ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 139

\[1)\ x_{n} = \sqrt[3]{\frac{n^{2} + 2n + 3}{4n^{2} - n + 2}}:\]

\[\lim_{x \rightarrow \infty}\sqrt[3]{\frac{n^{2} + 2n + 3}{4n^{2} - n + 2}} =\]

\[= \lim_{x \rightarrow \infty}\sqrt[3]{\frac{1 + \frac{2}{n} + \frac{3}{n^{2}}}{4 - \frac{1}{n} + \frac{2}{n^{2}}}} = \sqrt[3]{\frac{1}{4}} = \frac{\sqrt[3]{2}}{2}.\]

\[2)\ x_{n} = \frac{\sqrt[3]{8n^{3} + 2n^{2} + 1}}{n}:\]

\[\lim_{x \rightarrow \infty}\frac{\sqrt[3]{8n^{3} + 2n^{2} + 1}}{n} =\]

\[= \lim_{x \rightarrow \infty}\sqrt[3]{8 + \frac{2}{n} + \frac{1}{n^{3}}} = \sqrt[3]{8} = 2.\]

\[= \frac{6n - 4}{\sqrt{3n^{2} + 4n + 1} + \sqrt{3n^{2} - 2n + 5}};\]

\[\lim_{x \rightarrow \infty}x_{n} =\]

\[= \lim_{x \rightarrow \infty}\frac{6 - \frac{4}{n}}{\sqrt{3 + \frac{4}{n} + \frac{1}{n^{2}}} + \sqrt{3 - \frac{2}{n} + \frac{5}{n^{2}}}} =\]

\[= \frac{6}{2\sqrt{3}} = \sqrt{3}.\]

\[4)\ x_{n} = \sqrt[3]{n^{3} + 2n} - n =\]

\[= n\sqrt[3]{1 - \frac{2}{n^{2}}} - n:\]

\[\lim_{x \rightarrow \infty}\left( n\sqrt[3]{1 - \frac{2}{n^{2}}} - n \right) =\]

\[= n\sqrt[3]{1 - 0} - n = 0.\]