Question

Let $f\left(x\right)=x^3\{\sqrt{x^2+\sqrt{x^4+1}}-x\sqrt{2}\}$. Then $\underset{x\to \infty }{lim}f\left(x\right)$ is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$Wehavef\left(x\right)=\frac{x^3\{x^2+\sqrt{x^4+1}-2x^2\}}{\sqrt{x^2+\sqrt{x^4+1}}+x\sqrt{2}}=\frac{x^3\{\sqrt{x^4+1}-x^2\}}{\sqrt{x^2+\sqrt{x^4+1}}+x\sqrt{2}}=\frac{x^3(x^4+1-x^4)}{[\sqrt{x^2+\sqrt{x^4+1}}+x\sqrt{2}][\sqrt{x^4+1}+x^2]}=\frac{x^3}{[\sqrt{x^2+\sqrt{x^4+1}}+x\sqrt{2}][\sqrt{x^4+1}+x^2]}=\frac{1}{[\sqrt{1+\sqrt{1+\frac{1}{x^4}}}+\sqrt{2}][\sqrt{1+\frac{1}{x^4}}+1]}=\frac{1}{\left(\sqrt{1+\sqrt{1}+\sqrt{2}}\right)(\sqrt{1}+1)}=\frac{1}{2\sqrt{2}\left(2\right)}=\frac{1}{4\sqrt{2}}$