(Tricky) Absolute Value Inequalities


by vertciel
Tags: absolute, inequalities, tricky
vertciel
vertciel is offline
#1
Feb13-12, 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 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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Mark44
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#2
Feb13-12, 01:20 PM
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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.
vertciel
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#3
Feb13-12, 01:52 PM
P: 63
Thank you for your response, Mark44.

Could you please explain the red box?

SammyS
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#4
Feb13-12, 02:13 PM
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(Tricky) Absolute Value Inequalities


Quote Quote by vertciel View Post
Thank you for your response, Mark44.

Could you please explain the red box?
It looks like that's exactly what he did !


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