proof of De l’Hôpital’s rule

Let $x_0\in ℝ$, $I$ be an interval containing $x_0$ and let $f$ and $g$ be two differentiable functions defined on $I\setminus \{x_0\}$ with $g^{\prime }(x)\ne 0$ for all $x\in I$. Suppose that

$$\underset{x\to x_0}{lim}f(x)=0,\underset{x\to x_0}{lim}g(x)=0$$

and that

$$\underset{x\to x_0}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}=m.$$

We want to prove that hence $g(x)\ne 0$ for all $x\in I\setminus \{x_0\}$ and

$$\underset{x\to x_0}{lim}\frac{f(x)}{g(x)}=m.$$

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 $g(x)\ne 0$ for all $x\in I\setminus \{x_0\}$. If by contradiction $g(\overline{x})=0$ since we also have $g(x_0)=0$, by Rolle’s Theorem we get that $g^{\prime }(\xi )=0$ for some $\xi \in (x_0,\overline{x})$ 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 $x_n^{\prime }$ such that

$$\frac{f(x_n)}{g(x_n)}=\frac{f(x_n)-f(x_0)}{g(x_n)-g(x_0)}=\frac{f^{\prime }(x_n^{\prime })}{g^{\prime }(x_n^{\prime })}.$$

But as $x_n\to x_0$ and since $x_n^{\prime }\in (x_0,x_n)$ we get that $x_n^{\prime }\to x_0$ and hence

$$\underset{n\to \mathrm{\infty }}{lim}\frac{f(x_n)}{g(x_n)}=\underset{n\to \mathrm{\infty }}{lim}\frac{f^{\prime }(x_n)}{g^{\prime }(x_n)}=\underset{x\to x_0}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}=m.$$

Since this is true for any given sequence $x_n\to x_0$ we conclude that

$$\underset{x\to x_0}{lim}\frac{f(x)}{g(x)}=m.$$

Title	proof of De l’Hôpital’s rule
Canonical name	ProofOfDeLHopitalsRule
Date of creation	2013-03-22 13:23:31
Last modified on	2013-03-22 13:23:31
Owner	paolini (1187)
Last modified by	paolini (1187)
Numerical id	10
Author	paolini (1187)
Entry type	Proof
Classification	msc 26A24
Classification	msc 26C15