Answer
$\dfrac{1}{3}$
Work Step by Step
We know that $\tan^{-1} y= y-\dfrac{y^3}{3}+\dfrac{y^5}{5}-....$
$\lim\limits_{y \to 0}\dfrac{y-\tan^{-1} y}{y^3}=\lim\limits_{y \to 0} \dfrac{[y-(y-\dfrac{y^3}{3}+\dfrac{y^5}{5}-....)]}{y^3}\\=\lim\limits_{y \to 0} \dfrac{\dfrac{y^3}{3}- \dfrac{y^5}{5}+...}{y^3}\\= \lim\limits_{y \to 0} \dfrac{1}{3}-\lim\limits_{y \to 0} \dfrac{y^2}{5}+\lim\limits_{y \to 0} \dfrac{y^4}{7}-....]$
So, $\lim\limits_{y \to 0}\dfrac{y-\tan^{-1} y}{y^3}=\dfrac{1}{3}$
