${\lim }_{x\to 0}\frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x}$ is equal to:

Use the identity $1-\cos 2x=2{\sin }^{2}x$.
$$\underset{x\to 0}{\lim }\frac{2{\sin }^{2}x(3+\cos x)}{x\tan 4x}$$
Divide and multiply by appropriate terms to use standard limits:
$$=\underset{x\to 0}{\lim }\left[2{\left(\frac{\sin x}{x}\right)}^2\cdot \frac{x^2}{x\cdot \frac{\tan 4x}{4x}\cdot 4x}\cdot (3+\cos x)\right]$$
$$=2\cdot (1{)}^2\cdot \frac{1}{1\cdot 4}\cdot (3+1)=2\cdot \frac{1}{4}\cdot 4=2$$
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