Đáp án:
0
Giải thích các bước giải:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^3} + 1} - 1}}{{{x^2} + x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt {{x^3} + 1} - 1} \right)\left( {\sqrt {{x^3}} + 1} \right)}}{{{x^2} + x}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^3} + 1 - 1}}{{x\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^3}}}{{x\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}}}{{x + 1}} = \dfrac{{{0^2}}}{{0 + 1}} = 0
\end{array}$
