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. 

Alternatively, your limit can be written as $\lim_{x \to 0}
\frac{1- (\cos^{2}x-\sin^{2}x)}{x^{2}} = \lim_{x \to 0}  \frac{2\sin^{2}x}{x^{2}}$. Since $\lim_{x \to 0} \frac{ \sin x}{x} = 1$, ( a standard limit which might have been given in your course already- you could use L'Hopital, but you really need to know this limit to prove that $\sin x$ is differentiable at $0$ anyway), your limit is $2$.
