If $x_n \to 0$, show that $\sqrt{x_n} \to 0$

Let $x_n \geq 0$ for all $n \in N$. If $(x_n) \to 0$, show that $(\sqrt{x_n}) \to 0$

Unless I am missing something, this seems simple to me. Since $x_n$ converges for an arbitrary $\epsilon^2>0$, we have $|x_n| <\epsilon^2 $ for $n\geq N$. Taking square roots here we have $\sqrt{x_n}<\epsilon$ for all $n\geq N$ which is what we needed to show. Is that all? If so, then why $x_n \geq 0$ is important?

Solution 1:

Unless you are obligated to do it from first principles, you could also appeal to the continuity of $f(x)=\sqrt{x}$ to deduce the result.

Solution 2:

$f:\mathbb{R}^+\to\mathbb{R}^+, x\mapsto\sqrt{x}$ is continuous so $x_n\to 0\Leftrightarrow f(x_n)\to f(0)$