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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]

By the definition of little o notation, the required quotient isAs $ϵ\to 0^{+}$, write this as $\log ϵ/(1/ϵ)$. L'Hôpital's rule applies to this infinite-over-infinite ratio and givesTherefore ${ϵ}^2\log ϵ=o(ϵ)$. The logarithm is negative on this approach, which is consistent with a quotient tending to zero from below; little o notation concerns the limiting ratio, not its sign.^[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

Both numerator and denominator have a double zero at one. Applying L'Hôpital's rule twice givesAlternatively $x^3-3x+2=(x-1{)}^2(x+2)$ and $1-\sin (\pi x/2)\sim {\pi }^2(x-1{)}^2/8$, giving the same result from a Taylor expansion.

The numerator and denominator both vanish at zero. The fundamental theorem of calculus differentiates the numerator, while the denominator derivative is nonzero near zero. L'Hôpital's rule therefore gives

L'Hôpital's rule applies to a quotient $f(x)/g(x)$ of differentiable functions near a limiting point, excluding that point itself, provided $g^{\prime }(x)\ne 0$ there. If the quotient has an indeterminate form $0/0$ or $\infty /\infty$, and the limit of $f^{\prime }(x)/g^{\prime }(x)$ exists as a finite or infinite value $L$, thenThe rule also applies to one-sided limits and to limits at infinity, under the corresponding hypotheses. Existence of the derivative-ratio limit and the indeterminate-form hypotheses are required; an arbitrary quotient cannot be replaced by a quotient of derivatives.

Write the product as a quotient with indeterminate form $(-\infty )/(+\infty )$. By L'Hôpital's rule,