## Formal definition of limits of sequences

Suppose that $a_n=\dfrac{n^2-2n+1}{2n^2+4n-1}$

1) Find $L=\lim_{n \to \infty} a_n$

I found this to be $\frac{1}{2}$

2) For each positive number $\epsilon$ find a number $N$ such that:

$|a_n-L|<\epsilon$ whenever $n>N$

So far I've did this

$\left|\dfrac{n^2-2n+1}{2n^2+4n-1}-\dfrac{1}{2}\right|<\epsilon$

$\left|\dfrac{2n^2-4n+2-(2n^2+4n-1)}{4n^2+8n-2}\right|<\epsilon$

$\left|\dfrac{-8n+1}{4n^2+8n-2}\right|<\epsilon$

Not too sure what to do next?