Tính \(\mathop {\lim }\limits_{x \to 0} \frac{{\sq...

B

- Hướng dẫn giải

Ta có \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + ax} .\sqrt[3]{{1 + bx}} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + ax} .\left( {\sqrt[3]{{1 + bx}} - 1 + 1} \right) - 1}}{x}\)

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