145. Find $\lim _{x \rightarrow 0} \frac{1-\cos 8 x}{2 x^{2}}$.
Solution. Transform the numerator to the form $1-\cos 8 x=2 \sin ^{2} 4 x$. Next, we find

$$
\begin{gathered}
\lim _{x \rightarrow 0} \frac{1-\cos 8 x}{2 x^{2}}=\lim _{x \rightarrow 0} \frac{2 \sin ^{2} 4 x}{2 x^{2}}=\lim _{x \rightarrow 0}\left(\frac{\sin 4 x}{x} \cdot \frac{\sin 4 x}{x}\right)= \\
=\lim _{x \rightarrow 0} \frac{\sin 4 x}{x} \cdot \lim _{x \rightarrow 0} \frac{\sin 4 x}{x}=4 \cdot 4=16
\end{gathered}
$$