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?

Don't worry about $\epsilon$.
Instead,
try to get the difference
in a simple form
by assuming $n$
is as large as you need.
Then getting $n$
is much simpler.
Using your calculations:
$\begin{array}\\
\left|\dfrac{6n^3+5n-1}{2n^3+2n+8}-3\right| 
&= \left|\dfrac{6n^3+5n-1 - 6n^3 -6n -24}{2n^3+2n+8}\right|\\
&= \left|\dfrac{-n -25}{2n^3+2n+8}\right|\\
&= \left|\dfrac{n +25}{2n^3+2n+8}\right|\\
&\le \left|\dfrac{2n}{2n^3}\right|
\qquad\text{if } n \ge 25\\
&= \left|\dfrac{1}{n^2}\right|\\
&\lt \epsilon
\qquad\text{if } n \gt \frac1{\sqrt{\epsilon}}\\
\end{array}
$
Therefore the difference
is within $\epsilon$
if
$n \gt \max(25, \frac1{\sqrt{\epsilon}})$.
You don't have to get
the best possible $n$ - 
just showing one exists
is good enough.
