Question

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

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

Answer 1

Answer: (B)

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

Let $\frac{x}{2}=A$ or, $x=2^{n}A$ (1)

As $x\to \infty A\to 0$

$\underset{A\to 0}{lim}\sin A=A$ (2)

From trignometry

$\cos A.\cos 2A.\cos 4A.....nterms$

$=\frac{\sin 2^{n}A}{2^n\sin A}$

$\underset{n\to \infty }{lim}\frac{\sin 2^{n}A}{2^{n}A}$ (From(2))

$\underset{n\to \infty }{lim}\frac{\sin 2^{n}A}{2^{n}A}$

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