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Proof l'hopital for infinity over infinity

  1. Oct 28, 2011 #1

    georg gill

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    i wonder about this proof for l'hopital for infinity over infinity:

    http://planetmath.org/encyclopedia/ProofOfLHopitalsRuleForInftyinftyForm.html [Broken]

    how is this proved:

    http://bildr.no/view/1011658
     
    Last edited by a moderator: May 5, 2017
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  3. Oct 28, 2011 #2

    Stephen Tashi

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    [tex] \lim_{x \rightarrow a } \frac{ 1 - \frac{g(c)}{g(x)} }{ 1 - \frac{f(c)}{f(x)}} = 1 [/tex]

    I think this because [itex] g(c) [/itex] is a finite number and [itex] lim_{x \rightarrow a} g(x) = \infty [/itex] or [itex] - \infty [/itex]. So the numerator approaches [itex] 1 - 0 [/itex]. Likewise the denominator approaches [itex] 1 - 0 [/itex].
     
  4. Oct 28, 2011 #3

    georg gill

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    The only thing one should proof is just that f(c) and g(c) is not infinite I guess.
     
  5. Oct 28, 2011 #4

    Stephen Tashi

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    The planet math main article on l'Hospitals rule doesn't clearly state the hypotheses of the rule. (It even misspells the word "existence".) You are correct that the proof you linked ought to justify the assumption that f(c) and g(c) are finite.

    I think the key to doing that is to interpret the hypothesis that [itex] lim_{x \rightarrow a} \frac{g'(x)}{f'(x)}[/itex] exists. (The Wikipedia version of the rule says "exists or is infinite".) As I interpret this hypothesis, it is not assumption that f'(a) and g'(a) exists as individual functions. I think we must try to argue that if the limit of the quotient exists at x = a then there must be an open interval containing a where both f'(x) ang g'(x) exist everywhere in the interval except possibly at x = a. The functions f and g are differentiable at those points, hence continuous and hence finite. We can pick c to be in this interval.
     
  6. Oct 30, 2011 #5

    georg gill

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    Quote from proof:

    This is because f(x) og g(x) were assumed to approach [tex]\pm\infty[/tex] when x is close enough to a, they will exceed the fixed value f(c), g(c) and 0.

    I find the word exceed a bit strange if f(x) or g(x) goes to [tex]-\infty[/tex] it will have lover value then f(c) or g(c) I guess.

    But one thing that would make it clear even though I can struggle a bit with the words chosen would be if it is so that a fixed value can not be [tex]\pm\infty[/tex] Is it so that [tex]\pm\infty[/tex] is not a defined as a value?
     
    Last edited: Oct 30, 2011
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