Question

The value of the limit $\underset{x\to 0}{\lim }\frac{2\sin x-\sin 2x}{x^3}$ is

Answer 1

$\underset{x\to 0}{\lim }\frac{2\sin x-\sin 2x}{x^3}=\underset{x\to 0}{\lim }\frac{2\sin x-2\sin x\cos x}{x^3}=\underset{x\to 0}{\lim }\frac{2\sin x\left(1-\cos x\right)}{x^3}=2\underset{x\to 0}{\lim }\frac{\sin x}{x}\times \frac{\left(}{1-\cos x}=2\times 1\times \underset{x\to 0}{\lim }\frac{2{\sin }^2\left(\frac{x}{2}\right)}{x^2}=\underset{x\to 0}{\lim }\frac{{\sin }^2\left(\frac{x}{2}\right)}{\frac{x^2}{4}}={\left(\underset{\frac{x}{2}\to 0}{\lim }\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)}^2=1$