Answer
See the proof below.
Work Step by Step
Since $ m\neq 0$, then we have
\begin{align*}
\lim _{x \rightarrow 0}\frac{\cos mx-1}{x^2}&= \lim _{x \rightarrow 0}m^2 \frac{\cos m x-1}{(mx)^2}\\
&= - m^2\lim _{mx \rightarrow 0} \frac{1-\cos mx}{(mx)^2}\\
&=- \frac{m^2}{2}.\\
\end{align*}
Where we used the fact that $\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}=\frac{1}{2}$.