Answer
$-1$
Work Step by Step
The Taylor series for $\cos x $ can be defined as: $\cos x=1-\dfrac{x^2}{2!}+\dfrac{ x^4}{4!}-....$
Now, $\lim\limits_{y \to 0} \dfrac{y^2}{\cos y-\cos h y}=\lim\limits_{y \to 0} \dfrac{y^2}{(1-y^2/2+y^4/4!)-(1-y^2/2!+y^4/4!)}$
or, $=\lim\limits_{y \to 0} \dfrac{y^2}{-1-\dfrac{2y^4}{6!}-.......}$
or, $=\dfrac{1}{-1-1-0-....}$
or, $=-1$
