Limit of a sequence by definition I need to prove by definition that the limit of sequence :
$$a_n = \frac{5n^3-3n^2+1}{4n^3+n+2}$$
is $\dfrac54$ which means I need to show that :
$$\left|\frac{5n^3-3n^2+1}{4n^3+n+2} - \frac54\right| < \varepsilon$$
I tried for hours to solve it but could not,
Can anyone help please?
Thanks

Let $\varepsilon > 0$. Pick $n_0 \in \mathbb{N}$ such that $n_0 > \frac{23}{16\varepsilon}$.
For $n \ge n_0$ we have:
\begin{align}
\left|\frac{5n^3-3n^2+1}{4n^3+n+2} - \frac54\right| &= \left|\frac{4(5n^3-3n^2+1)-5(4n^3+n+2)}{4(4n^3+n+2)}\right|\\
&= \left|\frac{-12n^2-5n-6}{4(4n^3+n+2)}\right|\\
&= \frac{12n^2+5n+6}{4(4n^3+n+2)}\\
&\le \frac{12n^2+5n^2+6n^2}{4(4n^3 + 0 + 0)}\\
&= \frac{23n^2}{16n^3}\\
&= \frac{23}{16}\cdot\frac1n\\
&< \varepsilon
\end{align}
Therefore $$\lim_{n\to\infty} a_n = \frac54$$
