Tìm giới hạn \(A = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4x + 1} - \sqrt[3]{{2x + 1}}}}{x}\)

* Hướng dẫn giải

\(\begin{array}{l}
A = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4x + 1}  - \sqrt[3]{{2x + 1}}}}{x}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4x + 1}  - 1}}{x} - \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{2x + 1}} - 1}}{x}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{4x}}{{x\left( {\sqrt {4x + 1}  + 1} \right)}} - \mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{x\left[ {\sqrt[3]{{{{\left( {2x + 1} \right)}^2}}} + \sqrt[3]{{2x + 1}} + 1} \right]}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{4}{{\left( {\sqrt {4x + 1}  + 1} \right)}} - \mathop {\lim }\limits_{x \to 0} \frac{2}{{\left[ {\sqrt[3]{{{{\left( {2x + 1} \right)}^2}}} + \sqrt[3]{{2x + 1}} + 1} \right]}}\\
 = 2 - \frac{2}{3} = \frac{4}{3}
\end{array}\)