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Sequence limit help

  1. Sep 20, 2004 #1
    --------------------------------------------------------------------------------

    Hello all

    For the sequence (n^2 + n -1 )/ (3n^2 +1) I am trying to find

    lim (n^2 + n -1)/ (3n^2 +1) = 1/3
    n -> 00


    My solution"

    = 1 - (2n^2 - n +2/ 3n^2 +1) = 1- r(sub-n)

    We need to prove that r(sub-n) approaches 2/3.

    Hence 2n^2 - n + 2 < 2n^3
    3n^2 + 1 > 3n^2

    The remainder r sub n is 2n/3. Is this right? How would I find the remainder?

    Also how do I find an N such that n > N and the difference between f(x) and L is less than 1/ 10, 1/100, and 1/1000.

    Any help would be appreciated.

    Thanks
     
  2. jcsd
  3. Sep 20, 2004 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Please do not post the same question repeatedly (and how did it become "homework" in three hours?).

    You prove that "r(sub-n)" approaches 2/3 as n-> infinity exactly the same way you would prove (n^2+ n-1)/(3n^2+1)-> 1/3: divide both numerator and denominator by n^2.
     
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