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Proof of a limit.

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

    2. Relevant equations

    Lim{a}=L

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



    3. The attempt at a solution

    So here's the book's solution.

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

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

    they found an upper bound for the sequence.

    (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 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

    n/(n^(2)) <(3n+4)/(n^(3)-6)<epslion. Now I choose 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/epslion}

    So far, I'm convinced that my solution is the most simple. But is it correct?
     
  2. jcsd
  3. Aug 12, 2008 #2
    [tex]\epsilon[/tex] is "epsilon" and you can simply write "<or = " as "<=".

    Notice that your sequence is monotonically increasing. Let [tex]S= \left\{ a_{n}|n \in N \right\}[/tex] so [tex]\epsilon > 0, [/tex] you need to find an index N such that

    [tex]|a_{n}- l| < \epsilon[/tex]

    which is the same as

    [tex]a_{n} \leq l < l + \epsilon [/tex]

    Notice that l is an upper bound for the set S, so

    [tex] \implies l - \epsilon < a_{n} < l+ \epsilon [/tex]
    So, l is the least upper bond or supS . Thus, there is an index N such that [tex]l - \epsilon < a_{N}[/tex]; but, the sequence is mononotically increasing so

    [tex]l - \epsilon < a_{N} \leq a_{n}[/tex]

    In Short: Since your sequence is monotonically increasing, then it converges if and only if it is bounded above.
     
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