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?