Prove that $\sqrt{x}$ is continuous on its domain $[0, \infty).$ Prove that the function $\sqrt{x}$ is continuous on its domain $[0,\infty)$.
Proof.
Since $\sqrt{0} = 0, $ we consider the function $\sqrt{x}$ on $[a,\infty)$ where $a$ is real number and $s \neq 0.$ Let $\delta=2\sqrt{a}\epsilon.$ Then, $\forall x \in dom,$ and $\left | x-x_0\right | < \delta \Rightarrow \left| \sqrt{x}-\sqrt{x_0}\right| = \left| \frac{x-x_0}{ \sqrt{x}+\sqrt{x_0}} \right| < \left|\frac{\delta}{2\sqrt{a}}\right|=\epsilon.$
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Prove that $\sqrt{x}$ is continuous on its domain $[0,\infty[$
I have used the following definition and therom:
Definition 1:
A sequence of real numbers {$x_n$} is said to converge to a real number  $a \in R$ if and only if for every $\epsilon>0$ there is an N $\in$ N such that
$n \geq N$  implies $|x_n-a|<\epsilon$
Therom 1
Suppose that E is a nonempty subset of R, that a $\in$ E, and that f: E$\rightarrow$ R. Then the following statemants are equivalent:


*

*f is continuous at a $\in$ E.

*If $x_n$ converges to a and $x_n$ $\in$ E, then $f(x_n) \rightarrow f(a) $as $n \rightarrow \infty$


Proof
If $x_n \rightarrow a$ implies that $f(x_n) \rightarrow f(a)$ as $n\rightarrow \infty$
this means that the function f must be continuous. (This is just another way of stating the $\epsilon$ and $\delta$ statement)
Supose that as  $n \rightarrow \infty$ then $x_n \rightarrow x$ (x plays the role of a)
We need to prove that if $x_n \rightarrow x$ then $\sqrt{x_n} \rightarrow \sqrt{x}$ as $n \rightarrow \infty$. If this is true then $\sqrt{x}$ is continuous.
Case 1: x=0
(We cannot devide by zero, so we have to split up the proof, which will be clear in case 2.)
If x=0. We let $\epsilon > 0$ and choose N such that $n \geq N$ implies $|x_n-x|<\epsilon^2 \Leftrightarrow \sqrt{x_n-x} = \sqrt{x_n}<\epsilon$ for $n \geq N$.
So $\sqrt{x_n} \rightarrow 0$ when $n \rightarrow \infty$
(Remember that $\epsilon$ can be made aberitaryly small, but given a large enough N, $\sqrt{x_n}$ will be smaller thus it aproaches 0.)
Case 2: x > 0
$f(x_n)-f(x)=\sqrt{x_n}-\sqrt{x}=(\sqrt{x_n}-\sqrt{x})\cdot (\frac{\sqrt{x_n}+\sqrt{x}}{\sqrt{x_n}+\sqrt{x}})=\frac{|x_n-x|}{\sqrt{x_n}+\sqrt{x}}$
$\sqrt{x_n}\geq 0$ so we can write
$\sqrt{x_n}-\sqrt{x} \leq \frac{|x_n-x|}{\sqrt{x}}$
When $n \rightarrow 0$ then $x_n \rightarrow x$ meaning the righthandsite converges to 0.
$\sqrt{x_n}-\sqrt{x} \leq 0$
So it follows from the Squeeze Therom that $\sqrt{x_n} \rightarrow \sqrt{x}$ as $n \rightarrow \infty$
And therefore $\sqrt{x}$ is continuous on the domain [0,$\infty$[
