# Proof of the Square Root Rule for Sequences

$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$
Square Root Rule
Suppose that $(a_n)$ is a convergent sequences with $a_n\to a$. If $a_n\geq 0$ for all $n$, then $\sqrt{a_n} \to \sqrt{a}$.

Proof
Since $a_n\geq 0$ we know that $a$ is non-negative and hence $\sqrt{a}$ exists. We have two cases, $a\neq 0$ and $a=0$.

For $a\neq 0$ we have, given $\varepsilon >0$, that there exists $N$ such that $|a_n-a|<\sqrt{a} \varepsilon$ for all $n>N$.
We have $\left( \sqrt{a_n} – \sqrt{a} \right) \left( \sqrt{a_n} + \sqrt{a} \right) =a_n-a$ and so, for $n>N$,
$\left| \sqrt{a_n} – \sqrt{a} \right| = \left| \frac{a_n-a}{\sqrt{a_n} + \sqrt{a} } \right| \leq \left| \frac{a_n-a}{\sqrt{a} } \right| = \frac{1}{\sqrt{a}} \left| a_n-a \right| < \frac{1}{\sqrt{a}} \sqrt{a} \varepsilon = \varepsilon .$ Hence, $\sqrt{a_n} \to \sqrt{a}$. Now we assume $a=0$. Given $\varepsilon >0$ there exists $N$ such that $|a_n-0|<\varepsilon ^2$ for all $n>N$. Thus $|\sqrt{a_n}-0|=\sqrt{a_n} <\sqrt{\varepsilon ^2}=\varepsilon$.