Tính giới hạn \(E=\lim \limits_{x \rightarrow 7} \frac{\sqrt[3]{4 x-1}-\sqrt{x+2}}{\sqrt[4]{2 x+2}-2}\)
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Lời giải:
Báo saiTa có \({\rm{E}} = \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt[3]{{4x - 1}} - \sqrt {x + 2} }}{{\sqrt[4]{{2x + 2}} - 2}} = \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt[3]{{4x - 1}} - 3}}{{\sqrt[4]{{2x + 2}} - 2}} - \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt {x + 2} - 3}}{{\sqrt[4]{{2x + 2}} - 2}} = A - B\)
\(\begin{array}{*{20}{l}} {A = \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt[3]{{4x - 1}} - 3}}{{\sqrt[4]{{2x + 2}} - 2}} = \mathop {\lim }\limits_{x \to 7} \frac{{2(\sqrt[4]{{2x + 2}} + 2)\left( {\sqrt[4]{{{{(2x + 2)}^2}}} + 4} \right)}}{{\left( {\sqrt[3]{{{{(4x - 1)}^2}}} + 3\sqrt[3]{{4x - 1}} + 9} \right)}} = \frac{{64}}{{27}}}\\ {B = \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt {x + 2} - 3}}{{\sqrt[4]{{2x + 2}} - 2}} = \mathop {\lim }\limits_{x \to 7} \frac{{(\sqrt[4]{{2x + 2}} + 2)\left( {\sqrt[4]{{{{(2x + 2)}^2}}} + 4} \right)}}{{2(\sqrt {x + 2} + 3)}} = \frac{8}{3}} \end{array}\)
Vậy \(\mathrm{E}=\mathrm{A}-\mathrm{B}=\frac{64}{27}-\frac{8}{3}=\frac{-8}{27}\)