1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook