Question

The value of limit $\underset{x\to \infty }{lim}{x}^3(\sqrt{x^2+\sqrt{1+x^4}}-x\sqrt{2})$ is

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{4\sqrt{2}}$
• D. $\frac{1}{8\sqrt{2}}$

Answer 1

The correct option is C $\frac{1}{4\sqrt{2}}$

$L=\underset{x\to \infty }{lim}{x}^3(\sqrt{x^2+\sqrt{1+x^4}}-x\sqrt{2})$
$=\underset{x\to \infty }{lim}{x}^3\left(\frac{x^2+\sqrt{1+x^4}-2x^2}{\sqrt{x^2+\sqrt{1+x^4}}+x\sqrt{2}}\right)$
$=\underset{x\to \infty }{lim}{x}^3\left(\frac{\sqrt{1+x^4}-x^2}{\sqrt{x^2+\sqrt{1+x^4}}+x\sqrt{2}}\right)$
$=\underset{x\to \infty }{lim}{x}^3(\frac{\sqrt{1+x^4}-x^2}{\sqrt{x^2+\sqrt{1+x^4}}+x\sqrt{2}}\times \frac{\sqrt{1+x^4}+x^2}{\sqrt{1+x^4}+x^2})$
$=\underset{x\to \infty }{lim}\left(\frac{x^3}{(\sqrt{x^2+\sqrt{1+x^4}}+x\sqrt{2})(\sqrt{1+x^4}+x^2)}\right)$
$=\underset{x\to \infty }{lim}\frac{1}{(\sqrt{1+\sqrt{\frac{1}{x^4}+1}}+\sqrt{2})(\sqrt{1+\frac{1}{x^4}}+1)}$
$=\frac{1}{2\sqrt{2}\times 2}$
$=\frac{1}{4\sqrt{2}}$