Đáp án:
\[4\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4x + 1} - \sqrt[3]{{6x + 1}}}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {4x + 1} - \left( {2x + 1} \right)} \right) + \left[ {\left( {2x + 1} \right) - \sqrt[3]{{6x + 1}}} \right]}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{4x + 1 - 4{x^2} - 4x - 1}}{{\sqrt {4x + 1} + 2x + 1}} + \frac{{8{x^3} + 12{x^2} + 6x + 1 - 6x - 1}}{{2x + 1 + \sqrt[3]{{6x + 1}}}}}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{ - 4}}{{\sqrt {4x + 1} + 2x + 1}} + \frac{{8x + 12}}{{2x + 1 + \sqrt[3]{{6x + 1}}}}} \right]\\
= \frac{{ - 4}}{{\sqrt 1 + 1}} + \frac{{12}}{{1 + \sqrt[3]{1}}} = \frac{{ - 4}}{2} + \frac{{12}}{2} = 4
\end{array}\)
