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Math Help - Limit of ln(cosx)/x^2

  1. #1
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    Limit of ln(cosx)/x^2

    Can any help me with this. I need to solve this without L'Hopital's rule. \lim ( \ln ( \cos ( x ) ) / x^2 ) when x->0

    Thanks in advance
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Limit of ln(cosx)/x^2

    Quote Originally Posted by Bokas View Post
    Can any help me with this. I need to solve this without L'Hopital's rule. \lim ( \ln ( \cos ( x ) ) / x^2 ) when x->0

    Thanks in advance
    We start from the 'infinite product'...

    \cos x= \prod_{n=0}^{\infty} (1-\frac{ 4 x^{2}}{\pi^{2} (2n+1)^{2}}) (1)

    ... and from (1) we obtain...

    \ln \cos x = \sum_{n=0}^{\infty} \ln \{1-\frac{ 4 x^{2}}{\pi^{2} (2n+1)^{2}}\} (2)

    Now if we compute the coefficient of the x^{2} term of the McLaurin expansion of (2) we obtain...

    a_{2}= -\sum_{n=0}^{\infty} \frac{4}{\pi^{2} (2n+1)^{2}}= -\frac{4}{\pi^{2}}\ \frac{\pi^{2}}{8} = -\frac{1}{2} (3)

    ... so that is...

    \lim_{x \rightarrow 0} \frac{\ln \cos x}{x^{2}}= -\frac{1}{2} (4)

    Kind regards

    \chi \sigma
    Last edited by chisigma; November 30th 2011 at 08:13 PM. Reason: error of sign in (3)...
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  3. #3
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    Re: Limit of ln(cosx)/x^2

    Thank you very much. You saved me
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