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