$\epsilon$-$\delta$ limit proof that $\lim_{n\to \infty}\frac{n^2-n+2}{3n^2+2n-4}=\frac{1}{3}$ I need to prove that  $$\lim_{n\to \infty}\frac{n^2-n+2}{3n^2+2n-4}=\frac{1}{3}$$ using the epsilon definition. 
I'm having specific trouble understanding how to make it less than epsilon once I've simplified the equation. 
 A: Note that
$$
\begin{align}
\left|\frac{n^2-n+2}{3n^2+2n-4} - \frac{1}{3}\right|&=
\frac{1}{3}\left|\frac{3n^2-3n+6}{3n^2+2n-4} - 1\right|\\&= 
\left|\frac{-5n+10}{3(3n^2+2n-4)}\right|
\end{align}
$$
Given an $\epsilon>0$, we need to find an $n$ such that this expression is less than $\epsilon$.
A: $$
\left|\frac{n^2-n+2}{3n^2+2n-4} - \frac 1 3\right| = \left|\frac{3(n^2-n+2)}{3(3n^2+2n-4)} - \frac{3n^2+2n-4}{3(3n^2+2n-4)}\right|
$$
$$
= \left|\frac{-5n+10}{3(3n^2+2n-4)}\right| \le \frac{\text{some constant}}{n}
$$
Massage the whole thing a bit to figure out what "constant" can serve in this role, and then the whole thing is ${}<\varepsilon$ if $n$ is big enough.
PS:
We have
$$
\left|\frac{-5n+10}{3(3n^2+2n-4)}\right|.
$$
$$
3n^2 + 2n - 4.
$$
Let's try to show that this is $>n^2$ if $n$ is big enough.  We want
\begin{align}
3n^2+2n - 4 & > n^2 \\[6pt]
2n^2 + 2n - 4 & > 0 \\[6pt]
2\left(n^2+n+\frac 1 4 \right) - 4 - \frac 1 2 & > 0 \tag{completing the square} \\[6pt]
\left(n+\frac 1 2 \right)^2 & > \frac 9 4 \\[6pt]
n & > 1.
\end{align}
So we have
$$
\left|\frac{-5n+10}{3(3n^2+2n-4)}\right| < \frac{|-5n+10|}{n^2}\text{ if } n \ge 2,
$$
and
$$
|-5n+10| \le |-5n|+|10| = 5n+10 = 5(n+1). 
$$
Can we prove $\dfrac{5(n+1)}{n^2} \le \dfrac 6 n \text{ if }n>\text{something?}$
We would need $5(n+1)\le 6n$.  That yields $n\ge 6$.
We need $n\ge 2$ and $n\ge 6$.
So if $n\ge 6$ then the fraction above is $<\dfrac 6 n$.
(You should check details and adjust them as needed . . .)
Then if $n>\dfrac 6 \varepsilon$, you've got it.
A: Note that
$$\frac{n^2-n+2}{3n^2+2n-4}-\frac{1}{3}=\frac{3n^2-3n+6-3n^2-2n+4}{9n^2+6n-12}=\frac {-5n+10}{9n^2+6n-12}$$ is the expression you need to control.
Then you can make very simple estimates - ones which simplify the expression as much as possible. So if  $n\gt 2$ we have $9n^2+6n-12\gt9n^2$ and $5n-10\lt 9n$. That allows you to bound the absolute value of the difference by $\frac 1n$, which is all you need.
So - further to your edit and comments, you need $N\gt 2$ for the estimate to work, and you need $N$ big enough that $\epsilon \gt \frac 1N$ i.e. $N\gt \max(\frac 1{\epsilon},2)$, then for all $n\ge N$ your expression differs from $\frac 13$ by less than $\epsilon$.
A: just use l'hopitales rule differentiate the numerator to get 2n-1 then denominator to get 6n+2
and divide to get  2n-1/6n+2  seeing that this doesn't work do the same steps as the previous   and differentiate to get 2/6= 1/3 
