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