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

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- Jun 2nd 2011, 12:49 PMgaganvjlim 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 - Jun 2nd 2011, 01:39 PMTKHunny
$\displaystyle x\cdot ln(csc(x)) = x\cdot [-ln(sin(x))] = \frac{-ln(sin(x))}{\frac{1}{x}}$

I'm starting to see it.

There is a reason why you studied trigonometry and algebra. Now is the time to REMEMBER! - Jun 2nd 2011, 01:43 PMmr fantastic
limit as x approaches 0 of (Cosec[x])^x - Wolfram|Alpha

Click on Show steps. - Jun 2nd 2011, 09:41 PMchisigma
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

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