\(\mathop {\lim }\limits_{x...

A

- Hướng dẫn giải

$\mathop {\lim }\limits_{x \to {1^ - }} \sqrt {\frac{{1 - {x^3}}}{{3{x^2} + x}}}  = 0$

Vì $\mathop {\lim }\limits_{x \to {1^ - }} \left( {1 - {x^3}} \right) = 0,\mathop {\lim }\limits_{x \to {1^ - }} \left( {3{x^2} + x} \right) = 4$