Prove Limit of Sequence: (n+6)/(n^2-6)=0

[torquerotates]

Homework Statement

prove Lim(n->infinity) (n+6)/(n^(2)-6)=0

Homework Equations

lim{a}=a for n->infinity

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

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.

 

[Borek]

But n+6<n

 

[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)

how about now?