<s>[INST] Find the limit of $(\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$ Could you tell me how to find $\lim_{n \rightarrow \infty} (\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$ ?
 [/INST] As $x\rightarrow 0$ , $\sin(x) \rightarrow x$ 
So $ \sin(\frac{1}{n}) \rightarrow \frac{1}{n}$ as $n \rightarrow \infty$ 
The product $ [ \sin(\frac{1}{n}) \cdot \sin(\frac{2}{n}) \cdot ... \cdot \sin(\frac{n}{n}) ] \rightarrow \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdot...\frac{n}{n} =
\frac{n!}{n^n}$ 
As $n! = \sqrt{2 \cdot \pi \cdot n} \cdot (\frac{n}{e})^n  $ 
So $\frac{n!}{n^n} = \sqrt{2\pi n} \cdot (\frac{1}{e})^n $
$\lim_{ n \rightarrow \infty}$ of $[ \sin(\frac{1}{n}) \cdot \sin(\frac{2}{n}) \cdot ... \cdot \sin(\frac{n}{n}) ] ^ \frac{1}{n} \rightarrow \frac{1}{e}$
Since $\sin(\frac{m}{n})$ is always positive  if $0< m \le n$ and less than $\frac{m}{n}$ .
 The limit will be less than (1/e)
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