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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.