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Confusing limit

  1. Dec 8, 2004 #1
    My book tolds me that: [tex]\lim_{n\rightarrow\infty} \frac{n(n+1)}{2n^2}= \frac{1}{2}[/tex]. I don't get it. Maybe this with infinity, I dunno... Please, help!
     
  2. jcsd
  3. Dec 8, 2004 #2

    quasar987

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    Because

    [tex] \frac{n(n+1)}{2n^2} = \frac{n^2+n}{2n^2} = \frac{n^2}{2n^2} +\frac{n}{2n^2} = \frac{1}{2} + \frac{1}{2n} [/tex]
     
  4. Dec 8, 2004 #3
    Ok, I think I got it now. So it's based on that 1/n equals 0, when n approaches infinity and 2 * infinity = infinity, right?
     
  5. Dec 8, 2004 #4
    yeah, the 1/(2n)=(1/2)(1/n) so you can factor the constant out of the limit, ie lim (1/(2n)) = (1/2) * lim (1/n) --> 0 as n --> infinity
     
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