Question

$\displaystyle\lim _{n \rightarrow \infty}\left(\sin \frac{\pi}{2 n} \cdot \sin \frac{2 \pi}{2 n} \cdot \sin \frac{3 \pi}{2 n} \ldots \ldots . \cdot \sin \frac{(n-1) \pi}{n}\right)^{1 / n}$ is equal to :

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

Answer: (C)

$A=\left[\displaystyle\lim _{n \rightarrow \infty}\left(\sin \frac{\pi}{2 n} \sin \frac{2 \pi}{2 n} \ldots \ldots \sin \frac{(n-1)}{2 n}\right)\right]^{1 / n}$
$\Rightarrow \ln A=\frac{1}{n} \displaystyle\sum_{r=1}^{2 (n-1)} \ln \sin \frac{r \pi}{2 n}=\int\limits_0^2 \ell n \sin \left(\frac{\pi x}{2}\right) d x$
put $ \frac{\pi x}{2}=t$
$\Rightarrow \ln A=\frac{2}{\pi} \int\limits_0^\pi \ln (\sin t) d t=\frac{4}{\pi} \int\limits_0^{\pi / 2} \ln (\sin t) d t$
$\Rightarrow \ln A=-2 \ln 2 \Rightarrow A=\frac{1}{4}$