lim-x-x-3-x-2-x-4-1-x-2-

Question Number 120277 by bemath last updated on 30/Oct/20

\lim_{x \to \infty} x^{3} {\sqrt{x^{2} + \sqrt{x^{4} + 1}} - x \sqrt{2}} ?

Commented by benjo_mathlover last updated on 30/Oct/20

\lim_{x \to \infty} x^{4} {\sqrt{1 + \sqrt{1 + \frac{1}{x^{4}}}} - \sqrt{2}}

s e t t i n g \frac{1}{x} = z \land z \to 0

\lim_{z \to 0} \frac{\sqrt{1 + \sqrt{1 + z^{4}}} - \sqrt{2}}{z^{4}} \times \frac{\sqrt{1 + \sqrt{1 + z^{4}}} + \sqrt{2}}{\sqrt{1 + \sqrt{1 + z^{4}}} + \sqrt{2}}

\lim_{z \to 0} \frac{\sqrt{1 + z^{4}} - 1}{z^{4}} \times \frac{1}{2 \sqrt{2}} = \frac{\sqrt{2}}{4} \times \lim_{z \to 0} \frac{z^{4}}{z^{4} (\sqrt{1 + z^{4}} + 1)}

= \frac{\sqrt{2}}{4} \times \frac{1}{2} = \frac{\sqrt{2}}{8}

Answered by Dwaipayan Shikari last updated on 30/Oct/20

x^{3} (\sqrt{x^{2} + \sqrt{x^{4} + 1}} - x \sqrt{2})

= x^{3} (\sqrt{x^{2} + x^{2} \sqrt{1 + \frac{1}{x^{4}}}} - x \sqrt{2})

= x^{3} (x \sqrt{1 + \sqrt{1 + \frac{1}{x^{4}}}} - x \sqrt{2})

= x^{4} (\sqrt{1 + 1 + \frac{1}{2 x^{4}}} - \sqrt{2})

= \sqrt{2} x^{4} (\sqrt{1 + \frac{1}{4 x^{4}}} - 1)

= \sqrt{2} x^{4} (1 + \frac{1}{8 x^{4}} - 1)

= \frac{1}{4 \sqrt{2}}

Answered by bemath last updated on 30/Oct/20

\lim_{x \to \infty} x^{3} {\frac{x^{2} + \sqrt{x^{4} + 1} - 2 x^{2}}{\sqrt{x^{2} + \sqrt{x^{4} + 1}} + x \sqrt{2}}} =

\lim_{x \to \infty} \frac{x^{3} {\sqrt{x^{4} + 1} - x^{2}}}{\sqrt{x^{2} + \sqrt{x^{4} + 1}} + x \sqrt{2}} =

\lim_{x \to \infty} \frac{x^{3} {x^{4} + 1 - x^{4}}}{x (\sqrt{1 + \sqrt{1 + \frac{1}{x^{4}}}} + \sqrt{2}) (\sqrt{x^{4} + 1} + x^{2})} =

\lim_{x \to \infty} \frac{x^{3}}{x^{3} (\sqrt{1 + \sqrt{1 + \frac{1}{x^{4}}}} + \sqrt{2}) (\sqrt{1 + \frac{1}{x^{4}}} + 1)} =

\lim_{x \to \infty} \frac{1}{(\sqrt{1 + \sqrt{1 + \frac{1}{x^{4}}}} + \sqrt{2}) (\sqrt{1 + \frac{1}{x^{4}}} + 1)} =

\frac{1}{(\sqrt{2} + \sqrt{2}) \times 2} = \frac{1}{4 \sqrt{2}}

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