Cho \({u_n} = \frac{{1 + a + {a^2} + \cdots + {a^n}}}{{1 + b + {b^2} + \cdots + {b^n}}}\) với \(a,b\) là các số thực thoả mãn \(|a| < 1,|b| < 1\). Tính \(\mathop {\lim }\limits_{n \to + \infty } {u_n}\).

a) \(\lim {\left( {\frac{2}{3}} \right)^n} = 0\,\) \(\left( {{\rm{do}}\,\frac{2}{3} < 1} \right)\)

b) \(\lim \frac{1}{{{{(\sqrt 2 )}^n}}} = \lim {\left( {\frac{1}{{\sqrt 2 }}} \right)^n} = 0\,\left( {{\rm{do}}\,\frac{1}{{\sqrt 2 }}\, < 1} \right)\)

c) \(\lim \frac{1}{{{n^3}}} = 0\)

d) \(\lim 4 = 4\)

a) \(\lim {(\sqrt 3 )^n} = + \infty (\)do \(\sqrt 3 > 1)\)

b) \(\lim {\pi ^n} = + \infty (\) do \(\pi > 1)\)

c) \(\lim \left( {{n^3} + 2{n^2} - 4} \right) = \lim {n^3} \cdot \left( {1 + \frac{2}{n} - \frac{4}{{{n^3}}}} \right) = + \infty \).

Vì \(\left\{ {\begin{array}{*{20}{l}}{\lim {n^3} = + \infty }\\{\lim \left( {1 + \frac{2}{n} - \frac{4}{{{n^3}}}} \right) = 1 > 0}\end{array}} \right.\)

d) \(\lim \left( { - {n^4} + 5{n^3} - 4n} \right) = \lim {n^4} \cdot \left( { - 1 + \frac{5}{n} - \frac{4}{{{n^3}}}} \right) = - \infty \).

Vì \(\left\{ {\begin{array}{*{20}{l}}{\lim {n^4} = + \infty }\\{\lim \left( { - 1 + \frac{5}{n} - \frac{4}{{{n^3}}}} \right) = - 1 < 0}\end{array}} \right.\)