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.

$\frac{n^2}{n^2+1}=1-\frac{1}{n^2+1}$
As $n \to \infty$,
$\because(1-\frac{1}{n^2+1})^\frac{n-1}{n+1} > 1^\frac{n-1}{n+1} = 1$
$(1-\frac{1}{n^2+1})^\frac{n-1}{n+1} < (1-\frac{1}{n^2+1})^\frac{n}{n}=1-\frac{1}{n^2+1}$
$\because \lim \limits_{n \to \infty} 1 = 1$
$\lim \limits_{n \to \infty} (1-\frac{1}{n^2+1}) = 1$
$\therefore$ By Squeeze Theorem, $\lim \limits_{n \to \infty} \frac{n^2}{n^2+1} = 1$
