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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