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. 
 A: You are correct, but there is no need for a second L'Hospital, because: $$\lim_{x \to 0}2\frac{\sin 2x}{2x} = 2 \lim_{x \to 0}\frac{\sin 2x}{2x} = 2\lim_{u \to 0}\frac{\sin u}{u} = 2 \cdot 1 =2.$$
A: 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
$4\cdot [1/(1+1)] = 2.$
A: 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 $2$.
A: $$\lim_{x\to 0} \left(\frac{1-\cos(2x)}{x^2}\right)=$$
$$\lim_{x\to 0} \left(\frac{\frac{d}{dx}\left(1-\cos(2x)\right)}{\frac{d}{dx}x^2}\right)=$$
$$\lim_{x\to 0} \left(\frac{2\sin(2x)}{2x}\right)=$$
$$\lim_{x\to 0} \left(\frac{\sin(2x)}{x}\right)=$$
$$\lim_{x\to 0} \left(\frac{\frac{d}{dx}\sin(2x)}{\frac{d}{dx}x}\right)=$$
$$\lim_{x\to 0} \left(\frac{2\cos(2x)}{1}\right)=$$
$$\lim_{x\to 0} \left(2\cos(2x)\right)=$$
(Since $2\cos(2x)$ is a continuous function of $x$):
$$2\cos(2\times0)=2\cos(0)=2\times1=2$$
A: $\lim_{x\to 0} \left(\frac{1-\cos(2x)}{x^2}\right) = \lim_{x\to 0} \left(\frac{1-\cos(2x)}{(2x)^2}\frac{4x^2}{x²}\right) =\frac{1}{2}\times 4 = 2$
A: Using series is probably overkill here, but since the Taylor series of $\cos x$ is $1 - \frac{x^2}{2} + O(x^4)$, $\cos 2x = 1 - 2x^2 + O(x^4)$.
Therefore we have:
$$\lim_{x\to 0} \frac{1-\cos 2x}{x^2} = \lim_{x\to 0} \frac{1-(1-2x^2)}{x^2} = \boxed{2}.$$
