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Math Help - intermediate limit - L'Hopital's rule?

  1. #1
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    intermediate limit - L'Hopital's rule?

    Hi, can anyone explain how

    lim x->0 (x+( (e^(2x)) - (e^(3x)) / 3x+( e^(2x) )- (e^(5x)))

    Can be 5/21?
    When I try using L'Hopital I get -4/-18.
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  2. #2
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    Re: intermediate limit - L'Hopital's rule?

    I assume you mean:

    L=\lim_{x\to0}\frac{x+e^{2x}-e^{3x}}{3x+e^{2x}-e^{5x}}

    We see we have the indeterminate form 0/0, so application of L'H˘pital's rule gives:

    L=\lim_{x\to0}\frac{1+2e^{2x}-3e^{3x}}{3+2e^{2x}-5e^{5x}}

    We still have the indeterminate form 0/0, so application of L'H˘pital's rule again gives:

    L=\lim_{x\to0}\frac{4e^{2x}-9e^{3x}}{4e^{2x}-25e^{5x}}=\frac{4-9}{4-25}=\frac{5}{21}
    Thanks from topsquark
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    Re: intermediate limit - L'Hopital's rule?

    Thanks a lot!
    Last edited by Erka; November 28th 2012 at 01:35 AM.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: intermediate limit - L'Hopital's rule?

    Glad to help and welcome to the forums!
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