Question

Найти предел:
(в знаменателе первый корень кубический)​

Answer 1

Объяснение:

$\lim\limits _{n \to \infty}\frac{\sqrt{n^3+n^2-4}-\sqrt[5]{n^6}}{\sqrt[3]{n^5+2n}+\sqrt[4]{n^6+3n^4+2}}=\Big [\, \sqrt{n^3}=n^{3/2}\; ,\; \sqrt[5]{n^6}=n^{6/5}\; ,\; \sqrt[3]{n^5}=n^{5/3}\; ,\\\\\sqrt[4]{n^6}=n^{3/2}\; ,\; \; 5/3>3/2>6/5\; \; \to \; \; delim\; na\; n^{5/3}\; \Big ]\\\\\frac{\sqrt{n^3+n^2-4}}{n^{5/3}}=\sqrt{\frac{n^3+n^2-4}{n^{10/3}}}=\sqrt{n^{-1/3}+n^{-4/3}-4\cdot n^{-10/3}}=\sqrt{\frac{1}{n^{1/3}}+\frac{1}{n^{4/3}}-\frac{4}{n^{10/3}}}=\\\\=\sqrt{\frac{1}{\sqrt[3]{n}}+\frac{1}{\sqrt[3]{n^4}}-\frac{1}{\sqrt[3]{n^{10}}}}\; \to \; \sqrt{0+0-0}=0$

$\frac{\sqrt[5]{n^6}}{n^{5/3}}=\frac{n^{6/5}}{n^{5/3}}=n^{-7/15}=\frac{1}{\sqrt[15]{n^7}}\; \to \; \frac{1}{\infty }\; \to \; 0\\\\\frac{\sqrt[3]{n^5+2n}}{n^{5/3}}=\sqrt[3]{\frac{n^5}{n^{5}}+2\cdot \frac{n}{n^5}}=\sqrt[3]{1+\frac{2}{n^4}}\; \to \; \; \sqrt[3]{1+0}\; \to \sqrt[3]{1}=1\\\\\frac{\sqrt[4]{n^6+3n^4+2}}{n^{5/3}}=\sqrt[4]{\frac{n^6}{n^{20/3}}+3\cdot \frac{n^4}{n^{20/3}}+\frac{2}{n^{20/3}}}=\sqrt[4]{\frac{1}{n^{2/3}}+\frac{3}{n^{{8/3}}}+\frac{2}{n^{20/3}}}\; \to \; 0$

$\lim\limits _{n \to \infty}\frac{\sqrt{n^3+n^2-4}-\sqrt[5]{n^6}}{\sqrt[3]{n^5+2n}+\sqrt[4]{n^6+3n^4+2}}=\lim\limits _{n \to \infty}\frac{\sqrt{\frac{1}{\sqrt[3]{n}}+\frac{1}{\sqrt[3]{n^4}}\; -\; \frac{1}{\sqrt[3]{n^{10}}}}-\frac{1}{\sqrt[15]{n^7}}}{\sqrt[3]{1+\frac{2}{n^4}}\; +\; \sqrt[4]{\frac{1}{\sqrt[3]{n^{2}}}+\frac{3}{\sqrt[3]{n^{{8}}}}+\frac{2}{\sqrt[3]n^{20}}}}}=\\\\\\=\frac{0-0}{1+0}=\frac{0}{1}=0$