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