Limit till infinity

Let $L = \lim_{n \rightarrow \infty} \sqrt [n] {\sin{\dfrac{\pi}{2n}}\sin{\dfrac{2\pi}{2n}}\cdots\sin{\dfrac{(n-1)\pi}{2n}}}$ Taking Logarithm on both sides, $\ln{L}= \lim_{n \rightarrow \infty} \dfrac{1}{n}\displaystyle \sum_{r=1}^{n-1}\ln { \left(\sin{\dfrac{r\pi}{2n}} \right)}$ Using, Riemann Sums, the above can be approximated to, $\ln{L} = \displaystyle \int_{0}^{1} \ln{\left(\sin{\left(\dfrac{\pi x}{2} \right)}\right)}\text{d}x \\ = -\ln{2}$ So, $L= e^{-\ln{2}} = \boxed{\large { \dfrac{1}{2}}}$

Let 1,a1,a2,.....,a(n-1) be the roots of z^n=1
(z^n)-1=(z-1)(z-a1)...(z-a(n-1))
So,(z-a1)(z-a2)....(z-a(n-1))=z^(n-1)+z^(n-2)+...+z+1
Set z=1, (1-a1)(1-a2)...(1-a(n-1))=n
I1-a1II1-a2I...I1-a(n-1)I=n
Prod of r=1 to n-1((1-cos $\frac{2rpi}{n}$ )^2+(sin $\frac{2rpi}{n}$ ^2))^0.5=n
Prod of r=1 to n-1 2sin $\frac{rpi}{n}$ =n
(sin $\frac{pi}{n}$ )...(sin $\frac{(n-1)pi}{n}$ )= $\frac{n}{2^(n-1)}$
From here use lim n^(1/n)=1 to conclude the limit= $\frac{1}{2}$