Evaluate $\lim\limits_{x \to 0} \; \dfrac{\sin^{-1}x- \tan^{-1}x}{x^3}$
$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{\sin^{-1}x - \tan^{-1}x}{x^3} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\sin^{-1}x - \tan^{-1}x\right)}{\dfrac{d}{dx}\left(x^3\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{1}{\sqrt{1-x^2}}-\dfrac{1}{1+x^2}}{3x^2} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\dfrac{1}{\sqrt{1-x^2}}-\dfrac{1}{1+x^2}\right)}{\dfrac{d}{dx}\left(3x^2\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{-\dfrac{1}{2}\times \left(1-x^2\right)^{-3/2}\times \left(-2x\right)+\dfrac{2x}{\left(1+x^2\right)^2}}{6x} & \\ & & \\ & = \dfrac{1}{6} \; \lim\limits_{x \to 0} \; \left\{\dfrac{1}{\left(1-x^2\right)^{3/2}} + \dfrac{2}{\left(1+x^2\right)^2} \right\} & \\ & & \\ & = \dfrac{1}{6} \times 3 = \dfrac{1}{2} \end{aligned}$