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.
 A: $\frac{n^2}{n^2+1}=1-\frac{1}{n^2+1}$
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, $\lim \limits_{n \to \infty} \frac{n^2}{n^2+1} = 1$
A: 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 $1$. 
A: 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:
$$\lim_{n\to\infty}\left(\frac{n^2}{n^2+1}\right)^{\frac{n-1}{n+1}}=\left(\lim_{n\to\infty}\frac{n^2}{n^2+1}\right)^{\lim_{n\to\infty}\frac{n-1}{n+1}}=1^1=1.$$
A: 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) $
A: 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.
