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A limit proof

  1. Nov 3, 2012 #1
    Prove that the limit of (3n+5)/2(n+1)^2 is 0 when n goes to infinity.

    Attempt: I need to find an N such that for any €>0, (3n+5)/2(n+1)^2<€ holds for every n with n>N.

    Then I made some manipulation;

    (3n+5)/2(n+1)^2 < (3n+5)/(2n^2 +4n) < (3n+6)/(2n^2 +4n) = (n+3)/(n^2 +2n) < (n+3)/n^2

    Then what? Please help.
     
  2. jcsd
  3. Nov 3, 2012 #2

    tiny-tim

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    hi bedi! :smile:

    (try using the X2 button just above the Reply box :wink:)

    = n/n2 + 3/n2 ? :smile:

    (btw, you could have started (3n+3)/2(n+1)2 + 2/2(n+1)2 :wink:)
     
  4. Nov 3, 2012 #3
    Alright, then should I ignore 3/n2?
     
  5. Nov 3, 2012 #4

    tiny-tim

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    nooo, you should prove that its limit is 0 ! :smile:
     
  6. Nov 3, 2012 #5
    So I choose N such that 1/N<ε, which is permitted by the Archimedean property. Hence 1/n<1/N<ε. This proves that the limit of the first term is 0. To show that the limit of 3/n2 is also 0, I can use the same argument, can't I?
     
  7. Nov 3, 2012 #6

    tiny-tim

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