Доказать что последовательность [latex][x_{n}]=[ \frac{3n+5}{n-1}] [/latex] имеет предел A=3

[latex]\{x_{n}\}=\{\frac{3n+5}{n-1}\}\\\\lim_{n\to \infty }\frac{3n+5}{n-1}=lim_{n\to \infty }\frac{3+\frac{5}{n}}{1-\frac{1}{n}}=[\, \frac{3+0}{1-0}\, ]=3[/latex]   [latex]\forall \varepsilon \ \textgreater \ o[/latex]  [latex] \exists N , n\ \textgreater \ N : |\frac{3n+5}{n-1}-3|\ \textless \ \varepsilon [/latex]   [latex]|\frac{3n+5-3n+3}{n-1}|\ \textless \ \varepsilon \\\\|\frac{8}{n-1}|\ \textless \ \varepsilon \\\\\frac{8}{n-1}\ \textless \ \varepsilon \\\\8\ \textless \ \varepsilon (n-1)\\\\n\ \textgreater \ 1+\frac{8}{\varepsilon }\\\\N=1+\frac{8}{\varepsilon }[/latex]