Reference: G B Thomas, Calculus and Analytic Geometry, 9th Indian Edition.
The stronger form of l’Hopital’s rule is as follows:
Suppose that and the functions f and g are both differentiable on an open interval that contains the point . Suppose also that at every point in except possibly . Then,
…call this I, provided the limit on the right exists.
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 and differentiable through out and suppose also that through out . Then, there exists a number c in at which
(Note this becomes the ordinary mean value theorem when ).
Proof of Cauchy’s Mean Value theorem:
We apply the ordinary mean value theorem twice. First, we use it to show that . Because if , then the ordinary Mean Value theorem says that
for some c between a and b. This cannot happen because in .
We next apply the Mean Value Theorem to the function
This function is continuous and differentiable where f and g are, and note that . Therefore, by the ordinary mean value theorem, there is a number c between a and b for which . In terms of f and g, this says
or which is equation II above.
Proof of the stronger form of L’Hopital’s Rule:
We first establish equation I for the case . The method needs almost no change to apply to the case , and the combination of these two cases establishes the result.
Suppose that x lies to the right of . Then, and we can apply Cauchy’s Mean Value theorem to the closed interval from to x. This produces a number c between x and such that
As x approaches , c approaches as it lies between x and . Therefore,
This establishes l’Hopital’s Rule for the case where approaches from right. The case where x approaches from the left is proved by applying Cauchy’s Mean Value Theorem to the closed interval when .