<s>[INST] $\epsilon$-$\delta$ 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. 
 [/INST] $$
\left|\frac{n^2-n+2}{3n^2+2n-4} - \frac 1 3\right| = \left|\frac{3(n^2-n+2)}{3(3n^2+2n-4)} - \frac{3n^2+2n-4}{3(3n^2+2n-4)}\right|
$$
$$
= \left|\frac{-5n+10}{3(3n^2+2n-4)}\right| \le \frac{\text{some constant}}{n}
$$
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.
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