As $x_{n+1}=(1-\frac{1}{2n})x_{n}+\frac{1}{2n}x_{n-1}$, we can get that:
$\begin{align*} x_{n+1}-x_{n}&=-\frac{1}{2n}(x_{n}-x_{n-1})\\ x_{n}-x_{n-1}&=-\frac{1}{2(n-1)}(x_{n-1}-x_{n-2})\\ &\vdots\\ x_{2}-x_{1}&=-\frac{1}{2}(x_{1}-x_{0})\\ \end{align*}$
then,by computation $x_{n+1}-x_{n}=\frac{(-\frac{1}{2})^{n}}{n!}(b-a)$
$\begin{align*} x_{n}&=\sum_{k=0}^{n-1}(x_{k+1}-x_{k})+x_{0}\\ &=a+(b-a)\sum_{k=0}^{n-1}\frac{(-\frac{1}{2})^{k}}{k!}\\ \end{align*}$
$\begin{align*} \lim_{n\rightarrow\infty}x_{n} &=a+(b-a)\sum_{k=0}^{\infty}\frac{(-\frac{1}{2})^{k}}{k!}\\ &=a+(b-a)e^{-\frac{1}{2}}\\ &=(1-e^{-\frac{1}{2}})a+e^{-\frac{1}{2}}b \end{align*}$