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Math Help - L'Hopital's Rule Help

  1. #1
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    L'Hopital's Rule Help



    Can someone please explain to me the red circle? I don't understand how the previous step results in 1/infinity divided by infinity which then equals to zero..?

    Also, I don't understand anything in the picture below, so can someone also explain each step to me?



    Thank you!
    Last edited by azncocoluver; March 21st 2011 at 12:49 PM.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    {2ln(x)*(1/x)}/{2x} = {ln(x)}/{x}*(1/x) --->(x-->infty) 0*0=0

    In second problem:

    n*sin(n)=sin(1/n)/{1/n} --->(n-->infty) sin(0)/(0)=0/0 {sinx is continuous function}

    We ca use L'Hopital's rule! (WHY?)

    {sin(1/n)}'/{1/n}'={cos(1/n)*(-1/n^2)}/{(-1/n^2}=cos(1/n) --->(n-->infty) cos(0)=1 {cosx is continuous function}
    Last edited by Also sprach Zarathustra; March 21st 2011 at 01:04 PM.
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  3. #3
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    Thank you! I understand now, I just have one more question. In the first picture, 1/x is circled and infinity is plugged in for x. Why isn't ln x circled and infinity plugged in, as well?
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  4. #4
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    Quote Originally Posted by azncocoluver View Post
    Thank you! I understand now, I just have one more question. In the first picture, 1/x is circled and infinity is plugged in for x. Why isn't ln x circled and infinity plugged in, as well?
    It may be simplest to apply the rule a second time

    \displaystyle\frac{lnx\left(\frac{1}{x}\right)}{x}  =\frac{lnx}{x^2}

    Applying the rule again leads to the circled part.
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