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Limit as n approaches infinity

  1. Apr 14, 2013 #1
    1. The problem statement, all variables and given/known data

    What is the limit of the given equations as n approaches infinity?

    (1 + 3n-1)/3n
     
  2. jcsd
  3. Apr 14, 2013 #2

    Dick

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    You are supposed to show an attempt to solve it. Try it. Break it into two fractions.
     
  4. Apr 14, 2013 #3
    lim 1/3n + lim 3n-1/3n
    n→∞ n→∞

    the second equation:

    you can cancel the 3n-1
    and the second equations turns out to be the same as the first so that equals:

    lim 1/3n
    n→∞


    so in approaches e?
     
  5. Apr 14, 2013 #4

    Dick

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    e? No, what does 1/3^n approach? Just think about it. And 3^(n-1)/3^n isn't the same as the first term. What is it?
     
  6. Apr 14, 2013 #5
    1/3n approaches 1/ a really big number.
    1 over a big number equals 0.




    (3n-1)/(3n)

    thinking about this equation and plugging in values like 2, 3, 4, 5, ect.

    you get

    31/32

    32/33

    33/34

    they all equal 1/3

    so 1/3?
     
  7. Apr 14, 2013 #6

    Dick

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    Yes. But you didn't need to plug numbers in. 3^n=3^(n-1)*3. So 3^(n-1)/3^n=1/3.
     
  8. Apr 14, 2013 #7
    so it the answer is zero?

    1/3 x 0
     
  9. Apr 14, 2013 #8

    HallsofIvy

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    Where did you get that "0" you are multiplying?
     
  10. Apr 14, 2013 #9
    the limit as x approaches infinity of 1/3n
     
  11. Apr 14, 2013 #10

    Dick

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    It's 0+1/3, isn't it?
     
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