Question

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

Answer 1

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

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

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

$=2\times 1\times \frac{1}{2}=1$.