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Epsilon delta to N & M Definition

  1. Oct 8, 2008 #1
    Okay for a simple finite limit: e.g.
    lim (3x) = 3
    x->1

    in the end I say:

    "Therefore for every |x - 3| < delta, there exists an epsilon such that |3x-3| < epsilon"

    Hence I can make delta really really small and the y bounds of epsilon will constrain the limit.



    So lets come to the example I saw in an article
    lim (SQRT(x)) = INF
    x-INF

    Okay so:
    x > N - x is greater than any positive integer
    Match N with M^2
    x > M^2
    SQRT(x) > M

    Okay so how will I make my statement?
     
  2. jcsd
  3. Oct 8, 2008 #2
    so you are trying to prove that:

    [tex] \lim_{x\rightarrow \infty}\sqrt{x}=\infty[/tex] right?

    What we want to prove is that [tex] \forall N>0, \exists M>0[/tex] such that whenever

    [tex] x>M, \sqrt{x}>N[/tex]


    By observation we have, as you pointed out:[tex]\sqrt{x}>N=> x>N^2[/tex]

    so our statement would be

    [tex] \forall N>0, \exists M=N^2>0[/tex] such that whenever

    [tex]x>M=N^2=>\sqrt{x}>N[/tex]
     
  4. Oct 8, 2008 #3
    Yes. That is what I wanted. Still I think the best way for me to get this is convince myself by trying to prove a false limit (I obviously should not be able to...)
     
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