Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 8 Exercise 8.3
Solution: Let $\epsilon>0$. Since $\lim s_n=0$, there exists $N>0$ such that
$$
|s_n-0|=s_n < \epsilon^2
$$ for all $n>N$. Therefore $n>N$ implies that
$$
|\sqrt{s_n}-0|=\sqrt{s_n}<\sqrt{\epsilon^2}=\epsilon
$$ for all $n>N$ as desired. Thus $\lim s_n=0$.