Tìm 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 sai\(\begin{array}{l} \text { Ta có: } E=\lim\limits _{x \rightarrow 7} \frac{\sqrt[3]{4 x-1}-\sqrt{x+2}}{\sqrt[4]{2 x+2}-2}=\lim \limits _{x \rightarrow 7} \frac{\sqrt[3]{4 x-1}-3}{\sqrt[4]{2 x+2}-2}-\lim\limits _{x \rightarrow 7} \frac{\sqrt{x+2}-3}{\sqrt[4]{2 x+2}-2}=A-B \\ A=\lim\limits _{x \rightarrow 7} \frac{\sqrt[3]{4 x-1}-3}{\sqrt[4]{2 x+2}-2}=\lim\limits _{x \rightarrow 7} \frac{2(\sqrt[4]{2 x+2}+2)\left(\sqrt[4]{(2 x+2)^{2}}+4\right)}{\left(\sqrt[3]{(4 x-1)^{2}}+3 \sqrt[3]{4 x-1}+9\right)}=\frac{64}{27} \\ B=\lim\limits _{x \rightarrow 7} \frac{\sqrt{x+2}-3}{\sqrt[4]{2 x+2}-2}=\lim \limits _{x \rightarrow 7} \frac{(\sqrt[4]{2 x+2}+2)\left(\sqrt[4]{(2 x+2)^{2}}+4\right)}{2(\sqrt{x+2}+3)}=\frac{8}{3} \\ E=A-B=\frac{64}{27}-\frac{8}{3}=\frac{-8}{27} \end{array}\)