Answer
$\dfrac{1}{3}$
Work Step by Step
Taylor series for $\tan^{-1} y $ can be defined as: $\tan^{-1} y= y-\dfrac{y^3}{3}+\dfrac{y^5}{5}-....$
Now, $\lim\limits_{y \to 0}\dfrac{y-\tan^{-1} y}{y^3}=\lim\limits_{y \to 0} \dfrac{[y-(y-\dfrac{y^3}{3}+\dfrac{y^5}{5}-....)]}{y^3}$
or, $=\lim\limits_{y \to 0} \dfrac{\dfrac{y^3}{3}- \dfrac{y^5}{5}+...}{y^3}$
or, $=\lim\limits_{y \to 0} [\dfrac{1}{3}-\dfrac{y^2}{5}+\dfrac{y^4}{7}-....]$
or, $=\dfrac{1}{3}$
