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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 $

RM777
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  • The idea is that the choice of $\delta$ has to depend on the value of $x_0$. – Stan Tendijck Jan 13 '19 at 15:48
  • So, you cannot find a uniform $\delta$ that works for all $x_0$ since the 'optimal' value of $\delta$ will become smaller and smaller as $x_0$ grows. – Stan Tendijck Jan 13 '19 at 15:50
  • I changed the Question. I forgot to do it before I post it – RM777 Jan 13 '19 at 15:52
  • Why are you taking the limit as $x\to \infty$? What is $x_0$? – BigbearZzz Jan 13 '19 at 15:53
  • You are Right I will eddit it – RM777 Jan 13 '19 at 15:54
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    You seem to be assuming that $x_0$ is fixed, but it's not. To get $|x^2-x_0^2|>\varepsilon$ as $\delta$ gets small, you have to take both $x$ and $x_0$ progressively larger. – saulspatz Jan 13 '19 at 16:04
  • your $\delta$ depends on $x_0$. What you have proved is $f(x)=x^2$ is continuous. Have a look at https://math.stackexchange.com/questions/503093/proving-fx-x2-is-not-uniformly-continuous-on-the-real-line – skylark Jan 13 '19 at 16:05

2 Answers2

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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 $(-\infty,\infty)$.

xpaul
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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.

Matematleta
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