\(\mathop {\lim }\limits_{x \to {1^ - }} \sqrt {\frac{{1 - {x^3}}}{{3{x^2} + x}}} \)

* Hướng dẫn giải

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

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