proof of De l’Hôpital’s rule

Let x0, I be an interval containing x0 and let f and g be two differentiable functions defined on I{x0} with g(x)0 for all xI. Suppose that


and that


We want to prove that hence g(x)0 for all xI{x0} and


First of all (with little abuse of notation) we suppose that f and g are defined also in the point x0 by f(x0)=0 and g(x0)=0. The resulting functions are continuousMathworldPlanetmathPlanetmath in x0 and hence in the whole interval I.

Let us first prove that g(x)0 for all xI{x0}. If by contradictionMathworldPlanetmathPlanetmath g(x¯)=0 since we also have g(x0)=0, by Rolle’s Theorem we get that g(ξ)=0 for some ξ(x0,x¯) which is against our hypotheses.

Consider now any sequence xnx0 with xnI{x0}. By Cauchy’s mean value Theorem there exists a sequence xn such that


But as xnx0 and since xn(x0,xn) we get that xnx0 and hence


Since this is true for any given sequence xnx0 we conclude that

Title proof of De l’Hôpital’s rule
Canonical name ProofOfDeLHopitalsRule
Date of creation 2013-03-22 13:23:31
Last modified on 2013-03-22 13:23:31
Owner paolini (1187)
Last modified by paolini (1187)
Numerical id 10
Author paolini (1187)
Entry type Proof
Classification msc 26A24
Classification msc 26C15