What's the limit of the string? 
Find $$\lim_{n\to\infty} (x_n\sqrt{n})^{\sqrt{n^2-1}},$$ where $ x_{n+1} = \frac{x_n}{\sqrt{1+x_n^2}}$ and $x_1 = 2$.

I showed $x_n \to 0$, $x_n\sqrt{n} \to 1$, but i don't know how to solve limit properly.

Later EDIT: In the initial post there was an error, the old computation was using $y_0=2$, instead of $y_{\color{red}1}={\color{red}4}$. 
Let the sequence $(y_n)$ be given by $y_n=x_n^2$. Then we have the Möbius transformation action giving the recursion:
$$
y_n =\begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix}\cdot y_{n-1}\ .
$$
Here, the action of the $2\times 2$-matrix with entries $a,b,c,d$ on an element in $z\in \Bbb R$ (or $\Bbb C$) is given by
$$
\begin{bmatrix} a & b\\ c&d \end{bmatrix}\cdot  z
=\frac {az+b}{cz+d}\ .
$$
This gives
$$
\begin{aligned}
y_n &= 
\begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix}^{n-1}\cdot y_1
=
\begin{bmatrix} 1 & 0\\ n-1 & 1 \end{bmatrix}\cdot 4
\\
&=
\frac {4}{4n-3}\text{ . So we have:}
\\
x_n &=
\sqrt{\frac {4}{4n-3}}\ .
\end{aligned}
$$
(In case the argument using Möbius transformations is to heavy, just use induction to show the above formula.)
So we have to compute the limit of
$$
(x_n\sqrt{n})^{\sqrt{n^2-1}}
=
\left(\sqrt{\frac {4n}{4n-3}}\right)^{\sqrt{n^2-1}}
=
\left(1+\frac 3{4n-3}\right)^{\frac 12\sqrt{n^2-1}}
=
\left(1+\frac 3{4n-3}\right)^{(4n-3)\frac 1{2(4n-3)}\sqrt{n^2-1}}
=
\left[\ \left(1+\frac 3{4n-3}\right)^{(4n-3)}\ \right]^{\frac 1{2(4n-3)}\sqrt{n^2-1}}
\ .
$$
The limit is now easy to compute, 
the $[\dots]$ expression goes to $e^3$, and its exponent to $\frac 18$, so
the limit is $\exp\frac 38$.
