= Solution
First $g(b)\ne g(a)$: otherwise <Rolle's theorem> would produce a point with $g'=0$, contrary to the hypothesis. Define
$$
h(t)=[f(b)-f(a)][g(t)-g(a)]-[g(b)-g(a)][f(t)-f(a)].
$$
It is <continuous> on $[a,b]$, <differentiable> inside, and vanishes at both endpoints. <Rolle's theorem> gives a $\xi\in(a,b)$ with
$$
[f(b)-f(a)]g'(\xi)-[g(b)-g(a)]f'(\xi)=0.
$$
The two denominators are nonzero, so this proves <Cauchy's mean value theorem>:
$$
\boxed{\frac{f'(\xi)}{g'(\xi)}=\frac{f(b)-f(a)}{g(b)-g(a)}.}
$$
When $f(a)=g(a)=0$, apply that theorem on $[a,x]$ for each $a<x\leq b$. <Rolle's theorem> again ensures $g(x)\ne0$, and some $\xi_x\in(a,x)$ satisfies $f(x)/g(x)=f'(\xi_x)/g'(\xi_x)$. As $x\downarrow a$, the trapped point $\xi_x$ also tends to $a$. Therefore
$$
\boxed{\frac{f(x)}{g(x)}\longrightarrow\ell\quad\text{as }x\downarrow a.}
$$
This proves the stated endpoint case of <L'Hôpital's rule>. It does not assume <continuity> of the derivatives; only the indicated derivative-ratio limit is used. The same trapping argument also works for an extended infinite limit if that convention is allowed.
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