Prove that if $f(x) = \frac{\sqrt{x} - \sqrt{a}}{x-a}$, then $f$ has a limit at a.

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

$$ \left|\frac{\sqrt{x} - \sqrt{a}}{x-a}-\frac1{2\sqrt a}\right|<\frac{|x-a|}{3a^\frac32}<\epsilon$$

$x,a >0$.

$\epsilon >0$ be given.

$\left|\dfrac{\sqrt x-\sqrt a}{(\sqrt x+\sqrt a)(\sqrt x-\sqrt a)}-\dfrac{1}{2\sqrt a}\right|=$

$\left|\dfrac{1}{\sqrt x+\sqrt a}-\dfrac{1}{2\sqrt a}\right|=$

$\left|\dfrac{\sqrt a-\sqrt x}{2\sqrt a(\sqrt x+\sqrt a)}\right|=$

$\dfrac{\left|x-a\right|}{2\sqrt a(\sqrt x+\sqrt a)^2}<$

$\dfrac{\left|x-a\right|}{a^{3/2}}.$

Choose $\delta=\epsilon a^{3/2}$;