 Solution

First $g(b)\ne g(a)$: otherwise Rolle's theorem would produce a point with $g^{\prime }=0$, contrary to the hypothesis. DefineIt is continuous on $[a,b]$, differentiable inside, and vanishes at both endpoints. Rolle's theorem gives a $\xi \in (a,b)$ withThe two denominators are nonzero, so this proves Cauchy's mean value theorem:When $f(a)=g(a)=0$, apply that theorem on $[a,x]$ for each $a<x\le b$. Rolle's theorem again ensures $g(x)\ne 0$, and some ${\xi }_x\in (a,x)$ satisfies $f(x)/g(x)=f^{\prime }({\xi }_x)/g^{\prime }({\xi }_x)$. As $x\downarrow a$, the trapped point ${\xi }_x$ also tends to $a$. ThereforeThis 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.