A) 4
B) 3
C) 2
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
We have, \[\underset{x\to 0}{\mathop{\lim }}\,\frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}x(3+\cos x)}{x\times \frac{\tan 4x}{4x}\times 4x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}x}{{{x}^{2}}}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{(3+\cos x)}{4}\times \frac{1}{\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan 4x}{4x}}\] \[=2\times \frac{4}{4}\times 1\left[ \because \underset{\theta \to 0}{\mathop{\lim }}\,\frac{\sin \theta }{\theta }=1\text{and }\underset{\theta \to 0}{\mathop{\lim }}\,\frac{\tan \theta }{\theta }=1 \right]\]\[=2\]
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