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Math Help - l'Hopital's rule

  1. #1
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    l'Hopital's rule

    Find the limit by using the l' Hopital's rule!
    \lim_{x\to0}\frac{\cos\!\left(x\right)-1}{2x^2}


    So far i got to
    \lim_{x\to0}\frac{\-sin\!\left(x}{4x}

    What do you do next

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  2. #2
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    Quote Originally Posted by Emmeyh15@hotmail.com View Post
    Find the limit by using the l' Hopital's rule!
    \lim_{x\to0}\frac{\cos\!\left(x\right)-1}{2x^2}


    So far i got to
    \lim_{x\to 0}\frac{\-sin\!\left(x}{4x}

    What do you do next

    \lim_{x\to0}\frac{\cos\!\left(x\right)-1}{2x^2}=\lim_{x\to 0}\frac{-\sin(x)}{4x}=\text{ use L'Hopital again } = \lim_{x\to0}\frac{-\cos(x)}{4} = -\frac{1}{4}

    Tell your teacher that it is Bernoulli's rule actually.
    Last edited by courteous; January 23rd 2009 at 08:31 AM. Reason: Bernoulli
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  3. #3
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    Quote Originally Posted by courteous View Post
    \lim_{x\to0}\frac{\cos\!\left(x\right)-1}{2x^2}=\lim_{x\to 0}\frac{-\sin(x)}{4x}=\text{ use L'Hopital again } = \lim_{x\to0}\frac{-\cos(x)}{4} = -\frac{1}{4}

    Tell your teacher that it is Bernoulli's rule actually.
    But is you call it Bernoulli's rule most (if not all) people won't know what your talking about.

    Here's an interesting site.

    Calculus I Notes, Section 4-5
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