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. 

$$\lim_{x\to 0} \left(\frac{1-\cos(2x)}{x^2}\right)=$$
$$\lim_{x\to 0} \left(\frac{\frac{d}{dx}\left(1-\cos(2x)\right)}{\frac{d}{dx}x^2}\right)=$$
$$\lim_{x\to 0} \left(\frac{2\sin(2x)}{2x}\right)=$$
$$\lim_{x\to 0} \left(\frac{\sin(2x)}{x}\right)=$$
$$\lim_{x\to 0} \left(\frac{\frac{d}{dx}\sin(2x)}{\frac{d}{dx}x}\right)=$$
$$\lim_{x\to 0} \left(\frac{2\cos(2x)}{1}\right)=$$
$$\lim_{x\to 0} \left(2\cos(2x)\right)=$$
(Since $2\cos(2x)$ is a continuous function of $x$):
$$2\cos(2\times0)=2\cos(0)=2\times1=2$$
