Evaluate Left And Right Limits Of $f(x)=\frac{x}{\sqrt{1-\cos2x}}$ At $0$ Evaluate Left And Right Limits Of $f(x)=\frac{x}{\sqrt{1-\cos2x}}$ At $0$
The graph of $f(x)=\frac{x}{\sqrt{1-\cos2x}}$ appears to have a jump discontinuity at $0$ and I want to calculate the left and right limits of $f(x)$ to show there is a discontinuity at $0$.

I can't figure out how to manipulate the function in order to give different left and right limits.
Here's one of my attempts at trying to manipulate the funtion in to something more familiar to me:
$\lim_{x \to 0}\frac{x}{\sqrt{1-\cos2x}}$
(divide numerator and denominator by $(2x)^2)$
$ =\lim_{x \to 0}\frac{\frac{x}{4x^2}}{\sqrt{\frac{-(\cos2x-1)}{2x}}}$
$= \lim_{x \to 0}\frac{\frac{1}{4x}}{\sqrt{\frac{-(\cos2x-1)}{2x}}}$
Now I was thinking that I can apply $\lim_{x \to 0}\frac{\cos(\theta)-1}{\theta} =0$ but it doesn't help me at all.
Any ideas?
 A: A start: Use $\cos 2x=1-2\sin^2 x$. One needs to be careful when finding the square root of $2\sin^2 x$. It is $\sqrt{2}|\sin x|$. 
A: An alternate approach would be to multiply by $\sqrt{1+\cos 2x}$ on the top and bottom to get 
$\;\;\;\displaystyle\frac{x\sqrt{1+\cos2x}}{\sqrt{1-\cos^{2}2x}},$  and then use the identities $1-\cos^{2}\theta=\sin^{2}\theta$ and $\sqrt{t^2}=|t|$ 
to simplify the denominator.
(Notice that this method will work for $\cos nx$ for any integer $n\ge1$.)
A: $$
\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$$
$$\displaylines{
  1 - \cos \left( {2x} \right) = 2\sin ^2 x \cr 
  \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {1 - \cos \left( {2x} \right)} }} = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {2\sin ^2 x} }} \cr 
   = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt 2 \left| {\sin x} \right|}} \cr 
   = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt 2 \frac{{\left| {\sin x} \right|}}{x}}} \cr}
$$
$$
 \Rightarrow \left\{ \begin{array}{l}
 \mathop {\lim }\limits_{x \to 0^ +  } \frac{1}{{\sqrt 2 \frac{{\left| {\sin x} \right|}}{x}}} =  + \frac{1}{{\sqrt 2 }} \\ 
 \mathop {\lim }\limits_{x \to 0^ -  } \frac{1}{{\sqrt 2 \frac{{\left| {\sin x} \right|}}{x}}} =  - \frac{1}{{\sqrt 2 }} \\ 
 \end{array} \right.
$$
