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Thread: lim as x --> 0 of [cosec(x)]^x

  1. #1
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    lim as x --> 0 of [cosec(x)]^x

    Lim (cosecx)^x
    x-> 0


    how to solve this ? I know the initial step is to take log on both sides, but couldn't proceed further.

    Thanks.


    the answer to the problem is 1
    Last edited by mr fantastic; Jun 2nd 2011 at 01:44 PM. Reason: Re-titled.
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  2. #2
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    $\displaystyle x\cdot ln(csc(x)) = x\cdot [-ln(sin(x))] = \frac{-ln(sin(x))}{\frac{1}{x}}$

    I'm starting to see it.

    There is a reason why you studied trigonometry and algebra. Now is the time to REMEMBER!
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  3. #3
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    Quote Originally Posted by gaganvj View Post
    Lim (cosecx)^x
    x-> 0


    how to solve this ? I know the initial step is to take log on both sides, but couldn't proceed further.

    Thanks.


    the answer to the problem is 1
    limit as x approaches 0 of (Cosec[x])^x - Wolfram|Alpha

    Click on Show steps.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by gaganvj View Post
    Lim (cosecx)^x
    x-> 0


    how to solve this ? I know the initial step is to take log on both sides, but couldn't proceed further.

    Thanks.


    the answer to the problem is 1
    Because is $\displaystyle (\sin x)^{-x}= e^{-x\ \ln \sin x}$ what we have to valuate is...

    $\displaystyle \lim_{x \rightarrow 0} x\ \ln \sin x$ (1)

    From the 'infinite product'...

    $\displaystyle \sin x = x\ \prod_{n=1}^{\infty} (1-\frac{x^{2}}{n^{2}\ \pi^{2}})$ (2)

    ... we derive...

    $\displaystyle x\ \ln \sin x = x\ \{\ln x + \sum_{n=1}^{\infty} \ln (1-\frac{x^{2}}{n^{2}\ \pi^{2}})\}$ (3)

    ... so that is...

    $\displaystyle \lim_{x \rightarrow 0}x\ \ln \sin x =0 $ (4)

    What is interesting is the fact that the limit (4) is the same for $\displaystyle x \rightarrow 0+$ and $\displaystyle x \rightarrow 0-$ so that the function $\displaystyle (\sin x)^{-x}$ is continous in x=0...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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