I want to prove $\displaystyle \lim_{n \rightarrow \infty}n^3 - 4n^2 -100n = \infty$ using the definition $\forall c \exists N \in \mathbb{N} ,s.t.\ a_n>c\ \forall n\geq N$.
I'm not having the best luck in thinking about this one. I think that I cannot use the limit sum properties if the sequence goes to infinity. So I can't look at the parts. I know that like convergence I assume that I'm given $c$ and show that I can find an N that makes the inequality hold. But the function $n^3 - 4n^2 -100n$ has a sign change around n = 12. I'm not sure how that affects things if it does at all.
Anyways. If I get $c$ from someone and look at $N > (c+13)$. Then I say $n^3 - 4n^2 -100n > N^3 - 4N^2 -100N = (c+13)^3 - 4(c+13)^2 -100(c+13)$ and then $(c+13)^3 - 4(c+13)^2 -100(c+13)=c^3 + 35 c^2+303 c+ 221 > c$ Does this work or have I screwed up somewhere?