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

• February 2nd 2013, 09:42 AM
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?
• February 2nd 2013, 10:24 AM
emakarov
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.
• February 2nd 2013, 12:18 PM
Hartlw
Re: Limit x->0 (2/x)^(sin(x))
Quote:

Originally Posted by emakarov
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