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Math Help - líHopitalís Rule Help..

  1. #1
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    líHopitalís Rule Help..

    Hey guys having trouble solving this . Use líHopitalís rule to find:

    \lim_{x\to0}\frac{cosx-1}{x^2}

    any help much appreciated
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    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by jvignacio View Post
    Hey guys having trouble solving this . Use l’Hopital’s rule to find:

    \lim_{x\to0}\frac{cosx-1}{x^2}

    any help much appreciated
    Differentiate numerator and denominator

    \frac{d~}{dx} (cos(x) - 1 ) = -sin(x)

    \frac{d}{dx} (x^2) = 2x <br />

    What is

    (\frac{-1}{2})\lim_{x\to0}\frac{sin(x)}{x} ?

    You can do the same for finding this limit or just use the formula (which??)
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    Quote Originally Posted by ADARSH View Post
    Differentiate numerator and denominator

    \frac{d~}{dx} (cos(x) - 1 ) = -sin(x)

    \frac{d}{dx} (x^2) = 2x <br />

    What is

    (\frac{-1}{2})\lim_{x\to0}\frac{sin(x)}{x} ?

    You can do the same for finding this limit or just use the formula (which??)
    sorry what do u do after finding the derivative of the numerator and the denominator?
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by jvignacio View Post
    sorry what do u do after finding the derivative of the numerator and the denominator?

    1)Find derivatives separately

    2) Divide the numerator's derivative by denominator's derivative

    3)If on putting the limit you get a definite answer than game over!!

    4) If not than go to step 1


    ---------------------------------------------------
    I strongly recommend reading examples here

    The answer of this question will be -1/2 (after you follow above steps)

    Adarsh
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  5. #5
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    Quote Originally Posted by ADARSH View Post
    1)Find derivatives separately

    2) Divide the numerator's derivative by denominator's derivative

    3)If on putting the limit you get a definite answer than game over!!

    4) If not than go to step 1

    ---------------------------------------------------
    I strongly recommend reading examples here

    The answer of this question will be -1/2 (after you follow above steps)

    Adarsh
    are you able to do the working out for this question?

    im confused how to divide the derivatives of the original fraction


    \lim_{x\to0}\frac{-sin}{2x}
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  6. #6
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by jvignacio View Post
    are you able to do the working out for this question?

    im confused how to divide the derivatives of the original fraction

    Denominator =D Numerator =N

    Derivative of Numerator
    =  \frac{d~}{dx} (cos(x) - 1 ) = -sin(x)...(Also called N'(x))

    Derivative of Denominator
    = \frac{d}{dx} (x^2) = 2x...(Also called D'(x) )

    According to the l'Hopital's rule




    So using this

    Your Limit now becomes







    Once again applying the same rule this limit becomes


    (\frac{-1}{2}) lim_{x\to 0} cos(x)

    Now put the limit

    \implies (\frac{-1}{2}) \times (1)
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