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How is this asymptotic relation?

  1. Jul 30, 2007 #1
    1. The problem statement, all variables and given/known data
    e^((ln(x))^2)=O(1)?
    as x->0

    2. Relevant equations
    g(x)=O(f(x)) => |g(x)|<K|f(x)| as x->0 in this case but can approach any value as desired.
    K<infinity

    3. The attempt at a solution
    We can try numbers for small x but non zero that satisfy the inequality however are we allowed to manipulate infinities? Somehow I think that the equality shouldn't stand.

    I know that x^2 does not equal O(x) when x->infinity but does equal when x->0

    We can look at it this way, g(x)=O(f(x)) => |g(x)|<K|f(x)| as x->0, K<infinity => lim x->0 |g(x)/f(x)| is finite.
     
    Last edited: Jul 30, 2007
  2. jcsd
  3. Jul 30, 2007 #2

    Dick

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    Science Advisor
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    The limit is infinity as x->0. How can this be O(1)?
     
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