Limit of square root of sequence given value of limit of original sequence

Question

Suppose that $x_n\ge0$ is a sequence of real numbers such that $\forall n\in \mathbb{N}$ then $\lim_{n\to\infty}x_n=x_0\in\mathbb{R}$. Show that $\lim_{n\to\infty}\sqrt{x_n}=\sqrt{x_0}$.

In this context, a sequence $x_n$ converging to a constant $\alpha$ is given by:

$$\exists\alpha\in\mathbb{R}\forall\epsilon>0\exists N\in\mathbb{N}\forall n\ge N:|x_n-\alpha|<\epsilon$$

However, I'm struggling to prove this in a more abstract form; when I have a simpler algebraic expression fo $x_n$, I know the process, but I'm confused how to show it in general using the above definition.

Answer 1

The case that $x_{0}=0$ is left to OP.

For the case that $x_{0}>0$, find some $N_{1}$ such that $|x_{n}-x_{0}|<\dfrac{x_{0}}{2}$ for all $n\geq N_{1}$, then $x_{n}\geq x_{0}-\dfrac{x_{0}}{2}>\dfrac{x_{0}}{2}$ for all such $n$.

Given $\epsilon>0$, find some $N_{2}$ such that $|x_{n}-x_{0}|<\dfrac{3\sqrt{x_{0}}}{2}\cdot\epsilon$ for all $n\geq N_{2}$.

For all $n\geq N_{1}+N_{2}$, then \begin{align*} \left|\sqrt{x_{n}}-\sqrt{x_{0}}\right|&=\dfrac{|x_{n}-x_{0}|}{\sqrt{x_{n}}+\sqrt{x_{0}}}\\ &<\dfrac{|x_{n}-x_{0}|}{\dfrac{\sqrt{x_{0}}}{\sqrt{2}}+\sqrt{x_{0}}}\\ &<\dfrac{|x_{n}-x_{0}|}{\left(\dfrac{1}{2}+1\right)\sqrt{x_{0}}}\\ &=\dfrac{2}{3\sqrt{x_{0}}}\cdot|x_{n}-x_{0}|\\ &<\dfrac{2}{3\sqrt{x_{0}}}\cdot\dfrac{3\sqrt{x_{0}}}{2}\cdot\epsilon\\ &=\epsilon. \end{align*}

Answer 2

In order to show $$\lim_{n\to\infty}\sqrt{x_n}=\sqrt{x_0}$$

Note that for large enough $ n$, we have $$|\sqrt{x_n}-\sqrt{x_0}|=\frac {|x_n -x_0|}{|\sqrt{x_n}+\sqrt{x_0}|}\le$$

$$\frac {|x_n -x_0|}{(3/2) \sqrt{x_0}} <\epsilon$$

Answer 3

An option for $x \ge 0. $

$f(x) =√x$ is continuos at $x_0 \gt 0.$

$\iff$

For every sequence $x_n , x_n \ge 0$ , with

$\lim_{n \rightarrow \infty}x_n=x_0$ we have:

$\lim_{n \rightarrow \infty} f(x_n)=f(x_0).$

Show that $f$ is continuos at $x_0 \gt 0$ using $\epsilon,\delta$ definition.

Let $\epsilon >0$ be given.

Choose $\delta <\epsilon √x_o$, then

$|x-x_0|\lt \delta$ implies

$|√x-√x_0| = $

$\dfrac {|x-x_0|}{√x+√x_0}\lt \dfrac{\delta}{√x_0}= \epsilon.$