1. Limited time only! Sign up for a free 30min personal 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!

Infinite limits on a sequence

  1. Apr 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Well, Hi again. Here I got some limits to infinity that I don't know how to solve. The statement just ask me to calculate those limits, if exists, for the next sequences.

    [tex]\displaystyle\lim_{n \to{+}\infty}{(\sqrt[n]{n+1})}[/tex]

    [tex]\displaystyle\lim_{n \to{+}\infty}{\displaystyle\frac{\sin(n)}{n+1}}[/tex]


    [tex]\displaystyle\lim_{n \to{+}\infty}{\displaystyle\frac{1-\cos(5n)}{n^2}}[/tex]

    I don't pretend that someone solve all of these for me, but I need some orientation on which way to look. I'll be thanked for any help on any of those sequences.

    Edit: Deleted the ones that I could solve. Someone gave me some tips, and I could solve some of the limits that I have posted before. As there wasn't any responses to the topic I thought that it was unnecessary to repost giving the advertisement.
     
    Last edited: Apr 21, 2010
  2. jcsd
  3. Apr 21, 2010 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hints: Think about logarithms on the first and think about how big sines and cosines can be on the others.
     
  4. Apr 21, 2010 #3
    I resolved the trigonometric ones. Thanks for the hints, I'll keep working on the first.

    I established on the last two that sine and cosine are between 1 and -1, so both tend to zero.

    On the first, is this valid? [tex](\sqrt[n]{n+1})=\displaystyle\frac{log (n+1)}{n}\[/tex]
     
    Last edited: Apr 21, 2010
  5. Apr 21, 2010 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, those two quantities aren't equal are they? But you could call

    [tex] y = \left({n+1}\right)^{\frac 1 n}[/tex]

    and then say

    [tex] \ln(y) = \frac{log (n+1)}{n}[/tex]

    and work with ln(y).
     
  6. Apr 21, 2010 #5
    Thanks, I'll remember that, because it looks like the legs of a woman (y) :D

    How did you realize that applying the ln to the expression gives the same function?

    ln(y) tends to infinite really fast. How do I get the relation between what happens with ln(y) and y?
     
    Last edited: Apr 21, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook