# (Tricky) Absolute Value Inequalities

1. Feb 13, 2012

### vertciel

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!

Last edited: Feb 13, 2012
2. Feb 13, 2012

### Staff: 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.

3. Feb 13, 2012

### vertciel

Thank you for your response, Mark44.

Could you please explain the red box?

4. Feb 13, 2012

### SammyS

Staff Emeritus
It looks like that's exactly what he did !