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Theorem – weak L'Hôpital's rule – Let $$I$$ &sube; R, u &isin; $$I$$, and f, g: $$I$$ \ {u} → R be differentiable functions such that &exist;limu g' &isin; R, and  limu g' &ne; 0. If then limu(f/g) also exists, and
 * 1) $$\exists\lim\limits_{u} f=0,\quad\exists\lim\limits_{u} g=0$$ and
 * 2) $$\exists\lim\limits_{u}\frac{f'}{g'}\in \mathbf{R}$$
 * $$\lim\limits_{u}\frac{f}{g}=\lim\limits_{u}\frac{f'}{g'}$$

Proof. 1) u is an accumulation point of he domain of g. Since, limu g' &ne; 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 &isin; $$I$$ there is an x'&isin; $$I$$ between x and u such that
 * $$\frac{g(x)-g(u)}{x-u}=g'(x')$$

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