Prove: $s_n \to s \implies \sqrt{s_n} \to \sqrt{s}$ Prove: $s_n \to s \implies \sqrt{s_n} \to \sqrt{s}$ by the definition of the limit. $s \geq 0$ and $s_n$ is a sequence of non-negative real numbers.
This is my preliminary computation:
$|\sqrt{s_n} - \sqrt{s}| < \epsilon$
multiply by the conjugate:
$|\dfrac{s_n - s}{\sqrt{s_n}+\sqrt{s}}| < \epsilon$
Thus we can use the fact that $|\sqrt{s_n} - \sqrt{s}| < 
\dfrac{|s_n - s|}{\sqrt{s}} < \epsilon$
After this I am lost...
 A: If both $s$ and $s_n$ are non-negative
$$
|\sqrt{s}-\sqrt{s_n}|^2 \le |\sqrt{s}-\sqrt{s_n}||\sqrt{s} + \sqrt{s_n}|.
$$
Step by Step :)
Since both $s$ and $s_n$ are non-negative
$$
|\sqrt{s}-\sqrt{s_n}| \le |\sqrt{s} + \sqrt{s_n}|
$$
this is clear because the result of substracting a non-negative number from another is always less than the result of adding it, then
$$
|\sqrt{s}-\sqrt{s_n}|^2 \le |\sqrt{s}-\sqrt{s_n}| \cdot |\sqrt{s}+\sqrt{s_n}| =
|s - s_n|
$$
and you are done!
A: ADD You got to
$$\left| {\frac{{{s_n} - s}}{{\sqrt {{s_n}}  + \sqrt s }}} \right| < \frac{{\left| {{s_n} - s} \right|}}{{\sqrt s }}$$
Since $s_n\to s$, for every $\epsilon >0$ there is an $n_0$ for wich $$\left| {{s_n} - s} \right| < \varepsilon \sqrt s $$
whenever $n\geq n_0$ (i.e. $\varepsilon \sqrt s$ is also an $\epsilon'>0$). Then, for this $n_0$,
$$\left| {\frac{{{s_n} - s}}{{\sqrt {{s_n}}  + \sqrt s }}} \right| < \frac{{\left| {{s_n} - s} \right|}}{{\sqrt s }} < \frac{{\varepsilon \sqrt s }}{{\sqrt s }} = \varepsilon $$
which means $\sqrt {s_n}\to\sqrt s$.

You're almost done. You arrived at
$$\left|\dfrac{s_n - s}{\sqrt{s_n}+\sqrt{s}}\right| $$
You know that $s_n\to s$, so you can make $|s_n-s|$ as small as you wish. Now, we need to know how to handle $\sqrt{s_n}+\sqrt{s}$. Since $s_n\to s$, there is an $n_0$ for wich
$$|s-s_n|<3s/4$$
Since $s_n>0$,$s\geq 0$, then 
$$s-s_n\leq|s-s_n|<3s/4$$ 
This means that
$$s_n>s/4$$
then
$$2\sqrt {{s_n}}  > \sqrt s $$
or, since $\sqrt s>0$
$$\eqalign{
  & 2\sqrt {{s_n}}  + 2\sqrt s  > 3\sqrt s   \cr 
  & \sqrt {{s_n}}  + \sqrt s  > \frac{{3\sqrt s }}{2}  \cr 
  & \frac{1}{{\sqrt {{s_n}}  + \sqrt s }} < \frac{2}{{3\sqrt s }} \cr} $$
Again, since $s_n\to s$, there is an $n_1$ for wich
$$|s-s_n|<\epsilon \frac{3\sqrt s}{{2 }}$$
Then, taking $n\geq \max\{n_0,n_1\}$ we have
$$\left| {\frac{{{s_n} - s}}{{\sqrt {{s_n}}  + \sqrt s }}} \right| < \frac{2}{{3\sqrt s }}\left| {{s_n} - s} \right| < \frac{2}{{3\sqrt s }}\frac{{3\sqrt s }}{2}\varepsilon  = \varepsilon $$
A: Since $s_n\to s$, for any $\epsilon$, we can find an $N$ so that for all $n\ge N$ we have $|s_n-s|<\epsilon\sqrt{s}$.  Then
$$
|\sqrt{s_n}-\sqrt{s}|=\left|\frac{s_n-s}{\sqrt{s_n}+\sqrt{s}}\right|\le\left|\frac{s_n-s}{\sqrt{s}}\right|
$$
to get that
$$
|\sqrt{s_n}-\sqrt{s}|\le\epsilon
$$
A: well, you can actually take $\epsilon={\sqrt{s}}\epsilon'$ for any $\epsilon'>0$ given that for all $n>N, \epsilon'>|s_n-s|$
