# (Tricky) Absolute Value Inequalities

by vertciel
Tags: absolute, inequalities, tricky
 P: 63 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!
 Mentor P: 21,249 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.
 P: 63 Thank you for your response, Mark44. Could you please explain the red box?
Emeritus