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
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 (Cosec[x])^x - Wolfram|Alpha
Click on Show steps.
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
$\displaystyle \chi$ $\displaystyle \sigma$