I'm trying to work out this limit: $$ \lim_{n \to \infty} \left(\frac{n^2 - 2n + 1}{n^2-4n+2}\right)^n = e^2 $$  I have managed to show that it is equal to: $$\lim_{n \to \infty}\left (1+\frac{2n-1}{n^2-4n+2}\right)^n$$ but I have no idea how to continue from here, help would be appreciated.
I cant use lhopital's rule here is my attempt:  Limit of $(\frac{n^{2}-2n+1}{n^{2}-4n+2})^{n}$ =$e^{\lim_{n\to \infty} n\times \ln(\frac{1 -2-n +\frac{1}{n^{2}}}{1 -\frac{4}{n} +\frac{2}{n^{2}}})}$ I got stuck here
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