lim(n->бесконечность) n*(sqrt((n^2)+2)-sqrt((n^2)-2)). Created by zhekalisovenko. matematika-ru

lim(n->бесконечность) n*(sqrt((n^2)+2)-sqrt((n^2)-2))...

August 2022 1 6 Report

lim(n->бесконечность) n*(sqrt((n^2)+2)-sqrt((n^2)-2))

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animaldk $\lim\limits_{n\to\infty}n(\sqrt{n^2+2}-\sqrt{n^2-2})=\lim\limits_{n\to\infty}\frac{n(\sqrt{n^2+2}-\sqrt{n^2-2})(\sqrt{n^2+2}+\sqrt{n^2-2})}{\sqrt{n^2+2}+\sqrt{n^2-2}}\\\\=\lim\limits_{n\to\infty}\frac{n\left[(\sqrt{n^2+2})^2-(\sqrt{n^2-2})^2]}{\sqrt{n^2(1+\frac{2}{n^2})}+\sqrt{n^2(1-\frac{2}{n^2})}}=\lim\limits_{n\to\infty}\frac{n(n^2+2-n^2+2)}{n\sqrt{1+\frac{2}{n^2}}+n\sqrt{1-\frac{2}{n^2}}}$

$=\lim\limits_{n\to\infty}\frac{n\cdot4}{n\left(\sqrt{1+\frac{2}{n^2}}+\sqrt{1-\frac{2}{n^2}}\right)}=\lim\limits_{n\to\infty}\frac{4}{\sqrt{1+\frac{2}{n^2}}+\sqrt{1-\frac{2}{n^2}}}=\frac{4}{\sqrt1+\sqrt1}\\\\=\frac{4}{1+1}=\frac{4}{2}=2$

1 votes Thanks 1

zhekalisovenko Спасибо тебе огромное. А я на втором равно застрял :3

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najti uravnenie kasatelnoj k krivoj y 3x pri x 2

vypolnit raschyoty 3 i 4

1-x) )...

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