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Prove $x^2+x$ is uniformly continuous in $(0,1)$ using the $\epsilon , \delta$ method. My try:

Let $\epsilon>0$ and $\delta=\frac{\epsilon}3$. Then pick $x,y\in(0,1)$ s.t. $|x-y|<\delta$, so we have $|x^2+x-y^2-y|=|(x^2-y^2)+(x-y)|\leq|x^2-y^2|+|x-y|.$

Note that $|x^2-y^2|=|(x-y)(x+y)|$ and $x,y\in(0,1)$ so $(x+y)>0$$\space$ and $<2$. So it's equal to $|x-y|(x+y)<2\delta$.

Coming back we have: $|x^2-y^2|+|x-y|=|x-y|(x+y)+|x-y|<3\delta=\epsilon.$

It's this okay? How could I write it better? Am I wrong somewhere? Thanks.

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    $\begingroup$ Yes. Your argument is OK and I think you've written it as well as possible already. $\endgroup$ – stressed out Jan 29 at 18:01
  • $\begingroup$ That is, x(x+1) is uniformly continuous. It is suffficient to show that the product of uniformly continuous functions is uniformly continuous. $\endgroup$ – Jacob Wakem Jan 29 at 18:14
  • $\begingroup$ @Alephnull But it's not possible to show that (unless you also assume the two functions are bounded, or something along those lines). For example, $f(x) = g(x) = x$ is uniformly continuous on $\mathbb{R}$ but their product isn't. $\endgroup$ – Daniel Schepler Jan 29 at 20:43
  • $\begingroup$ It is well-known that x^2 is uniformly continuous. You can use the same delta (or is it epsilon?) . $\endgroup$ – Jacob Wakem Jan 30 at 17:00
  • $\begingroup$ x^2+x is between x^2 and (x+1)^2 . It is well-known x^2 is uniformlycontinuous and thus by graph-similarity (x+1)^2 is uniformly continuous. Thus x^2+x is uniformly continuous. $\endgroup$ – Jacob Wakem Feb 8 at 20:16
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You are correct. The same result can be obtained in a easier way by using the Mean Value Theorem: if $f(x)=x^2+x$ then for $x,y\in (0,1)$ there is $t\in (0,1)$ such that $$|f(x)-f(y)|=|f'(t)||x-y|=|2t+1||x-y|\leq 3|x-y|.$$ More generally a differentiable function whose derivative is bounded in an interval $I$ is also uniformly continuous in $I$.

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    $\begingroup$ @MayureshL Here we are talking about uniform continuity. $\endgroup$ – Robert Z Jan 29 at 18:08
  • $\begingroup$ Robert.Gives the result from a different angle.Nice! $\endgroup$ – Peter Szilas Jan 29 at 18:13
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Your proof is fine. A shortcut would be to note that $f$ is continuous on the compact set $[0,1]$, and so uniformly continuous there; hence, on $(0,1),$ too.

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