Question

Evaluate the limit: $\underset{x\to 0}{lim}\frac{2\sin x-\sin 2x}{x^3}$

Answer 1

Solution:
We have,

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

It becomes $\frac{0}{0}$ form.

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

$[\because \sin 2x=2\sin x\cos x]$

$=\underset{x\to 0}{lim}\frac{2\sin x(1-\cos x)}{x^3}$

$[\because 1-\cos x=2{\sin }^2\frac{x}{2}]$

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

$=\underset{x\to 0}{lim}4\times \frac{\sin x}{x}\times \frac{\sin \frac{x}{2}}{\frac{x}{2}}\times \frac{\sin \frac{x}{2}}{\frac{x}{2}}\times \frac{1}{4}$

$=4\times 1\times 1\times \frac{1}{4}$

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

$=1$

$\therefore \underset{x\to 0}{lim}\frac{2\sin x-\sin 2x}{x^3}=1$