The value of $$\lim_{x \to 0} \frac{(1 - \cos 2x)(3 + \cos x)}{x \tan 4x}$$ is equal to

We have to evaluate the limit

$$L=\lim_{x \to 0}\frac{(1-\cos 2x)(3+\cos x)}{x \tan 4x}\, .$$

First, recall the trigonometric identity

$$1-\cos\theta = 2\sin^2\!\left(\frac{\theta}{2}\right).$$

Applying it with $$\theta=2x$$, we obtain

$$1-\cos 2x = 2\sin^2 x.$$

Substituting this into the numerator gives

$$L=\lim_{x \to 0}\frac{2\sin^2 x\,(3+\cos x)}{x \tan 4x}.$$

Now, write the tangent in terms of sine and cosine:

$$\tan 4x=\frac{\sin 4x}{\cos 4x}\; \Longrightarrow\; x\tan 4x=\frac{x\sin 4x}{\cos 4x}.$$

Hence the limit becomes

$$L=\lim_{x \to 0}\frac{2\sin^2 x\,(3+\cos x)\,\cos 4x}{x\sin 4x}.$$

Separate the fraction into convenient pieces:

$$L=\lim_{x \to 0}2\,(3+\cos x)\,\cos 4x\;\frac{\sin^2 x}{x\sin 4x}.$$

Rewrite the last factor as a product:

$$\frac{\sin^2 x}{x\sin 4x}= \left(\frac{\sin x}{x}\right)\!\left(\frac{\sin x}{\sin 4x}\right).$$

We now evaluate each individual limit using standard small-angle limits.

1. The fundamental limit $$\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1.$$

2. For $$\displaystyle\lim_{x\to 0}\frac{\sin x}{\sin 4x},$$ we note that for small angles $$\sin x\sim x$$ and $$\sin 4x\sim 4x$$, so

$$\lim_{x\to 0}\frac{\sin x}{\sin 4x}=\lim_{x\to 0}\frac{x}{4x}=\frac14.$$

3. Finally, $$\displaystyle\lim_{x\to 0}\cos 4x=1$$ and $$\displaystyle\lim_{x\to 0}\cos x=1.$$

Putting all these limits together, we get

$$L = 2\;\bigl(3+1\bigr)\;(1)\;\bigl(1\bigr)\;\left(\frac14\right) = 2 \times 4 \times \frac14 = 2.$$

Hence, the correct answer is Option B.