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L'Hospital's rule and indeterminate forms

  1. Jun 11, 2015 #1
    I want to make sure I understand the conditions required for L'Hospital's Rule to work.
    $$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$$
    If ##\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}## exists.
    Should ##f## and ##g## be differentiable at ##a##? Or just around ##a##?
    Also, would it work if ##g'(a) = 0##?
  2. jcsd
  3. Jun 11, 2015 #2


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    It's not clear what you mean by a function being differentiable "just around a".

    For univariate functions, a derivative either exists at a given abscissa or it doesn't. I'm not aware of anything in between.

    If g'(a) = 0, then it is possible that the first application of L'Hopital's rule leads to another indeterminate form. You can apply L'Hopital's rule serially until either a limit is reached or it becomes clear that no convergent limit will ever be obtained. Then you have to move on.
  4. Jun 12, 2015 #3


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    By "just around a" I believe that MohamedRady97 means "in some neighborhood of a but not necessarily at x= a".
    For example, f(x)= x if x< 1, f(x)= 2x- 1 if x> 1 is differentiable "around x= 1" but not at 1.

    The answer to his question is "yes, that is correct". The definition of "limit" is such that what happens in a neighborhood of a, NOT at x= a itself, is all that is taken into consideration when taking a limit.
  5. Jun 13, 2015 #4
    See this. Theorem have special conditions.
  6. Jun 13, 2015 #5


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    as suggested above you do not have enough hypotheses to make the theorem true. you also need something like the limits of f and g are both = 0, or both = infinity. look in a book.
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