Prove that if $f(x) = \frac{\sqrt{x} - \sqrt{a}}{x-a}$, then $f$ has a limit at a. Prove that if $f(x) = \frac{\sqrt{x} - \sqrt{a}}{x-a}$, then $f$ has a limit at a.
I have written a very basic level proof here:
Given $\epsilon > 0$, we want $\delta > 0$ such that $|(\frac{1}{\sqrt{x} + \sqrt{a}}) - (\frac{1}{2\sqrt{a}})| < \epsilon$ whenever $0<|x - a|<\delta$. So for $\delta = g(\epsilon)$, $0<|x - a|<\delta \implies |(\frac{1}{\sqrt{x} + \sqrt{a}}) - (\frac{1}{2\sqrt{a}})| < \epsilon$.
I tried and failed to find a way to represent delta in terms of epsilon. Saying, $\delta = g(\epsilon)$, is wrong. I saw some examples where they set $\delta=1$ and the used the $min$ function. I tried this but was not successful.
Edit: I need to find an answer using the epsilon-delta form. just solving for the limit is not sufficient.
Anyone have any tips?

We have that for $a> 0$
$$f(x) = \frac{\sqrt{x} - \sqrt{a}}{x-a}= \frac{\sqrt{x} - \sqrt{a}}{x-a} \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}}=$$
$$=\frac{x-a}{(x-a)(\sqrt{x} + \sqrt{a})}=\frac{1}{\sqrt{x} + \sqrt{a}} \to \frac1{2\sqrt a}$$
As an alternative assuming wlog $|x-a|<\frac a 2$
$$ \left|\frac{\sqrt{x} - \sqrt{a}}{x-a}-\frac1{2\sqrt a}\right|=\left|\frac{1}{\sqrt{x} + \sqrt{a}}-\frac1{2\sqrt a}\right|=\left|\frac{2\sqrt a-\sqrt{x} - \sqrt{a}}{2\sqrt a(\sqrt{x} + \sqrt{a})}\right|=$$
$$=|x-a|\left|\frac{\sqrt a-\sqrt{x} }{2\sqrt a(\sqrt{x} + \sqrt{a})(x-a)}\right|<|x-a|\left|\frac{\sqrt a }{2\sqrt a(3 \sqrt{a})\frac a 2}\right|=\frac{|x-a|}{3a^\frac32}$$
therefore assuming $\delta=3\epsilon a^\frac32$ we have
$$ \left|\frac{\sqrt{x} - \sqrt{a}}{x-a}-\frac1{2\sqrt a}\right|<\frac{|x-a|}{3a^\frac32}<\epsilon$$
