# User:Mozo/plans

Theorem – weak L'Hôpital's rule – Let $I$R, u$I$, and f, g: $I$ \ {u} → R be differentiable functions such that ∃limu g' ∈ R, and limu g' ≠ 0. If

1. $\exists\lim\limits_{u} f=0,\quad\exists\lim\limits_{u} g=0$ and
2. $\exists\lim\limits_{u}\frac{f'}{g'}\in \mathbf{R}$

then limu(f/g) also exists, and

$\lim\limits_{u}\frac{f}{g}=\lim\limits_{u}\frac{f'}{g'}$

Proof. 1) u is an accumulation point of he domain of g. Since, limu g' ≠ 0.

2) Both f and g extend to an $I$R differentiable function, by setting f(u)=0 and g(u)=0. Indeed, by Lagrange Theorem, for every x$I$ there is an x'∈ $I$ between x and u such that

$\frac{g(x)-g(u)}{x-u}=g'(x')$

Hence,

$g'(u)=\lim\limits_{x\to u}\frac{g(x)-g(u)}{x-u}=\lim\limits_{x'\to u}g'(x')=\lim\limits_{u}g'\in \mathbf{R}$