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(Tricky) Absolute Value Inequalities 
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#1
Feb1312, 12:49 PM

P: 63

Hello everyone,
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 about x_{0} vs x. 2. Also, how does the above imply the case involving the orange; what "else" is there? Thank you very much! 


#2
Feb1312, 01:20 PM

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P: 21,407

The inequality x  x_{0} < ε^{2} doesn't specify whether x is to the right of x_{0} or to the left of it. That's the reason for the two inequalities.



#3
Feb1312, 01:52 PM

P: 63

Thank you for your response, Mark44.
Could you please explain the red box? 


#4
Feb1312, 02:13 PM

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(Tricky) Absolute Value Inequalities



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