1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    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!

Homework Help: L'Hôpital's rule

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data

    I have

    [tex] L = \lim_{x\rightarrow 0} \Big( {\frac{\cos(1.92x)-1} {e^{2.33x} - 1 -2.33x}} \Big)[/tex]

    I'm meant to use L'Hôpital's rule finding the limit, maybe twice.

    3. The attempt at a solution

    So, there's clearly something I have misunderstood, and hoping you might tell me what it is.

    [tex] L = \lim_{x\rightarrow 0} \Big( {\frac{(-sin 1.92x)1.92} {2.33e^{2.33x}-2.33}}\Big)[/tex]

    Getting rid of "1.92" gives

    [tex] \lim_{x\rightarrow 0} \Big( {\frac{-sin1.92x} {1.21e^{2.33x}-1.21}}\Big) [/tex]
    Then since [tex]{\frac {0} {0}}[/tex], I'll use the rule once more:

    [tex] \lim_{x\rightarrow 0} {\frac{-cos 1.92x} {2.82e^{2.33x}}} = -0.35 [/tex]
    What am I doing wrong?
  2. jcsd
  3. Nov 6, 2011 #2
    there is 1.92 missing in the numerator.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook