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!

Limits help

  1. Nov 14, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the limit as n tends to infinity of xn = (n^2 + exp(n))^(1/n)

    2. Relevant equations

    maybe use ( 1 + c/n )^n tends to exp(c)

    3. The attempt at a solution

    I know that inside the barckets are both inceasing and the 1/n makes it decrease but how do i find out which is stronger and what the limit is?
     
  2. jcsd
  3. Nov 14, 2007 #2

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    Try finding the limit of the ln of the function.

    Tip: ln(a^b) = b ln(a)
     
  4. Nov 14, 2007 #3
    i already tried that and i cant see how it helps, (1/n)ln(n^2 + exp(n)) has the same problem...
     
  5. Nov 14, 2007 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Try l'Hopital's rule, if you know that.
     
  6. Nov 14, 2007 #5
    i have a feeling im not allowed to use it, is there another way?
     
  7. Nov 14, 2007 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    What you really need to know is that n^2/exp(n)->0 as n->infinity. There are a variety of ways to show that - try to think of one. Once you done that then ln(exp(n)+n^2)=ln(exp(n)*(1+n^2/exp(n))=ln(exp(n))+ln(1+n^2/exp(n)) etc.
     
  8. Nov 14, 2007 #7
    [tex](n^2+e^n)^{1/n} = e(1+n^2/e^n)^{1/n}[/tex] now it is trivial but the fact that [tex](1+x_n/n)^{1/n} \to e^x[/tex] if [tex]x_n\to x[/tex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Limits help
  1. Help with Limits (Replies: 5)

  2. Help with limits (Replies: 4)

  3. Help with a limit (Replies: 3)

  4. Help with a limit. (Replies: 1)

  5. Limit help (Replies: 3)

Loading...