Question

The value of:
$\underset{n\to 0}{lim}\cos \left(\frac{x}{2}\right)\cos \left(\frac{x}{4}\right)\cos \left(\frac{x}{8}\right)......\cos \left(\frac{x}{2^n}\right)$ is

Answer 1

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

$=2\times 2\sin \frac{x}{4}\cos \frac{x}{4}\cos \frac{x}{2}$

$=4\times 2\sin \frac{x}{8}\cos \frac{x}{8}\cos \frac{x}{4}\cos \frac{x}{2}$

$=8\times 2\sin \frac{x}{16}\cos \frac{x}{16}\cos \frac{x}{8}\cos \frac{x}{4}\cos \frac{x}{2}$

$..,n$ times

$=2^n\sin \frac{x}{2^n}\times \cos \frac{x}{2}\cos \frac{x}{4}\cos \frac{x}{8}\cos \frac{x}{16}...\cos \frac{x}{2^n}$

$\Rightarrow \cos \frac{x}{2}\cos \frac{x}{4}\cos \frac{x}{8}\cos \frac{x}{16}...\cos \frac{x}{2^n}=\frac{\sin x}{2^n\sin \frac{x}{2^n}}$

${lim}_{n\to 0}\cos \frac{x}{2}\cos \frac{x}{4}\cos \frac{x}{8}...\cos \frac{x}{2^n}={lim}_{n\to 0}\frac{\sin x}{2^n\sin \frac{x}{2^n}}$

Just apply limit n=0

Anaswer will be $1$