Уравнение и граница

[Hephaestus]

Да се пресметне границата [tex]\lim_{n \to \infty}n^{p}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})[/tex],
където [tex]p[/tex] е най-големият корен на уравнението [tex]\frac{x^{3} - 2x^{2} + x + 4}{x+1} = \frac{7}{4}sin\left(\frac{\pi x}{3}\right)[/tex].

[Добромир Глухаров]

$\frac{x^3-2x^2+x+4}{x+1}=\frac{7}{4}sin\left(\frac{\pi x}{3}\right)$

$f(x)=\frac{x^3-2x^2+x+4}{x+1}=x^2-3x+4=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}$

$f(x)\geq\frac{7}{4}$

$\varphi(x)=\frac{7}{4}sin\left(\frac{\pi x}{3}\right)\leq\frac{7}{4}$

$\Rightarrow f(x)=\varphi(x)=\frac{7}{4}$ за $x=\frac{3}{2}$

$p=\frac{3}{2}$

$\lim_{n\to\infty}n^p(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})=\lim_{n\to\infty}n^{\frac{3}{2}}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})=\lim_{n\to\infty}n\sqrt{n}\frac{(\sqrt{n+1}+\sqrt{n-1})^2-(2\sqrt{n})^2}{\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}}=$

$=\lim_{n\to\infty}n\sqrt{n}\frac{2n+2\sqrt{n^2-1}-4n}{\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}}=\lim_{n\to\infty}n\cdot\frac{1}{\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}+2}\cdot2(\sqrt{n^2-1}-n)=$

$=\lim_{n\to\infty}\frac{2}{\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}+2}\cdot\lim_{n\to\infty}\frac{n(n^2-1-n^2)}{\sqrt{n^2-1}+n}=\frac{2}{4}\cdot\lim_{n\to\infty}\frac{-1}{\sqrt{1-\frac{1}{n^2}}+1}=\frac{1}{2}\cdot\left(-\frac{1}{2}\right)=-\frac{1}{4}$