TITLE: How does one prove L'Hôpital's rule?
QUESTION [20 upvotes]: L'Hôpital's rule  can be stated as follows:

Let $f, g$ be differentiable real functions defined on a deleted one-sided neighbourhood$^{(1)}$ of $a$, where $a$ can be any real number or $\pm \infty$. Suppose that both $f,g$ converge to $0$ or that both $f,g$ converge to $+\infty$ as $x \to a^{\pm}$ ($\pm$ depending on the side of the deleted neighbourhood). If
  $$\frac{f'(x)}{g'(x)} \to L,$$
  then 
  $$\frac{f(x)}{g(x)} \to L,$$
  where $L$ can be any real number or $\pm \infty$. 

This is an ubiquitous tool for computations of limits, and some books avoid proving it or just prove it in some special cases. Since we don't seem to have a consistent reference for its statement and proof in MathSE and it is a theorem which is often misapplied (see here for an example), it seems valuable to have a question which could serve as such a reference. This is an attempt at that.
$^{(1)}$E.g., if $a=1$, then $(1,3)$ is such a neighbourhood.

REPLY [2 votes]: By definition, $f'(a) = \lim_\limits{x\to a} \frac {f(x) - f(a)}{x-a}$
if $f'(a), g'(a)$ exist and $g'(a) \ne 0$
$\frac {f'(a)}{g'(a)} = \lim_\limits{x\to a} \dfrac {\frac {f(x) - f(a)}{x-a}}{\frac {g(x) - g(a)}{x-a}} =  \lim_\limits{x\to a}\frac {f(x) - f(a)}{g(x) - g(a)}$
if $f(a), g(a) = 0$
$\frac {f'(a)}{g'(a)} = \lim_\limits{x\to a}\frac {f(x) }{g(x)}$