Question

The value of $\underset{n\to \infty }{lim}\cos \frac{x}{2}\cos \frac{x}{2^2}...\cos \frac{x}{2^n}$ is

• A. $1$
• B. $\frac{\sin x}{x}$
• C. $\frac{x}{\sin x}$
• D. none of these

Answer 1

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

Let $y=\cos \frac{x}{2}\cos \frac{x}{2^2}...\cos \frac{x}{2^n}$
$=\frac{1}{2\sin x/2^n}(\cos \frac{x}{2}\cos \frac{x}{2^2}...\cos \frac{x}{2^{n-2}}\sin \frac{x}{2^{n-2}})$
$⋮$
$=\frac{1}{2\sin x/2^n}\sin x$
Hence $\underset{n\to \infty }{lim}\frac{2\sin x/2^n}{\sin 2\sin x/2^n}\frac{\sin x}{x}=1.\frac{\sin x}{x}=\frac{\sin x}{x}$