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Math Help - The Pinching Theorem; Trig Limits

  1. #1
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    The Pinching Theorem; Trig Limits

    Can anyone help me evaluate this limit?

     \lim_{x\to 0} \frac{x^2}{1-\cos2x}

    I can see that the numerator approaches 0 faster and that you can say that this is equal to  (x)(\frac{x}{1-\cos2x}) but i'm not sure where to go from there. Oh and the answer is 1/2 and i cant use L'hospital's theorem.

    Thanks!
    Last edited by discobob; January 24th 2010 at 07:47 AM.
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  2. #2
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    Hello, discobob!

    Do you have to use the "Squeeze Theorem"?


     \lim_{x\to 0} \frac{x^2}{1-\cos2x}

    We have: . \frac{x^2}{1-\cos2x} \;=\;\frac{x^2}{2\sin^2\!x}\;=\;\frac{1}{2}\left(\  frac{x}{\sin x}\right)^2


    Therefore: . \lim_{x\to0}\,\left[\frac{1}{2}\left(\frac{x}{\sin x}\right)^2\right] \;=\; \frac{1}{2}\cdot 1^2 \;=\;\frac{1}{2}

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  3. #3
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    Oh, ok. I just really need to study my trig Identities. I understand how you do it. Thanks a lot Soroban!
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