Evaluate $\;$ $\lim\limits_{y \rightarrow 0} \dfrac{y - \tan^{-1}y}{y - \sin y}$
$\begin{aligned} \lim\limits_{y \rightarrow 0} \dfrac{y - \tan^{-1}y}{y - \sin y} \;\;\; \left[\dfrac{0}{0} \text{ form}\right] & = \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{d}{dy} \left[y - \tan^{-1}y\right]}{\dfrac{d}{dy} \left[y - \sin y\right]} \;\;\; \left[\text{by L'Hopital's rule}\right] \\\\ & = \lim\limits_{y \rightarrow 0} \dfrac{1 - \dfrac{1}{1 + y^2}}{1 - \cos y} \;\;\; \left[\dfrac{0}{0} \text{ form}\right] \\\\ & = \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{d}{dy} \left[1 - \dfrac{1}{1 + y^2}\right]}{\dfrac{d}{dy} \left[1 - \cos y\right]} \;\;\; \left[\text{by L'Hopital's rule}\right] \\\\ & = \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{1}{\left(1 + y^2\right)^2} \times 2y}{\sin y} \;\;\; \left[\dfrac{0}{0} \text{ form}\right] \\\\ & = 2 \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{d}{dy} \left[\dfrac{y}{\left(1 + y^2\right)^2}\right]}{\dfrac{d}{dy} \left[\sin y\right]} \;\;\; \left[\text{by L'Hopital's rule}\right] \\\\ & = 2 \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{\left(1 + y^2\right)^2 \times 1 - y \times 2 \left(1 + y^2\right) \times 2y}{\left(1 + y^2\right)^4}}{\cos y} \\\\ & = 2 \lim\limits_{y \rightarrow 0} \dfrac{\left(1 + y^2\right)^2 - 4 y^2 \left(1 + y^2\right)}{\left(1 + y^2\right)^4 \cos y} \\\\ & = 2 \times \left[\dfrac{1 - 0}{1 \times 1}\right] \\\\ & = 2 \end{aligned}$