Câu hỏi 58

\(\frac{{\tan ax}}{{\sqrt {1 + bx}  - \sqrt[3]{{1 + cx}}}} = a.\frac{{\tan ax}}{{ax}}.\frac{x}{{\sqrt {1 + bx}  - \sqrt[3]{{1 + cx}}}}\)

Lại có:  \(\mathop {\lim }\limits_{x \to 0} \frac{{\tan ax}}{{ax}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{ax.\cos ax}} = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos ax}} = 1\) ;

Xét: \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + ax}} - 1}}{x}\;\left( {a\; \ne 0;\;n \in {\mathbb{N}^*}} \right)\) .

Đặt \(\sqrt[n]{{1 + ax}} = y \Leftrightarrow {y^n} = 1 + ax \Rightarrow x = \frac{{{y^n} - 1}}{a};\,\,\mathop {\lim }\limits_{x \to 0} y = 1\)                                                                                      \(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + ax}} - 1}}{x} = a\mathop {\lim }\limits_{y \to 1} \frac{{y - 1}}{{{y^n} - 1}} = a\mathop {\lim }\limits_{y \to 1} \frac{{y - 1}}{{\left( {y - 1} \right)\left( {{y^{n - 1}} + ... + y + 1} \right)}} = a\mathop {\lim }\limits_{y \to 1} \frac{1}{{\left( {{y^{n - 1}} + ... + y + 1} \right)}} = \frac{a}{n}\\\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + bx}  - \sqrt[3]{{1 + cx}}}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {1 + bx}  - 1} \right) - \left( {\sqrt[3]{{1 + cx}} - 1} \right)}}{x} = \frac{b}{2} - \frac{c}{3} = \frac{{3b - 2c}}{6}\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\tan ax}}{{\sqrt {1 + bx}  - \sqrt[3]{{1 + cx}}}} = \frac{{6a}}{{3b - 2c}}.\end{array}\)

Do đó hệ thức liên hệ là: \(\frac{{6a}}{{3b - 2c}} = \frac{1}{2} \Leftrightarrow \frac{a}{{3b - 2c}} = \frac{1}{{12}}\)

Chọn D.