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L'Hoptial's rule

  1. Mar 11, 2014 #1
    1. The problem statement, all variables and given/known data

    I have an equation in which a term equals zero and in this case the whole equations equals zero. I know it is possible to use the L'Hopital's rule to calculate the equation but having I'm having a little trouble.

    2. Relevant equations

    The equation I would like to is the following:

    [tex]k = \frac{pA}{x}*[(\frac{V}{V-Ax})^y - 1][/tex]

    I have to solve this when x = 0;


    3. The attempt at a solution

    Plugging in 0 for x, results in 0, which should not be the answer. I figure it is possible to use the L'Hopitals rule, but currently only my denominator has the x term which would be approaching zero. But in order to use the rule I would have to manipulate the equation for both the numerator and denominator to approach zero?
     
  2. jcsd
  3. Mar 11, 2014 #2

    vanhees71

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    Instead of using de L'Hospital's rule you can (equivalently) expand the expression in the square brackets in powers of [itex]x[/itex]. This should help!
     
  4. Mar 11, 2014 #3
    But if I would still like to use L'Hopital's rule, how should I proceed?
     
  5. Mar 11, 2014 #4

    BvU

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    Differentiate the numerator wrt x

    Not true. It results in 0/0 !
     
  6. Mar 11, 2014 #5

    Ray Vickson

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    Your function is ##f(x)## is of the form
    [tex] f(x) = \frac{N(x)}{x}, \;\; N(x) = pA\left[ \left(\frac{V}{V-Ax}\right)^y -1\right] [/tex]
    When you use l'Hospital's rule (not l'Hopital!) you compute
    [tex] \lim_{x \to 0} f(x) = \frac{N'(0)}{1},[/tex]
    so all you are really doing is just taking the first term of the series expansion of ##N(x)##. l'Hospital and series expansion are really the same thing.
     
    Last edited: Mar 11, 2014
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