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How to evaluate the following limit? $$ \lim_{n\to \infty }(\frac{n^{2}+1}{n-1})^{-2n^{2}} $$

I have tried the following: $$ \begin{align} \lim_{n\rightarrow \infty }(\frac{n^{2}+1+2n-2n}{n-1})^{-2n^{2}} &= \lim_{n\rightarrow \infty}(\frac{(n-1)^{2}+2n}{n-1})^{-2n^{2}}\\ &= \lim_{n\rightarrow \infty}(\frac{(n-1)^{2}}{n-1}+\frac{2n}{n-1})^{-2n^{2}}\\ &= \lim_{n\to \infty }({n-1}+\frac{2n}{n-1})^{-2n^{2}} \end{align} $$
I know that: $$ \lim_{n\rightarrow \infty }(1+\frac{1}{n})^{n}=e $$ But I can't come to that form in my equation. Can you help me, please?

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$$\lim_{n\to\infty}\left(\dfrac{n^2+1}{n-1}\right)^{-2n^2}=\dfrac1{\lim_{n\to\infty}\left(n+1+\dfrac2{n-1}\right)^{2n^2}}=?$$

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Note that: $$\left(\frac{n^{2}+1}{n-1}\right)^{-2n^{2}}=\left(\frac{n-1}{n^{2}+1}\right)^{2n^{2}}<\left(\frac1n\right)^{2n^2}\to 0.$$

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We have that

$$\left(\frac{n^{2}+1}{n-1}\right)^{-2n^{2}}=n^{-2n^2}\left(\frac{n+\frac1n}{n-1}\right)^{-2n^{2}}=n^{-2n^2}\left(\frac{n-1+1+\frac1n}{n-1}\right)^{-2n^{2}}=$$

$$=n^{-2n^2}\left(1+\frac{n+1}{n(n-1)}\right)^{-2n^{2}}=n^{-2n^2}\left[\left(1+\frac{n+1}{n(n-1)}\right)^{\frac{n(n-1)}{n+1}}\right]^{\frac{-2n^{2}(n+1)}{n(n-1)}}=$$$$=\frac1{n^{2n^2}\left[\left(1+\frac{n+1}{n(n-1)}\right)^{\frac{n(n-1)}{n+1}}\right]^{\frac{2n^{2}(n+1)}{n(n-1)}}}\sim \frac{1}{n^{2n^2}e^{2n}}$$

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  • $\begingroup$ Great solution, thank you very much. $\endgroup$ – violettagold Nov 23 '18 at 13:16
  • $\begingroup$ @violettagold It can be simplified a lot, I did in that way in order to use the knwn limit for $e$ but it is not necessary of course. $\endgroup$ – gimusi Nov 23 '18 at 13:50
  • $\begingroup$ Yes, I have noticed. I had simplified my answer at the end. $\endgroup$ – violettagold Nov 23 '18 at 15:27

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