proof of l’Hôpital’s rule for / form

This is the proof of L’Hôpital’s Rule ( in the case of the indeterminate form ±/. Compared to the proof for the 0/0 case (, more complicated estimates are needed.

Assume that


where a and m are real numbers. The case when a or m is infinite only involves slight modifications to the argumentsMathworldPlanetmath below.

Given ϵ>0. there is a δ>0 such that


whenever 0<|ξ-a|<δ.

Let c and x be points such that a-δ<c<x<a or a<x<c<a+δ. (That is, both c and x are within distance δ of a, but x is always closer.) By Cauchy’s mean value theorem, there exists some ξx in between c and x (and hence 0<|ξx-a|<δ) such that


We can assume the values f(x), g(x), f(x)-f(c), g(x)-g(c) are all non-zero when x is close enough to a, say, when 0<|x-a|<δ for some 0<δ<δ. (So there is no division by zero in our equations.) This is because f(x) and g(x) were assumed to approach ±, so when x is close enough to a, they will exceed the fixed values f(c), g(c), and 0.

We write

f(x)g(x) =f(x)f(x)-f(c)g(x)-g(c)g(x)f(x)-f(c)g(x)-g(c)

Note that


but ξx is not guaranteed to approach a as x approaches a, so we cannot just take the limit xa directly. However: there exists 0<δ′′<δ so that


whenever 0<|x-a|<δ′′. Then

|f(x)g(x)-m| =|(f(ξx)g(ξx)-m)+f(ξx)g(ξx)(1-g(c)/g(x)1-f(c)/f(x)-1)|

for 0<|x-a|<δ′′.

This proves



  • 1 Michael Spivak, Calculus, 3rd ed. Publish or Perish, 1994.
Title proof of l’Hôpital’s rule for / form
Canonical name ProofOfLHopitalsRuleForinftyinftyForm
Date of creation 2013-03-22 15:40:15
Last modified on 2013-03-22 15:40:15
Owner stevecheng (10074)
Last modified by stevecheng (10074)
Numerical id 7
Author stevecheng (10074)
Entry type Proof
Classification msc 26A06