TITLE: Show that if $x_n \to x$ then $\sqrt{x_n} \to \sqrt{x}$
QUESTION [7 upvotes]: Show that if $x_n \to x$ then $\sqrt{x_n} \to \sqrt{x}$ I have been stuck on this for a while. I tried $$|\sqrt{x_n} - \sqrt{x}| = \frac{|x_n - x|}{|\sqrt{x_n} + \sqrt{x}|},$$ and then I at least can get the top to be as small as I want, so I have $$\frac{\epsilon}{|\sqrt{x+\epsilon} + \sqrt{x}|},$$ but I get stuck here at choosing the N, and I don't know if my first step in breaking down the absolute value is legitimate. Please help.

REPLY [2 votes]: Let $\epsilon > 0$.  Since $(x_n)\rightarrow x$, then for some $N\in \mathbb{N}$, we have,
$$|x_n-x|<\epsilon \sqrt{x} \hspace{10pt} \text{when}\hspace{10pt} n\geq N$$
This implies 
$$\frac{|x_n-x|}{\sqrt{x}}<\epsilon \hspace{10pt} \text{when}\hspace{10pt} n\geq N$$
Notice,
\begin{align*}
|\sqrt{x_n}-\sqrt{x}|&=|\sqrt{x_n}-\sqrt{x}|\left(\frac{\sqrt{x_n}+\sqrt{x}}{\sqrt{x_n}+\sqrt{x}}\right)\\
&=\frac{|x_n-x|}{\sqrt{x_n}+\sqrt{x}}\\
&\leq  \frac{|x_n-x|}{\sqrt{x}} 
\end{align*}
So, when $n\geq N$,
$$|\sqrt{x_n}-\sqrt{x}|\leq\frac{|x_n-x|}{\sqrt{x}}<\epsilon $$
Hence, $(\sqrt{x_n})\rightarrow \sqrt{x}$.