<s>[INST] 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.
 [/INST] Strating from @dan_fulea nice solution
$$x_n=
\sqrt{\frac {4}{4n-3}}$$we could have nice approximation of
$$y_n=\big(x_n\sqrt{n}\big)^{\sqrt{n^2-1}}=\left(1+\frac 3{4n-3}\right)^{\frac 12\sqrt{n^2-1}}$$ Take logarithms
$$\log(y_n)={\frac 12\sqrt{n^2-1}}\,\log\left(1+\frac 3{4n-3}\right)$$
Now, use Taylor series for large values of $n$
$$\sqrt{n^2-1}=n-\frac{1}{2 n}-\frac{1}{8 n^3}+O\left(\frac{1}{n^5}\right)$$
$$\log\left(1+\frac 3{4n-3}\right)=\frac{3}{4 n}+\frac{9}{32 n^2}+\frac{9}{64 n^3}+O\left(\frac{1}{n^4}\right)$$
$$\log(y_n)=\frac{3}{8}+\frac{9}{64 n}-\frac{15}{128 n^2}+O\left(\frac{1}{n^3}\right)$$
$$y_n=e^{\log(y_n)}=e^{3/8}\left(1+\frac{9}{64 n}-\frac{879}{8192 n^2}+O\left(\frac{1}{n^3}\right) \right)$$
Just for the fun, using your pocket calculator for $n=10$; the exact value is $1.47381$ whle le above approximation gives $1.47389$.
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