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Question Number 131789 by bemath last updated on 08/Feb/21

$$\mathrm{Let}\phi =\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^4+3({\mathrm{a}}^2-\sqrt{{\mathrm{a}}^4+{\mathrm{x}}^4})}{{\mathrm{x}}^8};\mathrm{a}>0$$ $$\mathrm{If}\phi \mathrm{is}\mathrm{finite}\mathrm{then}$$ $$\left(\mathrm{a}\right)\mathrm{a}=\frac{3}{2}\left(\mathrm{b}\right)\mathrm{a}=\sqrt{\frac{3}{2}}\left(\mathrm{c}\right)\phi =\frac{1}{3}\left(\mathrm{d}\right)\phi =\frac{1}{9}$$

Answered by Dwaipayan Shikari last updated on 08/Feb/21

$$\underset{x\to 0}{\lim }\frac{x^4+3(a^2-a^2(1+\frac{x^4}{2a^4}-\frac{x^8}{8a^8}))}{x^8}=\frac{x^4-\frac{3x^4}{2a^2}+\frac{3x^8}{8a^6}}{x^8}=\phi$$ $$\Rightarrow x^4-\frac{3x^4}{2a^2}=0\Rightarrow 1=\frac{3}{2a^2}\Rightarrow a=\pm \sqrt{\frac{3}{2}}$$ $$\phi =\frac{3}{8a^6}=\frac{3}{8}.\frac{8}{27}=\frac{1}{9}$$

Answered by EDWIN88 last updated on 08/Feb/21

$$\phi =\underset{x\to 0}{\lim }\frac{({\mathrm{x}}^4+{3\mathrm{a}}^2)-3\sqrt{{\mathrm{a}}^4+{\mathrm{x}}^4}}{{\mathrm{x}}^8}=$$ $$\phi =\underset{x\to 0}{\lim }\frac{{({\mathrm{x}}^4+{3\mathrm{a}}^2)}^2-9({\mathrm{a}}^4+{\mathrm{x}}^4)}{{\mathrm{x}}^8(({\mathrm{x}}^4+{3\mathrm{a}}^2)+3\sqrt{{\mathrm{a}}^4+{\mathrm{x}}^4}))}=$$ $$\phi =\frac{1}{{6\mathrm{a}}^2}\times \underset{x\to 0}{\lim }\frac{{\mathrm{x}}^8+{6\mathrm{a}}^2{\mathrm{x}}^4+{9\mathrm{a}}^4-{9\mathrm{a}}^4-{9\mathrm{x}}^4}{{\mathrm{x}}^8}$$ $$\phi =\frac{1}{{6\mathrm{a}}^2}\times \underset{x\to 0}{\lim }\frac{{\mathrm{x}}^8+{\mathrm{x}}^4({6\mathrm{a}}^2-9)}{{\mathrm{x}}^8}$$ $$\mathrm{since}\mathrm{the}\mathrm{value}\mathrm{of}\phi \mathrm{is}\mathrm{finite}\mathrm{it}\mathrm{follows}\mathrm{that}$$ $${6\mathrm{a}}^2-9\mathrm{must}\mathrm{be}\mathrm{is}\mathrm{zero};\mathrm{a}=\sqrt{\frac{3}{2}}$$ $$\mathrm{then}\phi =\frac{1}{6\left(\frac{9}{6}\right)}=\frac{1}{9}$$

Commented byliberty last updated on 08/Feb/21

$$\mathrm{Hallo}\mathrm{Herr}\mathrm{wie}\mathrm{geht}\mathrm{es}\mathrm{dir}$$

Commented byEDWIN88 last updated on 08/Feb/21

$$$$ $$\mathrm{Hallo}\mathrm{Herr}\mathrm{auch}.\mathrm{Gute}\mathrm{Nachrichten}$$

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