# Homework Help: Limit of a sequence

1. Jul 29, 2008

### torquerotates

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. Jul 29, 2008

### Staff: Mentor

:surprised

3. Jul 29, 2008

### torquerotates

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)