Limit proof by the definition 
If $X=(x_n)$ is a positive sequence which converges to $x$, then
  $(\sqrt {x_n})$ converges to $\sqrt x$.

The hint my book provided is: $$\sqrt x_n -\sqrt x = \frac{x_n-x}{\sqrt x_n +\sqrt x}$$
which I would never even guessed to try, so maybe I am not totally understanding what the limit defintion is trying to convey.
My attempt: 
We know that $\lim x_n=x$. So, given any $\epsilon>0$ $\exists$ an $N$ such that $|x_n−x|<\epsilon$ for all $n\ge N.$ 
Then for $\lim \sqrt {x_n}=\sqrt x$. Given any $\epsilon>0$ there is an $N$ such that $|\sqrt x_n -\sqrt x|<\epsilon$, $\forall n\ge N$ $$|\sqrt x_n -\sqrt x|=\frac{|x_n-x|}{|\sqrt x_n +\sqrt x|}< \frac{\epsilon}{|\sqrt x_n +\sqrt x|}.$$
Which I don't quite think is correct because I feel like I am missing a lot of information.
 A: You're on a good track. Now all you need to do is bound $\sqrt {x_n}+\sqrt x$ from below. Say you obtain a constant $\frac 1 \eta$ such that $$\frac 1 \eta<\sqrt {x_n}+\sqrt x$$ for each $n$. Then you'll have for each $n$ that
$$ \frac{1}{\sqrt {x_n}+\sqrt x}<\eta$$
which means 
$$\frac {|x_n-x|}{\sqrt {x_n}+\sqrt x}< \eta{\epsilon} $$
Thus, given $\epsilon$ you could take $N$ such that $n\geq N$ implies $|x-x_n|<\epsilon/\eta$ which would leave you with 
$$\frac {|x_n-x|}{\sqrt {x_n}+\sqrt x}< \eta{\epsilon/\eta}=\epsilon$$
and the proof would be complete.
Can you do that? 
Thomas has a nice solution when $\lim x_n\neq 0$. If $\lim x_n=0$, we'd have to prove that $\sqrt {x_n}<\epsilon$ for $n$ large enough, i.e $x_n<\epsilon^2$ for $x$ large enough. And that is not hard! =)
ADD You say "...which I would never even guessed to try, so maybe I am not totally understanding what the limit defintion is trying to convey." The "trick" the book used might not be too obvious at first, but you'll get used to them later. It has nothing to do with understanding the definition of limit wants to convey. What you might want to note is that since we know we can make $|x-x_n|$ small, it would be good to write $|\sqrt x-\sqrt{x_n}|$ in terms of $|x-x_n|$, and work with that. For example, say $x_n\to x$. If you want to prove $x_n^2\to x^2$, you want $$|x^2-x_n^2|$$ small. Yet again, we have $$|x^2-x_n^2|=|x-x_n|\cdot|x+x_n|$$
One part we can make small, the other we can "control" in some way, so we know all in all the product will be small. I think I told you this before: practice, practice, practice.
A: First assume that $x\neq 0$.
Given $x$ you can pick $N$ such that when $n\geq N$ then 
$$
\lvert x_n - x\lvert < \epsilon\sqrt{x}.
$$
So then
$$
\lvert\sqrt x_n -\sqrt x\rvert = \frac{|x_n-x|}{|\sqrt x_n +\sqrt x|} = \frac{|x_n-x|}{\sqrt x_n +\sqrt x} \leq \frac{\lvert x_n - x\rvert}{\sqrt{x}} <  \epsilon.
$$
Here you are just using that $\sqrt{x_n} \geq 0$ for all $n$.
For completion let us now consider the case where $x = 0$. Then you have again $\epsilon > 0$ given. You pick $N$ such that $\lvert x_n\rvert = x_n< \epsilon^2$. So then
$$
\lvert \sqrt{x_n}\rvert = \sqrt{x_n} < \sqrt{\epsilon^2} = \epsilon.
$$
