Question

Evaluate:

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

Answer 1

${lim}_{x\to 0}\frac{2\sin x-\sin 2x}{x^3}$ $(\frac{0}{0}\to \text{indeterminate form})$

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

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

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

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

$={lim}_{x\to 0}4\left(\frac{\sin x}{x}\right)\left[{\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)}^2\right].\frac{1}{4}$

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