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Rolle's theorem states that a continuous function $f:[a,b]\to \mathbb{R}$, differentiable on $(a,b)$ with $a<b$ and $f(a)=f(b)$, has $f^{\prime }(c)=0$ for some $c\in (a,b)$. If $f$ is constant this is immediate. Otherwise the extreme value theorem supplies a maximum or minimum with value different from the common endpoint value, so one such extremum occurs in $(a,b)$. At an interior maximum, the difference quotients for positive increments are nonpositive and those for negative increments are nonnegative. Their common limit, the derivative, must be zero. At a minimum the inequalities reverse, with the same conclusion. This proves Rolle's theorem.

For the two-function identity, write $\Delta f=f(b)-f(a)$ and $\Delta g=g(b)-g(a)$, and set $\varphi (x)=g(x)\Delta f-f(x)\Delta g$. This is a continuous function on the closed interval and is differentiable inside it. Its endpoint values agree:They need not be zero, but Rolle's theorem only requires equality. At the resulting point $c$,which proves the Cauchy mean value theorem identityFor the final endpoint limit, first state the denominator convention explicitly. In the usual zero-over-zero version of L'Hopital rule, $g^{\prime }(x)\ne 0$ on some whole interval $(0,\delta )$; this is understood if the displayed derivative quotient is to be a function defined throughout a deleted right neighborhood of zero. Define $f(0)=g(0)=0$, making both functions continuous at zero. For $0<x<\delta$, Rolle's theorem implies $g(x)\ne 0$, since otherwise $g$ would have equal zero values at $0$ and $x$ and a zero derivative between them. Apply the proved Cauchy mean value theorem on $[0,x]$ to obtain $c_x\in (0,x)$ withGiven $\epsilon >0$, the hypothesis on the derivative quotient supplies $\eta >0$ such that its distance from $\ell$ is less than $\epsilon$ whenever $0<t<\eta$. If $0<x<\min (\eta ,\delta )$ then $0<c_x<x<\eta$, so the same bound holds for $f(x)/g(x)$. ThereforeNo continuity of the derivatives is needed.

The PDF does not explicitly state the nonvanishing condition on $g^{\prime }$. If “the limit exists” permits a limit only along the natural domain where $g^{\prime }\ne 0$, the assertion is false, even with $g>0$ everywhere. The distinction concerns derivative quotient limits on partial domains. Here is a concrete counterexample to that weaker reading. Put $x_n=2^{-n}$ and $h_n=4^{-n}$ for $n\ge 0$. On each interval $[x_n/2,x_n]$, split off the plateau $[3x_n/4,x_n]$. On the plateau setOn the transition interval $[x_n/2,3x_n/4]$ setExtend $g=1,f=0$ for $x\ge 1$. The endpoints fit because $S(0)=0$, $S(1)=1$, while both $S^{\prime }$ and $q^{\prime }$ vanish at zero and one and $q$ also vanishes there. Thus $f,g$ are differentiable for all $x>0$ and tend to zero at zero. The derivative $g^{\prime }$ is nonzero exactly inside the transition intervals, where $f^{\prime }=0$, so the derivative quotient has natural-domain limit zero. But $f(x_n)/g(x_n)=0$, whereas $f(7x_n/8)/g(7x_n/8)=1$. Hence $f/g$ has no limit. The preceding qualified proof is the intended valid form of L'Hopital rule.