# Uniform continuous proofs

1. Oct 27, 2013

### Lee33

1. The problem statement, all variables and given/known data
Show that the function $x^2$ is not uniformly continuous on $\mathbb{R}$

2. Relevant equations

Delta - Epsilon Definition:

$\exists \epsilon > 0, \ \forall \delta >0, \exists x \in S [|x-x_0|< \delta \text{and} |x^2 - x_0^2| \ge \epsilon ].$

3. The attempt at a solution

I am confused on how they got $\epsilon$ and $\delta$. It seems they found it through thin air.

What I did was:

Let $\delta >0$ and $|x-x_0| < \delta$ then for a given $\epsilon >0$ we have that $|f(x) - f(x_0)| = |x^2 - x_0^2| = |x - x_0||x+x_0| = \delta|x+x_0| = \delta|x-x_0 +2x_0| = \delta[|x-x_0| + |2x_0|] = \delta^2 + \delta|2x_0| ...$

I googled this and find out they chose $\epsilon = 1$ and $x= x_0 + \delta/2$ and $x_ 0 = 1/\delta$. I am not sure how they got that? And this happens a lot in real analysis, they choose an $\epsilon$ and $\delta$ that I have no idea how they got. It's like they got it from thin air. I know those $\epsilon$ and $\delta$ work but they never provide information on how they obtained it.

Can anyone help me rigorously show how to solve problems like this?