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For $a<b$ and real-valued functions $f,g$ that are continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, there exists $\xi \in (a,b)$ withApply Rolle's theorem to the linear combination $h(x)=[f(b)-f(a)]g(x)-[g(b)-g(a)]f(x)$, which has equal endpoint values. The displayed product identity permits zero derivatives. For the ratio form, additionally assume $g^{\prime }(x)\ne 0$ for every $x\in (a,b)$. Then Rolle's theorem ensures $g(b)\ne g(a)$, so both denominators below are nonzero:Neither derivative needs to be continuous, and $f^{\prime }$ may vanish. Choosing $g(x)=x$ recovers the mean value theorem; the ratio form also gives the zero-over-zero endpoint case of L'Hôpital's rule. ^[ref]

Now impose the stated zero endpoint values and nonvanishing assumptions near $a$. For each $x\in (a,a+\delta )$, applying the proved identity on $[a,x]$ gives some $c_x\in (a,x)$ such thatAs $x\downarrow a$, every possible $c_x$ also tends to $a$. Explicitly, for any $\epsilon >0$, choose a right neighborhood of $a$ where $|f^{\prime }/g^{\prime }-L|<\epsilon$. If $x$ lies in it, so does $c_x$, and therefore $|f(x)/g(x)-L|<\epsilon$. This proves the endpoint zero-over-zero form of L'Hôpital's rule:No continuity of either derivative is required, only their ratio limit and the given nonvanishing conditions.

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.

Set $F(h)=f(a+h)+f(a-h)-2f(a)$ and $G(h)=h^2$. Both vanish at zero. For $h\ne 0$, their derivative ratio isTwice differentiability at $a$ makes both quotients tend to $f^{\prime \prime }(a)$. Apply the endpoint L'Hôpital's rule proved in part (iii) for $h\downarrow 0$. For the other side use the reflected positive variable $t=-h$; $F$ and $G$ are even, so their quotient is unchanged. Thus the two-sided limit isThis is the second-order central difference limit. The proof only uses differentiability of $f^{\prime }$ at $a$ and differentiability nearby; it does not incorrectly assume $f^{\prime \prime }$ is continuous.

As ${\lambda }_1\to {\lambda }_2=\lambda$, L'Hôpital's rule gives $b\to t{e}^{\lambda t}$, soThe quadratic becomesFor $t\ge 0$, its maximum root is