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.

Using $x=e^{\ln x}$ we take the log. So the limit is the exponential of the limit of the log, which is the limit of $\frac{n-1}{n+1}\ln(1-\frac{1}{n^2+1}$. That fraction goes to zero, so the log is asymptotic to it, hence the limit is the limit of $\frac{n-1}{n+1}\frac{1}{n^2+1}$, which is easily seen to be 0. Hence yout limit is 1.
Sorry for the wordy answer but typing the passages in MathJax from my mobile... I decided to spare myself such an ordeal :). If I remember, I will edit this tomorrow.
There is, however, a far easier way: the base and exponent both tend to 1, and $1^1$ is no indeterminate form, it is precisely 1.
Update
Here are the above passages in symbols:
$$\lim_{n\to\infty}\left(\frac{n^2}{n^2+1}\right)^{\frac{n-1}{n+1}}=\lim_{n\to\infty}e^{\ln\left(\left(\frac{n^2}{n^2+1}\right)^{\frac{n-1}{n+1}}\right)}=e^{\lim_{n\to\infty}\frac{n-1}{n+1}\ln\frac{n^2}{n^2+1}}=e^A,$$
and:
\begin{align*}
A={}&\lim_{n\to\infty}\frac{n-1}{n+1}\ln\left(1-\frac{1}{n^2+1}\right)=\lim_{n\to\infty}\frac{n-1}{n+1}\left(-\frac{1}{n^2+1}\right)={} \\
{}={}&-\left(\lim_{n\to\infty}\frac{n-1}{n+1}\right)\cdot\left(\lim_{n\to\infty}\frac{1}{n^2+1}\right)=-1\cdot0=0,
\end{align*}
hence, the desired limit is:
$$e^A=e^0=1.$$
Faster way:
$$\lim_{n\to\infty}\left(\frac{n^2}{n^2+1}\right)^{\frac{n-1}{n+1}}=\left(\lim_{n\to\infty}\frac{n^2}{n^2+1}\right)^{\lim_{n\to\infty}\frac{n-1}{n+1}}=1^1=1.$$
