Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 310

\[1)\ f(x) = \frac{x + 1}{x} = 1 + \frac{1}{x}.\]

\[Ответ:\ x = 0;\ y = 1.\]

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

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

\[= \lim_{x \rightarrow \infty}\frac{1}{1 + \frac{4}{x}} = \frac{1}{1 + 0} = 1;\]

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

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

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

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

\[= \lim_{x \rightarrow \infty}\frac{- 4}{1 + \frac{4}{x}} = - \frac{4}{1 + 0} = - 4.\]

\[Ответ:\ \ x = - 4;\ y = x - 4.\]

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

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

\[= \lim_{x \rightarrow \infty}\frac{1 - \frac{2}{x} + \frac{3}{x^{2}}}{1 + \frac{2}{x}} = \frac{1 - 0 + 0}{1 + 0} = 1;\]

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

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

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

\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) = \lim_{x \rightarrow \infty}\frac{3 - 4x}{x + 2} =\]

\[= \lim_{x \rightarrow \infty}\frac{\frac{3}{x} - 4}{1 + \frac{2}{x}} = \frac{0 - 4}{1 + 0} = - 4.\]

\[Ответ:\ \ x = - 2;\ y = x - 4.\]

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

\[\lim_{x \rightarrow \infty}\frac{f(x)}{x} = \lim_{x \rightarrow \infty}\frac{x^{2}}{x^{2} + 6x + 9} =\]

\[= \lim_{x \rightarrow \infty}\frac{1}{1 + \frac{6}{x} + \frac{9}{x^{2}}} = \frac{1}{1 + 0 + 0} = 1;\]

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

\[= \lim_{x \rightarrow \infty}\left( \frac{x^{3}}{x^{2} + 6x + 9} - x \right) =\]

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

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

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

\[= \lim_{x \rightarrow \infty}\frac{- 6 - \frac{9}{x}}{1 + \frac{6}{x} + \frac{9}{x^{2}}} =\]

\[= \frac{- 6 - 0}{1 + 0 + 0} = - 6.\]

\[Ответ:\ \ x = - 3;\ y = x - 6.\]