Question

Evaluate: $\underset{x\to 0}{lim}\frac{\sec 4x-\sec 2x}{\sec 3x-\sec x}$

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is A $\frac{3}{2}$

Given,

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

$\sec x=\frac{1}{\cos x}$

$=\underset{x\to 0}{lim}\frac{\frac{1}{\cos 4x}-\frac{1}{\cos 2x}}{\frac{1}{\cos 3x}-\frac{1}{\cos x}}$

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

$={lim}_{x\to 0}\frac{-2\sin 3x\sin (-x)}{-2\sin \left(2x\right)\sin (-x)}\frac{\cos 3x\cos x}{\cos 4x\cos 2x}$

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

as $\cos 0=1$ and ${lim}_{x\to 0}\frac{\sin x}{x}=1$

$=\frac{3x}{2x}=\frac{3}{2}$