Functions Question 316
Question: The value of $ \underset{n\to \infty }{\mathop{\lim }},\cos ( \frac{x}{2} )\cos ( \frac{x}{4} )\cos ( \frac{x}{8} )…\cos ( \frac{x}{2^{n}} ) $ is
Options:
A) 1
B) $ \frac{\sin x}{x} $
C) $ \frac{x}{\sin x} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We know that $ \cos A\cos 2A\cos 4A….\cos {2^{n-1}}A=\frac{\sin 2^{n}A}{2^{n}\sin A} $ Taking $ A=\frac{x}{2^{n}}, $ we get $ \cos ,( \frac{x}{2^{n}} ),\cos ,( \frac{x}{{2^{n-1}}} ),…\cos ( \frac{x}{4} )\cos ,( \frac{x}{2} )=\frac{\sin x}{2^{n}\sin ( \frac{x}{2^{n}} )} $
$ \therefore ,\underset{n\to \infty }{\mathop{\lim }},\cos ,( \frac{x}{2} )\cos ,( \frac{x}{4} )…\cos ,( \frac{x}{{2^{n-1}}} ),\cos ,( \frac{x}{2^{n}} ) $ $ =\underset{n\to \infty }{\mathop{\lim }}\frac{\sin x}{2^{n}\sin ,( \frac{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} $ .
BETA
