac{1}{g(a)g(a+h)} \right) =  \\ = -\lim_{h \to 0} \frac{g(a+h)-g(a)}{h} \cdot   \lim_{h \to 0} \frac{1}{g(a)g(a+h)} = -g'(a) \cdot \frac{1}{[g(a)]^2} = -\frac{g'(a)}{[g(a)]^2}.
$$
Can you recognize the limit of the left hand side if $h$ tends to $0$?
