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

Let f and g be differentiable functions on the interval [a,b], with a < b.
Consider the following function:

h(x) = (f(x) - f(a)(g(b) - g(a)) - (g(x) - g(a))(f(b) - f(a).

Then if we evaluate h at x = a and at x = b, we get:

h(a) = h(b) = 0.

So by Rolle's theorem there exists a point $\xi$ in the interval (a, b), such that $h'(\xi) = 0$.
Then we have $0 = h'(\xi) = f'(\xi)(g(b) - g(a)) - g'(\xi)(f(b) - f(a))$.
In particular, if g' is non-zero in the open interval (a,b) and if $g(b) \ne g(a)$, we have, for some $\xi$ in the interval (a,b):

$$\frac{f'(\xi)}{g'(\xi)} = \frac{f(b) - f(a)}{g(b) - g(a)}.$$

This is called Cauchy's Mean Value Theorem.
In the special case that g(x) = x, so g'(x) = 1, this reduces to the ordinary mean value theorem.
Now consider the case that both f(a) and g(a) vanish and replace b by a variable x.
Then we have, provided f(a) = g(a) = 0 and $g'(x) \ne 0$ in an interval around a, except possibly at x = a:

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

where $x \ne a$, but x is near a and then $\xi(x)$ (which depends on x) is a point between x and a.
Now suppose $\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)} = L$ exists.
Then $\lim_{x\rightarrow a} \frac{f'(\xi(x))}{g'(\xi(x))} = L$ also, so we get L'Hopital's rule:

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

It must be emphasized that this rule only has been shown to hold when both f(a) and g(a) vanish.
If now f and g are twice differentiable and f'(a) and g'(a) both vanish and if g''(x) is non-zero near x = a, except possibly at x = a and then if $\lim_{x\rightarrow a} \frac{f''(x)}{g''(x)}$ exists, we may apply L'Hopital's rule twice to give:

$$\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow ... ...rac{f'(x)}{g'(x)} = \lim_{x\rightarrow a}\frac{f''(x)}{g''(x)}.$$

Finally if g and its first n-1-derivatives vanish at a, and if g and its first n derivatives do not vanish in an interval around a, except possibly at a, then we have the iterated L'Hopital's rule:

$$\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a}\frac{f^{(n)}(x)}{g^{(n)}(x)}.$$

Examples:

$$\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = \lim_{x\rightarrow 0} \frac{\cos(x)}{1} = 1,$$

$$\lim_{x\rightarrow 0}\frac{\sin(x) - x}{x^3} = \lim_{x\righta... ...2} = \lim_{x\rightarrow 0} \frac{-\sin(x)}{6x} = -\frac{1}{6}.$$

L'Hopital's rule is also valid for other cases, although we do not prove these:

• $\frac{0}{0}$; a finite; this is the standard L'Hopital we proved above: and applies to the case that f(a) = g(a) = 0.
• $\frac{0}{0}$, $a = \pm \infty$; here $\lim_{x\rightarrow \pm \infty} f(x) = 0$, and $\lim_{x\rightarrow \pm \infty} g(x) = 0$.
• $\frac{\infty}{\infty}$; a finite; here $\lim_{x\rightarrow a} f(x) = \pm \infty$, and $\lim_{x\rightarrow a} g(x) = \pm\infty$.
• $\frac{\infty}{\infty}$; $a = \pm \infty$; here $\lim_{x\rightarrow \pm \infty} f(x) = \pm \infty$, and $\lim_{x\rightarrow \pm \infty} g(x) = \pm\infty$.

Examples:

$$\lim_{x\rightarrow \infty} x^3e^{-x} = \lim_{x\rightarrow \in... ... \frac{6x}{e^x} = \lim_{x\rightarrow \infty} \frac{6}{e^x} = 0.$$

$$\lim_{x\rightarrow \infty} x(e^{\frac{1}{x}} - 1) = \lim_{x\r... ...= \lim_{x\rightarrow \infty} (- e^{\frac{1}{x}}) = - e^0 = - 1.$$

$$\lim_{x\rightarrow \infty} \frac{(\tan^{-1}(x) - \frac{\pi}{2... ...\lim_{x\rightarrow \infty} - \frac{1}{1 + \frac{1}{x^2}} = - 1.$$

$$\lim_{x\rightarrow \infty} x(\ln(x + 5) - \ln(x)) = \lim_{x\rightarrow \infty} \frac{\ln(\frac{x + 5}{x})}{\frac{1}{x}}$$

$$= \lim_{x\rightarrow \infty} \frac{\ln(1 + \frac{5}{x})}{\f... ...}} = \lim_{x\rightarrow \infty} 5\frac{1}{1 + \frac{5}{x}} = 5.$$

Sometimes the calculation goes easier if we replace x by $\frac{1}{x}$.
Then a limit to $\infty$ is replaced by a limit to 0⁺ and a limit to $-\infty$ is replaced by a limit to 0^- and vice-versa.
Using this trick the last example reads:

$$\lim_{x\rightarrow \infty} x(\ln(x + 5) - \ln(x)) = \lim_{x\rightarrow \infty} \frac{\ln(\frac{x + 5}{x})}{\frac{1}{x}}$$

$$= \lim_{x\rightarrow \infty} \frac{\ln(1 + \frac{5}{x})}{\f... ...{x} = \lim_{x\rightarrow 0^+} \frac{(\frac{5}{1 + 5x})}{1} = 5.$$

Next we can handle some exponent limits by first taking logarithms and then using L'Hopital.
Consider $\lim_{x\rightarrow \infty}(1 + \frac{1}{x})^x$.
Let $y = (1 + \frac{1}{x})^x$, then $\ln(y) = x\ln(1 + \frac{1}{x})$ and we have:

$$\lim_{x\rightarrow \infty} \ln(y) = \lim_{x\rightarrow \infty... ... \frac{1}{x}) = \lim_{x\rightarrow 0^+}(\frac{1}{x}\ln(1 +x))$$

$$= \lim_{x\rightarrow 0^+}(\frac{\ln(1 +x)}{x}) = \lim_{x\rightarrow 0^+}(\frac{(\frac{1}{1 + x})}{1}) = 1.$$

So $\lim_{x\rightarrow \infty}(1 + \frac{1}{x})^x = \lim_{x\rightarrow \infty}e^{\ln(y)} = e^{\lim_{x\rightarrow \infty}(\ln(y))} = e^1 = e$.

Finally, although again we do no prove this, L'Hopital can be used to show that limits do not exist. For example:

$$\lim_{x \rightarrow 0}\frac{\sin(x) - x}{x^4} = \lim_{x \rightarrow 0}\frac{\cos(x) - 1}{4x^3}$$

$$= \lim_{x \rightarrow 0}\frac{-\sin(x)}{12x^2} = \lim_{x \rightarrow 0}\frac{-\cos(x)}{24x},$$

which does not exist, since the numerator goes to -1 and the denominator goes to zero.