# lim as x --> 0 of [cosec(x)]^x

• June 2nd 2011, 12:49 PM
gaganvj
lim 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
• June 2nd 2011, 01:39 PM
TKHunny
$x\cdot ln(csc(x)) = x\cdot [-ln(sin(x))] = \frac{-ln(sin(x))}{\frac{1}{x}}$

I'm starting to see it.

There is a reason why you studied trigonometry and algebra. Now is the time to REMEMBER!
• June 2nd 2011, 01:43 PM
mr fantastic
Quote:

Originally Posted by gaganvj
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

limit as x approaches 0 of &#40;Cosec&#91;x&#93;&#41;&#94;x - Wolfram|Alpha

Click on Show steps.
• June 2nd 2011, 09:41 PM
chisigma
Quote:

Originally Posted by gaganvj
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

Because is $(\sin x)^{-x}= e^{-x\ \ln \sin x}$ what we have to valuate is...

$\lim_{x \rightarrow 0} x\ \ln \sin x$ (1)

From the 'infinite product'...

$\sin x = x\ \prod_{n=1}^{\infty} (1-\frac{x^{2}}{n^{2}\ \pi^{2}})$ (2)

... we derive...

$x\ \ln \sin x = x\ \{\ln x + \sum_{n=1}^{\infty} \ln (1-\frac{x^{2}}{n^{2}\ \pi^{2}})\}$ (3)

... so that is...

$\lim_{x \rightarrow 0}x\ \ln \sin x =0$ (4)

What is interesting is the fact that the limit (4) is the same for $x \rightarrow 0+$ and $x \rightarrow 0-$ so that the function $(\sin x)^{-x}$ is continous in x=0...

Kind regards

$\chi$ $\sigma$