Evaluate: $\lim\limits_{x \to 0} \dfrac{\tan 2x - \sin 2x}{x^3}$
$\begin{aligned} \lim\limits_{x \to 0} \dfrac{\tan 2x - \sin 2x}{x^3} & = \lim\limits_{x \to 0} \dfrac{\dfrac{\sin 2x}{\cos 2x}- \sin 2x}{x^3} \\ & = \lim\limits_{x \to 0} \dfrac{\sin 2x - \sin 2x \cos 2x}{x^3 \cos 2x} \\ & = \lim\limits_{x \to 0} \dfrac{\sin 2x \left(1-\cos 2x\right)}{x^3 \cos 2x} \\ & = \lim\limits_{x \to 0} \dfrac{\sin 2x \times 2 \sin^2 x}{x^3 \cos 2x} \;\; \left[\text{Note: }1- \cos 2\theta = 2 \sin^2 \theta\right] \\ & = 2 \lim\limits_{2x \to 0} \dfrac{\sin 2x}{2x} \times 2 \lim\limits_{x \to 0} \left(\dfrac{\sin x}{x}\right)^2 \times \lim\limits_{x \to 0} \dfrac{1}{\cos 2x} \\ & \left[\text{Note: As }x \to 0, \; 2x \to 0\right] \\ & = 2 \times 1 \times 2 \times 1^2 \times 1 \\ & = 4 \end{aligned}$