Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 311

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

\[\lim_{x \rightarrow \infty}\frac{f(x)}{x} = \lim_{x \rightarrow \infty}\frac{(x + 3)^{3}}{x(x - 1)^{2}} =\]

\[= \lim_{x \rightarrow \infty}\frac{x^{3} + 9x^{2} + 9x + 27}{x^{3} - 2x^{2} + x} = \frac{1}{1} = 1;\]

\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]

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

\[= \lim_{x \rightarrow \infty}\left( \frac{x^{3} + 9x^{2} + 9x + 27}{x^{2} - 2x + 1} - x \right);\]

\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]

\[= \lim_{x \rightarrow \infty}\frac{11x^{2} + 8x + 27}{x^{2} - 2x + 1} = \frac{11}{1} = 11.\]

\[Ответ:\ \ x = 1;\ y = x + 11.\]

\[2)\ f(x) = \sqrt{x^{2} + 4x + 3};\]

\[\lim_{x \rightarrow \infty}\frac{f(x)}{x} = \lim_{x \rightarrow \infty}\frac{\sqrt{x^{2} + 4x + 3}}{x} =\]

\[= \pm \lim_{x \rightarrow \infty}\sqrt{1 + \frac{4}{x} + \frac{3}{x^{2}}} = \pm \sqrt{1} = \pm 1;\]

\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]

\[= \lim_{x \rightarrow \infty}\left( \sqrt{x^{2} + 4x + 3} \mp x \right) =\]

\[= \lim_{x \rightarrow \infty}\frac{\left( x^{2} + 4x + 3 \right) - x^{2}}{\sqrt{x^{2} + 4x + 3} \pm x};\]

\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]

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

\[= \lim_{x \rightarrow \infty}\frac{4 + \frac{3}{x}}{\pm \sqrt{1 + \frac{4}{x} + \frac{3}{x^{2}}} \pm 1};\]

\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) = \frac{4}{\pm \sqrt{1} \pm 1} =\]

\[= \frac{4}{\pm 2} = \pm 2.\]

\[Ответ:\ \ y = x + 2;\ y = - x - 2.\]