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Limit of square root function

  1. Nov 17, 2009 #1
    Could someone help me with this? I feel it should likely be easy, but I'm baffled anyway:

    1. The problem statement, all variables and given/known data

    Prove the validity of the limit lim x -> x_0 sqrt(x) = sqrt(x_0)


    2. Relevant equations

    use definition of limit


    3. The attempt at a solution

    Generally confused. I start with |sqrtx - sqrtx0| < epsilon
     
  2. jcsd
  3. Nov 17, 2009 #2

    Mark44

    Staff: Mentor

    You want to show that for any delta > 0, there is an epsilon so that |sqrt(x) - sqrt(x_0)| < eps when |x - x_0| < delta.

    |sqrt(x) - sqrt(x_0)| = |sqrt(x) - sqrt(x_0)| *|sqrt(x) + sqrt(x_0)| /|sqrt(x) + sqrt(x_0)| = |x - x_0|/ *|sqrt(x) + sqrt(x_0)|

    Without loss of generality, you can assume that delta is reasonably small, say less than 1. That puts x within 1 unit of x_0.

    Can you work with |sqrt(x) + sqrt(x_0)| to find max and min values for it?
     
  4. Nov 18, 2009 #3
    Okay...this is starting to make more sense now. Thanks for the tips. When I get it I'll try to upload what I have.
     
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