Answer:
Option D
Explanation:
Let I= $\lim_{x \rightarrow 0}\frac{(1-\cos 2x)}{x^{2}}\frac{(3+\cos x)}{1}.\frac{x}{\tan 4x}$
= $\lim_{x \rightarrow 0}\frac{2\sin^{2}x}{x^{2}}.\frac{3+\cos x}{1}.\frac{x}{\tan 4x}$
= $2\lim_{x \rightarrow 0}\left(\frac{\sin^{}x}{x^{}}\right)^{2}\lim_{x \rightarrow 0}.(3+\cos x).\lim_{x \rightarrow 0}\frac{4x}{4\tan 4x}$
= $2.(1)^{2}.(3+\cos 0^{o}).\frac{1}{4}(1)$
= $2.1.(3+1).\frac{1}{4}=2.4.\frac{1}{4}=2$
