Epsilon delta proof for $\lim_\limits{x\to a} \sqrt{x} = \sqrt{a}$ where $a>0$

Question

$f(x) = \sqrt{x}$. 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 $\sqrt a - \sqrt x \lt \epsilon$

Answer 1

No, you don't choose $\varepsilon$. It can be any number greater than $0$. Then you take $\delta=\varepsilon{\sqrt a}$. Then\begin{align}|x-a|<\delta\iff&|x-a|<\varepsilon{\sqrt a}\\\iff&\bigl|\sqrt x-\sqrt a\bigr|\left(\sqrt x+\sqrt a\right)<\varepsilon\sqrt a\\\implies&\bigl|\sqrt x-\sqrt a\bigr|\sqrt a<\varepsilon\sqrt a\\\iff&\bigl|\sqrt x-\sqrt a\bigr|<\varepsilon.\end{align}Now, start all over again when $a=0$; what I wrote above doesn't work in that case, since $\delta$ would be $0$ then.

Answer 2

For any $\varepsilon >0$ you must find apropriate $\delta$

$$\sqrt{x} > \sqrt{a} \implies \sqrt{x}+\sqrt{a}>2\sqrt{a} \implies {1\over \sqrt{x}+\sqrt{a}}<{1\over 2\sqrt{a}}$$

So take $\delta = 2\sqrt{a}\cdot\varepsilon$

Answer 3

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 $

$\dfrac{\delta}{√a} \le\epsilon.$