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

$|f(n)|:=|\dfrac{5n^3-3n^2+1}{4n^3+n+2} -5/4| =$
$|\dfrac{-12n^2 -5n-6}{4(4n^3+n+2)}| \lt$
$|\dfrac{12n^2 +5n+6}{16n^3}| \lt$
$|\dfrac{12n^2+5n^2+6n^2}{n^3}| =$
$|\dfrac{23n^2}{n^3}| = |\dfrac{23}{n}|.$
Let $\epsilon >0$ be given:
Choose $M \gt 23/\epsilon,$ with $M$, real.
There is a $n_0 \gt M$.(Archimedes)
Then for $n \ge n_0:$
$|f(n)| \lt 23/n \le 23/n_0 \lt 23/M $
$\lt \epsilon.$
