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  1. #1
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    Homework

    Use l' Hopital's Rule to find the limit. Show your work.

    Lim cosx-1/2x^2 is 0/0 form
    x->0
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Emmeyh15@hotmail.com View Post
    Use l' Hopital's Rule to find the limit. Show your work.

    Lim cosx-1/2x^2 is 0/0 form
    x->0
    Please a bit clearer on your notation.

    \lim_{x\to0}\frac{\cos\!\left(x\right)-1}{2x^2}

    Is this what you mean?
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  3. #3
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    Yes sorry i didnt know how to make that arrow!!!
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Emmeyh15@hotmail.com View Post
    Yes sorry i didnt know how to make that arrow!!!
    In applying L'Hopitals rule, you need to differentiate both numerator and denominator of the limit. Thus we see that

    \lim_{x\to0}\frac{\cos\!\left(x\right)-1}{2x^2}=\lim_{x\to0}\frac{-\sin\!\left(x\right)}{4x}

    Again, this limit has the indeterminate form \frac{0}{0}

    So, I leave it for you to apply the rule one more time.

    Does this make sense?
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