L'Hôpital's rule

A standard one-sided version is as follows. Let $f,g$ be differentiable on $(a,b)$, with $g^{\prime }(x)\ne 0$ there. Suppose either $f(x),g(x)\to 0$ as $x\downarrow a$, or $|g(x)|\to \infty$ there. Ifthen $f(x)/g(x)\to L$. The denominator is nonzero sufficiently near the endpoint. Corresponding versions apply at other finite endpoints and at infinity. The quotient of derivatives must have a limit; differentiating numerator and denominator is not an unrestricted quotient identity.

In the zero-over-zero case, extend both functions continuously to the endpoint by zero and use the Cauchy mean value theorem to write $f(x)/g(x)=f^{\prime }({\xi }_x)/g^{\prime }({\xi }_x)$ for an intermediate ${\xi }_x$. For example, $\sin x/x\to 1$ as $x\to 0$. The rule is sufficient, not necessary: $(x+\sin x)/x\to 1$ as $x\to \infty$, while its derivative ratio $1+\cos x$ has no limit. Without the required indeterminate setting, it can give a wrong answer: $(x+1)/x$ diverges at zero although its derivative ratio is one.

The usual zero-over-zero L'Hopital rule requires the denominator's derivative to be nonzero throughout a sufficiently small deleted neighborhood. A limit taken only where that derivative is nonzero is insufficient. An explicit counterexample uses $x_n=2^{-n}$, $h_n=4^{-n}$, $S(u)=3u^2-2u^3$ and $q(u)=16u^2(1-u{)}^2$. On $[3x_n/4,x_n]$ put $g=h_n$ and $f=h_{n}q((x-3x_n/4)/(x_n/4))$. On $[x_n/2,3x_n/4]$ put $g=h_{n+1}+(h_n-h_{n+1})S((x-x_n/2)/(x_n/4))$ and $f=0$. Extend $g=1,f=0$ beyond one. Values and first derivatives agree at the joins, both functions tend to zero, and $g$ is positive. Where $g^{\prime }\ne 0$, it lies inside a transition interval and $f^{\prime }=0$. Thus $f^{\prime }/g^{\prime }$ has natural-domain limit zero, but $f/g$ equals zero at $x_n$ and one at $7x_n/8$, so has no limit.