Tính được các giới hạn sau, khi đó:

a) \(\lim {\left( {\frac{2}{3}} \right)^n} = 0\)

b) \(\lim \frac{1}{{{{(\sqrt 2 )}^n}}} = - \infty \)

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

d) \(\lim 4 = 0\)

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

Do \(\mathop {\lim }\limits_{n \to + \infty } \frac{1}{{{3^n}}} = 0\) nên \(\mathop {\lim }\limits_{n \to + \infty } \left( {{u_n} - 2} \right) = 0\). Vậy \(\mathop {\lim }\limits_{n \to + \infty } {u_n} = 2\).