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Limits and log

  1. Oct 21, 2007 #1
    Please help, i take out n^2 top and bottom so end up with 0 as demominator....

    Find lim (n to infinity) xn

    xn = (n^2 + log n)/(2n^3 - 1)^(1/2)


    ...?
     
  2. jcsd
  3. Oct 21, 2007 #2
    You won't get a 0 in the denominator. You forgot to divide the 1 by n^2.
     
  4. Oct 21, 2007 #3
    if i divide the bottom by n^2 i get (1/n + 1/n^4)
     
  5. Oct 21, 2007 #4
    sorry... (1/n - 1/n^4)^(1/2)
     
  6. Oct 21, 2007 #5
    damn, sorry again, i mean (2/n - 1/n^4)^(1/2)
     
  7. Oct 21, 2007 #6
    which becomes 0 as n tends to infinity?
     
  8. Oct 22, 2007 #7

    Gib Z

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    [tex]\lim_{n\to \infty} x_n = \lim_{n\to\infty} \left( \frac{ 1+ \frac{\log n}{n^2} }{ \sqrt{ \frac{2n^3-1}{n^4}}} \right)[/tex]

    Consider separately, what is the numerator tending towards? How about the denominator?
     
    Last edited: Oct 22, 2007
  9. Oct 22, 2007 #8
    Rationalize the denominator

    Try rationalizing the denominator first. Then things get much easier.
     
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