Answer
Please see below.
$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline
x & -0.1 & -0.01 & - 0.001 & 0 & 0.001 & 0.01 & 0.1 \\ \hline
\frac{\cos x -1}{2x^2} & -0.24979174 & -0.24999792 & -0.24999998 & \text{undefined} & -0.24999998 & -0.24999792 & -0.24979174 \\ \hline
\end{array}$$
Work Step by Step
Looking at the graph, we find that when $x$ approaches $0$ from the left and right, the function approaches $-\frac{1}{4}=-0.25$. So, we can conclude that$$\lim_{x \to 0}\frac{\cos x-1}{2x^2}=-\frac{1}{4}=-0.25 \, .$$
The table confirms our result.
Now, we find the limit analytically. Since by direct substitution we get the indeterminate form $\frac{0}{0}$, we should use the trigonometric relation $\cos x= 1- 2 \sin ^2 \frac{x}{2}$ as folllows.$$\lim_{x \to 0}\frac{\cos x-1}{2x^2}=\lim_{x \to 0}\frac{-\sin ^2 \frac{x}{2}}{x^2}=\lim_{x \to 0} \left ( - \frac{1}{4} \right ) \left ( \frac{\sin \frac{x}{2}}{\frac{x}{2}} \right )^2= \lim_{u \to 0} \left ( - \frac{1}{4} \right ) \left ( \frac{\sin u}{u} \right )^2= (- \frac{1}{4}) (1)^2=-\frac{1}{4}$$(In finding the limit we have used Thereom 1.9 (1), $\lim_{x \to 0} \frac{\sin x}{x}=1$, also $u=\frac{x}{2}$).
