A) \[\frac{2}{5}\]
B) \[\frac{3}{5}\]
C) \[\frac{3}{2}\]
D) \[\frac{3}{4}\]
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{1-{{\cos }^{3}}x}{x\sin x\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2(1-{{\cos }^{3}}x)}{x\sin 2x}\] [Using LHospital Rule] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2[-3{{\cos }^{2}}x(-\sin x)]}{\sin 2x+x+\cos 2x.2}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{6{{\cos }^{2}}x\sin x}{\sin 2x+2x\cos 2x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{6[-2\cos x{{\sin }^{2}}x+{{\cos }^{3}}x]}{2\cos 2x+2[-x\sin 2x.2+\cos 2x]}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{6[-2\cos x{{\sin }^{2}}x+{{\cos }^{3}}x]}{2\cos 2x-4x\sin 2x+2\cos 2x}\] \[=\frac{6}{2+2}\] \[=\frac{3}{2}\]
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