Understanding convergence of $a_n := \frac{2n^3 +n^2 +3}{n^3-4}$ We proved by definition that the sequence $a_n := \frac{2n^3 +n^2 +3}{n^3-4}$ converges.
Let $\epsilon > 0$. Choose $N := \lceil \frac{24}{\epsilon} \rceil + 2 $.
Then for all $n \geq N$ it holds, that
$$|a_n - 2| = \big |\frac{2n^3+n^2+3}{n^3-4} - \frac{2n^3-8}{n^3-4} \big | = \big | \frac{n^2+11}{n^3-4} \big | = \frac{n^2+11}{n^3-4} \leq \frac{n^2+11}{n^3 - \frac{1}{2}n^3} \leq \frac{12n^2}{\frac{1}{2} n^3} = \frac{24}{n} < \epsilon $$
What I don't understand is how we get to $$- \frac{2n^3-8}{n^3-4} $$
Could someone explain? Thanks

I like to do these by dividing out the highest power of the variable.  You get $\dfrac{2n^3+n^2+3}{n^3-4}=\dfrac{2+1/n+3/n^3}{1-4/n^3}\to2/1=2$.
