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Math Help - Limits(trigonometric)

  1. #1
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    Unhappy Limits(trigonometric)

    Find:
    lim (lncosx)/x^2
    x-->0
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  2. #2
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    Quote Originally Posted by matsci0000 View Post
    Find:
    lim (lncosx)/x^2
    x-->0

    When the x-->0 the function is 0/0.
    So use L'hopitals rule.
    For this function, use it twice.
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  3. #3
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    Quote Originally Posted by matsci0000 View Post
    Find:
    lim (lncosx)/x^2
    x-->0
    ln(cos(0))=ln(1) so the logarithm with base e that equals 1 is just equal to zero. This means that the numerator and the denominator approach zero, so use l-hopitals rule.

    \lim_{x->0}\frac{ln(\cos(x))}{x^2}=\lim_{x->0}\frac{-\sin(x)}{2x\cos(x)}

    No just apply the rule a second time. It's pretty easy from here.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by matsci0000 View Post
    Find:
    lim (lncosx)/x^2
    x-->0
    Notice that for x within a sufficently small neighborhood of 0 that \ln\left(\cos(x)\right)=\ln\left(\sqrt{1-\sin^2(x)}\right)=\frac{1}{2}\ln\left(1-\sin^2(x)\right). Also, it is a well known fact that \sin(x)\underset{x\to0}{\sim}x and \ln\left(1-x\right)\underset{x\to0}{\sim}-x. So combining these two we can see that \ln\left(1-\sin^2(x)\right)\underset{x\to0}{\sim}\ln\left(1-x^2\right)\underset{x\to0}{\sim}-x^2. So we may conclude that \lim_{x\to0}\frac{\ln\left(\cos(x)\right)}{x^2}=\l  im_{x\to0}\frac{\tfrac{1}{2}\left(-x^2\right)}{x^2}=\frac{-1}{2}.
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