# Cauchy’s Mean Value Theorem and the Stronger Form of l’Hopital’s Rule

Reference: G B Thomas, Calculus and Analytic Geometry, 9th Indian Edition.

The stronger form of l’Hopital’s rule is as follows:

Suppose that $f(x_{0})=g(x_{0})=0$ and the functions f and g are both differentiable on an open interval $(a,b)$ that contains the point $x_{0}$. Suppose also that $g^{'} \neq 0$ at every point in $(a,b)$ except possibly $x_{0}$. Then, $\lim_{x \rightarrow x_{0}} \frac{f(x)}{g(x)} = \lim_{x \rightarrow x_{0}}\frac{f^{'}(x)}{g^{'}(x)}$…call this I, provided the limit on the right exists.

Remarks:

The proof of the stronger from of l’Hopital’s rule in based on Cauchy’s mean value theorem, a mean value theorem that involves two functions instead of one. We prove Cauchy’s theorem first and then show how it leads to l’Hopital’s rule.

Cauchy’s Mean Value Theorem:

Suppose the functions f and g are continuous on $[a,b]$ and differentiable through out $(a,b)$ and suppose also that $g^{'} \neq 0$ through out $(a,b)$. Then, there exists a number c in $(a,b)$ at which $\frac{f^{'}(c)}{g^{'}(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}$.

(Note this becomes the ordinary mean value theorem when $g(x)=x$).

Proof of Cauchy’s Mean Value theorem:

We apply the ordinary mean value theorem twice. First, we use it to show that $g(b) \neq g(a)$. Because if $g(b)=g(a)$, then the ordinary Mean Value theorem says that $g^{'}(c) = \frac{g(b)-g(a0}{b-a}=0$ for some c between a and b. This cannot happen because $g^{'}(x) \neq 0$ in $(a,b)$.

We next apply the Mean Value Theorem to the function $F(x) = f(x)-f(a) - \frac{f(b)-f(a)}{g(b)-g(a)}(g(x)-g(a))$

This function is continuous and differentiable where f and g are, and note that $F(b)=F(a)=0$. Therefore, by the ordinary mean value theorem, there is a number c between a and b for which $F^{'}(c)=0$. In terms of f and g, this says $F^{'}(c) = f^{'}(c) - \frac{f(b)-f(a)}{g(b)-g(a)}(g^{'}(c)) = 0$

or $\frac{f^{'}(c)}{g^{'}(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}$ which is equation II above.

Proof of the stronger form of L’Hopital’s Rule:
We first establish equation I for the case $\lim x \rightarrow x_{0}^{+}$. The method needs almost no change to apply to the case $\lim x \rightarrow x_{0}^{-}$, and the combination of these two cases establishes the result.

Suppose that x lies to the right of $x_{0}$. Then, $g^{'}(x) \neq 0$ and we can apply Cauchy’s Mean Value theorem to the closed interval from $x_{0}$ to x. This produces a number c between x and $x_{0}$ such that $\frac{f^{'}(c)}{g^{'}(c)} = \frac{f(x)-f(x_{0})}{g(x)-g(x_{0})}$

But, $f(x_{0})=g(x_{0})=0$ so

That $\frac{f^{'}(c)}{g^{'}(c)}= \frac{f(x)}{g(x)}$

As x approaches $x_{0}$, c approaches $x_{0}$ as it lies between x and $x_{0}$. Therefore, $\lim_{x \rightarrow x_{0}^{+}} \frac{f(x)}{g(x)} = \lim_{x \rightarrow x_{0}^{+}} \frac{f^{'}(c)}{g^{'}(c)} = \lim_{x \rightarrow x_{0}^{+}} \frac{f^{'}(x)}{g^{'}(x)}$.

This establishes l’Hopital’s Rule for the case where approaches $x_{0}$ from right. The case where x approaches $x_{0}$ from the left is proved by applying Cauchy’s Mean Value Theorem to the closed interval $[x,x_{0}]$ when $x < x_{0}$.

QED.

Regards,

Nalin Pithwa

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