Question

The value of $ \underset{n\to \infty }{\mathop{\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 }{\mathop{\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 }{\mathop{\lim }}\,\frac{\sin \,x}{{{2}^{n}}\,\sin (x/{{2}^{n}})} $
$=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\frac{\sin \,x}{x}.\frac{(x/{{2}^{n}})}{\sin (x/{{2}^{n}})} $
$=\frac{\sin \,\,x}{x} $