How Do I Find $x_n$ First?

We have $x_nx_{n+1}=2n$ and $x_{n+1}x_{n+2}=2(n+1)$ . Dividing the second by the first, we obtain $\frac{x_{n+2}}{x_{n}}=\frac{n+1}{n}$ Using the above and the given initial condition and iterating $\sim n$ times, we have $x_{2n+1}=\frac{2^{2n}}{\binom{2n}{n}}, x_{2n}=\frac{4n\binom{2n}{n}}{2^{2n}}.$ Using Stirling's formula , we have the following asymptotic relation $\binom{2n}{n}\asymp \frac{2^{2n}}{\sqrt{n\pi}}$ Substituting it, we obtain $\lim_{n\to \infty}\frac{|x_{2n+1}-x_{2n}|}{\sqrt{2n}}=\frac{1}{\sqrt{2}}(\sqrt{\pi}-\frac{1}{\sqrt{\pi}})\approx 0.342$

I used a spreadsheet to observe the behavior of the numbers and the equation. The limit converged fairly slowly, so I averaged two adjacent values of the equation and came out with $\boxed{.342}$ .