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Limit using L'hospital's rule

  1. Jan 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Calculate the limit as x tends to 0:


    2. Relevant equations

    3. The attempt at a solution

    I have to use L'hospital's rule but i dont know how to start this particular question.
    I tried making the substitution y=1/x^2
    but didnt get anywhere.
    Any help or hints would be v much appreciated.
    Thank you
  2. jcsd
  3. Jan 4, 2009 #2


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    My only suggestion is is to use logarithmic differentiation and L'hospital's rule to solve the problem.
  4. Jan 4, 2009 #3
    yes that sounds good.
    but what do you mean by logarithmic differentiation??

  5. Jan 4, 2009 #4


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    Let y = (sin(x)/x)^(1/x^2), than log(y) = log(sin(x)/x)/(x^2). Since we may think of y as y = e^f(x), we need only find the limit as f(x) tends to zero. Does that make sense? As a side note, log designates the natural logarithm or logarithm base e.
  6. Jan 4, 2009 #5
    so f(x)= log(sin(x)/x)/(x^2)

    we need to find this limit??
    doesnt this make things more diificult? i dont know mayb i dont get it. :s
  7. Jan 4, 2009 #6


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    If you're required to use L'hospital's rule I think you may need to find that limit. You may need to apply the rule several times.
  8. Jan 4, 2009 #7
    but, log(sin(x)/x)/(x^2) is not a fraction so how do we apply L'Hospitals?
    What would be the first step
  9. Jan 4, 2009 #8
    Assuming the problem is to find

    [tex] y = \lim_{x \rightarrow 0} \left( \frac{sin(x)}{x} \right)^{\frac{1}{x^2}} [/tex]

    Then, we can take ln(.) of both sides to get

    [tex] ln|y| = \lim_{x \rightarrow 0} \frac{1}{x^2} ln \left( \frac{sin(x)}{x} \right) [/tex]

    Now we know that the limit inside the ln(.) tends to 1 and ln(1) = 0 so we effectively have a case of 0/0 and so we can use L'Hopital's.

    Can you take it from here?
  10. Jan 4, 2009 #9


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    It is a fraction: f(x) = log(sin(x)/x) and g(x) = x^2. Therefore f(x)/g(x) = log(sin(x)/x)/x^2.
  11. Jan 4, 2009 #10
    This is a pretty nasty limit but I'm pretty sure there's a trick. Plus knowing the answer makes me think the trick is to use Maclaurin if you know it and play around with derivative of sums.
    Last edited: Jan 4, 2009
  12. Jan 4, 2009 #11
    ok so if i do:
    and use L'hospital's,
    i get:
    ln(y)=lim (2xcosx-sinx)/sinx

    using L'hospitals rule again, i get:

    ln(y)= lim (cosx-2xsinx)/(cosx)

    substitutin x=0 into the limit, i get:

    ln(y) = 1

    ?? did i make a mistake somehere?
  13. Jan 4, 2009 #12
    Yes. What's the derivative of

  14. Jan 4, 2009 #13


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    Correct me if I'm wrong but shouldn't the first application of L'hosptial's rule yield: (xcos(x) - sin(x))/(2x^2sin(x))?
  15. Jan 4, 2009 #14
    after simplifying it will be:
    (xcosx - sinx)/(xsinx)

    Oh sorry i made a mistake,
    after applying the first L'hospitals, i should get:
    ln(y)=lim (2xcosx-2sinx)/sinx

    so applying again, i should get:
    ln(y)=lim (-2xsinx)/cosx

    so do i have to apply L'hospitals rule again?
  16. Jan 4, 2009 #15


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    I still don't agree with the answer on your first application; however, assuming you've done the steps correctly, you can no longer apply L'hospital's rule at the step you've arrived at.

    d(ln(sin(x)/x))/dx = (x/sin(x))((xcos(x)-sin(x))/x^2) correct?
  17. Jan 4, 2009 #16
    That is correct. Why can't you apply L'Hopital's here? When you "plug in" x = 0, you get 0/0 again.
  18. Jan 4, 2009 #17


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    His or her final step: -2xsin(x)/cosx is not in the form 0/0, that's what I was pointing to - he or she asked if another application of the rule was needed. Sorry for any confusion. Moreover, While he or she could substitute 0 into this step and obtain an answer, the result is not correct.
  19. Jan 4, 2009 #18
    the first time i applied L'hospital's, i differentiate ln(sinx / x) divided by derivative of x^2

    so: = (x/sinx)((xcosx-sinx)/x^2) cancel out an x to get: (1/sinx)((xcosx-sinx)/x)
    divide this by 2x to give:.........oh ur right!!!!! after the first time, you would get:
    ln(y)= (xcosx-sinx)/(2x^2sinx) ...like you said.
    then after the second time, you would get:


  20. Jan 4, 2009 #19
    oh then after another time:

  21. Jan 4, 2009 #20
    Yes that's the last step, and now since you are trying to find y... how do you get y = ?
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