Answer:
Option C
Explanation:
We have
$\lim_{x \rightarrow 0}\frac{(1-\cos 2x)(3+\cos x)}{x \tan4x}$
= $\lim_{x \rightarrow 0}\frac{(2\sin ^{2}x(3+\cos x))}{x \times \frac{\tan 4x}{4x}\times 4x}$
= $\lim_{x \rightarrow 0}\frac{2\sin ^{2}x}{x^{2}}\times\lim_{x \rightarrow 0}\frac{(3+\cos x)}{4} \times\frac{1}{\lim_{x \rightarrow 0}\frac{\tan 4x}{4x}}$
$=2\times \frac{4}{4}\times1$ $\left( \therefore \lim_{\theta \rightarrow 0}\frac{\sin \theta}{\theta}=1 and \lim_{\theta \rightarrow 0}\frac{\tan \theta}{\theta}=1\right)$
=2
