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Math Help - Limit x->0 (2/x)^(sin(x))

  1. #1
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    Limit x->0 (2/x)^(sin(x))

    Can somebody please show me how one computes

    \lim_{x \rightarrow 0^+}  \left(\frac 2x \right)^{\sin (x)}

    If we assign it to the variable y and then take the logarithm of both sides, we should make it more easier. But after using some log laws and simplifications, I end up with

    \ln y = \ln (2) \cd \sin (x) - \sin (x) \cd \ln (x) \, .

    Now, because of ln(x), we need to use L'Hôspitals Rule but that isn't getting me anywhere since ln(x) keeps showing up. Can somebody help?
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  2. #2
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    Re: Limit x->0 (2/x)^(sin(x))

    If you know that x\ln x\to0 as x\to0^+ and \sin x=x+o(x) around 0, then this is easy.
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    Re: Limit x->0 (2/x)^(sin(x))

    Quote Originally Posted by emakarov View Post
    If you know that x\ln x\to0 as x\to0^+ and \sin x=x+o(x) around 0, then this is easy.
    Personally I think its more straight-forward (and clearer) to not invoke the series expansion for sin x and use L”Hopitals rule directly to get:

    lim –sinxlnx = lim( –cosxlnx –sinx/x) = +infinity -> y -> infinity. (x -> 0+).

    As a matter of interest, if your going to use sinx app x, might as well do it at the beginning to find lim y = (2/x)^x and log y = xln2 – xlnx to get same result.

    Not finding fault, just giving a different perspective.

    Edit: As a matter of further interest:
    lim (2/x)^x = lim (2^x) lim (1/x)^x = lim (1/x)^x
    Last edited by Hartlw; February 2nd 2013 at 12:29 PM.
    Thanks from topsquark
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