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.

For all $n$, we have $0 < \frac{n^2}{n^2 + 1} < 1.$
For $n > 1$, we have $\frac{n-1}{n+1} > 0.$
Therefore, for $n > 1$,
$$ \left(\frac{n^2}{n^2 + 1}\right)^{(n-1)/(n+1)} < 1.$$
On the other hand, $\frac{n^2 + 1}{n^2} > 1$ for all $n$, and for $n > -1$ we have $\frac{2}{n+1} > 0$, so
\begin{align}
\left(\frac{n^2}{n^2 + 1}\right)^{(n-1)/(n+1)}
&= \left(\frac{n^2}{n^2 + 1}\right) \cdot
   \left(\frac{n^2}{n^2 + 1}\right)^{-2/(n+1)} \\
&= \left(\frac{n^2}{n^2 + 1}\right) \cdot
   \left(\frac{n^2 + 1}{n^2}\right)^{2/(n+1)} \\
&> \frac{n^2}{n^2 + 1}. \\
\end{align}
In short, for any $n > 1$,
$$ \frac{n^2}{n^2 + 1} < \left(\frac{n^2}{n^2 + 1}\right)^{(n-1)/(n+1)} < 1.$$
Finding the limit should now be simple enough.
