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Math Help - calculate the limit

  1. #1
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    calculate the limit

    \lim_{x\rightarrow 0}\left ( \frac{1-\textup{cos}(1-\textup{cosx})}{\textup{x}^4} \right )
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  2. #2
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    I would suggest using L'Hopital's rule, four times.
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  3. #3
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    YA i know that....but is there any shorter way
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  4. #4
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    L'Hospital's Rule is nearly always the shortest way...
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  5. #5
    Member Mathelogician's Avatar
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    Of course, series expansion would be easier than hospital's rule many times!Indeed some complicated limits which are impossible to be solved with hospital's rule could be solved using series expansion!
    1/8-x^2/48-x^4/960+O(x^6)
    See:
    lim ((1-cos(1-cosx))/x^4 as x->0 - Wolfram|Alpha
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  6. #6
    Senior Member DeMath's Avatar
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    Quote Originally Posted by banku12 View Post
    \lim_{x\rightarrow 0}\left ( \frac{1-\textup{cos}(1-\textup{cosx})}{\textup{x}^4} \right )
    Try using this \sin^2\theta=\dfrac{1-\cos2\theta}{2}

    \lim\limits_{x\to0}\frac{1-\cos(1-\cos{x})}{x^4}=\lim\limits_{x\to0}\frac{1-\cos\left(2\sin^2\frac{x}{2}\right)}{x^4}=2\lim\li  mits_{x\to0}\frac{\sin^2\sin^2\frac{x}{2}}{x^4}=

    =\frac{1}{8}\lim\limits_{x\to0}\frac{\sin^4\frac{x  }{2}}{\frac{x^4}{16}}\frac{\sin^2\sin^2\frac{x}{2}  }{\sin^4\frac{x}{2}}=\frac{1}{8}\cdot1^4\cdot1^2=\  frac{1}{8}.
    Last edited by DeMath; August 12th 2010 at 07:02 AM. Reason: typo
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  7. #7
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    thanks
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