1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to find the bounds of this siquence

  1. Dec 2, 2008 #1
    (n^2+2)^0.5 - (n)^0.5

    i thought of doing a limit where n->infinity

    but here i get undefined form and even if i whould get some finite limite
    it will only be one bound

    and i cant do limit n->-infinity because its a sequence must be positive??
     
  2. jcsd
  3. Dec 2, 2008 #2

    Mark44

    Staff: Mentor

    It's not bounded, so you're going to have a tough time finding a bound for it. Of the two terms, the first is dominant and is approximately n. n grows large more quickly than sqrt(n).

    [tex]\lim_{n \rightarrow \infty} \sqrt{n^2 + 2} - \sqrt{n} = \infty[/tex]
     
  4. Dec 2, 2008 #3
    Factor out a [tex]\sqrt(n^2)[/tex] to see that it increases beyond all positive bounds.

    [tex]\lim_{n \rightarrow \infty} \sqrt(n^2 +2) - \sqrt(n) = \lim_{n \rightarrow \infty}\sqrt(n^2)(\sqrt(1 + 2/n^2) - \sqrt(\frac{1}{n})) [/tex]

    In the second bracket, the first term converges to 1 and the second term converges to 0. So we now have [tex]\lim_{n \rightarrow \infty} \sqrt(n^2)*1 = \infty[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?