# 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 $\delta = 16\epsilon$.
• Forgot to factor out the $x$ at the end. Im still new to this style of proof. So I want my $|x|<\delta$ to get into the form of $|\frac{1}{2+\sqrt{4-x}}-\frac{1}{4}|$? So I got tried found a common denominator and got $\frac{1-\sqrt{4-x}}{4+2\sqrt{4-x}}$ but I dont see how i can get $\delta$ into this form. Unless I am thinking of this wrong. – ThisLax Apr 26 '15 at 2:47
• Last question. Looking at your work I see that you are finding a large enough $\epsilon$ so that the problem becomes easier to solve for $x$. Is that what you were doing there? More specifically is that a recommended thought process when solving these types of proofs. – ThisLax Apr 26 '15 at 2:57
• no. Fix $\epsilon$ without constrains, and then look for a convenient $\delta$ – mookid Apr 26 '15 at 2:59