Epsilon and Delta proof of $\lim_{x\to0} \frac{2-\sqrt{4-x}}{ x}$ I need to prove $\lim_{x\to0} \frac{2-\sqrt{4-x}}{ x}$
I first found the limit to be $\frac{1}{4}$ by using l'hopital's rule.
By definition i need to find a $\delta > 0$ for every $\epsilon >0$ 
Then i will have $|x-0|<\delta$ and 
$$|\frac{2-\sqrt{4-x}}{ x}-\frac{1}{4}|<\epsilon$$
I have tried multiple ways to simplify, but I can't seem to get it in the form of just $x$. And I am a bit confused on how to pick my delta in this case. 
Any help would be much appreciated. 

It gets much simple when you write
$$
\frac{2−\sqrt{4−x}}x = \frac{(2-\sqrt{4−x})(2+\sqrt{4−x})}{x(2+\sqrt{4−x})}
=\frac1{2+\sqrt{4−x}}
$$
then:
\begin{align}
\left|
        \frac1{2+\sqrt{4−x}} - \frac 14
\right|
&=   \frac{|2 - \sqrt{4-x}|}{4(2+\sqrt{4−x})} 
\le \frac{|2 - \sqrt{4-x}|}8 
\\ &=  
\frac{|2 - \sqrt{4-x}|(2 + \sqrt{4-x})}{8(2 + \sqrt{4-x})} 
\\ &= \frac{|x|}{8(2 + \sqrt{4-x})} 
\le \frac{|x|}{16}
\end{align}
so just take $\delta = 16\epsilon$.
