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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?
     
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