Tính giới hạn: $L = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} - \sqrt[3]{{1 + 3x}}}}{{{x^2}}}$
Giải:
Ta có: \[L = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} - \sqrt[3]{{1 + 3x}}}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {1 + 2x} - \left( {x + 1} \right)} \right) - \left( {\sqrt[3]{{1 + 3x}} - \left( {x + 1} \right)} \right)}}{{{x^2}}}\]\[ = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} - \left( {x + 1} \right)}}{{{x^2}}} - \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{1 + 3x}} - \left( {x + 1} \right)}}{{{x^2}}}\]\[ = \mathop {\lim }\limits_{x \to 0} \frac{{ - {x^2}}}{{{x^2}\left( {\sqrt {1 + 2x} + x + 1} \right)}} - \mathop {\lim }\limits_{x \to 0} \frac{{ - {x^3} - 3{x^2}}}{{{x^2}\left[ {\sqrt[3]{{{{\left( {1 + 3x} \right)}^2}}} + \sqrt[3]{{1 + 3x}}\left( {x + 1} \right) + {{\left( {x + 1} \right)}^2}} \right]}}\]\[ = \mathop {\lim }\limits_{x \to 0} \frac{{ - 1}}{{\sqrt {1 + 2x} + x + 1}} + \mathop {\lim }\limits_{x \to 0} \frac{{x + 3}}{{\sqrt[3]{{{{\left( {1 + 3x} \right)}^2}}} + \sqrt[3]{{1 + 3x}}\left( {x + 1} \right) + {{\left( {x + 1} \right)}^2}}}\]\[ = \frac{3}{{1 + 1 + 1}} - \frac{1}{{1 + 1}} = \frac{1}{2}\]
Vậy $L = \frac{1}{2}$
