Tính giới hạn \(I = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} .\sqrt[3]{{1 + 4x}} - 1}}{x}\)

Chỉ được phép điền số 0, nguyên âm, nguyên dương và phân số dạng a/b

Đáp án:

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

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

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

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

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

\( = \dfrac{{4.1}}{{1 + 1 + 1}} + \dfrac{2}{{1 + 1}} = \dfrac{7}{3}\)