Epsilon delta proof for $\lim_\limits{x\to a} \sqrt{x} = \sqrt{a}$ where $a>0$ $f(x) = \sqrt{x}$. For right limit, $x-a > 0$. Let $x-a < \delta$ then $(\sqrt{x}-\sqrt{a}) \lt \frac{\delta}{\sqrt{x}+\sqrt{a}}$. Then chose we must choose $\epsilon $ so that $\epsilon > \frac{\delta}{\sqrt{x}+\sqrt{a}}$. Is this correct? How would I conclude the proof?
Edit - Follow up
From answers, I try a proof for the same limit but this time Left side:
For left side limit, $x-a < 0$ so $a-x > 0$. Let $a-x < \delta$ then $\sqrt{a}-\sqrt{x} \lt \frac{\delta}{\sqrt{a}+\sqrt{x}} \lt \frac{\delta}{\sqrt{a}}$. 
So let me begin the proof by saying to choose $\delta = \epsilon \sqrt a$ then $a-x \lt \delta $ so $(\sqrt a - \sqrt x )(\sqrt a + \sqrt x) \lt \epsilon \sqrt a$ so $\sqrt a - \sqrt x \lt \epsilon$

For  $x >0$, $a>0$:
$|√x-√a| =\dfrac {|x-a|}{√x+√a}\lt$
$\dfrac{|x-a|}{√a}.$
Let $\epsilon >0$ be given.
Choose $\delta \le \epsilon √a.$
Then
$|x-a| \lt \delta$ implies
$|√x-√a| \lt \dfrac{|x-a|}{√a} \lt $
$\dfrac{\delta}{√a} \le\epsilon.$
