Cauchy's Mean Value Theorem and L'Hopital's Rule

Consider the problem of the limit of a quotient,

$$\frac{f(x)}{g(x)}$$

as x tends to a. There are really four possibilities if f and g separately have limits as x tends to a. Both limits can be different from , and then the limit of the quotient is the quotient of the limits. The limit of f can be while the limit of g is not , and then the limit of the quotient is still the quotient of the limits. The limit of f can be different from while the limit of g is , and then the limit of the quotient is undefined. Finally, both limits can be , and then we are never sure what happens. Sometimes the limit of the quotient is defined, as in the case of derivatives, and sometimes not. We shall give below another tool to try in such cases, called L'Hopital's Rule. It is based on the following rather mysterious looking version of Rolle's Theorem, called Cauchy's Mean Value Theorem (CMVT).

Theorem 8238 (Cauchy's Mean Value Theorem)

Suppose that $F:[a,b]\rightarrow(-\infty,\infty)$ and $G:[a,b]\rightarrow (-\infty,\infty)$ are continuous functions which are differentiable on (a,b). Then for some $z\in(a,b)$

F'(z)(G(b) - G(a)) = G'(z)(F(b)-F(a)).

Note that when G(x) = x we get the conclusion of the ordinary Mean Value Theorem.

The CMVT is no harder to prove than the Mean Value Theorem, as it relies on the same trick. Define $R:[a,b]\rightarrow(-\infty,\infty)$ given by

R(u) = (F(u)-F(a))(G(b)-G(a))-(F(b)-F(a))(G(u)-G(a))

R satisfies the hypotheses of Rolle's Theorem:

• R is continuous on [a,b];
• R is differentiable on (a,b);
• R(a) = 0 = R(b).

Hence for some $z\in(a,b)$ we have R'(z) = 0. We have

R'(u) = (F'(u)-0)(G(b)-G(a))-(F(b)-F(a))(G'(u)-0) = F'(u)(G(b)-G(a))-(F(b)-F(a))G'(u)

so plugging in z for u gives

0 = F'(z)(G(b)-G(a))-(F(b)-F(a))G'(z)

which is the same as the conclusion of the CMVT.

QED

What has this got to do with finding the behaviour of

$$\frac{f(x)}{g(x)}$$

as x tends to a in the case where

$$\begin{array} {rcl} \lim_{x\rightarrow a}f(x) & = & 0\\ \lim_{x\rightarrow a}g(x) & = & 0?\end{array}$$

If we define f(a) = 0 and g(a) = 0 then f and g are continuous at a. If we know that f and g are differentiable at all points near a then f and g are continouous on some closed interval containing a. Hence for each $x\neq a$ there is (by using the CMVT on an interval with endpoints x and a) a number z_x between x and a so that

f'(z_x)(g(x)-g(a)) = g'(z_x)(f(x) - f(a)).

Since g(a) = f(a) = 0 we have

f'(z_x)g(x) = g'(z_x)f(x),

so if dividing by is not a problem we have

$$\frac{f(x)}{g(x)} = \frac{f'(z_x)}{g'(z_x)}$$

and as x tend to a, so does z_x. Hence, if

$$\lim_{z\rightarrow a}\frac{f'(z)}{g'(z)} = L$$

then

$$\lim_{x\rightarrow a}\frac{f(x)}{g(x)} = L$$

also. The precise statement of what we know is called L'Hopital's Rule:

Theorem 8294 (L'Hopital's Rule)

Suppose that $f:(a,b)\rightarrow(-\infty,+\infty)$ and $g:(a,b)\rightarrow(-\infty,+\infty)$ are differentiable and $g'(x) \neq 0$ for all $x\in(a,b)$, where $-\infty \leq a < b \leq +\infty$. Suppose that

$$\frac{f'(x)}{g'(x)} \rightarrow A\;\;{\rm as}\;\; x\rightarrow a.$$

If

$$f(x)\rightarrow 0 \;\;{\rm and}\;\;g(x)\rightarrow 0 \;\;{\rm as}\;\;x\rightarrow a$$

or

$$g(x)\rightarrow +\infty \;\;{\rm as}\;\;x\rightarrow a$$

then

$$\frac{f(x)}{g(x)} \rightarrow A\;\;{\rm as}\;\; x\rightarrow a.$$

The analogous statements are true if $g(x)\rightarrow +\infty$ is replaced with $g(x) \rightarrow -\infty$, or if $x \rightarrow b$, or if $A = -\infty$or if $A = -\infty$. Notice also that as stated, the rule is for right and left handed limits. It also applies to ordinary limits.

As a practical rule it is best to use L'Hopital's Rule as a last resort. Also, in the case of limits of ratios of the form

$$\frac{f(x)-f(a)}{x-a}$$

L'Hopital's rule is of no theoretical importance as such limits define the derivative of f and would, therefore, be needed to apply L'Hopital's rule. However, it does provide a rather mindless way to evaluate such limits if one has forgotten how the derivative formulas were derived.