Find the limit $\lim_{x\to 0}\frac{1-\cos 2x}{x^2}$ 
Find the limit $$\lim_{x\to 0}\frac{1-\cos 2x}{x^2}$$

This is what I did: $$\lim_{x\to 0}\frac{1-\cos 2x}{x^2} = \frac{0}{0}$$
Then, if we apply L'hopital's, we get: $$\lim_{x\to 0}\frac{2\sin 2x}{2x} =\frac{0}{0}.$$ 
Once again, using L'hopital's rule, we get:$$\lim_{x\to 0}\frac{4\cos 2x}{2} = 2\cos 2x = 2.$$ 
Can someone please tell me what I did wrong here? Thanks. 
Update: Thanks everyone for your wonderful answers. I have found out the reason for taking a point off of my work. It is because I didn't use the correct expression. For example, since $$\lim_{x\to 0} \frac{f(x)}{g(x)} = \frac{0}{0},$$ I didn't write the correct term $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = \dots$$ and instead I equated everything when I was applying L'Hopital's rule. So, I thought I should've mention it here. Thanks again. 

Without L'Hopital: Multiply top and bottom by $1+\cos 2x$ to get
$$\frac{1-\cos^2 2x}{x^2(1+\cos 2x)}= \frac{\sin^2 2x}{x^2(1+\cos 2x)}.$$
It's easily seen that $(\sin^2 2x)/x^2 \to 4.$ Thus the limit is
$4\cdot [1/(1+1)] = 2.$
