Answer
$-\dfrac{1}{6}$
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, $\lim\limits_{y \to 0}\dfrac{\tan^{-1} y -\sin y}{y^3 \cos y}=\lim\limits_{y \to 0} \dfrac{(y-\dfrac{y^3}{3!}+\dfrac{y^5}{5!}-....)-(y-\dfrac{y^3}{3}+\dfrac{y^5}{5}-....)}{y^3}$
or, $=\lim\limits_{y \to 0} \dfrac{\dfrac{-y^3}{6}+ \dfrac{23y^5}{5!}+...}{y^3 \cos y}$
or, $=\lim\limits_{y \to 0} \dfrac{(-\dfrac{1}{6}+\dfrac{23y^2}{5!}-...)}{\cos y}$
or, $=-\dfrac{1}{6}$
