Answer
$$
\lim _{x \rightarrow 0} \frac{\cos x-1+\frac{1}{2} x^{2}}{x^{4}} =\frac{1}{24}
$$
Work Step by Step
$$
\lim _{x \rightarrow 0} \frac{\cos x-1+\frac{1}{2} x^{2}}{x^{4}}
$$
Since
$$
\lim _{x \rightarrow 0} ( \cos x-1+\frac{1}{2} x^{2})=0
$$
and
$$
\lim _{x \rightarrow 0} ( x^{4})=0
$$
the limit is an indeterminate form of type $\frac{0}{0}$ so we can apply l’Hospital’s Rule:
$$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\cos x-1+\frac{1}{2} x^{2}}{x^{4}} & \stackrel{H}{=} \lim _{x \rightarrow 0} \frac{-\sin x+x}{4 x^{3}} \rightarrow \frac{0}{0} \\
&\stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0} \frac{-\cos x+1}{12 x^{2}}
\rightarrow \frac{0}{0} \\
& \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0} \frac{\sin x}{24 x} \rightarrow \frac{0}{0} \\
& \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0} \frac{\cos x}{24} \\
&=\frac{1}{24}
\end{aligned}
$$