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