Answer
$-\dfrac{1}{24}$
Work Step by Step
Taylor series for $\cos t $ can be defined as: $\cos t=1-\dfrac{t^2}{2}+\dfrac{t^4}{4}-\dfrac{t^6}{6}+....$
Now, $\lim\limits_{t \to 0}\dfrac{1-\cos t-\dfrac{t^2}{2}}{t^4}=\lim\limits_{t \to 0} \dfrac{(1-\dfrac{t^2}{2})-(1-\dfrac{t^2}{2}+\dfrac{t^4}{4}-\dfrac{t^6}{6}+....)}{t^4}$
or, $=\lim\limits_{t \to 0} \dfrac{-\dfrac{t^4}{4!}+\dfrac{t^6}{6!}+...}{t^4}$
or, $=\lim\limits_{x \to 0} [-\dfrac{1}{4!}+\dfrac{t^2}{6!}+....]$
or, $=-\dfrac{1}{24}$
