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L'Hopital's rule

  1. Jul 6, 2010 #1
    1. The problem statement, all variables and given/known data
    lim as x -> 0 sin(e^(x^2)-1)/e^(cosx)-e
    See Link Below.

    I saw this on the net and tried to solve it. I'm wondering if I'm correct by any means or did I mess up somewhere along the line.

    2. Relevant equations

    3. The attempt at a solution
    http://img291.imageshack.us/img291/1174/hopitalsrule.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 6, 2010 #2
    The second differentiation of the numerator isn't correct. It shouldn't be just one term but two.
    (f*g)' = fg' + gf'
  4. Jul 6, 2010 #3
    You did it well.

  5. Jul 6, 2010 #4
    I think you're right Bohrok (because I have a tendency of messing up).

    @Bohrok: Do you mean the second differentiation should be


    also, would the negative signs go away since we have both in the numerator and denominator? could we reduce the answer to 0 or do we keep it as 0/e?
  6. Jul 6, 2010 #5


    Staff: Mentor

    I agree with Bohrok and disagree with njama. Let's take it a step at a time. Your first differentiation of the numerator was correct, but the second one wasn't, and your latest try doesn't look right either.

    What do you get for
    [tex]\frac{d}{dx} 2xe^{x^2}~cos(e^{x^2} - 1)[/tex] ?

    Also, you should NOT leave an answer as 0/e.
  7. Jul 6, 2010 #6
    Thank you for clearing that up, so the limit as x->0 is 0

    I get:
  8. Jul 6, 2010 #7


    Staff: Mentor

    Close, but you are missing a pair of necessary parentheses and one exponent is incorrect, according to my work.
    Edit: the 3 exponent below should be 2.

    [tex]2xe^{x^2}*2xe^{x^2} = 4x^2e^{2x^2}[/tex]
    Last edited: Jul 6, 2010
  9. Jul 6, 2010 #8
    I'm glad I finally got it because it took me a few tries to get it :/ which is typical.
    I don't understand how it's 4x3.
    And thank you for telling me that it's 4x^2e^(2x^2) because I didn't know how to simplify it :shy:
  10. Jul 6, 2010 #9


    Staff: Mentor

    You're right - that exponent shouldn't be 3. It took me several times checking to find what I did wrong.

    [tex]d/dx(2xe^{x^2}~cos(e^{x^2} - 1)) = 2xe^{x^2}(-sin(e^{x^2} - 1) \cdot 2xe^{x^2}) + d/dx(2xe^{x^2}) \cdot cos(e^{x^2} - 1)[/tex]
    [tex]=-4x^2e^{2x^2}sin(e^{x^2} - 1) + (2e^{x^2} + 2x \cdot 2xe^{x^2})cos(e^{x^2} - 1)[/tex]
    [tex]= (2e^{x^2} + 4x^2e^{x^2})cos(e^{x^2} - 1) - 4x^2e^{2x^2} sin(e^{x^2} - 1)[/tex]
  11. Jul 6, 2010 #10
    Great! Thank you for the help... I actually added the parenthesis on my paper but somehow didn't do that when I wrote it on the computer. :| It took me a whole bunch of tries to get this right (even when I had an idea on what to do), I need to work on my math skills which I'm hoping to get better over the summer. Hopefully I'll finally gain full understanding on math once I practice and learn the rules.
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