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Homework Help: Limits question L'Hopitals rule

  1. Nov 15, 2007 #1
    1. The problem statement, all variables and given/known data
    lim (sin(x)/x)^(1/X^2)

    2. Relevant equations
    for the life of me i cannot get the correct solution

    3. The attempt at a solution
    Ive tried taking the log of both sides etc and working from there then applying l'hopitals rule until i get a result but the answer i always get is e^(-1/2) but the answer is e^(-1/6) any help would be much appreciated
  2. jcsd
  3. Nov 15, 2007 #2


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    Homework Helper

    e^(-1/6) is correct. And that's the way to do it alright. But there is no way to tell why you get e^(-1/2) unless you show us what you did.
  4. Nov 15, 2007 #3
    sorry ill give more info
    lim (sin(x)/x)^(1/X^2)

    let y = (sin(x)/x)^(1/X^2)
    ln y = (ln(sin(x)/x))^(1/X^2)
    ln y = (ln(sin(x)/x))/(X^2)
    ln y = (ln sin(x) - ln(x))/(X^2)
    apply l'hopital's rule
    = (cos(x)/sin(x) - 1/y)/2X
    apply l'hotital's rule again
    = (-sin(x)/cos(x) -1/1)/2
    and from there i get
    = -1/2
    y = e^(-1/2)
    Last edited: Nov 15, 2007
  5. Nov 15, 2007 #4


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    How did y get over on the right side?
    Applying L'Hopital's rule gives you
    [tex]\frac{\frac{cos(x)}{sin(x)}- 1/x}}{2x}[/tex]
    so I assume the "1/y" was "1/x". That reduces to
    [tex]\frac{xcos(x)- sin(x)}{2x^2sin(x)[/tex]
    You are also differentiating incorrectly. The derivative of cos(x)/sin(x) is NOT (cos(x))'/(sin(x))' and the derivative of 1/x is NOT1/(x)'!

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