User:Mozo/plans

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Theorem – weak L'Hôpital's rule – Let IR, uI, 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 IR differentiable function, by setting f(u)=0 and g(u)=0. Indeed, by Lagrange Theorem, for every xI 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}