## (Tricky) Absolute Value Inequalities

Hello everyone,

I'm posting here since I'm only having trouble with an intermediate step in proving that

$$\sqrt{x} \text{ is uniformly continuous on } [0, \infty]$$.

By definition, $$|x - x_0| < ε^2 \Longleftrightarrow -ε^2 < x - x_0 < ε^2 \Longleftrightarrow -ε^2 + x_0 < x < ε^2 + x_0$$

1. How does this imply the inequality in red?

$$\text{ Since } ε > 0 \text{ then } x_0 - ε^2 < x_0$$

However, I do not know more about x0 vs x.

2. Also, how does the above imply the case involving the orange; what "else" is there?

Thank you very much!

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 Mentor The inequality |x - x0| < ε2 doesn't specify whether x is to the right of x0 or to the left of it. That's the reason for the two inequalities.
 Thank you for your response, Mark44. Could you please explain the red box?

Mentor

## (Tricky) Absolute Value Inequalities

 Quote by vertciel Thank you for your response, Mark44. Could you please explain the red box?
It looks like that's exactly what he did !