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?
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.
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 $x\to 0.$