Tính giới hạn \(A=\lim\limits _{x \rightarrow 1} \frac{\sqrt[3]{7 x+1}-\sqrt{5 x-1}}{x-1}\)
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Lời giải:
Báo saiTa có:
\(A{\rm{ }} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[3]{{7x + 1}} - 2 - (\sqrt {5x - 1} - 2)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[3]{{7x + 1}} - 2}}{{x - 1}} - \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {5x - 1} - 2}}{{x - 1}} = I - J\)
\(\begin{array}{l} {\rm{I}} = \mathop {\lim }\limits_{x \to 1} \frac{{7({\rm{x}} - 1)}}{{({\rm{x}} - 1)\left( {\sqrt[3]{{{{(7{\rm{x}} - 1)}^2}}} + 2\sqrt[3]{{7{\rm{x}} - 1}} + 4} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{7}{{\sqrt[3]{{{{(7{\rm{x}} - 1)}^2}}} + 2\sqrt[3]{{7{\rm{x}} - 1}} + 4}} = \frac{7}{{12}}\;\\ {\rm{J}} = \mathop {\lim }\limits_{x \to 1} \frac{{5({\rm{x}} - 1)}}{{({\rm{x}} - 1)(\sqrt {5{\rm{x}} - 1} + 1)}} = \mathop {\lim }\limits_{x \to 1} \frac{5}{{\sqrt {5{\rm{x}} - 1} + 1}} = \frac{5}{3} \end{array}\)
Vậy \(A=-\frac{2}{3}\)