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Question Number 94528 by i jagooll last updated on 19/May/20

Commented by i jagooll last updated on 19/May/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{x^2\{\sqrt[4]{1-\frac{2}{x}}-1+\frac{2}{x^2}\}}{x^2\{1+\frac{1}{x^2}-\sqrt[4]{16+\frac{7}{x^7}}\}}=0$$$$$$

Answered by abdomathmax last updated on 20/May/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{{({\mathrm{x}}^8-{2\mathrm{x}}^7)}^{\frac{1}{4}}+2-{\mathrm{x}}^2}{{\mathrm{x}}^2+1-{({16\mathrm{x}}^8+7\mathrm{x})}^{\frac{1}{4}}}{\lim }_{\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)?$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2{(1-\frac{2}{\mathrm{x}})}^{\frac{1}{4}}+2-{\mathrm{x}}^2}{{\mathrm{x}}^2+1-{2\mathrm{x}}^2{(1+\frac{7}{16\mathrm{x}})}^{\frac{1}{4}}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{{\mathrm{x}}^2(1-\frac{1}{2\mathrm{x}})+2-{\mathrm{x}}^2}{{\mathrm{x}}^2+1-{2\mathrm{x}}^2(1+\frac{7}{64\mathrm{x}})}=\frac{{\mathrm{x}}^2-\frac{\mathrm{x}}{2}+2-{\mathrm{x}}^2}{{\mathrm{x}}^2+1-{2\mathrm{x}}^2-\frac{7\mathrm{x}}{32}}$$$$=\frac{-\frac{\mathrm{x}}{2}+2}{-{\mathrm{x}}^2-\frac{7\mathrm{x}}{32}+1}\Rightarrow {\lim }_{\mathrm{x}\to \mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)={\lim }_{\mathrm{x}\to \mathrm{\infty }}\frac{-\mathrm{x}}{-{2\mathrm{x}}^2}$$$$={\lim }_{\mathrm{x}\to \mathrm{\infty }}\frac{1}{2\mathrm{x}}=0$$$$$$

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