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Limit property

  1. Jul 18, 2011 #1
    The problem statement, all variables and given/known data
    if lim f(n) = s , prove that lim (f(n))^{1/3} = s^{1/3}. How do you know that as n approaches infinity, f(n) and s have the same sign. n is just an index in this case and f is not a function but a sequence.

    The attempt at a solution
    so we know that |f(n) - s| < epsilon, using a^3 - b^3 I can factor:
    |f(n)^{1/3} - s^{1/3}|*|f(n)^{2/3} + (f(n)*s)^{1/3} + s^{1/3}| < epsilon. Divide both sides and we get |f(n)^{1/3} - s^{1/3}| < epsilon/(|f(n)^{2/3} + (f(n)*s)^{1/3} + s^{1/3}|) which proves the first part. I don't know how to argue that as n approaches infinity, f(n) and s have the same sign.
     
  2. jcsd
  3. Jul 18, 2011 #2

    vela

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    You still have work to do to finish the proof of the first part.

    You should have no problems showing f(n) and s have the same sign if you understand the definition of the limit. Suppose s>0. You can find an interval around s that only contains positive numbers, right? So...
     
  4. Jul 18, 2011 #3
    what am i missing from the first part of the proof?
     
  5. Jul 18, 2011 #4

    vela

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    To prove a limit using the definition, you assume ε>0, and then show that for the appropriate choice of N, n>N implies that |f(n)-L|<ε. So first, you need to find N.
     
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