Maybe there is a misunderstanding but $x^2$ is continious but not uniformly continous. Because it is not uniformly continious
$\exists_{\epsilon>0}\forall_{\delta>0}\exists_{x,x_o\in\mathbb{R}}:|x-x_0|<\delta,|x^2-x_0^2|\geq \epsilon $
which would be equivalent to
$\lim_{x\rightarrow x_0}|x^2-x_0^2|\neq 0$.
However I can Chose an arbitrary $\epsilon$.
And I would get $|x^2-x_0^2|\leq\epsilon$ For all $x=x_0 +a$ with the property that
$|(x_0+a)^2-x_0^2|\leq\epsilon \stackrel{cmplt. \square}{\iff}a<\sqrt{2x_0}-\sqrt{2\sqrt{2}x_0a-\epsilon}$.
Where is my mistake and how can I prove the Statement in the Question: I.e:
$\exists_{\epsilon>0}\forall_{\delta>0}\exists_{x,x_o\in\mathbb{R}}:|x-x_0|<\delta,|x^2-x_0^2|\geq \epsilon $
Best Answer
Fix $\epsilon_0=1$. For $\forall \delta\in(0,1)$, for $|a|<\delta$, take $x_0$ such that $|x_0|=\frac{1}{|a|}>1$ $$ |(x_0+a)^2-x_0^2|=|a^2+2ax_0|=|a||a+2x_0|\ge|a|(2|x_0|-|a|)\ge|a|(\frac{2}{|a|}-|a|)\ge1=\epsilon_0.$$ Therefore $f(x)=x^2$ is not uniformly continuous in $(-\infty,\infty)$.