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Lim as x goes to a of sqrtx = sqrta

  1. Sep 20, 2009 #1
    this is a theorem that has a proof. in the proof they use a constant c so that epsilon is not in terms of x when you go to prove it. can someone explain this proof and also how and why you get/use a constant c to prove this theorm?
  2. jcsd
  3. Sep 20, 2009 #2
    I would have to see this proof to explain it.
    With the typical epsilon/delta proof of a limit, to prove that lim x->a f(x) = L you need to prove that for EVERY epsilon, a delta exists such that if a-delta < x < a+delta, then |f(x)-L| < epsilon.
    epsilon is arbtrary and not in terms of x or anything else, and delta will be a funtion of epsilon that you will have to choose to make |f(x)-L| small enough.
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