Question

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

• A. $\frac{x}{\sin x}$
• B. $\frac{x}{\cos x}$
• C. $\frac{(\sin x)}{x}$
• D. $\frac{(cosx)}{x}$

Answer 1

Answer: (C)

We know that $\cos A\cos 2A\cos 4A....\cos 2^{n-1}A=\frac{\sin 2^{n}A}{2^n\sin A}$ Put $A=\frac{x}{2^n},$ we get $\cos \left(\frac{x}{2^n}\right)\cos \left(\frac{x}{2^{n-1}}\right)....\cos \left(\frac{x}{4}\right)\cos \left(\frac{x}{2}\right)$
$=\frac{\sin x}{2^n\sin (x/2^n)}$
$\therefore$ $\underset{n\to \infty }{\lim }\cos \left(\frac{x}{2}\right)\cos \left(\frac{x}{4}\right).....\cos \left(\frac{x}{2^{n-1}}\right)\cos \left(\frac{x}{2^n}\right)$
$=\underset{n\to \infty }{\lim }\frac{\sin x}{2^n\sin (x/2^n)}$
$=\underset{n\to \infty }{\lim }\frac{\sin x}{x}.\frac{(x/2^n)}{\sin (x/2^n)}$
$=\frac{\sin x}{x}$