$\epsilon$-$\delta$ 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. 

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 $\epsilon$.
