Prove that if a sequence $\{a_{n}\}$ converges then $\{\sqrt a_{n}\}$ converges to the square root of the limit.

There are two possibilities:

1) $a = 0$. Let $\epsilon > 0$, then there exists a $N$ such that if $n > N$, then $|a_n - 0| < \epsilon^2$. Thus: $|\sqrt{a_n} - 0| < \epsilon$. This shows that $\sqrt{a_n} \rightarrow 0$.

2) $a > 0$. Given $\epsilon > 0$, there exists a $N$ such that if $n > N$, then $|a_n - a| < \epsilon\sqrt{a}$. Thus: $|\sqrt{a_n} - \sqrt{a}| = \dfrac{|a_n - a|}{\sqrt{a_n} + \sqrt{a}} < \dfrac{|a_n - a|}{\sqrt{a}} < \dfrac{\epsilon\sqrt{a}}{\sqrt{a}} = \epsilon$. This shows that $\sqrt{a_n} \rightarrow \sqrt{a}$.