Let-lim-x-0-x-4-3-a-2-a-4-x-4-x-8-a-gt-0-If-is-finite-then-a-a-3-2-b-a-3-2-c-1-3-d-1-9-

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

Let φ = \lim_{x \to 0} \frac{x^{4} + 3 (a^{2} - \sqrt{a^{4} + x^{4}})}{x^{8}}; a > 0

If φ is finite then

(a) a = \frac{3}{2} (b) a = \sqrt{\frac{3}{2}} (c) φ = \frac{1}{3} (d) φ = \frac{1}{9}

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

\lim_{x \to 0} \frac{x^{4} + 3 (a^{2} - a^{2} (1 + \frac{x^{4}}{2 a^{4}} - \frac{x^{8}}{8 a^{8}}))}{x^{8}} = \frac{x^{4} - \frac{3 x^{4}}{2 a^{2}} + \frac{3 x^{8}}{8 a^{6}}}{x^{8}} = φ

\Rightarrow x^{4} - \frac{3 x^{4}}{2 a^{2}} = 0 \Rightarrow 1 = \frac{3}{2 a^{2}} \Rightarrow a = \pm \sqrt{\frac{3}{2}}

φ = \frac{3}{8 a^{6}} = \frac{3}{8} . \frac{8}{27} = \frac{1}{9}

Answered by EDWIN88 last updated on 08/Feb/21

φ = \lim_{x \to 0} \frac{(x^{4} + {3 a}^{2}) - 3 \sqrt{a^{4} + x^{4}}}{x^{8}} =

φ = \lim_{x \to 0} \frac{{(x^{4} + {3 a}^{2})}^{2} - 9 (a^{4} + x^{4})}{x^{8} ((x^{4} + {3 a}^{2}) + 3 \sqrt{a^{4} + x^{4}}))} =

φ = \frac{1}{{6 a}^{2}} \times \lim_{x \to 0} \frac{x^{8} + {6 a}^{2} x^{4} + {9 a}^{4} - {9 a}^{4} - {9 x}^{4}}{x^{8}}

φ = \frac{1}{{6 a}^{2}} \times \lim_{x \to 0} \frac{x^{8} + x^{4} ({6 a}^{2} - 9)}{x^{8}}

since the value of φ is finite it follows that

{6 a}^{2} - 9 must be is zero; a = \sqrt{\frac{3}{2}}

then φ = \frac{1}{6 (\frac{9}{6})} = \frac{1}{9}

Commented by liberty last updated on 08/Feb/21

Hallo Herr wie geht es dir

Commented by EDWIN88 last updated on 08/Feb/21

Hallo Herr auch . Gute Nachrichten

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