# proof of De l’Hôpital’s rule

Let $x_{0}\in\mathbb{R}$, $I$ be an interval containing $x_{0}$ and let $f$ and $g$ be two differentiable functions defined on $I\setminus\{x_{0}\}$ with $g^{\prime}(x)\neq 0$ for all $x\in I$. Suppose that

 $\lim_{x\to x_{0}}f(x)=0,\quad\lim_{x\to x_{0}}g(x)=0$

and that

 $\lim_{x\to x_{0}}\frac{f^{\prime}(x)}{g^{\prime}(x)}=m.$

We want to prove that hence $g(x)\neq 0$ for all $x\in I\setminus\{x_{0}\}$ and

 $\lim_{x\to x_{0}}\frac{f(x)}{g(x)}=m.$

First of all (with little abuse of notation) we suppose that $f$ and $g$ are defined also in the point $x_{0}$ by $f(x_{0})=0$ and $g(x_{0})=0$. The resulting functions are continuous   in $x_{0}$ and hence in the whole interval $I$.

Let us first prove that $g(x)\neq 0$ for all $x\in I\setminus\{x_{0}\}$. If by contradiction   $g(\bar{x})=0$ since we also have $g(x_{0})=0$, by Rolle’s Theorem we get that $g^{\prime}(\xi)=0$ for some $\xi\in(x_{0},\bar{x})$ which is against our hypotheses.

Consider now any sequence $x_{n}\to x_{0}$ with $x_{n}\in I\setminus\{x_{0}\}$. By Cauchy’s mean value Theorem there exists a sequence $x^{\prime}_{n}$ such that

 $\frac{f(x_{n})}{g(x_{n})}=\frac{f(x_{n})-f(x_{0})}{g(x_{n})-g(x_{0})}=\frac{f^% {\prime}(x^{\prime}_{n})}{g^{\prime}(x^{\prime}_{n})}.$

But as $x_{n}\to x_{0}$ and since $x^{\prime}_{n}\in(x_{0},x_{n})$ we get that $x^{\prime}_{n}\to x_{0}$ and hence

 $\lim_{n\to\infty}\frac{f(x_{n})}{g(x_{n})}=\lim_{n\to\infty}\frac{f^{\prime}(x% _{n})}{g^{\prime}(x_{n})}=\lim_{x\to x_{0}}\frac{f^{\prime}(x)}{g^{\prime}(x)}% =m.$

Since this is true for any given sequence $x_{n}\to x_{0}$ we conclude that

 $\lim_{x\to x_{0}}\frac{f(x)}{g(x)}=m.$
Title proof of De l’Hôpital’s rule ProofOfDeLHopitalsRule 2013-03-22 13:23:31 2013-03-22 13:23:31 paolini (1187) paolini (1187) 10 paolini (1187) Proof msc 26A24 msc 26C15