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.
 A: 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}
A: $$\lim_{x\rightarrow 0}{\cos(x)}\sim 1-\frac{x^2}{2}$$
That's a formula . Derived from the below expression.
$$\cos(x) = 1-{1 \over 2!} (x)^2+{1 \over 4!} (x)^4- \ldots$$
Now, use it and get the answer.
$$\cos(3x) = 1-\frac{9x^2}{2}$$
A: Top and bottom both go to zero, so use L'Hospitals' rule, take the derivative of the top and bottom, and you get 
$\frac{-3\sin(3x)}{2x}$.  Top and bottom still go to zero so take derivatives again to get $\frac{-9\cos(3x)}{2}$  Now the limit is $-9/2$.
A: As Andre Nicolas already said in comment, you can do algebraic manipulations to get the limit
If you are not familiar with Taylor expansion of $\cos x$, which are easily applied here, you have to somehow manage to remove the terms which cause the trouble:
We divide and multiply by $\cos(3x) +1$ and try to convert it into a function of $\sin$
$$\frac{\cos(3x)-1}{x^2}\times\frac{\cos(3x)+1}{\cos(3x)+1}$$
$$= \frac{-\sin^2 3x}{x^2 (\cos(3x) + 1)}$$
Now we must do some adjustments to use: $\lim_{x\to 0}\dfrac{\sin x}{x} = 1$
So as $x\to 0$, $3x \to 0$. WE need $3^2$ in the numerator, so divide and multiply by $3^2$
$$=\frac{-9}{\cos(3x)+1}\times \left(\frac{\sin 3x}{3x}\right)^2=-\frac{9}{2}$$
A: $$\lim_{x \to 0}\frac{\cos(3x)-1}{x^2}=$$
$$-9 \cdot \lim_{x\to 0} \frac{1-\cos(3x)}{(3x)^2} \stackrel{t=3x}{=} \ldots$$
