Question

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

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{\tan \left(2x\right)-\sin \left(2x\right)}{x^3}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{{\displaystyle \frac{\sin 2x}{\cos 2x}}-\sin 2x}{x^3}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\sin 2x\left\{1-\cos 2x\right\}}{\mathrm{co}S2x\times x^3}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\sin 2x\times 2{\sin }^{2}x}{\cos 2x\times x^3}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\sin 2x}{2x}\times \frac{2}{\cos 2x}\times 2{\left(\frac{\sin x}{x}\right)}^2\right] \\ =\frac{2\times 2}{\cos 0} \\ =4 \end{gathered}$$