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Find limit as n -> infinity

  1. Jan 17, 2008 #1
    1. The problem statement, all variables and given/known data

    Find limit as n -> infinity

    [ (n+1)^2 ] / [ [tex]\sqrt{}3+5n^2+4n^4[/tex] ]

    2. Relevant equations

    L'Hopital won't do the job, I think.

    3. The attempt at a solution

    It's something really small I'm just completely missing.
     
  2. jcsd
  3. Jan 17, 2008 #2
    divide the numerator and denominator by n^4
     
  4. Jan 17, 2008 #3
    I found an example that said exactly that, but I guess I don't see clearly how, first of all,

    the radical / n^4 comes out pretty.
     
  5. Jan 17, 2008 #4
    you have, (n+1)^2/(sqrt(3) + 5n^2 + 4n^4), or

    (n^2 + 2n + 1)/(sqrt(3) + 5n^2 + 4n^4), so dividing num and denom by n^4,

    (1/n^2 + 2/n^3 + 1/n^4)/(sqrt(3)/n^4 + 5/n^2 + 4) and now the limit as n->inf is ...
     
  6. Jan 17, 2008 #5
    Oh, I'm sorry. In the denominator, everything is actually under the radical. It doesn't matter, however, because the numerator comes out to zero. Thanks for the help.
     
  7. Jan 17, 2008 #6
    [tex]\sqrt{3}+5n^2+4n^4[/tex] OR [tex]\sqrt{3+5n^2+4n^4}[/tex] ?

    if the last one, then divide num and denom by n^2
     
    Last edited: Jan 17, 2008
  8. Jan 17, 2008 #7
    It's the last one, everything under the radical.

    My problem is I'm not seeing how I would divide something like

    [tex]\sqrt{3+5n^2+4n^4}[/tex] / n^4, or / any number, for that matter. I'm not seeing how to simplify that radical into individual terms.
     
  9. Jan 17, 2008 #8
    You should not simplify the radical, just put n^2 INSIDE the radical... and remember that n^2 becomes n^4 when inside radical...
     
  10. Jan 17, 2008 #9
    figured this would be your difficulty, here is an example
    sqrt(n^2 + 2n)/n^2 = sqrt(n^2 + 2n)/sqrt(n^4) = sqrt((n^2 + 2n)/n^4) = sqrt(1/n^2 + 2/n^3)

    try to see why it works, now mimic it for your problem.
     
  11. Jan 17, 2008 #10
    Ahh, I see it. I forgot about that property where you're allowed to simply "take out" multiplying/dividing terms. I see it now, thanks a bunch.
     
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