Prove by definition that $\frac{1-2n^3}{4n^3-2n^2-1} \to -\frac12$ $a_n = \dfrac{1-2n^3}{4n^3-2n^2-1}$
My attempt (of the scratch):
$\left| \dfrac{1-2n^3}{4n^3-2n^2-1} + \dfrac{1}{2}\right|=\left| \dfrac{2-4n^3+4n^3-2n^2-1}{8n^3-4n^2-2}\right|=\left|\dfrac{-2n^2+1}{8n^3-4n^2-2} \right|=\dfrac{2n^2-1}{8n^3-4n^2-2}<\dfrac{2n^2}{8n^3-4n^2-2}=\dfrac{n^2}{4n^3-2n^2-1}<\epsilon$
But I don't know what to do next to choose my $N \in \Bbb{N}$

For large $n$, we have $4n^3 - 2n^2 - 1 > n^3$, in which case $\frac{n^2}{4n^3-2n^2-1} < \frac{n^2}{n^3} = \frac{1}{n}$.
