A) 2
B) \[\frac{1}{2}\]
C) 4
D) 3
Correct Answer: A
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( 1-\cos 2x \right)\left( 3+\cos x \right)}{x\tan 4x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x\left( 3+\cos x \right)}{{{x}^{2}}\frac{\tan 4x}{4x}.4}\] \[=2\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}x}{{{x}^{2}}}.\underset{x\to 0}{\mathop{\lim }}\,\left( 3+\cos x \right).\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\frac{\tan 4x}{4x}}.\frac{1}{4}\]\[=2:1.4.\frac{1}{4}=2\]Code : D
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