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  $\frac{n^3}2≥6$.


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 $n > 9$, $0 < (n^3 - 64n),$ which is a lower bound for the denominator.  
For $n > 9$, $(n-8) > 1 \implies$ that the numerator is 
less than $3(n+8)(n-8) = 3(n^2 - 64).$
Therefore, the fraction is less than
$$\frac{3(n^2 - 64)}{n^3 - 64n} = \frac{3}{n}.$$
Therefore, choose $n$ such that $n > 9$ and 
$\frac{3}{n} < \epsilon \implies \frac{3}{\epsilon} < n.$
