Evaluate the limit or show that is does not exist $$\lim_{x \to 0}\frac{1-\cos(3x)}{2x^2}$$
So far I have noted
$$\frac{1-\cos(3x)}{2x^2} = \frac{1-\cos(3x)}{2x^2} \cdot \frac{1+\cos(3x)}{1+\cos(3x)}$$
and then used the identity $$\sin^2x = 1-\cos^2x$$ to reduce this to
$$\frac{\sin^2(9x)}{2x^2+\cos(9x)}.$$
Any tips on how to proceed?
 A: Given: $L=\lim_{x \to 0} \frac{1-\cos(3x)}{2x^2}$  
$\implies L=\lim_{x \to 0} \frac{1-\cos(3x)}{(3x)^2} \times \frac{(3x)^2}{2x^2}$  
$\implies L=\lim_{x \to 0} \frac{1}{2} \times \frac{9x^2}{2x^2}$   [Using $\lim_{x \to 0} \frac{1-\cos(x)}{x^2}=\frac{1}{2}$] 
$\implies \boxed{L= \frac{9}{4}}$   
EDIT: I have changed my solution based on the changes in your question.
A: You had the right idea, but your algebra is suspect. For example, $\cos 3x \cdot \cos 3x \ne \cos^2 9x$
$$\lim_{x\to 0} \frac{1-\cos 3x}{2x} = \frac{3}{2}\cdot\lim_{x\to 0} \frac{1-\cos 3x}{3x}$$
$$= \frac{3}{2}\cdot\lim_{3x\to 0} \frac{1-\cos 3x}{3x}$$
$$= \frac{3}{2}\cdot\lim_{\theta\to 0} \frac{1-\cos \theta}{\theta}$$
where $\theta = 3x$
$$= \frac{3}{2}\cdot\lim_{\theta\to 0} \frac{1-\cos \theta}{\theta}\cdot \frac{1+\cos \theta}{1+\cos \theta}$$
$$= \frac{3}{2}\cdot\lim_{\theta\to 0} \frac{1-\cos^2 \theta}{\theta}\cdot \frac{1}{1+\cos \theta}$$
$$= \frac{3}{2}\cdot\lim_{\theta\to 0} \frac{\sin^2 \theta}{\theta}\cdot \frac{1}{1+\cos \theta}$$
$$= \frac{3}{2}\cdot\lim_{\theta\to 0} \frac{\sin \theta}{\theta}\cdot \frac{\sin \theta}{1+\cos \theta}$$
$$= \frac{3}{2}\cdot\lim_{\theta\to 0} \frac{\sin \theta}{\theta}\cdot \lim_{\theta\to 0}\frac{\sin \theta}{1+\cos \theta}$$
$$= \frac{3}{2}\cdot\frac{0}{1+1}\cdot\lim_{\theta\to 0} \frac{\sin \theta}{\theta}$$
$$= 0\cdot\lim_{\theta\to 0} \frac{\sin \theta}{\theta}$$
$$= 0$$
Now all that is required is to prove that $\displaystyle\lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1$ (or at the very least, that it exists).
on $(0, \pi/2)$
$$0 \lt \sin \theta \lt \theta$$
$$0/\theta \lt \sin \theta / \theta \lt \theta/\theta$$
$$0 \lt \frac{\sin \theta}{\theta} \lt 1$$
thus $ 0 \le \displaystyle\lim_{\theta\to 0}\frac{\sin \theta}{\theta} \le 1$
and
$$0\cdot 0 \le 0\cdot \displaystyle\lim_{\theta\to 0}\frac{\sin \theta}{\theta} \le 0\cdot1$$
$$0 \le 0\cdot \displaystyle\lim_{\theta\to 0}\frac{\sin \theta}{\theta} \le 0$$
$$\therefore 0\cdot \displaystyle\lim_{\theta\to 0}\frac{\sin \theta}{\theta} = 0$$
A: You would have  have it here as follows if your algebra were correct.
$$\frac{\sin^2(9x)}{2x^2+\cos(9x)}.$$
As $x\to 0$, $\sin^2(9x) \to 0$ and $2x^2 + \cos(9x)\to 1.$
The limit of the ratio is zero.
However, the algebra leading up to your expression is flawed. I made the mistake
of not checking your preceding calculations. 
$$\frac{1-\cos(3x)}{2x^2} = \frac{1-\cos(3x)}{2x^2} \cdot \frac{1+\cos(3x)}{1+\cos(3x)} ={\sin^2(3x)\over 2x^2(1 + \cos(3x))} \sim {\sin^2(3x)\over 4x^2}\to {9\over 4} $$
as $x\to 0.$
