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Proof of a limit of a cubic like sequence.

  1. Aug 12, 2008 #1
    Now this is an example in my book that I don't really understand. It says prove that the limit of (4n^(3)+3n)/(n^(3)-6))=4 as n goes to infinity.

    Basically the definition of a limit of a sequence as n->infinity is as follows.
    Lim{a}=L

    For any epsilion>0 there is a N>0 such that n>N=> |{a}-L|<epsilion


    So here's the book's solution. I just don't know why they found upper bounds.

    |(4n^(3)+3n)/(n^(3)-6) - 4|=|(3n+4)/(n^(3)-6)|<epslion

    but if n>or=2, (3n+4)/(n^(3)-6)<epslion.

    (3n+24)<or =30n

    & (n^(3)-6)>or =(1/2)n^(3)

    So, (3n+4)/(n^(3)-6)<or= (30n)/( (1/2)n^(3))<epsion

    => 60/n^(2)<epslion
    => (60/epslion)^(-1/2)<n

    so this implies that we make N=max{2, (60/epslion)^(-1/2)}


    Now here's my much more simplistic solution.

    I never found a reason why they used an upper bound. Wouldn't a lower bound work just as well?


    (3n+4)/(n^(3)-6)<epslion. { still given that n>or=2}

    n/(n^(2)) <(3n+4)/(n^(3)-6)<epslion. Now I chose the denominator,n^2, arbitaritly just to make just sure that the denominator becomes smaller.

    So solving, 1/epslion<n. This implies that we make N=max{2, 1/epsilion}.

    See how simple this is. I didn't even use equal to or greater than/ less than. I just used a straightforward inequality. But is it correct? I have no clue why the book did it with upper bounds.
     
  2. jcsd
  3. Aug 12, 2008 #2
    (4n^(3)+3n)/(n^(3)-6))=4
    (4 * oo^3 + 3*oo)/(oo^3-6)) = 4
    (4 * oo^3)/(oo^3)
    4 = 4

    I don't see what would be wrong with just showing that kn^3/n^3 for all n is k, and that n^3 + n = n^3 as n approaches infinity.

    I could be wrong though.
     
  4. Aug 13, 2008 #3

    HallsofIvy

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    I'm not sure you can justify cancelling infinities like that. More rigorous is to divide by both numerator and denominator by the highest power of n:
    [tex]\frac{4n^3+ 3n}{n^3- 6}= \frac{4+ \frac{3}{n^2}}{1- \frac{6}{n^3}}[/tex]
    Since both 3/n2 and 6/n3[/sub] go to 0 as n goes to infinity, the limit is 4/1= 4.
     
  5. Aug 13, 2008 #4
    But I'm trying to understand it from a formal approach using epslion and a bound as they do in real analysis.
     
  6. Aug 13, 2008 #5
    For proving this limit I would use the same method as Hall, I would devide each term in the numerator and denominator by the highest power of x that occurs in the denominator, namely, n3. Just like he did.

    You must remember that [tex]lim_{x \rightarrow\pm \infty} \frac{1}{x} = 0[/tex] etc, and voila, you have limit = 4

    Why can't you try L'Hopital's rule, btw?
     
  7. Aug 13, 2008 #6
    I'm not allowed to do that because I'm self studying real analysis. I need to prove things from the most fundamental steps. I have to use the definition of a limit. Which is, for a sequence,

    Lim{a}=L as n-> infinity means,

    For any epsilion>0 there is a N>0 such that n>N=> |{a}-L|<epsilion
     
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