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

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?

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

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

We have that

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

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

• Great solution, thank you very much. – violettagold Nov 23 '18 at 13:16
• @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. – gimusi Nov 23 '18 at 13:50
• Yes, I have noticed. I had simplified my answer at the end. – violettagold Nov 23 '18 at 15:27