Proof of Limit Using Unusual Definition Let $\lbrace a_n \rbrace$ be a real sequence.  We say $\lim_{n\to\infty} a_n=\infty$ provided that:
$$\forall K>0, \exists N\in \mathbb{N} \forall n \ge N:a_n>K$$
Use this definition, as well as the Archimedean property, to prove that:
$$\lim_{n\to\infty}n^3-4n^2-99n=\infty$$
In this context, the Archimedean property is defined as:
$$\forall c\in\mathbb{R} \exists n\in\mathbb{N}:n>c$$
I'm a little lost on how to use this particular definition of the limit.  Perhaps if the roots were easier, it wouldn't be so difficult, but I'm not making progress even when I try simpler versions.  So how could you prove this using that definition in a fairly simple way?

Note that
$$
n^3-4n^2-99n=n(n^2-4n-99)
$$
and that $n^2-4n-99=n(n-4)-99>1$ as soon as $n>12$. Thus, for $n>12$ you have $n^3-4n^2-99n>n$.
