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Question Number 167593 by cortano1 last updated on 20/Mar/22

Answered by qaz last updated on 20/Mar/22

$$\mathrm{A}=\underset{\mathrm{x}\to 0}{\lim }\frac{\sqrt{17-2{(\mathrm{x}+2)}^2}\cdot \sqrt[3]{3{(\mathrm{x}+2)}^3+3}-9}{{\mathrm{x}}^2}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{\sqrt{9-{2\mathrm{x}}^2-8\mathrm{x}}\cdot \sqrt[3]{{3\mathrm{x}}^3+{18\mathrm{x}}^2+36\mathrm{x}+27}-9}{{\mathrm{x}}^2}$$$$=9\underset{\mathrm{x}\to 0}{\lim }\frac{\sqrt{1-\frac{8}{9}\mathrm{x}-\frac{2}{9}{\mathrm{x}}^2}\cdot \sqrt[3]{1+\frac{36}{27}\mathrm{x}+\frac{18}{27}{\mathrm{x}}^2+\frac{3}{27}{\mathrm{x}}^3}-1}{{\mathrm{x}}^2}$$$$=9\underset{\mathrm{x}\to 0}{\lim }\frac{\ln (\sqrt{1-\frac{8}{9}\mathrm{x}-\frac{2}{9}{\mathrm{x}}^2}\cdot \sqrt[3]{1+\frac{36}{27}\mathrm{x}+\frac{18}{27}{\mathrm{x}}^2+\frac{3}{27}{\mathrm{x}}^3})}{{\mathrm{x}}^2}$$$$=9\underset{\mathrm{x}\to 0}{\lim }\frac{\frac{1}{2}\ln (1-\frac{8}{9}\mathrm{x}-\frac{2}{9}{\mathrm{x}}^2)+\frac{1}{3}\ln (1+\frac{36}{27}\mathrm{x}+\frac{18}{27}{\mathrm{x}}^2+\frac{3}{27}{\mathrm{x}}^3)}{{\mathrm{x}}^2}$$$$=9\underset{\mathrm{x}\to 0}{\lim }\frac{1}{2\mathrm{x}}[\frac{-\frac{8}{9}-\frac{4}{9}\mathrm{x}}{2(1-\frac{8}{9}\mathrm{x}-\frac{2}{9}{\mathrm{x}}^2)}+\frac{\frac{36}{27}+\frac{36}{27}\mathrm{x}+\frac{9}{27}{\mathrm{x}}^2}{3(1+\frac{36}{27}\mathrm{x}+\frac{28}{27}{\mathrm{x}}^2+\frac{3}{27}{\mathrm{x}}^3)}]$$$$=9\underset{\mathrm{x}\to 0}{\lim }\frac{1}{2\mathrm{x}}\cdot (\frac{-4-2\mathrm{x}}{9-8\mathrm{x}-{2\mathrm{x}}^2}+\frac{12+12\mathrm{x}+{3\mathrm{x}}^2}{27+36\mathrm{x}+{28\mathrm{x}}^2+{3\mathrm{x}}^3})$$$$=9\underset{\mathrm{x}\to 0}{\lim }\frac{(-4-2\mathrm{x})(27+36\mathrm{x}+{28\mathrm{x}}^2+{3\mathrm{x}}^3)+(12+12\mathrm{x}+{3\mathrm{x}}^2)(9-8\mathrm{x}-{2\mathrm{x}}^2)}{2\mathrm{x}(9-8\mathrm{x}-{2\mathrm{x}}^2)(27+36\mathrm{x}+{28\mathrm{x}}^2+{3\mathrm{x}}^3)}$$$$=9\underset{\mathrm{x}\to 0}{\lim }\frac{(-4\cdot 36-2\cdot 27+12\cdot 9-12\cdot 8)\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right)}{2\cdot 9\cdot 27\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right)}$$$$=-\frac{31}{9}$$

Answered by cortano1 last updated on 20/Mar/22

$$\mathrm{A}=\underset{x\to 2}{\lim }\frac{(17-{2\mathrm{x}}^2)\sqrt[3]{{({3\mathrm{x}}^3+3)}^2}-81}{[\sqrt{17-{2\mathrm{x}}^2}\sqrt[3]{{3\mathrm{x}}^3+3}+9]{(\mathrm{x}-2)}^2}$$$$\mathrm{A}=\frac{1}{18}\underset{x\to 2}{\lim }\frac{-4\mathrm{x}\sqrt[3]{{({3\mathrm{x}}^3+3)}^2}+(17-{2\mathrm{x}}^2)\frac{2.\left({9\mathrm{x}}^2\right)}{3\sqrt[3]{{3\mathrm{x}}^3+3}}}{2(\mathrm{x}-2)}$$$$\mathrm{A}=\frac{1}{36}\underset{x\to 2}{\lim }\frac{\frac{(17-{2\mathrm{x}}^2){6\mathrm{x}}^2}{\sqrt[3]{{3\mathrm{x}}^3+3}}-4\mathrm{x}\sqrt[3]{{({3\mathrm{x}}^3+3)}^2}}{(\mathrm{x}-2)}$$$$\mathrm{A}=\frac{1}{36}\underset{x\to 2}{\lim }\frac{(17-{2\mathrm{x}}^2){6\mathrm{x}}^2-4\mathrm{x}({3\mathrm{x}}^3+3)}{(\mathrm{x}-2)\sqrt[3]{{3\mathrm{x}}^3+3}}$$$$\mathrm{A}=\frac{1}{108}\underset{x\to 2}{\lim }\frac{{102\mathrm{x}}^2-{24\mathrm{x}}^4-12\mathrm{x}}{(\mathrm{x}-2)}$$$$\mathrm{A}=\frac{1}{108}\underset{x\to 2}{\lim }\frac{204\mathrm{x}-{96\mathrm{x}}^3-12}{1}$$$$\mathrm{A}=\frac{408-768-12}{108}=-\frac{31}{9}$$

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