This one is worth doing

$\left(1-\frac{1}{\frac{n(n+1)}{2}}\right) = \frac{n^2+n-2}{n\left(n+1\right)}=\frac{n-1}{n} \cdot\frac{n+2}{n+1}$

$x_{n} = \displaystyle \prod_{n=1}^{n}\left(\frac{n-1}{n} \cdot \frac{n+2}{n+1}\right)^2$

$\implies \frac{\large \ln x_{n}}{2} = \displaystyle \sum_{n=2}^{n} \ln \left(\frac{n-1}{n} \cdot \frac{n+2}{n+1}\right) = \displaystyle \sum_{n=2}^{n}\left( \ln \frac{n-1}{n} - \ln\frac{n+2}{n+1} \right)$

$= \ln \frac{1}{2}+\ln\frac{2}{3}-\ln \frac{n}{n+1} - \ln\frac{n+1}{n+2}$

$\lim_{n \to \infty} x_{n} = 2 \lim_{n \to \infty}\left( \ln \frac{1}{2}+\ln\frac{2}{3}-\ln \frac{n}{n+1} - \ln \frac{n+1}{n+2}\right)=2 \left[ \ln \left(\frac{1}{2}\times \frac{2}{3}\right)- \lim_{n \to \infty} \left(\ln\frac{n}{n+1}\cdot \frac{n+1}{n+2}\right)\right]$

$=2\left( \ln \frac{1}{3}-\lim_{n \to \infty}\ln\frac{n}{n+2}\right)$

Since as $n \to \infty$ , $\frac{n}{n+2} \to 1$ and thus $\ln \frac{n}{n+2} \to 0$

Putting this our expression equals $2\ln\left(\frac{1}{3}\right)=\ln\left(\frac{1}{3}\right)^2= \ln\left(\frac{1}{9}\right)$

$\implies \lim_{n \to \infty} x_{n} =\boxed{\frac{1}{9}}$