Tính giới hạn $\ L=\underset{x\to 1}{\mathop{\lim }}\,\frac{\left( 1-\sqrt{x} \right)\left( 1-\sqrt[3]{x} \right)\left( 1-\sqrt[4]{x} \right)\left( 1-\sqrt[5]{x} \right)}{{{\left( 1-x \right)}^{5}}}.$
Giải:
$\ L=\underset{x\to 1}{\mathop{\lim }}\,\frac{\left( 1-\sqrt{x} \right)\left( 1-\sqrt[3]{x} \right)\left( 1-\sqrt[4]{x} \right)\left( 1-\sqrt[5]{x} \right)}{{{\left( 1-x \right)}^{5}}}=\underset{x\to 1}{\mathop{\lim }}\,\frac{1-\sqrt{x}}{1-x}.\underset{x\to 1}{\mathop{\lim }}\,\frac{1-\sqrt[3]{x}}{1-x}.\underset{x\to 1}{\mathop{\lim }}\,\frac{1-\sqrt[4]{x}}{1-x}.\underset{x\to 1}{\mathop{\lim }}\,\frac{1-\sqrt[5]{x}}{1-x}.$
$\ =\underset{x\to 1}{\mathop{\lim }}\,\frac{1}{1+\sqrt{x}}.\underset{x\to 1}{\mathop{\lim }}\,\frac{1}{1+\sqrt[3]{x}+\sqrt[3]{{{x}^{2}}}}.\underset{x\to 1}{\mathop{\lim }}\,\frac{1}{1+\sqrt[4]{x}+\sqrt[4]{{{x}^{2}}}+\sqrt[4]{{{x}^{3}}}}.\underset{x\to 1}{\mathop{\lim }}\,\frac{1}{1+\sqrt[5]{x}+\sqrt[5]{{{x}^{2}}}+\sqrt[5]{{{x}^{3}}}+\sqrt[5]{{{x}^{4}}}}.$
$\ =\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}=\frac{1}{5!}=\frac{1}{120}.$
TỔNG QUÁT: ${L_n} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {1 - \sqrt x } \right)\left( {1 - \sqrt[3]{x}} \right)...\left( {1 - \sqrt[n]{x}} \right)}}{{{{\left( {1 - x} \right)}^n}}} = \frac{1}{{n!}}$
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