Cho ${u_n} = \dfrac{{{n^2} - 3n}}{{1 - 4{n^2}}}$. Khi đó $\lim {u_n}$bằng?

$1.$    

$\dfrac{2}{3}.$

$- 1.$

$- \dfrac{1}{3}.$

$\lim \sqrt[3]{{{u_n}}} = L$

$\lim \sqrt {{u_n}}  = L$

$\lim \sqrt {{u_n}}  = \sqrt L$         

$\lim \sqrt[3]{{{u_n}}} = \sqrt[3]{L}$

$\lim \left( {{u_n}.{v_n}} \right) = 0$          

$\lim \left( {{u_n}.{v_n}} \right) =  + \infty$          

$\lim \left( {{u_n}.{v_n}} \right) =  - \infty$

$\lim \left( {{u_n}.{v_n}} \right) = 1$

$0.$     

$1.$    

$\dfrac{3}{5}.$         

$+ \infty .$

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

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

${u_n} = n$

${u_n} = \sqrt n$

$\lim n =  + \infty$    

$\lim \sqrt n  =  + \infty$      

$\lim \sqrt[3]{n} =  + \infty$

$\lim \dfrac{1}{n} =  + \infty$