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.

Given that $n^2/(n^2+1) < 1$, and $1>\frac{n-1}{n+1} >0$ for $n>1$, we can upper bound $\left(\frac{n^2}{n^2+1} \right)^{\frac{n-1}{n+1}} \leq 1^{\frac{n-1}{n+1}} \leq 1$. So, the limit is upper bounded by $1$. 
Now, note that we have the lower bound $\left(\frac{n^2}{n^2+1} \right)^{\frac{n-1}{n+1}} \geq \frac{n^2}{n^2+1}$ since we have (something smaller than 1)^(something smaller than 1). The limit of $\frac{n^2}{n^2+1}$ is $1$. So the limit is lower bounded by $1$.
The upper and lower bounds match, so its limit is $1$. 
