# Thread: Limit x->0 (2/x)^(sin(x))

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

Can somebody please show me how one computes

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

If we assign it to the variable $\displaystyle y$ 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

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

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?

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

If you know that $\displaystyle x\ln x\to0$ as $\displaystyle x\to0^+$ and $\displaystyle \sin x=x+o(x)$ around 0, then this is easy.

3. ## Re: Limit x->0 (2/x)^(sin(x)) Originally Posted by emakarov If you know that $\displaystyle x\ln x\to0$ as $\displaystyle x\to0^+$ and $\displaystyle \sin x=x+o(x)$ 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

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