Evaluate: $\lim\limits_{x \rightarrow 0} \left[\dfrac{x \cos x - \sin x}{x^2 \sin x}\right]$
$\begin{aligned} \lim\limits_{x \rightarrow 0} \left[\dfrac{x \cos x - \sin x}{x^2 \sin x}\right] \;\; \left[\dfrac{0}{0} \text{ form}\right] & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\dfrac{d}{dx} \left(x \cos x - \sin x\right)}{\dfrac{d}{dx} \left(x^2 \sin x\right)}\right] \left[\text{ by L'Hopital's rule}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\cos x - x \sin x - \cos x}{2x \sin x + x^2 \cos x}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{- x \sin x}{x \left(2 \sin x + x \cos x\right)}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{- \sin x}{2 \sin x + x \cos x}\right] \;\; \left[\dfrac{0}{0} \text{ form}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\dfrac{d}{dx} \left(- \sin x\right)}{\dfrac{d}{dx} \left(2 \sin x + \cos x\right)}\right] \left[\text{ by L'Hopital's rule}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\cos x}{2 \cos x + \sin x}\right] \\\\ & = \dfrac{1}{2 + 0} \\\\ & = \dfrac{1}{2} \end{aligned}$