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Math Help - Finding this limit without l'Hospital's rule

  1. #1
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    Finding this limit without l'Hospital's rule

    <br />
\lim_{x\to0} \frac{1-cos(2x)}{x^2} <br />

    We haven't covered l'Hospital's rule yet so Wolfram Alpha's not all that helpful, and even if I did know it it wouldn't have flown on the exam.

    I had misread it as cosine squared x, and that is in fact how I solved it on the test which I got no credit for, for some reason.

    Any hints?
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  2. #2
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    \lim_{x\to 0}\frac{1-cos(2x)}{x^{2}}

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

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

    \lim_{x\to 0}\frac{sin^{2}(2x)}{x^{2}}\cdot\lim_{x\to 0}\frac{1}{1+cos(2x)}

    \lim_{x\to 0}\left(\frac{sin(2x)}{x}\right)^{2}\cdot \lim_{x\to 0}\frac{1}{1+cos(2x)}

    As can be seen from the left most limit, we have 4. The right one is 1/2.

    And the limit is 2.
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  3. #3
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    Hello, mbacarella!

    \lim_{x\to0} \frac{1-\cos2x}{x^2}
    Here's one way . . . probably not the shortest solution.

    Multiply by \frac{1+\cos2x}{1+\cos2x}\!:

    . . \frac{1-\cos2x}{x^2}\cdot\frac{1+\cos2x}{1+\cos2x}

    . . =\;\frac{1-\cos^2\!2x}{x^2(1+\cos2x)}

    . . =\;\frac{\sin^2\!2x}{x^2(1+\cos2x)}

    . . = \;\frac{\sin^2\!2x}{x^2}\cdot\frac{1}{1+\cos2x}

    . . = \;4\cdot\frac{\sin^2\!2x}{4x^2}\cdot\frac{1}{1+\co  s2x}

    . . =\;4\cdot\left(\frac{\sin2x}{2x}\right)^2\cdot\fra  c{1}{1+\cos2x}


    \text{We have: }\;\lim_{x\to0}\left[4\cdot\left(\frac{\sin2x}{2x}\right)^2\cdot\frac{1  }{1+\cos2x}\right]

    . . . . . . =\;\; 4\cdot\underbrace{\left[\lim_{x\to0}\left(\frac{\sin2x}{2x}\right)^2\right]}_{\text{This is }1^2} \cdot \underbrace{\left[\lim_{x\to0}\left(\frac{1}{1+\cos2x}\right)\right]}_{\text{This is }\frac{1}{2}}

    . . . . . . = \;\;4\cdot1\cdot\frac{1}{2}

    . . . . . . =\;\;\boxed{2}



    Edit: too slow . . . again!
    .
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  4. #4
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    \frac{1-\cos 2x}{x^{2}}=\frac{2\sin^{2}x}{x^{2}}.
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  5. #5
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    Oh! Of course!

    Thank you very much.
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    If you know the series expansion it can also be done like this :

    Since \cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...

    then

    \cos 2x = 1-\frac{4x^2}{2!}+\frac{16x^4}{4!}-\frac{64x^6}{6!}+...

    so \frac{1-\cos 2x}{x^2} = \frac{4}{2!}-\frac{16x^2}{4!}+\frac{64x^4}{6!}+...

    taking the limit as x \rightarrow 0 on both sides, you get \frac{4}{2!}=4/2=2.
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