A) 1
B) \[\frac{\sin x}{x}\]
C) \[\frac{x}{\sin x}\]
D) None of these
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 \[\cos \,\left( \frac{x}{{{2}^{n}}} \right)\,\cos \,\left( \frac{x}{{{2}^{n-1}}} \right)\,...\cos \left( \frac{x}{4} \right)\cos \,\left( \frac{x}{2} \right)=\frac{\sin x}{{{2}^{n}}\sin \left( \frac{x}{{{2}^{n}}} \right)}\] \[\therefore \,\,\,\underset{n\to \infty }{\mathop{\lim }}\,\,\,\cos \,\left( \frac{x}{2} \right)\cos \,\left( \frac{x}{4} \right)...\cos \,\left( \frac{x}{{{2}^{n-1}}} \right)\,\cos \,\left( \frac{x}{{{2}^{n}}} \right)\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{\sin x}{{{2}^{n}}\sin \,\left( \frac{x}{{{2}^{n}}} \right)}=\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{\sin x}{x}\frac{(x/{{2}^{n}})}{\sin \,(x/{{2}^{n}})}=\frac{\sin x}{x}\].
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