Proving an expression for the limit of a certain sequence [closed]

Question

If $(a_{n})_{n=1}^{\infty}$ is a sequence converging to $L$, with $a_n \geq 0$ for all $n$, how can I prove that $L \geq 0$ and that $(\sqrt{a_n})_{n=1}^{\infty}$ converges to $\sqrt{L}$.

I was under the impression that I would have to use the epsilon proof. However, I need a bound that isn't directly given here to do so. I think that I can bound it myself knowing that $a_n$ converges? Otherwise, I am confused about how to accomplish this proof. I am still new to the limits of sequences in proofs. I feel as though this should be obvious and I may be overthinking it.

Any advice and help would be greatly appreciated on how to go about this problem.

Answer 1

You can treat them in two cases:

$(1) L=0$, you don't need a lower bound for $a_n$

$(2) L>0$, when $n>N$ is sufficiently large, you can have a lower bound $a_n>\frac{L}2$

Answer 2

Let's suppose that $L<0$. Then, $-L>0$ and for $\epsilon=-L>0$ we would have that there is $n_{0}\in{\mathbb{N}}$ such that for every $n\ge n_{0}$ we would have that: $a_n-L\leq |a_n-L|< -L$ and thus $a_n<0$ for $n\ge n_{0}$ a contradiction. Hence, we conclude that $L\ge 0$. As for the second part, maybe this slightly abstract usage of the definition will help you understand how to get there.