граница на функция

[Гост]

[tex]\lim_{x \to \infty}a= ?[/tex]
[tex]a= n^3 ( \sqrt{n^2+ \sqrt{n^4+1}}-n \sqrt{2})[/tex]

[Добромир Глухаров]

Два пъти рационализираме числителя:

$a=n^3\cdot\frac{\left(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2}\right)\left(\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}\right)}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}$

$a=n^3\cdot\frac{(n^2+\sqrt{n^4+1})-2n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=\frac{n^3}{n\left(\sqrt{1+\sqrt{1+\frac{1}{n^4}}}+\sqrt{2}\right)}\cdot\frac{(\sqrt{n^4+1}-n^2)(\sqrt{n^4+1}+n^2)}{\sqrt{n^4+1}+n^2}$

$a=\frac{\cancel{n^2}}{\sqrt{2}+\sqrt{1+\sqrt{1+\frac{1}{n^4}}}}\cdot\frac{\cancel{n^4}+1\cancel{-n^4}}{\cancel{n^2}\left(1+\sqrt{1+\frac{1}{n^4}}\right)}$

$\lim_{n\to\infty}a=\lim_{n\to\infty}\frac{1}{\left(\sqrt{2}+\sqrt{1+\sqrt{1+\frac{1}{n^4}}}\right)\left(1+\sqrt{1+\frac{1}{n^4}}\right)}=\frac{1}{\left(\sqrt{2}+\sqrt{1+\sqrt{1+0}}\right)\left(1+\sqrt{1+0}\right)}=\frac{1}{2.\sqrt{2}.2}=\frac{\sqrt{2}}{8}$