Phương pháp giải:

- Biến đổi biểu thức, đưa về dạng \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + nx}} - 1}}{x}\)

- Nhân liên hợp.

Lời giải chi tiết:

Ta có:

\(\begin{array}{l}\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\sqrt[4]{{1 + 4x}} - 1\\ = \sqrt {1 + 2x}  - \sqrt {1 + 2x}  + \sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}} - \sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}} + \sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\sqrt[4]{{1 + 4x}} - 1\\ = \left( {\sqrt {1 + 2x}  - 1} \right) + \sqrt {1 + 2x} \left( {\sqrt[3]{{1 + 3x}} - 1} \right) + \sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\left( {\sqrt[4]{{1 + 4x}} - 1} \right)\end{array}\)

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

Tính:

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sqrt {1 + 2x}  - 1}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {1 + 2x}  - 1} \right)\left( {\sqrt {1 + 2x}  + 1} \right)}}{{x\left( {\sqrt {1 + 2x}  + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{x\left( {\sqrt {1 + 2x}  + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{2}{{\sqrt {1 + 2x}  + 1}} = \frac{2}{{1 + 1}} = 1\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\frac{{\sqrt[3]{{1 + 3x}} - 1}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\frac{{\left( {\sqrt[3]{{1 + 3x}} - 1} \right)\left[ {{{\left( {\sqrt[3]{{1 + 3x}}} \right)}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}{{x.\left[ {{{\left( {\sqrt[3]{{1 + 3x}}} \right)}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}} \right)\\= \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\frac{{3x}}{{x.\left[ {{{\left( {\sqrt[3]{{1 + 3x}}} \right)}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{3\sqrt {1 + 2x} }}{{\left[ {{{\left( {\sqrt[3]{{1 + 3x}}} \right)}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}} \right) = \frac{{3.1}}{{1 + 1 + 1}} = 1\end{array}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\frac{{\sqrt[4]{{1 + 4x}} - 1}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\frac{{\left( {\sqrt[4]{{1 + 4x}} - 1} \right)\left[ {{{\left( {\sqrt[4]{{1 + 4x}}} \right)}^3} + {{\left( {\sqrt[4]{{1 + 4x}}} \right)}^2} + \sqrt[4]{{1 + 4x}} + 1} \right]}}{{x\left[ {{{\left( {\sqrt[4]{{1 + 4x}}} \right)}^3} + {{\left( {\sqrt[4]{{1 + 4x}}} \right)}^2} + \sqrt[4]{{1 + 4x}} + 1} \right]}}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\frac{{4x}}{{x\left[ {{{\left( {\sqrt[4]{{1 + 4x}}} \right)}^3} + {{\left( {\sqrt[4]{{1 + 4x}}} \right)}^2} + \sqrt[4]{{1 + 4x}} + 1} \right]}}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \frac{{4\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}}}{{{{\left( {\sqrt[4]{{1 + 4x}}} \right)}^3} + {{\left( {\sqrt[4]{{1 + 4x}}} \right)}^2} + \sqrt[4]{{1 + 4x}} + 1}} = \frac{{4.1.1}}{{1 + 1 + 1 + 1}} = 1\end{array}\)Vậy \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\sqrt[4]{{1 + 4x}} - 1}}{x} = 1 + 1 + 1 = 3\)

Chọn: D.