$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$

**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 $|\sqrt{a_n}-0|=\sqrt{a_n} <\sqrt{\varepsilon ^2}=\varepsilon$.

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