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

Can somebody please show me how one computes

If we assign it to the variable 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

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?

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

If you know that as and around 0, then this is easy.

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

Quote:

Originally Posted by

**emakarov** If you know that

as

and

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