Cho hàm số \(f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{\dfrac{{2x}}{{\sqrt {1 - x} }}}\,\,khi\,\,{x < 1}\\{\sqrt {3{x^2} + 1} }\,\,khi\,\,{x \ge 1}\end{array}} \right..\) Khi đó \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)\) là:

$\lim \dfrac{{2{n^2} - 3}}{{ - 2{n^3} - 4}}.$

$\lim \dfrac{{2{n^2} - 3}}{{ - 2{n^2} - 1}}.$

$\lim \dfrac{{2{n^2} - 3}}{{2{n^2} + 1}}.$

$\lim \dfrac{{2{n^3} - 3}}{{2{n^2} - 1}}.$

$ - \dfrac{1}{3}.$      

$0.$

$\dfrac{1}{3}.$         

Không tồn tại

\({u_n} = \sin n\)

\({v_n} = \cos n\)

\({x_n} = {\left( { - 1} \right)^n}\)

\({y_n} = \dfrac{1}{2}\)

\( + \infty \)

\( - \infty \)      

\(\dfrac{{1 - \sqrt[3]{3}}}{{\sqrt[4]{2} - 1}}\)        

$1$

\(S = \dfrac{{{u_1}}}{{1 - q}}\)

\(S = \dfrac{{{u_1}}}{{q - 1}}\) 

\(S = \dfrac{{1 - q}}{{{u_n}}}\)     

\(S = \dfrac{{{u_1}}}{{1 - {q^n}}}\)

$5$

$7$

$9$

$6$

\(0.\)

\(1.\)

\( + \infty .\)

\(2.\)