A) \[\frac{2}{5}\]
B) \[\frac{3}{5}\]
C) \[\frac{3}{2}\]
D) \[\frac{3}{4}\]
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{I-{{\cos }^{3}}x}{x\sin x\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{(1-\cos x)(1+\cos x+{{\cos }^{2}}x)}{x\sin x\cos x}\]\[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\left( \frac{x}{2} \right)}{x.2\sin \left( \frac{x}{2} \right)\cos \left( \frac{x}{2} \right)}\times \frac{(1+\cos x+{{\cos }^{2}}x)}{\cos x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \left( \frac{x}{2} \right)}{2\left( \frac{x}{2} \right)}\times \frac{1+\cos x+{{\cos }^{2}}x)}{\cos \left( \frac{x}{2} \right)\cos x}=\frac{1}{2}\times 3=\frac{3}{2}\]
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