- #1
vertciel
- 63
- 0
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!
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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