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Math Help - Limit

  1. #1
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    Limit

    Could anyone solve this:

    lim (x -> o) ((1-cos3x) / 7x^2)

    Thanks!
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  2. #2
    Math Engineering Student
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    Consider \lim_{u \to 0} \frac{{1 - \cos u}}<br />
{{u^2 }} = \frac{1}<br />
{2}.

    Substitute u=3x,

    \lim_{x \to 0} \frac{{1 - \cos 3x}}<br />
{{9x^2 }} = \frac{1}<br />
{2}.

    Multiply both sides by \frac97 to get the answer.
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  3. #3
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    Hello, clapaucius!

    Here's one way . . . there may be better ways.


    \lim_{x\to0} \frac{1- \cos3x}{7x^2}

    Multiply by \frac{1+\cos3x}{1+\cos3x}

    . . \frac{1-\cos3x}{7x^2}\cdot\frac{1+\cos3x}{1+\cos3x} \;=\;\frac{1-\cos^2\!3x}{7x^2(1+\cos3x)} \;=\;\frac{\sin^2\!3x}{7x^2(1 + \cos3x)}


    Multiply by \frac{9}{9}

    . . \frac{9}{9}\cdot\frac{\sin^2\!3x}{7x^2(1+\cos3x)} \;=\;\frac{9}{7(1+\cos3x)}\cdot\frac{\sin^2\!3x}{9  x^2} \;=\;\frac{9}{7(1+\cos3x)}\cdot\left(\frac{\sin3x}  {3x}\right)^2


    Hence: . \lim_{x\to0}\,\left[\frac{9}{7(1+\cos3x)}\cdot\left(\frac{\sin3x}{3x}\  right)^2\right] \;=\;\frac{9}{7(1 + \cos0)}\cdot(1^2) \;=\;\frac{9}{14}

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  4. #4
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    \frac{9}{14} is correct ... Thanks!!
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  5. #5
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    Use a Taylor 3rd Degree polynomial.
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