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Limit ln(n-1/n+1) as n->infinity

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi! I need help confirming the limit of ln(n-1/n+1) as n->infinity.

    If you multiply top and bottom of the quotient by 1/n you'd end up with ln(1) = 0, no? I must be missing something rather simple here because my hp50 won't even compute. Thanks!
    1. The problem statement, all variables and given/known data
     
  2. jcsd
  3. Oct 6, 2012 #2

    Dick

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    Yes, you did it correctly. I'm not sure why the hp50 has problems.
     
    Last edited: Oct 6, 2012
  4. Oct 6, 2012 #3
    Thank you! After getting your response I looked further into the problem with the hp and figured it out. I had a 1. instead of 1 (sans decimal), and it won't take a limit with the decimal because it's a "real" number. This was of much help.
     
  5. Oct 7, 2012 #4

    Ray Vickson

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    You should realize that what you have written is ln[n + 1 - 1/n], which has no limit. Did you really mean ln[(n-1)/(n+1)]? If so, use brackets!

    RGV
     
  6. Oct 7, 2012 #5
    Yes ln[(x-1)/(x+1)] is what I actually meant, thanks for correcting my mistake.
     
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