How to prove that $\sqrt x$ is continuous in $[0,\infty)$?

Question

I am trying to prove that

$\sqrt x$ is continuous in $[0,\infty)$.

I have started writing the following proof: Given $x_0 \in [0,\infty)$ and $\epsilon > 0$. We have to show that there exists a $\delta > 0$ such that for every $x \in (x_0 - \delta,x_0 +\delta)$, $\sqrt x \in (\sqrt x_0 - \epsilon, \sqrt x_0 + \epsilon)$.

So, $| \sqrt x - \sqrt x_0 | = \frac {| (\sqrt x - \sqrt x_0)(\sqrt x + \sqrt x_0) |}{| \sqrt x + \sqrt x_0 |} = \frac {| x - x_0 |}{| \sqrt x + \sqrt x_0 |}$

I am not sure how to continue from here.. I could take $M = max\{ x,x_0 \}$ and $\delta = 2M \epsilon$ but this is unnecessary if $x,x_0 > 1$.

Should I split to cases? $x,x_0 > 1$, $0 < x,x_0 < 1$ etc.. Should I dea with $x_0 = 0$ separately with a right neighborhood $[0,\delta)$?

I feel like I am over complicating this..

What is the simplest way to define $\delta$?

Thanks!!

Answer 1

For $x_{0}\neq 0$, we have $$ |\sqrt{x}-\sqrt{x_{0}}|=\frac{|x-x_{0}|}{\sqrt{x}+\sqrt{x_{0}}}\leq \frac{|x-x_{0}|}{\sqrt{x_{0}}}. $$ So you can take $\delta=\sqrt{x_{0}}\epsilon$.

For $x_{0}=0$, we need $\sqrt{x}<\epsilon$ for $x\in(0,\delta)$, so we can take $\delta=\epsilon^{2}$.

Answer 2

I will prove the stronger result that the function is uniformly continuous. If $x <\epsilon ^{2} /4$ and $x_o <\epsilon ^{2} /4$ then then $|\sqrt x -\sqrt {x_0}| <\epsilon /2+\epsilon/ 2=\epsilon$. Otherwise $|\sqrt x -\sqrt {x_0}| \leq \frac {2|x-x_0|} {\epsilon}$ from the inequality you have derived. Hence $|\sqrt x -\sqrt {x_0}|<\epsilon$ if $|x-x_0| <\epsilon^{2}/2$.

Answer 3

Inverse of a continuous function is continuous on the interval hence the $\sqrt x$ is continuous off course being the inverse of square function