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.
 A: You have to make sure that $\forall n\in\mathbb{N}^*$, $x_n\geq 0$. Otherwise the limit doesn't exist because the sequence $\{\sqrt{x_n}\}_{n=1}^{\infty}$ isn't even defined(well, in $\mathbb{R}$).
Suppose the statement above is true. Let $A=\lim\limits_{x\to\infty}x_n$ Then there are 2 cases:
(1)$A=0$. $\forall\epsilon>0$,$\exists N\in\mathbb{N}^*$ so that $\forall n\geq N$ there is $x_n<\epsilon^2$,that is, $\sqrt{x_n}<\epsilon$, so that $\lim\limits_{x\to\infty}\sqrt{x_n}=0=\sqrt{\lim\limits_{n\to\infty}x_n}$.
(2)Otherwise,$A>0$. $\forall\epsilon>0$,$\exists N\in\mathbb{N}^*$ so that $\forall n\geq N$ there is $|x_n-A|<\sqrt{A}\epsilon$.This means that $\epsilon>\frac{x_n}{\sqrt{A}}\geq|\frac{x_n-A}{\sqrt{x_n}+\sqrt{A}}|=|\sqrt{x_n}-\sqrt{A}|$, therefore $\lim\limits_{n\to\infty}\sqrt{x_n}=\sqrt{A}=\sqrt{\lim\limits_{n\to\infty}x_n}$.
This proof uses solely the definition of the limit of a sequence :)
A: It's wrong of course. 
For $x_n\geq0$ it's true because $f$ is a continuous function, where $f(x)=\sqrt{x}$. 
