# If $(an)\rightarrow L$, $a_n > 0$ for all $n \in \mathbb N$, and $L > 0$, then prove that $\sqrt{a_n}\rightarrow \sqrt L$.

To be honest, I don't even understand what the question is asking, and have no idea how to answer it. Any guidance would be great. I know convergent/divergent definitions, as well as basic limit laws, but not sure how I can apply any of them to prove this.

[Hint: for all $n \in \mathbb N$, $\sqrt{a_n} + \sqrt L > \sqrt L$]

• What does this have to do with number theory? – Omnomnomnom Dec 8 '14 at 4:03

Since $a_n \rightarrow L$ and $L > 0$, then for any $\epsilon > 0$ and $n$ sufficiently large
$$|a_n - L| < \epsilon|\sqrt{L}| \\ \implies |\sqrt{a_n}-\sqrt{L}||\sqrt{a_n} + \sqrt{L}| < \epsilon|\sqrt{L}| \\ \implies |\sqrt{a_n}-\sqrt{L}| < \frac{\epsilon|\sqrt{L}| }{|\sqrt{a_n} + \sqrt{L}|}\\ \implies |\sqrt{a_n}-\sqrt{L}| <\frac{\epsilon|\sqrt{L}| }{|\sqrt{L}|}= \epsilon$$