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