Answer
$\lim\limits_{x \to a} \sqrt{x}= \sqrt{a}~~~$ if $a \gt 0$
Work Step by Step
Suppose that $a \gt 0$
Let $\epsilon \gt 0$ be given.
Let $\delta = \sqrt{a}~\epsilon$
Suppose that $\vert x-a \vert \lt \delta$
Then:
$\vert \sqrt{x}-\sqrt{a} \vert = \frac{\vert x-a \vert}{\sqrt{x}+\sqrt{a}} \lt \frac{\delta}{\sqrt{x}+\sqrt{a}} \lt \frac{\delta}{\sqrt{a}} = \frac{\sqrt{a}~\epsilon}{\sqrt{a}}=\epsilon$
Therefore, $\lim\limits_{x \to a} \sqrt{x}= \sqrt{a}~~~$ if $a \gt 0$
