Tính giới hạn \(D = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {3x + 1} - 2}}{{\sqrt[3]{{3x + 1}} - 2}}\)
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Lời giải:
Báo sai\(\begin{array}{l} D = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {3x + 1} - 2}}{{\sqrt[3]{{3x + 1}} - 2}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {3x + 1} - 2} \right)\left( {\sqrt {3x + 1} + 2} \right).\left[ {{{\left( {\sqrt[3]{{3x + 1}}} \right)}^2} + 2.\sqrt[3]{{3x + 1}} + 4} \right]}}{{\left( {\sqrt[3]{{3x + 1}} - 2} \right)\left[ {{{\left( {\sqrt[3]{{3x + 1}}} \right)}^2} + 2.\sqrt[3]{{3x + 1}} + 4} \right]\left( {\sqrt {3x + 1} + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {3x - 3} \right).\left[ {{{\left( {\sqrt[3]{{3x + 1}}} \right)}^2} + 2.\sqrt[3]{{3x + 1}} + 4} \right]}}{{\left( {3x - 7} \right)\left( {\sqrt {3x + 1} + 2} \right)}} = \frac{{\left( {3.1 - 3} \right).\left[ {{{\left( {\sqrt[3]{{3.1}}} \right)}^2} + 2.\sqrt[3]{{3.1 + 1}} + 4} \right]}}{{\left( {3.1 - 7} \right)\left( {\sqrt {3.1} + 2} \right)}} = 0 \end{array}\)