Proof that $\lim\limits_{n\to \infty} \sqrt{x_n} = \sqrt{\lim\limits_{n\to \infty} x_n}$ It is asked to prove that $\lim\limits_{n\to \infty} \sqrt{x_n} = \sqrt{\lim\limits_{n\to \infty} x_n}$, and suggested to use the following two inequalities:
$$a+b\leq a+ 2\sqrt{a}\sqrt{b}+b$$
$$\sqrt{b}\sqrt{b}\leq \sqrt{a}\sqrt{b}$$
The second inequality holds iff $a\geq b \geq 0$.
I've tried different possibilities, but couldn't figure out how to either take the limit sign out of the square root, or take the limit sign into the square root. Would appreciate some hints, but not an entire solution please.

It's wrong of course. 
For $x_n\geq0$ it's true because $f$ is a continuous function, where $f(x)=\sqrt{x}$. 
