Phương pháp giải:

Tính \(\mathop {\lim }\limits_{x \to {0^{}}} f(x)\) để hàm số liên tục tại \(x = 0\) khi và chỉ khi  \(\mathop {\lim }\limits_{x \to 0} f(x) = f\left( 0 \right)\)

Lời giải chi tiết:

Ta có:

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

Để hàm số liên tục tại \(x = 0\)  thì  \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right)\)

\( \Leftrightarrow \frac{a}{2} + \frac{b}{3} = a + b \Leftrightarrow 3a + 2b = 6a + 6b \Leftrightarrow 3a + 4b = 0.\)

Chọn C.