Proving ${\lim\limits_ {n\to\infty}}\frac{6n^3+5n-1}{2n^3+2n+8} = 3$ I'm trying to show that $\exists \,\varepsilon >0\mid\forall n>N\in\mathbb{N}$ such that:
$$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| < \varepsilon$$
Let's take $\varepsilon = 1/2$:
$$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| = \left|\frac{6n^3+5n-1 - 6n^3 -6n -24}{2n^3+2n+8}\right| < \left|\frac{-n -25}{2n^3+2n+8}\right| = \left|-\left(\frac{n +25}{2n^3+2n+8}\right)\right|$$
Since $|-x| = |x|$:
$$\left|-\left(\frac{n +25}{2n^3+2n+8}\right)\right| = \left|\frac{n +25}{2n^3+2n+8}\right|<\varepsilon $$
This is the part which I have trouble with. Here, I always end up with my minimum-required index to be smaller than zero, which seems peculiar to me:
$$\left|\frac{n +25}{2n^3+2n+8}\right| < |n+25| = n+ 25 < \varepsilon  = 1/2$$
From here, I will get that my $N$ will be less than zero,  which means that all the elements of the sequence are inside of my given epsilon environment, but I know that's not true since $a_1 = 5/6 < 3 - \varepsilon $, so what I did do wrong? Is it because I completely removed the denominator? If so, why does that break the inequality?

Note that for $n\geq 1$ we have $n+25\leq n+25n=26n$ and $2n^3+2n+8\geq 2n^3$ so you have that $$\left|\frac{n+25}{2n^3+2n+8}\right|\leq \frac{26n}{2n^3}=\frac{13}{n^2}$$
Can you take it from here?
