Let $\lbrace a_n \rbrace$ be a sequence; then $\lim_{n\to\infty} a_n=\infty$ provided that:
$$\forall K>0, \exists N\in \mathbb{N} \forall n \ge N:a_n>K$$
Use this definition to prove that:
$\lim_{n\to\infty}\frac{n^3-6n^2+1}{n^2+5}=\infty$
I've tried about as many ways to get this to work as I can think of, but nothing ever ends up working out. I can't decide the lower bound for $N$ to choose so that I end up with the sequence always being greater than $K$. It's just a really annoying fraction to work with. Unfortunately I'm stuck and can't figure out the algebraic manipulation.
You want $\lim_{n\to\infty}\frac{n^3-6n^2+1}{n^2+5}=\infty $.
If $f(n) =\frac{n^3-6n^2+1}{n^2+5} $, then $n^3-6n^2+1 =n^3(1-6/n)+1 \gt n^3/2 $ for $n \gt 12$ and $n^2+5 =n^2(1+5/n^2) \lt 2n^2$ for $n \ge 3$.
Therefore, for $n \gt 12$, $f(n) =\frac{n^3-6n^2+1}{n^2+5} \gt \frac{n^3/2}{2n^2} =\frac{n}{4} $.
Therefore, to get $f(n) > K$ it is enough to choose $n > \max(12, 4K)$.
Note that the inequalities can be very crude; to prove divergence (or convergence if that is what holds) you do not need to get anywhere near the best possible bounds.