$\lim \left( {\dfrac{2}{n} + \dfrac{3}{{{n^2}}}} \right)$ bằng

${u_n} = \dfrac{n}{2}$

${u_n} = \dfrac{2}{n}$

${u_n} = n$

${u_n} = \sqrt n$

$\dfrac{1}{5}.$

$- \dfrac{4}{5}.$     

$\dfrac{4}{5}.$

$- \dfrac{1}{5}.$

$\lim \dfrac{{3{u_n} - 1}}{{{u_n} + 1}} = 3$

$\lim \dfrac{{3{u_n} - 1}}{{{u_n} + 1}} =  - 1$

$\lim \dfrac{{3{u_n} - 1}}{{{u_n} + 1}} = 2$

$\lim \dfrac{{3{u_n} - 1}}{{{u_n} + 1}} = 1$

$1.$

$- \dfrac{1}{4}.$

$\dfrac{4}{5}.$

$- \dfrac{3}{4}.$

${u_n} = \dfrac{1}{{\sqrt n }}$                  

${u_n} = \dfrac{1}{{\sqrt[3]{n}}}$

${u_n} = \dfrac{{\sqrt[3]{n}}}{2}$

${u_n} = 0$

$0.$    

$- \dfrac{1}{4}.$      

$\dfrac{3}{4}.$

$- \dfrac{3}{4}.$

$\lim {u_n} = 0$                   

$\lim {u_n} > \lim {v_n}$

$\lim {u_n} < \lim {v_n}$

$\lim {u_n} < 0$