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 $N \in \Bbb{N}$

Let $N \in \mathbb{N}$ s.t. $4N-2>n'+\delta$ for $n' \in \mathbb{N}$ and $\delta >0$, then $n \geq N$ implies $4n-2 \geq 4N-2 > n'+\delta$.
Let $N' \in \mathbb{N}$ s.t. $\frac{1}{N'^2}< \delta$ for $\delta >0$, then $n\geq N'$ implies $\frac{1}{n^2} \leq \frac{1}{N'^2} < \delta$.
If $n \geq \text{max}\{N, N'\}$ then $4n-2-\frac{1}{n^2} > n'+ \delta - \delta= n'$ and $\frac{n^2}{4n^3 -2n^2 -1} =\frac{1}{4n-2-\frac{1}{n^2}}<\frac{1}{n'}.$
Let $n'$ be such that $\frac{1}{n'}<\epsilon$.
