Tính $\mathop {\lim }\limits_{x \to  - \infty } x\sqrt {\dfrac{{3x + 2}}{{2{x^3} + {x^2} - 1}}} $ bằng?

\( - 1.\)

\(1.\)

\( + \infty .\)

\(0.\)

\(\mathop {\lim }\limits_{x \to  + \infty } {x^n} =  - \infty \)           

\(\mathop {\lim }\limits_{x \to  \pm \infty } {x^n} =  + \infty \)        

\(\mathop {\lim }\limits_{x \to  - \infty } {x^n} =  - \infty \)

\(\mathop {\lim }\limits_{x \to  - \infty } {x^n} =  + \infty \) 

\( + \infty \)     

\( - \infty \)

\(\dfrac{2}{3}\)

\(1\)

$0$

\( - \infty .\)

$1$

\( + \infty .\)

\( - \dfrac{3}{2}.\)

\( - \dfrac{2}{3}.\)

\(0.\)

\( + \infty .\)

\(\lim \left( {{u_n} - 2{v_n}} \right) = \dfrac{1}{3}\)

\(\lim \left( {2{u_n} - {v_n}} \right) = 4\)

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

\(\lim \left( {{u_n} + {v_n}} \right) = \dfrac{1}{3}\)

$0.$    

$ - \dfrac{1}{4}.$      

$\dfrac{3}{4}.$

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