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General Delta Epsilon Proof

  1. Feb 6, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove: [tex]\lim_{x \to a} \sqrt{x} = \sqrt{a},[/tex] if [tex] a>0[/tex]

    2. Relevant equations
    [tex]|f(x)-L| < \epsilon[/tex]
    [tex]|x-a| < \delta[/tex]

    3. The attempt at a solution
    [tex]|\sqrt{x}-\sqrt{a}| < \epsilon[/tex], when [tex]|x-a| < \delta [/tex]
    [tex]|x - 2\sqrt{x}\sqrt{a}-a| < \epsilon^2[/tex], when [tex]|x-a| < \delta[/tex]

    From here I don't know where to go. I don't see any obvious way to get delta into terms of epsilon.
  2. jcsd
  3. Feb 6, 2008 #2
    well [tex]|\sqrt{x}-\sqrt{a}| =|(\sqrt{x}-\sqrt{a})\frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}|[/tex] can you see what to do now?
  4. Feb 6, 2008 #3
    Yep. Thanks, it has been a long night.
  5. Feb 6, 2008 #4
    It is 3 in the morning here, lol !!!
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