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Question Number 153404 by liberty last updated on 07/Sep/21

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt[3]{8^x+3^x}-\sqrt{4^x-2^x}=?$$

Answered by EDWIN88 last updated on 07/Sep/21

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt[3]{8^x+3^x}-\sqrt[3]{8^x}+\sqrt{4^x}-\sqrt{4^x-2^x}=$$$$L_1=\underset{x\to \mathrm{\infty }}{\lim }\frac{(8^x+3^x)-8^x}{\sqrt[3]{{(8^x+3^x)}^2}+\sqrt[3]{8^x(8^x+3^x)}+4^x}$$$$L_1=\underset{x\to \mathrm{\infty }}{\lim }\frac{3^x}{\sqrt[3]{8^{2x}{(1+{\left(\frac{3}{8}\right)}^x)}^2}+\sqrt[3]{8^{2x}(1+{\left(\frac{3}{8}\right)}^x)}+4^x}$$$$L_1=\underset{x\to \mathrm{\infty }}{\lim }\frac{{\left(\frac{3}{4}\right)}^x}{\sqrt[3]{{(1+{\left(\frac{3}{8}\right)}^x)}^2}+\sqrt[3]{1+{\left(\frac{3}{8}\right)}^x}+1}$$$$L_1=\frac{0}{3}=0$$$$L_2=\underset{x\to \mathrm{\infty }}{\lim }\sqrt{4^x}-\sqrt{4^x-2^x}$$$$L_2=\underset{x\to 0}{\lim }{2}^x(1-\sqrt{1-{\left(\frac{1}{2}\right)}^x})$$$$L_2=\underset{X\to 0}{\lim }\frac{1-\sqrt{1-X}}{X}=\frac{1-(1-\frac{1}{2}X)}{X}=\frac{1}{2}$$$$ThenL=L_1+L_2=0+\frac{1}{2}=\frac{1}{2}$$

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