用数学归纳法证明：$ 2\leqslant x_n<x_{n+1}< 3 $．
（i）当 $ n=1 $ 时，$ x_1=2 $，直线 $ PQ_1 $ 的方程为\[ y-5={\dfrac{f\left(2\right)-5}{2-4}}\left(x-4\right),\]令 $ y=0 $，解得\[x_2={\dfrac{11}{4}} ,\]所以 $ 2\leqslant x_1<x_2< 3$．
（ii）假设当 $ n=k $ 时，结论成立，即 $ 2\leqslant x_k<x_{k+1}< 3 $，直线 $ PQ_{k+1} $ 的方程为\[ y-5={\dfrac{f\left(x_{k+1}\right)-5}{x_{k+1}-4}}\left(x-4\right),\]令 $ y=0 $，解得\[ x_{k+2}={\dfrac{3+4x_{k+1}}{2+x_{k+1}}}.\]由归纳假设知\[\begin{split} x_{k+2}&={\dfrac{3+4x_{k+1}}{2+x_{k+1}}}\\&=4-{\dfrac{5}{2+x_{k+1}}}\\&<4-{\dfrac{5}{2+3}}\\&=3;\\x_{k+2}-x_{k+1}&={\dfrac{\left(3-x_{k+1}\right)\left(1+x_{k+1}\right)}{2+x_{k+1}}}>0,\end{split} \]即 $ x_{k+1}<x_{k+ 2}$．
所以 $ 2\leqslant x_{k+1}<x_{k+2}<3 $，即当 $ n=k+1 $ 时，结论成立．
由（i）（ii）知对任意的正整数 $ n$，$2\leqslant x_n<x_{n+1}<3$．

由（1）及题意得\[ x_{n+1}={\dfrac{3+4x_n}{2+x_n}}. \]设 $ b_n=x_n-3 $，则\[ \begin{split} \dfrac{1}{b_{n+1}} &= \dfrac{5}{b_n}+1,\\ {\dfrac{1}{b_{n+1}}}+{\dfrac{1}{4}}&=5\left( {\dfrac{1}{b_n}}+{\dfrac{1}{4}} \right),\end{split} \]数列 $ \left\{{\dfrac{1}{b_n}}+{\dfrac{1}{4}}\right\} $ 是首项为 $ -{\dfrac{3}{4}} $，公比为 $ 5 $ 的等比数列．
因此\[ {\dfrac{1}{b_n}}+{\dfrac{1}{4}}=-{\dfrac{3}{4}}\cdot 5^{n-1}, \]即\[ b_n=-{\dfrac{4}{3\cdot 5^{n-1}+1}}, \]所以数列 $ \left\{x_n\right\} $ 的通项公式为\[ x_n=3-{\dfrac{4}{3\cdot 5^{n-1}+1}}. \]