A) \[\frac{x}{\sin \,x}\]
B) \[\frac{x}{\cos \,x}\]
C) \[\frac{(\sin x)}{x}\]
D) \[\frac{(cosx)}{x}\]
Correct Answer: C
Solution :
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 \[\cos \,\left( \frac{x}{{{2}^{n}}} \right)\cos \left( \frac{x}{{{2}^{n-1}}} \right)....\cos \left( \frac{x}{4} \right)\cos \left( \frac{x}{2} \right)\] \[=\frac{\sin \,x}{{{2}^{n}}\,\sin \,(x/{{2}^{n}})}\] \[\therefore \]\[\underset{n\to \infty }{\mathop{\lim }}\,\cos \left( \frac{x}{2} \right)\,\cos \left( \frac{x}{4} \right).....\cos \left( \frac{x}{{{2}^{n-1}}} \right)\,\cos \left( \frac{x}{{{2}^{n}}} \right)\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\sin \,x}{{{2}^{n}}\,\sin (x/{{2}^{n}})}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\frac{\sin \,x}{x}.\frac{(x/{{2}^{n}})}{\sin (x/{{2}^{n}})}\] \[=\frac{\sin \,\,x}{x}\]
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