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.

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