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Math Help - Limit question

  1. #1
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    Limit question

    Sorry, I have no idea how to solve this limit problem

    Use a CAS (Computer Algebra System) to evaluate the following limit:
    I got this answer but it's wrong

    cos(0)-sec(0)^2)/2
    Last edited by killasnake; November 3rd 2006 at 08:24 PM. Reason: adding more detail
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  2. #2
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    Well if you use something like Mathematica (software) it should give you 1/24 as the answer.

    To solve it analytically, the easiest way is to use L'Hopital's Rule.

    Here's what you start with.
    \lim_{x \to 0} \frac{\cos x - 1 + \frac{x^2}{2}}{x^4}

    If you substitute 0 in for x, then you get \frac{0}{0} which is criterion for using L'Hopitale.

    Derive the numerator and denominator with respect to x.
    The limit becomes:

    \lim_{x \to 0} \frac{-\sin x + x}{4x^3}

    Once again, direct substitution will yield \frac{0}{0}, so L'Hopitale can be applied again, giving the limit:

    \lim_{x \to 0} \frac{-\cos x + 1}{12x^2}

    Again, direct substitution gives \frac{0}{0}. Do L'Hopitale yet again to get the limit:

    \lim_{x \to 0} \frac{\sin x}{24x}

    You may recognize this form as \frac{1}{24} \cdot \left( \lim_{x \to 0} \frac{\sin x}{x} \right) = \frac{1}{24} \cdot (1) = \frac{1}{24}.

    (if you didn't recognize the form, applying L'Hopitale one last time would give 1/24 trivially).
    Hope that helps.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by killasnake View Post
    Sorry, I have no idea how to solve this limit problem



    I got this answer but it's wrong

    cos(0)-sec(0)^2)/2
    I dont know what you are doing with your CAS but see attachment.

    RonL
    Attached Thumbnails Attached Thumbnails Limit question-gash.jpg  
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