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

Reference: Thomas, Finney, 9th edition, Calculus and Analytic Geometry.

Continuing our previous discussion of “theoretical” calculus or “rigorous” calculus, I am reproducing below the proof of the finite limit case of the stronger form of l’Hopital’s Rule :

L’Hopital’s Rule (Stronger Form):

Suppose that

$f(x_{0})=g(x_{0})=0$

and that 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 at $x_{0}$ . Then,

$\lim_{x \rightarrow x_{0}}\frac{f(x)}{g(x)}=\lim_{x \rightarrow x_{0}}\frac{f^{x}}{g^{x}}$ ….call this equation I,

provided the limit on the right exists.

The proof of the stronger form of l’Hopital’s Rule is 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 that the functions f and g are continuous on $[a,b]$ and differentiable throughout $(a,b)$ and suppose also that $g^{'} \neq 0$ throughout $(a,b)$ . Then there exists a number c in $(a,b)$ at which

$\frac{f^{'}(c)}{g^{'}(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}$ …call this II.

The ordinary Mean Value Theorem is the case where $g(x)=x$ .

Proof of Cauchy’s Mean Value Theorem:

We apply the Mean Value Theorem twice. First we use it to show that $g(a) \neq g(b)$ . For if $g(b)$ did equal to $g(a)$ , then the Mean Value Theorem would give:

$g^{'}(c)=\frac{g(b)-g(a)}{b-a}=0$ 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 $F(b) = F(a)=0$ . Therefore, 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 II above. QED.

Proof of the Stronger Form of l’Hopital’s Rule:

We first prove I for the case $x \rightarrow x_{o}^{+}$ . The method needs no change to apply to $x \rightarrow x_{0}^{-}$ , and the combination of those two cases establishes the result.

Suppose that x lies to the right of $x_{o}$ . Then, $g^{'}(x) \neq 0$ and we can apply the Cauchy’s Mean Value Theorem to the closed interval from $x_{0}$ to x. This produces a number c between $x_{0}$ and x 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 $x_{0}$ because 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 x approaches $x_{0}$ from above. The case where x approaches $x_{0}$ from below is proved by applying Cauchy’s Mean Value Theorem to the closed interval $[x,x_{0}]$ , where $x< x_{0}$ . QED.

Mathematics Hothouse