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Homework Help: L'Hopital's Rule

  1. May 15, 2010 #1
    1. The problem statement, all variables and given/known data

    Evalutate the limit, as x approaches 0, log(coshx)/x2


    2. Relevant equations

    L'Hopital's rule


    3. The attempt at a solution

    I can get as far as (sinhx.cosh-1x)/2x by differentiating the top and bottom separately. I'm not sure how to do the next differentiation.

    Thanks for any help.
     
  2. jcsd
  3. May 15, 2010 #2
    Be careful about writing cosh-1 x; that means something quite different from what you mean.

    So you have quite rightly differentiated top and bottom, leaving you with:

    [tex] \frac{\frac{sinh x}{cosh x}}{2x} [/tex]

    Fix up that fraction so that it is in a more usual form (ie, move the cosh x to the denominator), and then apply L'Hopital's rule again.
     
  4. May 15, 2010 #3

    Mark44

    Staff: Mentor

    By cosh-1x, you really mean 1/cosh x, but this notation suggests the inverse cosh function. Don't write cosh-1x if you mean 1/cosh x.

    Your numerator is sinh(x)/cosh(x) = tanh(x). Since the numerator approaches zero, and 2x approaches zero, use L'Hopital's Rule another time.
     
  5. May 15, 2010 #4
    Here's what I've done. I've treated the numerator as being sinhx/coshx and differentiated it to get (1/cosh2x). I've treated denominator as being 2x and differentiated it to get 2. Therefore I end up with (1/cosh2x)/2 which gives 1/2 as x approaches 0. Is this the correct answer?
     
  6. May 15, 2010 #5
    Yup.
     
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