proof of l'Hôpital's rule that minimizes special-casing A simple form of l'Hôpital's rule looks like this: If $u$ and $v$ are functions with $u(0)=0$ and $v(0)=0$, the derivatives $\dot{v}(0)$ and $\dot{v}(0)$ are defined,
and the derivative $\dot{v}(0)\ne 0$,
then
\begin{align*}
  \lim_{x\rightarrow 0} \frac{u}{v} &= \frac{\dot{u}(0)}{\dot{v}(0)} \qquad .
\end{align*}
To me, the clearest way to arrive at this result uses a little nonstandard analysis: Since $u(0)=0$, and the derivative $d u/d x$ is defined at $0$, $u(d x)=d u$ is infinitesimal, and likewise for $v$. By the definition
of the limit, the limit is the standard part of
\begin{equation*}
  \frac{u}{v} = \frac{d u}{d v} = \frac{d u/d x}{d v/d x} \qquad ,
\end{equation*}
where by assumption the numerator and denominator are both defined (and finite, because the derivative
is defined in terms of the standard part). The standard part of a quotient like $p/q$ equals the quotient of the
standard parts, provided that both $p$ and $q$ are finite (which we've established), and $q \ne 0$ (which is
true by assumption). But the standard part of $d u/d x$ is the definition of the derivative $\dot{u}$, and
likewise for  $d v/d x$, so this establishes the result.
The generalizations to $x\rightarrow a$, where $a\ne 0$, and $x\rightarrow \infty$ are pretty trivial with the changes of variable $x\rightarrow x-a$ and $x\rightarrow 1/x$.
But there are a bunch of other cases of l'Hôpital's that seem to me to involve toxic doses of case-splitting. There are cases where you have to differentiate more than once, and cases where the indeterminate form is $\infty/\infty$ rather than $0/0$.
Is it possible to treat all of this in a unified way, possibly using ideas from projective geometry or inversions with respect to a circle in the complex plane?
 A: Last week I covered L'Hôpital's Rule in my Spivak Calculus course.  I ended up preferring the proof given in Rudin's Principles to Spivak's treatment, because (i) he does compile several different cases in a rather efficient way and (ii) he proves a stronger result in the case where $\lim_{x \rightarrow a} g(x) = \infty$, namely that there is no hypothesis needed on the limiting behavior of the numerator $f$.
I wrote this up in $\S$ 1 of these notes.  However I must confess that I found the proof itself not very interesting and ended up not covering it in class.
I must also confess that I don't really understand the OP's sketch proof.  In particular I do not recognize the hypothesis $g'(a) \neq 0$ that the OP is assuming: who says that $f$ and $g$ are even defined at $a$?  I am also slightly skeptical that NSA really helps here to give a shorter proof, but I would be very interested to be proven wrong about this.
Added: If you are willing to assume that $f$ and $g$ are defined and differentiable at $a$, that $f(a) = g(a) = 0$ and that $g'(a) \neq 0$, the proof becomes almost trivial:
$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{g(x)-g(a)} = \lim_{x \rightarrow a} \frac{ \frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}} = \frac{ \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a}}{\lim_{x \rightarrow a} \frac{g(x)-g(a)}{x-a}} = \frac{f'(a)}{g'(a)}$.
(However this version is inadequate for many of the standard applications of freshman calculus.)  So I presume we're talking about a stronger version than this?
