Validity of a Limit Proof I am trying to show that if $\displaystyle{\lim_{s\to\infty}} s_n = s$, $\displaystyle{\lim_{s\to\infty}} \sqrt{s_n} = \sqrt{s}$ for some sequence $s_n$. 
We must note that $s_n > 0$ for all $n$ for the limit to be defined in $\mathbb{R}$.
I've put together this proof, but I'm not sure if it is valid:
Let $\epsilon >0$. Since $s=\lim{s_n}$, there exists an $N$ such that $n > N$ implies $|s_n-s| < \epsilon^2$. Thus, for an arbitrary $n>N$, we have:
$\sqrt{|s^n - s|} < \epsilon$
Case 1 ($s_n > s$):
$\sqrt{s_n-s} < \epsilon$
$\sqrt{s_n} - \sqrt{s} < \sqrt{s_n-s} < \epsilon$
$|\sqrt{s_n} - \sqrt{s}| < \epsilon$
Case 2 ($s > s_n$):
$\sqrt{s-s_n} < \epsilon$
$\sqrt{s_n} - \sqrt{s} < \sqrt{s-s_n} < \epsilon$
$|\sqrt{s} - \sqrt{s_n}| < \epsilon$ $\blacksquare$
Couple of things that are worrying me:


*

*I'm not sure if the $\sqrt{a}-\sqrt{b} < \sqrt{a-b}$ actually holds up in this case. For proving it, I though $a-b > a-b-2\sqrt{ab}$ if $a, b > 0$, which is true in this case since $s, s_n > 0$.

*Does this whole casework idea make sense?

 A: Overall this looks fine. In the second-to-last step, you should have $\sqrt{s}-\sqrt{s_n}$ instead of $\sqrt{s_n}-\sqrt{s}$.
If $a>b>0$, $\sqrt{a}-\sqrt{b} < \sqrt{a-b}$. To see this, notice that
$$a-b = (\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}) \geq (\sqrt{a}-\sqrt{b})^2$$
and then take the square root of both sides.
A: I was going to post this as a comment, but it started to get a bit long.
So if both $a>b$, you're argument establishes that $\sqrt{a}-\sqrt{b}<\sqrt{a-b}$ (since $f(x)=x^2$ is strictly increasing on $[0,\infty)$). 
Then, Case 1 works fine, since $s_n>s$ implies that $\sqrt{s_n}-\sqrt{s}>0$ (since $f(x)=\sqrt{x}$ is also increasing on $[0,\infty)$) which combined with $\sqrt{s_n}-\sqrt{s}<\varepsilon$ gives you that $|\sqrt{s_n}-\sqrt{s}|<\varepsilon$. 
Case 2 requires minor tweaking since you haven't actually bound from below $\sqrt{s_n}-\sqrt{s}$. However, if you use the same argument as in Case 1 you get that 
$$0<\sqrt{s}-\sqrt{s_n}<\sqrt{s-s_n}<\varepsilon$$
which gives you the desired $|\sqrt{s}-\sqrt{s_n}|<\varepsilon$.
