Hello everyone,(adsbygoogle = window.adsbygoogle || []).push({});

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

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

By definition, [tex] |x - x_0| < ε^2 \Longleftrightarrow -ε^2 < x - x_0 < ε^2 \Longleftrightarrow -ε^2 + x_0 < x < ε^2 + x_0 [/tex]

1. How does this imply the inequality in red?

[tex] \text{ Since } ε > 0 \text{ then } x_0 - ε^2 < x_0 [/tex]

However, I do not know more aboutvsx_{0}.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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# (Tricky) Absolute Value Inequalities

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