Mathematics - Limits Question with Solution | TestHub

$\underset{x\to 0}{\lim }\frac{(1-\cos 2x\left) \right(3+\cos x)}{x\tan 4x}$

$= \underset{x\to 0}{\lim }\frac{2{\sin }^{2}x \cdot (3+\cos x)}{x\cdot \tan 4x}$

$= \underset{x\to 0}{\lim }\frac{2\cdot {\left(\frac{\sin x}{x}\right)}^2\cdot \left(3+\cos x\right)}{\frac{\tan 4x}{4x} \cdot 4}$

$= \frac{2\cdot {\left(1\right)}^2 \cdot (3+1)}{1 \cdot 4}$

$=2$.