Calculus - integration's application in solving limits

2) \lim_{n\rightarrow \infty }\left(sin\frac{\pi }{2n}sin\frac{2\pi }{2n}.........sin\frac{2(n-1)\pi }{2n} \right)^{1/n}

ans..........> 1/4

$\mathop{\lim}\limits_{n \to \infty} \sum_{r= 1}^{2n} . \frac{1}{n}\ .\frac{ \frac{r}{n}}{\sqrt{1 +\frac{r^2}{n^2} }}$

PUT rn = x

now integral of form

$\mathop{\lim}\limits_{n \to \infty} \sum_{r= 1}^{n} .\frac{1}{n } f (\frac{r}{n} ) =\int_{0}^{1} f ( x ) d x$

here n= 2 n hence integral reduces to

$=\int_{0}^{2} f ( x ) d x$

hence rewritin f ( x )

$=\int_{0}^{2} \frac{x}{\sqrt{1 + x ^2}} d x$

= √5 - 1

COPYIN FROM ARIHANT BY AMIT SIR

CONSIDER EXPRESSION AS A

AND TAKE LOG BOTH SIDES

log A = 1n log sin ( Ï€ / 2n . sin 2Ï€ / 2n. sin 2 ( n-1) Ï€/ 2n

take

r = 2 ( n-1 )

then do it !!!!!
u get log A = log 1/4

A= 1/4

integration's application in solving limits