Đáp án đúng: B
Giải chi tiết:\(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[m]{{1 + ax}}.\sqrt[n]{{1 + bx}} - 1}}{x}\,\,\,\,\left( {ab \ne 0,m,n \ge 2} \right)\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[m]{{1 + ax}} - 1} \right).\sqrt[n]{{1 + bx}}}}{x} + \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + bx}} - 1}}{x}\\ = {L_1} + {L_2}\end{array}\)
\(\begin{array}{l}{L_1} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[m]{{1 + ax}} - 1} \right).\sqrt[n]{{1 + bx}}\left( {{{\sqrt[m]{{1 + ax}}}^{m - 1}} + {{\sqrt[m]{{1 + ax}}}^{m - 2}} + ... + \sqrt[m]{{1 + ax}} + 1} \right)}}{{x\left( {{{\sqrt[m]{{1 + ax}}}^{m - 1}} + {{\sqrt[m]{{1 + ax}}}^{m - 2}} + ... + \sqrt[m]{{1 + ax}} + 1} \right)}}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{ax.\sqrt[n]{{1 + bx}}}}{{x\left( {{{\sqrt[m]{{1 + ax}}}^{m - 1}} + {{\sqrt[m]{{1 + ax}}}^{m - 2}} + ... + \sqrt[m]{{1 + ax}} + 1} \right)}}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{a.\sqrt[n]{{1 + bx}}}}{{{{\sqrt[m]{{1 + ax}}}^{m - 1}} + {{\sqrt[m]{{1 + ax}}}^{m - 2}} + ... + \sqrt[m]{{1 + ax}} + 1}}\\\,\,\,\,\,\, = \frac{{a.1}}{{1 + 1 + 1 + ... + 1\,\,\,\left( {m\,\,chu\,\,so\,\,1} \right)}} = \frac{a}{m}\end{array}\)
Chứng minh tương tự ta có \({L_2} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + bx}} - 1}}{x} = \frac{b}{n}\).
Vậy \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[m]{{1 + ax}}.\sqrt[n]{{1 + bx}} - 1}}{x} = \frac{a}{m} + \frac{b}{n}\)
Chọn B.
