Question

Evaluate: $\underset{n\to \infty }{lim}(\cos \frac{x}{2}\cos \frac{x}{4}\cos \frac{x}{8}...\cos \frac{x}{2^n})$

• A. $\frac{x}{\sin x}$
• B. $\frac{\sin x}{x}$
• C. $0$
• D. None of these

Answer 1

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

$\underset{n\to \infty }{lim}(\cos \frac{x}{2}\cos \frac{x}{4}\cos \frac{x}{8}...\cos \frac{x}{2^n})$

$\frac{2\sin \frac{x}{2^n}}{2\sin \frac{x}{2^n}}(\cos \frac{x}{2}\cos \frac{x}{4}\cos \frac{x}{8}....\cos \frac{x}{2^n})$

$=\frac{2\times (\cos \frac{x}{2}\cos \frac{x}{4}\cos \frac{x}{8}....\cos \frac{x}{2^{n-1}}\sin \frac{x}{2^{n-1}})}{2\times 2\sin \frac{x}{2^n}}$

$=\frac{2\cos \frac{x}{2}\sin \frac{x}{2}}{2^n\sin \frac{x}{2^n}}=\frac{x\sin x\frac{1}{2^n}}{x\sin \frac{x}{2^n}}$

$=\frac{\sin x}{x}\times \frac{\frac{x}{2^n}}{\sin \frac{x}{2^n}}$

${lim}_{n\to \infty }(\cos \frac{x}{2}\cos \frac{x}{4}\cos \frac{x}{8}....\cos \frac{x}{2^n})={lim}_{n\to \infty }\frac{\sin x}{x}\times \frac{\frac{x}{2^n}}{\sin \frac{x}{2^n}}[n\to \infty ;\frac{x}{2^n}\to 0]$

$=\frac{\sin x}{x}\times 1$

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