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Math Help - L'Hospital

  1. #1
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    L'Hospital

    calculate:

    \lim_{x \to 0} (cos2x)^\frac{1}{x^2}



    Answer:
    e^-4
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  2. #2
    Eater of Worlds
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    \lim_{x\to 0}(cos(2x))^{\frac{1}{x^{2}}}

    The limit is e^{-2}, not e^{-4}

    There are many ways to go about it.

    Let t=2x

    Then you get:

    \lim_{t\to 0}\left[(cos(t))^{\frac{1}{t^{2}}}\right]^{4}

    Rewrite as:

    \left[e^{\lim_{t\to 0}\frac{ln(cos(t))}{t^{2}}}\right]^{4}

    The inside limit is a 'famous' one which equals -1/2, which easily derived using L'Hopital. You can do it without, but it's easier with the ol' Hospital rule

    So, we have:

    \left[e^{-1/2}\right]^{4}=e^{-2}=\frac{1}{e^{2}}

    You wouldn't have to make the sub:

    Start out with:

    e^{\lim_{x\to 0}\frac{ln(cos(2x))}{x^{2}}}

    Now, use L'Hopital and get:

    e^{-\lim_{x\to 0}\frac{sin(2x)}{xcos(2x)}}

    I hope you can read that. Sure is tiny. I have tried to make it bigger to no avail.
    Last edited by galactus; June 23rd 2008 at 05:12 PM.
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  3. #3
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    thanks for help
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