Evaluate the following limits:

$=\lim _{x \rightarrow 0} \frac{\sec 4 x-\sec 2 x}{\sec 3 x-\sec x}$

$=\lim _{x \rightarrow 0} \frac{\frac{1}{\cos 4 x}-\frac{1}{\cos 2 x}}{\frac{1}{\cos 3 x}-\frac{1}{\cos x}}$

$=\lim _{x \rightarrow 0} \frac{(\cos 2 x-\cos 4 x) \times \cos x \times \cos 3 x}{(\cos x-\cos 3 x) \times \cos 2 x \times \cos 4 x}$

$=\lim _{x \rightarrow 0} \frac{2 \times \sin 3 x \times \sin x \times \cos x \times \cos 3 x}{2 \times \sin 2 x \times \sin x \times \cos 2 x \times \cos 4 x}$

$=\lim _{x \rightarrow 0} \frac{\sin 6 x \times \cos x}{\sin 4 x \times \cos 4 x} \times \frac{2}{2}$

$=2 \times \lim _{x \rightarrow 0} \frac{\sin 6 x \times \cos x}{\sin 8 x} \times \frac{8 x}{6 x} \times \frac{6}{8}$

$=\frac{3}{2} \times \lim _{x \rightarrow 0} \frac{\frac{\sin 6 x}{6 x} \times \cos x}{\frac{\sin 8 x}{8 x}}$

$=\frac{3}{2} \times \frac{1 \times 1}{1}$

$=\frac{3}{2}$

$\therefore \lim _{x \rightarrow 0} \frac{\sec 4 x-\sec 2 x}{\sec 3 x-\sec x}=\frac{3}{2}$