What is the procedure to solve for the $\lim\limits_{x\rightarrow 0} \frac{\cos(3x)-1}{x^2}$? What is the procedure to solve for the $\lim\limits_{x\rightarrow 0} \frac{\cos(3x)-1}{x^2}$?
My calculator tells me the answer is -9/2, but I don't know how to solve without substituting values of x.
I suspect there is some trigonometric identity involved.

From the trigonometric identity $\;\color{blue}{\cos(2\theta)=1-2\sin^2\theta}\;$ we have
\begin{align}
\lim_{x\to 0}\frac{\cos(3x)-1}{x^2}&=\lim_{x\to 0}\frac{1-2\sin^2(3x/2)-1}{x^2}\\
&=\lim_{x\to 0}\frac{-2\sin^2(3x/2)}{x^2}\\
&=-2\cdot\frac{9}{4}\cdot\lim_{x\to 0}\frac{\sin^2(3x/2)}{\frac{9}{4}x^2}\\
&=-2\cdot\frac{9}{4}\cdot\left(\lim_{x\to 0}\frac{\sin(3x/2)}{3x/2}\right)^2\\
&=-2\cdot\frac{9}{4}\cdot\left(1\right)^2\\
&=\boxed{\color{blue}{-\frac{9}{2}}}
\end{align}
