Limit It!

$\because \underset { n\rightarrow \infty }{ lim } n!{ x }_{ i }=\begin{cases} even\quad if\quad { x }_{ i }\quad is\quad rational\left( why? \right) \\ irrational\quad if\quad { x }_{ i }\quad is\quad irrational \end{cases}\\ \\ \& \quad \underset { m\rightarrow \infty }{ lim } \cos ^{ m }{ \left( \pi n!{ x }_{ i } \right) } =\begin{cases} 1\quad if\quad n!{ x }_{ i }\quad is\quad even\quad \\ 0\quad if\quad n!{ x }_{ i }\quad is\quad irrational \end{cases}\\ \\ \therefore \underset { m\rightarrow \infty }{ lim } \underset { n\rightarrow \infty }{ lim } \cos ^{ m }{ \left( \pi n!{ x }_{ i } \right) } =\begin{cases} 1\quad if\quad { x }_{ i }\quad is\quad rational\quad \\ 0\quad if\quad { x }_{ i }\quad is\quad irrational \end{cases}\\ \\ Now,\left( x-1 \right) \left( x-8 \right) \left( x-31 \right) =1\\ \\ or,{ \quad x }^{ 3 }-40{ x }^{ 2 }+287x-249=0\\ \\ By\quad Rational\quad root\quad theorem\quad every\quad ratinal\quad { sol }^{ n }\quad x=\cfrac { p }{ q } \quad \\ \\ will\quad satisfy\quad \\ \\ 1.p\quad is\quad integer\quad factor\quad of\quad -249=1\times 3\times 83\quad (const.\quad term)\\ \\ \& \quad 2.q\quad is\quad integer\quad factor\quad of\quad 1\quad (coeff.\quad of\quad { x }^{ 3 })\\ \\ so\quad possible\quad values\quad of\quad p=\pm 1,\pm 3,\pm 83,\pm 249\\ \\ \& \quad possible\quad values\quad of\quad q=\pm 1\quad \\ \\ but\quad none\quad of\quad \cfrac { p }{ q } \quad \left( \pm 1,\pm 3,\pm 83,\pm 249 \right) \quad satisfy\quad the\quad { eq }^{ n }\\ \\ it\quad means\quad root\quad are\quad irrational\quad so\quad all\quad limits\quad will\quad be\quad 0\\ \\ \Rightarrow 0+0+0=k+7\quad or\quad k=-7$