Real analysis, The limits of two related sequences I am currently working on som of the exercises in Abbot's Understanding Analysis, and I have gotten stuck on an exercise. The exercise is as follows:
If $(x_{n})\rightarrow x$, show that $\sqrt{(x_{n})}\rightarrow \sqrt{x}$
I have been stuck on this for a while and would appreciate some help.
I have tried using the definition of that $x_{n}\rightarrow x$, $\forall\epsilon >0\exists N\in\mathbb{N}: n\geq N\Rightarrow |x_{n}-x|<\epsilon$, but I can't find a proper way to manipulate this expression to get $|\sqrt{x_{n}}-\sqrt{x}|<\epsilon$
Thanks in advance.
 A: Hint:
Prove (or assert) that
$$
\left|\sqrt{x} - \sqrt{y}\right| \leq \sqrt{\left|x - y\right|}.
$$
It may be easier (more intuitive) to first prove, for positive $a$, $b$,
$$
a + b \leq (\sqrt a + \sqrt b)^2
$$
which implies 
$$
\sqrt{a+b} \leq \sqrt a + \sqrt b.
$$
A: How about some nonstandard analysis. Let $H$ be an infinite Hyperreal number.
$x_H\approx x$ (meaning they are infinitely close.) Now we must show that $\sqrt{x_H}\approx \sqrt x$. We know that $x-x_H \approx 0$.
$$(x-x_h)^2\approx0$$
$$x^2-2x*x_H+x_h^2\approx0$$
$$x^2+2x*x_H+x_H^2\approx4x*x_H$$
$$x+x_H\approx2\sqrt{x*x_H}$$
$$x-2\sqrt{x}\sqrt{x_H}+x_H\approx0$$
and
$$(\sqrt x-\sqrt{x_h})^2=x-2\sqrt{x}\sqrt{x_H}+x_H$$
$$(\sqrt{x}-\sqrt{x_h})^2\approx0$$
$$\sqrt x-\sqrt{x_H}\approx0$$
$$\sqrt x \approx \sqrt{x_H}$$
End of proof
A: Hint:
You can use the fact that $$\sqrt x_n -\sqrt x = \frac{x_n-x}{\sqrt x_n +\sqrt x}.$$ And apply the logic you have written down in your second last sentence. If you need further elaboration let me know.  
