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Prove that the lim as x goes to a of sqrtx = sqrta

  1. Sep 20, 2009 #1
    can i get some help on proving this limit?
     
  2. jcsd
  3. Sep 21, 2009 #2

    lurflurf

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    Homework Helper

    start with
    sqrt(x)-sqrt(a)=(x-a)/(sqrt(x)+sqrt(a))
     
  4. Sep 21, 2009 #3
    I think it would be very helpful if you guys used latex

    The question should be

    Proove that [tex]\lim {x \rightarrow a} \sqrt{x} = \sqrt{a}[/tex]

    And the first reply should be

    start with
    [tex]\sqrt{x}-\sqrt{a}=\frac{x-a}{\sqrt{x}+\sqrt{a}}[/tex]

    For basic things like this, latex only takes about a minute to learn and it VERY much easier to read :)
     
  5. Sep 21, 2009 #4

    HallsofIvy

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    In order to prove that "[itex]\lim_{x\to a}\sqrt{x}= \sqrt{a}[/itex], you need to prove that "Given [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]|x- a|< \delta[/itex] then [itex]|\sqrt{x}-\sqrt{a}|< \epsilon[/itex]".

    Start from the second inequality and show (using the above hint) that you can find such a [itex]\delta[/itex].
     
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