Question

$\underset{x\to 0}{lim}\cos \frac{x}{2}\cos \frac{x}{2^2}\cos \frac{x}{2^3}......\cos \frac{x}{2^n}$ is equal to

• A. $1$
• B. $-1$
• C. $\frac{\sin x}{x}$
• D. $\frac{x}{\sin x}$

Answer 1

The correct option is C $\frac{\sin x}{x}$

$\underset{x\to \infty }{lim}(\cos \frac{x}{2}\cos \frac{x}{2^2},\cos \frac{x}{2^3}.......\cos \frac{x}{2^x})$

$\sin 2\theta =2\sin \theta \cos \theta$ $2\theta =x$

$=\sin x=2\left(\sin \frac{x}{2}\right)\cos \frac{x}{2}$ $\theta =\frac{x}{2}$

$=2\left(2\sin \frac{x}{4}\cos \frac{x}{4}\right)\cos \frac{x}{2}$ $2\theta =\frac{x}{2}$

$\theta =\frac{x}{4}$

$=8\left(\left(\sin \frac{x}{8}\right)\cos \frac{x}{8}\cos \frac{x}{4}\cos \frac{x}{2}\right)$ $\frac{x}{4}=2\theta$

$\theta =\frac{x}{8}$

$=16\left(\sin \frac{x}{16}\cos \frac{x}{4}\cos \frac{x}{8}\cos \frac{x}{4}\cos \frac{x}{2}\right)\theta =\frac{x}{16}$

$=16\left(\sin \frac{x}{2^4}\cos \frac{x}{2^4}\cos \frac{x}{2^3}\cos \frac{x}{2^2}\cos \frac{x}{2}\right)$

$\sin x=\left(2^n\sin \frac{x}{2^x}\right)\cos \frac{x}{2x}\cos \frac{x}{2^{x-1}}.......\cos \frac{x}{2^3}\cos \frac{x}{2^2}\cos \frac{x}{2}$

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

$\underset{x\to \infty }{lim}\frac{\sin x}{2^n\sin \left(\frac{x}{2^x}\right)}\underset{x\to 0}{lim}\frac{\sin x}{x}=1$

$\overline{\infty }=0$

$\underset{x\to \infty }{lim}\frac{\sin x}{2^x\frac{\sin \left(\frac{x}{2^x}\right)}{\left(\frac{x}{2^x}\right)}\times \left(\frac{x}{2^x}\right)}=\frac{\sin x}{x}]\underset{x\to \infty }{lim}\frac{\frac{1}{\sin \left(\frac{x}{2^x}\right)}}{\frac{x}{2^x}}$

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