Hey guys, I've already seen how to prove this problem using continuity however I'd like to also understand how to do it another way

Suppose $\displaystyle a_n>0$ for all $\displaystyle n$ and $\displaystyle a_n\rightarrow 3$ show that $\displaystyle \sqrt{a_n}\rightarrow \sqrt{3}$.

How would I begin to show this by first showing algebraically that $\displaystyle |\sqrt{a_n}-\sqrt{3}|<|a_n-3|$?