L'Hôpital's rule

For an indeterminate quotient whose numerator and denominator both tend to zero or both diverge, suitable differentiability and denominator conditions allow its limit to be computed from the limit of the quotient of their derivatives.

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.