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!

Homework Help: Limit of a sequence

  1. Jul 29, 2008 #1
    1. The problem statement, all variables and given/known data
    prove Lim(n->infinity) (n+6)/(n^(2)-6)=0



    2. Relevant equations

    lim{a}=a for n->infinity

    For any epslion>0 there is a N>0 such that n>N => |{a}-a|<epslion

    3. The attempt at a solution

    For n>or=3, |(n+6)/(n^(2)-6)|<epslion becomes( dropping the absolute value sign) (n+6)/(n^(2)-6)< epslion

    But n+6<n and n^(2)-6>or equal (1/3)n^(2)

    so 3/n=n/((1/3)n^(2))<(n+6)/(n^(2)-6)<epslion

    both n>or=3 and n>3/epslion implies that I make N=Max(3,3/epslion)

    Is this correct? The book had a different way of doing it but since there are many ways of proving things, the book's solution doesn't help.
     
  2. jcsd
  3. Jul 29, 2008 #2

    Borek

    User Avatar

    Staff: Mentor

    :surprised
     
  4. Jul 29, 2008 #3
    lol ok. ( fixing it)

    For n>or=3, |(n+6)/(n^(2)-6)|<epslion becomes( dropping the absolute value sign) (n+6)/(n^(2)-6)< epslion

    But n+6>n and n^(2)-6< n^(2)

    so 1/(n)=n/(n^(2))<(n+6)/(n^(2)-6)<epslion

    both n>or=3 and n>1/epslion implies that I make N=Max(3,1/epslion)

    how about now?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...