Prove that $x^2$ is not uniformly continious

Question

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 $

Answer 1

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 $(-\infty,\infty)$.

Answer 2

Suppose toward a contradiction that such a $\delta$ exists and take $\epsilon=1$ (in fact, any fixed value of $\epsilon$ will do.) Then, with $x=x_0+\delta/2,\ $ we would have $|x-x_0|<\delta$ and $|(x_0+\delta/2)^2-x^2_0|=|x_0\delta+\delta^2/4|<1.$ All that remains to do is to take $x_0$ large enough to get a contradiction.