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