$\lim\limits_{n \to\infty}\dfrac{n^{2}-n+2}{3n^{2}+2n-4}=\dfrac{1}{3}.$ Change $n$'s by $x$ and find $M$. $\lim\limits_{n\to\infty}\dfrac{n^2-n+2}{3n^2+2n-4}=\dfrac{1}{3}$.
With epsilon definition I get my answer as $N=\left[ \dfrac{5}{9\varepsilon }\right] +1$. But then I thought that how can I evaluate this sequence, in functions $\delta$-$\varepsilon$ definition by substituting $n$ by $x$. But I can't make it.
So here's my question:
What can I do is I have function like this one?
$\lim\limits_{x\to+\infty}\dfrac{x^2-x+2}{3x^2+2x-4}=\dfrac{1}{3}$.
I'm thinking about that function. But how to find.
$\forall \varepsilon>0$ be given, then $\exists M>0$ such that $x>M$ implies $| f\left( x\right) -l|<\varepsilon$
For function:
$\left| \dfrac{x^{2}-x+2}{3x^{2}+2x-4}-\dfrac{1}{3}\right|=
\left| \dfrac{-5x+10}{9x^{2}+6x-12}\right|<\left| \dfrac{-5x+10}{9x^{2}}\right|$
And this is where I stuck. Because $x\in \mathbb{R}$, so I think we can't remove the brackets. Isn't it?
For y'all's helps I solved it thank you all!

We have to prove that
$\forall\;\varepsilon>0\;\;\exists\;M>0\;$ such that $\;x>M\;$ implies $\;\left|f\left(x\right)-l\right|<\varepsilon\;.$
For any $\;\varepsilon>0\;,\;$ let $\;M=\max\left\{2,\dfrac{5}{9\varepsilon}\right\}>0\;.$
Moreover,
$x>M\ge2\implies|-5x+10|=5x-10<5x\;,$
$x>M\ge2\implies|9x^2+6x-12|=9x^2+6x-12>9x^2\;.$
Hence,
$x>M\implies\left| \dfrac{-5x+10}{9x^{2}+6x-12}\right|<\dfrac{5x}{9x^2}=\dfrac{5}{9x}\;.$
Consequently,
$x>M\implies|f(x)-l|=\left|\dfrac{x^{2}-x+2}{3x^{2}+2x-4}-\dfrac{1}{3}\right|=\left| \dfrac{-5x+10}{9x^{2}+6x-12}\right|<\dfrac{5}{9x}<\dfrac{5}{9M}\le\dfrac{5}{9\frac{5}{9\varepsilon}}=\varepsilon\;.$
