ГДЗ по алгебре и начала математического анализа 11 класс Колягин Проверь себя II. Сложный вариант

\[\mathbf{1}.\]

\[x_{n} = \frac{2n^{4} + n^{2} - 4}{3n^{4} - n^{3} + 2};\]

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

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

\[= \frac{2 + 0 - 0}{3 - 0 + 0} = \frac{2}{3}.\]

\[Ответ:\ \ \frac{2}{3}.\]

\[\mathbf{2}.\]

\[\lim_{n \rightarrow 3}\frac{x^{2} - 6x + 9}{x - 3} =\]

\[= \lim_{n \rightarrow 3}\frac{(x - 3)^{2}}{x - 3} = \lim_{n \rightarrow 3}(x - 3) = 0.\]

\[Ответ:\ \ нет.\]

\[\mathbf{3.\ }\]

\[f(x) = 2\sqrt{x} - 3\ln(x + 2);\]

\[f^{'}(x) = 2 \bullet \frac{1}{2\sqrt{x}} - 3 \bullet \frac{1}{x + 2} = 0;\]

\[(x + 2) - 3\sqrt{x} = 0\]

\[x + 2 = 3\sqrt{x}\]

\[x^{2} + 4x + 4 = 9x\]

\[x^{2} - 5x + 4 = 0\]

\[D = 25 - 16 = 9\]

\[x_{1} = \frac{5 - 3}{2} = 1;\]

\[x_{2} = \frac{5 + 3}{2} = 4.\]

\[Ответ:\ \ 1;\ 4.\]

\[\mathbf{4}\text{.\ }\]

\[f(x) = \frac{1}{3}x^{3} - x^{2} + 5;\text{\ \ }\]

\[y = 3x - 2;\]

\[y^{'}(x) = \frac{1}{3} \bullet 3x^{2} - 2x = 3;\]

\[x^{2} - 2x - 3 = 0\]

\[D = 4 + 12 = 16\]

\[x_{1} = \frac{2 - 4}{2} = - 1;\ \]

\[x_{2} = \frac{2 + 4}{2} = 3;\]

\[f( - 1) = - \frac{1}{3} - 1 + 5 = \frac{11}{3};\]

\[f(3) = 9 - 9 + 5 = 5;\]

\[y = \frac{11}{3} + 3(x + 1) =\]

\[= 3x + 6\frac{2}{3} = 5 + 3(x - 3) =\]

\[= 3x - 4.\]

\[Ответ:\ \ y = 3x - 4;\ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }y = 3x + 6\frac{2}{3}.\]