Answer
$$-1$$
Work Step by Step
The Taylor series can be written as follows: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-... $ and
$\cos x=1-\dfrac{x^2}{2}+\dfrac{ x^3}{3}-...$
$$\lim\limits_{y \to 0} \dfrac{y^2}{\cos y-\cos (h y)}=\lim\limits_{y \to 0} \dfrac{y^2}{(1-\dfrac{y^2}{2}+\dfrac{y^4}{4!})-(1-\dfrac{y^2}{2!}+\dfrac{y^4}{4!})} \\ \\ \space =\dfrac{1}{-1-1-0-....} \space \\=-1$$
