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?

I do it like this:
$\forall n \ge 1, \; \dfrac{6n^3 + 5n - 1}{2n^3 + 2n + 8} = \dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}}; \tag 1$
since it is easy to see that
$\displaystyle \lim_{n \to \infty} (6 + 5n^{-2} - n^{-3}) = 6, \tag 2$
and
$\displaystyle \lim_{n \to \infty} (2 + 2n^{-2} + 8n^{-3} ) = 2, \tag 3$
we have
$\displaystyle \lim_{n \to \infty} \dfrac{6n^3 + 5n - 1}{2n^3 + 2n + 8} = \lim_{n \to \infty} \dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}} = \dfrac{\lim_{n \to \infty}( 6 + 5n^{-2} - n^{-3})}{\lim_{n \to \infty} ( 2 + 2n^{-2} + 8n^{-3})  } =   \dfrac{6}{2} = 3.  \tag 4$
Of course, such an answer, though perfectly rigorous and on point,  is bound to be less than satisfactory to readers who want, as the text of the question indicates, to see the $\epsilon$-$N$ mechanism operate in detail; what I have done here is simply invoke standard and elementary results on the behavior of limits; we may also cast things into $\epsilon$-$N$ form if we write
$\dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}} - 3$
$= \dfrac{ 6 + 5n^{-2} - n^{-3} - 3( 2 + 2n^{-2} + 8n^{-3})}{2 + 2n^{-2} + 8n^{-3}}$
$=\dfrac{5n^{-2} - n^{-3} - 6n^{-2} - 24n^{-3}}{2 + 2n^{-2} + 8n^{-3}} =  \dfrac{-n^{-2} - 25n^{-3}}{2 + 2n^{-2} + 8n^{-3}} ; \tag 5$
$\left \vert \dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}} - 3 \right \vert =  \left \vert \dfrac{-n^{-2} - 25n^{-3}}{2 + 2n^{-2} + 8n^{-3}} \right \vert; \tag 6$
it is evident via inspection of the right-hand side of this equation that, given any $\epsilon > 0$ there exists a sufficiently large $N \in \Bbb N$ that 
$n > N \Longrightarrow \left \vert \dfrac{-n^{-2} - 25n^{-3}}{2 + 2n^{-2} + 8n^{-3}} \right \vert < \epsilon, \tag 7$
which in light of (6) by definition implies
$\displaystyle \lim_{n \to \infty} \dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}} = 3, \tag 8$
completing an $\epsilon$-$N$ demonstration of the requisite limit.
It appears to me that our OP Fatima Ens' principal error lies in
confusing the roles of $\exists$ and $\forall$ in the definition of limit; the quantifying symbol string should read
$\forall \epsilon \exists N \mid \forall n > N \; \text{and so forth}; \tag 9$
if this correction is accepted and the remaining statements bought into accord with this (corrected) version, the chances of completing a successful proof are greatly enhanced.  
