Question

Evaluate the limit:

$\underset{x\to 0}{lim}\frac{\cos 3x-\cos 7x}{x^2}$

Answer 1

We have,

$\underset{x\to 0}{lim}\frac{\cos 3x-\cos 7x}{x^2}$

It becomes 0/0 form.

$\because \cos A-\cos B=-2\sin \frac{A+B}{2}\sin \frac{A-B}{2}$

$=\underset{x\to 0}{lim}\frac{-2\sin \left(\frac{3x+7x}{2}\right)\sin \left(\frac{3x-7x}{2}\right)}{x^2}$

$=\underset{x\to 0}{lim}\frac{-2\sin \left(5x\right)\sin (-2x)}{x^2}$

$=\underset{x\to 0}{lim}\frac{-2\sin 5x(-\sin 2x)}{x^2}[\therefore \sin (-\theta )=-\sin (\theta \left)\right]$

$=\underset{x\to 0}{lim}\frac{2\sin 5x}{5x}\times \frac{\sin 2x}{2x}\times 5\times 2$

$[\therefore \underset{x\to 0}{lim}\frac{\sin x}{x}=1]$

$=2\times 1\times 1\times 5\times 2$

$=20$

Therefore, $\underset{x\to 0}{lim}\frac{\cos 3x-\cos 7x}{x^2}=20$