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L'Hopitals Rule Conditions

  1. Feb 28, 2009 #1

    I'm slightly confused about one aspect of the conditions for applying L'Hopital's rule.

    N.b. apologies in advance for the lack of LaTeX.

    L'Hopitals rule:
    Let f,g:(a,b) → R be differentiable and let c ε (a,b) be such that f(c)=g(c)=0 and g'(x)≠0 for x≠c.
    Then lim(x→ c)[f(x)/g(x)] = lim(x→ c)[f'(x)/g'(x)], provided latter limit exists.

    I have lim(x→ 2)[(x²+x-6)/(x²-x-2)] = lim(x→ 2)[(2x+1)/(2x-1)]

    I compute this to be 5/3 (which is correct) by L'Hopital's rule.
    My point is that this function satisfies f(2)=g(2)=0, but does NOT satisfy g'(x)≠0 for x≠2, as the solution of g'(x)=2x-1=0 is x=0.5; i.e. g'(x)=0 for x=0.5, and 0.5≠c(=2).

    So why are we able to apply L'Hopital in this case?
  2. jcsd
  3. Feb 28, 2009 #2
    If take (a,b)=(1,3), then the requirements are satisfied. The point is that you need g'(x) nonzero only "close" to c, i.e. on some open interval containing c.
  4. Mar 1, 2009 #3
    I think you may have stated the conditions incorrectly. In particular, "g'(x)≠0 for x≠c", I believe, should be "g'(x)≠0 for x=c", since that is all that is necessary for the limit of f'(x)/g'(x) to exist (which seems to be consistent with the definitions I've checked from other sources, then if it still doesn't exist, you can apply it again, where it may eventually work).

    I might be wrong though. Where did you get the definition of L'Hospital's Rule from?
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