proving limits exist Prove $\lim\frac{ 4n^3+3n}{n^3-6}= 4$.
I basically need to determine how large $n$ has to be to imply
$$\frac{3n+24}{n^3-6}<\epsilon$$
the idea is to upper bound the numerator and lower bound the denominator.
For example, since $3n + 24 ≤ 27n$, it suffices for us to get $\frac{27n}{n3-6} < ε$.
it is inferred, thereafter, that all we need is $\frac{n^3}2 ≥6$ or $n^3 ≥12$ or $n>2$.
this is where I have a problem. I don't understand how we got to the point where all we need is  $\frac{n^3}2≥6$.

If $\frac{n^3}2\ge 6$ then
$$n^3-6\ge n^3-\frac{n^3}2=\frac{n^3}2 $$
is your desired monomial lower bound for the denominator.

In fact, we are allowed to be wasteful and may begin right away with, say, assuming $n>1000$.
That makes $3n+24<3.024 n$ and $n^3-6>0.999999994n^3$ and so the error term certainly $<\frac{3.025}{n^2}$. But that doesn't matter for the sole purpose of proving the limit where any other bound of the form $<\frac C{n^2}$ (or even weaker) is good enough.
