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?