Tính \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2{\rm{x}}} \cdot \sqrt[3]}} \cdot \sqrt[4]}} - 1}}{x}\).

Đáp án đúng là: D

Ta có: \(\sqrt {1 + 2{\rm{x}}} \cdot \sqrt[3]}} \cdot \sqrt[4]}} - 1\)

\( = \sqrt {1 + 2{\rm{x}}} - \sqrt {1 + 2{\rm{x}}} + \sqrt {1 + 2{\rm{x}}} .\sqrt[3]}} - \sqrt {1 + 2{\rm{x}}} \cdot \sqrt[3]}} + \sqrt {1 + 2{\rm{x}}} \cdot \sqrt[3]}} \cdot \sqrt[4]}} - 1\)\( = \left( {\sqrt {1 + 2{\rm{x}}} - 1} \right) + \sqrt {1 + 2{\rm{x}}} \cdot \left( {\sqrt[3]}} - 1} \right) + \sqrt {1 + 2{\rm{x}}} \cdot \sqrt[3]}}\left( {1 + \sqrt[4]}}} \right)\)

Suy ra: \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2{\rm{x}}} .\sqrt[3]}}.\sqrt[4]}} - 1}}{x}\)

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

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

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

Ta có:

\(\mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2{\rm{x}}} .\frac{{\sqrt[3]}} - 1}}{x}} \right)\)

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

\( = \mathop {\lim }\limits_{x \to 0} \left\{ {\sqrt {1 + 2{\rm{x}}} .\frac{{3{\rm{x}}}}{{x\left[ {{{\left( {\sqrt[3]}}} \right)}^2} + \sqrt[3]}} + 1} \right]}}} \right\}\)

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{3\sqrt {1 + 2{\rm{x}}} }}{{\left[ {{{\left( {\sqrt[3]}}} \right)}^2} + \sqrt[3]}} + 1} \right]}}} \right) = \frac = 1\)

Ta có: \(\mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2{\rm{x}}} .\sqrt[3]}}.\frac{{\sqrt[4]}} - 1}}{x}} \right)\)

\( = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2{\rm{x}}} .\sqrt[3]}}.\frac{{\left( {\sqrt[4]}} - 1} \right)\left[ {{{\left( {\sqrt[4]}}} \right)}^3} + {{\left( {\sqrt[4]}}} \right)}^2} + \left( {\sqrt[4]}}} \right) + 1} \right]}}{{x\left[ {{{\left( {\sqrt[4]}}} \right)}^3} + {{\left( {\sqrt[4]}}} \right)}^2} + \left( {\sqrt[4]}}} \right) + 1} \right]}}} \right)\)

\( = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2{\rm{x}}} .\sqrt[3]}}.\frac{{4{\rm{x}}}}{{x\left[ {{{\left( {\sqrt[4]}}} \right)}^3} + {{\left( {\sqrt[4]}}} \right)}^2} + \left( {\sqrt[4]}}} \right) + 1} \right]}}} \right)\)

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{4\sqrt {1 + 2{\rm{x}}} .\sqrt[3]}}}}{{\left[ {{{\left( {\sqrt[4]}}} \right)}^3} + {{\left( {\sqrt[4]}}} \right)}^2} + \left( {\sqrt[4]}}} \right) + 1} \right]}}} \right) = \frac = 1\)

Suy ra \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2{\rm{x}}} .\sqrt[3]}}.\sqrt[4]}} - 1}}{x} = 1 + 1 + 1 = 3\)

Vậy ta chọn đáp án D.