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Math Help - Limit

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    Limit

    \lim_{x\to 0}\frac{1-cos\;x}{x^2} without l'Hopital Rule.

    Thank you
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    Quote Originally Posted by Adonis View Post
    \lim_{x\to 0}\frac{1-cos\;x}{x^2} without l'Hopital Rule.

    Thank you
    Substitute the Maclaurin Series for cos x, simplify and take the limit:

    \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - ......

    Therefore you have \frac{1 - 1 + \frac{x^2}{2} - \frac{x^4}{24} + .....}{x^2} = \frac{1}{2} - \frac{x^2}{24} + ......

    As x --> 0 the limiting value is obvious .....
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    Quote Originally Posted by Adonis View Post
    \lim_{x\to 0}\frac{1-cos\;x}{x^2} without l'Hopital Rule.
    \frac{1-\cos x}{x^2} \cdot \frac{1+\cos x}{1+\cos x} = \frac{\sin^2 x}{x^2} \cdot \frac{1}{1+\cos x}.

    Now use the fact that \lim_{x\to 0}\frac{\sin x}{x} = 1.


    ---
    Using infinite series is like shooting a dead fly with a Grand Piano.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    [tex][snip]
    Using infinite series is like shooting a dead fly with a Grand Piano.
    True.

    But I've found most students can be taught to kill even a live fly using a Grand Piano. As opposed to teaching them how to pull a rabbit out of a hat (even when it's a fairly ordinary rabbit, as in the present case) .....

    By the way, I don't mean to be critical of the rabbit - I love magic - it's just that most students struggle to learn it. Even when they do, they usually pull out a dead fish .....

    My view, all comment welcome.
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  5. #5
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    Quote Originally Posted by Adonis View Post
    \lim_{x\to 0}\frac{1-cos\;x}{x^2} without l'Hopital Rule.

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    Use the equality:1-cosx=2sin(x/2)^2.
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  6. #6
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    Quote Originally Posted by curvature View Post
    Use the equality:1-cosx=2sin(x/2)^2.
    Now thats what I call a rabbit
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