   Next: Problems Up: No Title Previous: No Title

# 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 in the interval (a, b), such that .
Then we have .
In particular, if g' is non-zero in the open interval (a,b) and if , we have, for some in the interval (a,b): 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 in an interval around a, except possibly at x = a: where , but x is near a and then (which depends on x) is a point between x and a.
Now suppose exists.
Then also, so we get L'Hopital's rule: 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 exists, we may apply L'Hopital's rule twice to give: 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: Examples:  L'Hopital's rule is also valid for other cases, although we do not prove these:
• ; a finite; this is the standard L'Hopital we proved above: and applies to the case that f(a) = g(a) = 0.
• , ; here , and .
• ; a finite; here , and .
• ; ; here , and .
Examples:     Sometimes the calculation goes easier if we replace x by .
Then a limit to is replaced by a limit to 0+ and a limit to is replaced by a limit to 0- and vice-versa.
Using this trick the last example reads:  Next we can handle some exponent limits by first taking logarithms and then using L'Hopital.
Consider .
Let , then and we have:  So .

Finally, although again we do no prove this, L'Hopital can be used to show that limits do not exist. For example:  which does not exist, since the numerator goes to -1 and the denominator goes to zero.   Next: Problems Up: No Title Previous: No Title
George A. J. Sparling
2001-01-11