Question

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

Answer 1

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

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

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

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

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

$={lim}_{x\to 0}\frac{\sin 2x}{\frac{1}{2}\times 2x}\times \frac{2{\sin }^{2}x}{x^2}\times \frac{1}{\cos 2x}$ using multiple angle formula $\cos 2x=1-2{\sin }^{2}x$

$={lim}_{x\to 0}2\left(\frac{\sin 2x}{2x}\right)\times 2\frac{{\sin }^{2}x}{x^2}\times \frac{1}{\cos 2x}$ by rearranging the terms

$=2\times 1\times 2\times 1\times \frac{1}{\cos 0}$ since ${lim}_{\theta \to 0}\frac{\sin \theta }{\theta }=1$

$=4$ since $\cos 0=1$