If $b_n$ is a non-negative sequence and the limit is 0, then the limit of $\sqrt{b_n}$ is 0.

Question

My proof so far is as follows:

Choose $\epsilon > 0$. Since the limit of $b_n$ is 0, $|b_n - 0| = |b_n| = b_n < \epsilon$.

I'm not sure how to proceed here. I'm thinking of showing $b_n < \epsilon^2$, and from there claiming $\sqrt{b_n} < \epsilon$. I don't thinking this is right however, as if $\epsilon \in (0,1)$, $\epsilon^2 < \epsilon$ and so $b_n < \epsilon^2$ does not follow.

Answer 1

Assume that $\epsilon\in(0,1)$, consider the positive number $\epsilon^{2}$, then there is some $N$ such that $|b_{n}|<\epsilon^{2}$ for all $n\geq N$, then $\sqrt{b_{n}}<\epsilon$.

Answer 2

You have the right idea. Since $b_n \to 0$, there exists $N>0$ such that for all $n\ge N$, $b_n < \epsilon^2$. Then $$\sqrt{b_n} < \sqrt{\epsilon^2} = \epsilon$$