Answer
$\dfrac{1}{12}$
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, $\lim\limits_{t \to 0} (\dfrac{1}{2-2 \cos t}-\dfrac{1}{t^2}=\lim\limits_{t \to 0} \dfrac{t^2-2+2 \cos t }{2t^2 (1-\cos t)}$
or, $=\lim\limits_{t \to 0} \dfrac{t^2-2+2 (1-\dfrac{t^2}{2!}+\dfrac{ t^4}{4}-....) }{2t^2 (1-1+\dfrac{t^2}{2!}+\dfrac{ t^4}{4}-....)}$
or, $=\lim\limits_{t \to 0} \dfrac{2 (\dfrac{1}{4!}-\dfrac{ t^2}{6!}-....) }({1-\dfrac{2t^2}{4!}+...})$
or, $=\dfrac{1}{12}$
