Question

$\underset{x\to \frac{\pi }{2}}{lim}\left(\underset{x\to \infty }{lim}\cos \frac{x}{2}\cos \frac{x}{2^2}\cos \frac{x}{2^3}......\cos \frac{x}{2^n}\right)$ equals to

Answer 1

$\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}$

$\sin x=2^2\sin \frac{x}{2^2}\cos \frac{x}{2}\cos \frac{x}{2^2}$

$\sin x=2^3\sin \frac{x}{2^3}\cos \frac{x}{2}\cos \frac{x}{2^2}\cos \frac{x}{2^3}$

$...$

$....$

$sinx=2^n\sin \frac{x}{2^n}\prod _{k=1}^n\cos \frac{x}{2^k}$

$\prod _{k=1}^n\cos \frac{x}{2^k}=\frac{\sin x}{x}{\left(\frac{\sin 2^{-n}x}{2^{-n}x}\right)}^{-1}$

${lim}_{n\to \infty }\prod _{k=1}^n\cos \frac{x}{2^k}=\frac{\sin x}{x}$

Substituting this value in the given expression, we get

${lim}_{x\to \frac{\pi }{2}}\frac{\sin x}{x}$

$=\frac{\sin \frac{\pi }{2}}{\frac{\pi }{2}}$

$=\frac{2}{\pi }.$