how does one solve $\lim_{n\to \infty} (\frac{n^{2}-2n+1}{n^{2}-4n+2})^{n}$

Question

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

Answer 1

Lemma: If $g(n)>0$, $h(n)>0$ when $n \to \infty$ and $$\lim_{n \to \infty}g(n)=a, \lim_{n \to \infty}h(n)=b,$$ then $$\lim_{n \to \infty}g(n)^{h(n)}=a^b.$$

So

$$\lim_{n \to \infty}(\frac{n^2-2n+1}{n^2-4n+2})^n=\lim_{n \to \infty}(1+\frac{2n-1}{n^2-4n+2})^n=\lim_{n \to \infty}(1+\frac{2n-1}{n^2-4n+2})^{ {\frac{n^2-4n+2}{2n-1} (\frac{2n-1}{n^2-4n+2} n)}}=e^2$$.

Answer 2

Hint

After taking logarithm, observe that

$$n\ln(\frac{n^2-2n+1}{n^2-4n+2})=$$

$$n\ln(\color{red}{1}+\frac{2n-1}{n^2-4n+2})=$$

$$n\frac{2n-1 }{ n^2-4n+2}\frac{ \ln( \color{red}{1}+\frac{2n-1 }{n^2-4n+2 } ) }{ \frac{2n-1 }{ n^2-4n+2 } }.$$

Now, use the well-known limit

$$\lim_{X\to 0}\frac{\ln(\color{red}{1}+X)}{X}=1$$.

the limit is $e^2$.