Question

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

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{3\sin x-\sin \left(3x\right)}{x^3}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{3\sin x-\left(3\sin x-4{\sin }^{3}x\right)}{x^3}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{4{\sin }^{3}x}{x^3}\right] \\ =\underset{x\to 0}{\lim }\left[4{\left(\frac{\sin x}{x}\right)}^3\right] \\ =4\times 1=4 \end{gathered}$$