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