Tính $\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt[3]{{{x^3} + 1}} + x - 1} \right)$ bằng?

$5$

$7$

$9$

$6$

\(-2\)

\(5\)

\(9\)

\(10\)

$3$

$\sqrt 3 .$                   

$-3$

$\dfrac{1}{3}.$

$-3$

$-2$

$2$

$3$

$ - \dfrac{1}{3}.$      

$0.$

$\dfrac{1}{3}.$         

Không tồn tại

$\mathop {\lim }\limits_{x \to  - 1} \dfrac{{{x^2} + 3x + 2}}{{\left| {x + 1} \right|}} =  - 1$

$\mathop {\lim }\limits_{x \to  - 1} \dfrac{{{x^2} + 3x + 2}}{{\left| {x + 1} \right|}} =  0$

$\mathop {\lim }\limits_{x \to  - 1} \dfrac{{{x^2} + 3x + 2}}{{\left| {x + 1} \right|}} = 1$

Không tồn tại $\mathop {\lim }\limits_{x \to  - 1} \dfrac{{{x^2} + 3x + 2}}{{\left| {x + 1} \right|}}$

$\dfrac{1}{5}.$          

$\dfrac{2}{5}.$         

$\dfrac{1}{2}.$

$\dfrac{1}{3}.$