<s>[INST] Calculate the limit $\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$ $$\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$$
My attempt
\begin{align}
\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)} &=  \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-n+1)/(n+1)} \\
&= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{1 + (-2n)/(n+1)} \\
&= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)\left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)}
\end{align}
since $\left(\frac{n^2+1}{n^2} \right) \to 1$ when $n\to\infty$
$$\lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)\left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)} = \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)}$$
Also I tried to make similar simplifications into the braсkets but nothing happens and no proof that limit $= 1$ or whatever.
And here is a rule for the task. This is a limit of a sequence so no usage of functional simplifications and derives are allowed. If you have really beautiful solution for the task then post it anyway.
 [/INST] Hints:


*

*For $0 \lt k \lt 1$ and $0 \lt x \lt 1$ you have $x \lt x^k \lt 1$

*$ \displaystyle \lim_{n\to\infty} \bigg(\dfrac{n^2}{n^2 + 1} \bigg) = \lim_{n\to\infty} \bigg(1-\dfrac{1}{n^2 + 1} \bigg) $
</s>