Tính giới hạn \(\lim\limits _{x \rightarrow 0} \frac{\sqrt{1-x}-\sqrt[3]{1-x}}{x}\).
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Lời giải:
Báo saiTa có :
\(\begin{array}{l} \begin{array}{*{20}{l}} {{\rm{ }}\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - x} - \sqrt[3]{{1 - x}}}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - x} - 1}}{x} - \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{1 - x}} - 1}}{x}}\\ { = \mathop {\lim }\limits_{x \to 0} \frac{{(\sqrt {1 - x} - 1)(\sqrt {1 - x} + 1)}}{{x(\sqrt {1 - x} + 1)}} - \mathop {\lim }\limits_{x \to 0} \frac{{(\sqrt[3]{{1 - x}} - 1)\left( {\sqrt[3]{{{{(1 - x)}^2}}} + \sqrt[3]{{1 - x}} + \sqrt[3]{1}} \right)}}{{x\left( {\sqrt[3]{{{{(1 - x)}^2}}} + \sqrt[3]{{1 - x}} + \sqrt[3]{1}} \right)}}} \end{array}\\ \begin{array}{*{20}{l}} { = \mathop {\lim }\limits_{x \to 0} \frac{{ - x}}{{x(\sqrt {1 - x} + 1)}} - \mathop {\lim }\limits_{x \to 0} \frac{x}{{x\left( {\sqrt[3]{{{{(1 - x)}^2}}} + \sqrt[3]{{1 - x}} + \sqrt[3]{1}} \right)}}}\\ { = \mathop {\lim }\limits_{x \to 0} \frac{{ - 1}}{{(\sqrt {1 - x} + 1)}} - \mathop {\lim }\limits_{x \to 0} \frac{1}{{\left( {\sqrt[3]{{{{(1 - x)}^2}}} + \sqrt[3]{{1 - x}} + \sqrt[3]{1}} \right)}} = - \frac{1}{6}} \end{array} \end{array}\)