Question

Evaluate the limit:

$\underset{x\to 0}{lim}\frac{\sin (2+x)-\sin (2-x)}{x}$

Answer 1

We have,

$\underset{x\to 0}{lim}\frac{\sin (2+x)-\sin (2-x)}{x}$

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

$=\underset{x\to 0}{lim}\frac{2\cos \frac{(2+x+2-x)}{2}\sin \frac{(2+x-2+x)}{2}}{x}$

$=\underset{x\to 0}{lim}\frac{2\cos \frac{4}{2}\sin \frac{2x}{2}}{x}$

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

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

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

$=\left(2\cos 2\right)\times 1$

$=2\cos 2$

Therefore,

$\underset{x\to 0}{lim}\frac{\sin (2+x)-\sin (2-x)}{x}=2\cos 2$