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Epsilon-delta proof of sqrt of abs value of x

  1. Sep 26, 2011 #1
    Prove that the limit as x->0 of [itex]\sqrt{\left|x\right|}[/itex] = 0

    attempt:

    [itex]\sqrt{\left|x\right|}[/itex] < [itex]\epsilon[/itex] when |x|<[itex]\delta[/itex]

    -[itex]\epsilon[/itex] < [itex]\sqrt{\left|x\right|}[/itex] < [itex]\epsilon[/itex]

    [itex]\left|x\right|[/itex] < [itex]\epsilon[/itex][itex]^{2}[/itex]

    [itex]\delta[/itex]=[itex]\epsilon[/itex][itex]^{2}[/itex]

    This seems to work...but I just ignored the negative epsilon since I cant square it in my inequality...I dont know if you are allowed to just ignore this?
     
  2. jcsd
  3. Sep 26, 2011 #2
    [tex]
    \sqrt{|x|} < \epsilon \Leftrightarrow |x| < \epsilon^{2}
    [/tex]

    If you choose a [itex]\delta[/itex] such that:
    [tex]
    0 < \delta < \epsilon^2
    [/tex]
    you are done.
     
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