\(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {...

D

- Hướng dẫn giải

Phương pháp giải:

Sử dụng phương pháp nhân liên hợp để khử dạng \(\dfrac{0}{0}\) .

Giải chi tiết:

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