Đáp án:
$\begin{array}{l}
10)\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + 1} + \sqrt[3]{{{x^3} - 1}}}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + 1} - 1 + 1 + \sqrt[3]{{{x^3} - 1}}}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{x^2}}}{{\sqrt {{x^2} + 1} + 1}} + \frac{{{x^3}}}{{1 - \sqrt[3]{{{x^3} - 1}} + \sqrt[3]{{{{\left( {{x^3} - 1} \right)}^2}}}}}}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt {{x^2} + 1} + 1}} + \frac{x}{{1 - \sqrt[3]{{{x^3} - 1}} + \sqrt[3]{{{{\left( {{x^3} - 1} \right)}^2}}}}}\\
= \frac{1}{2}
\end{array}$
