ГДЗ по алгебре 11 класс Никольский Параграф 2. Предел функции и непрерывность Задание 19

\[\boxed{\mathbf{19.}}\]

\[\mathbf{а)\ }\lim_{x \rightarrow \infty}\frac{3x + 7}{2x + 1} = \lim_{x \rightarrow \infty}\frac{\frac{3x}{x} + \frac{7}{x}}{\frac{2x}{x} + \frac{1}{x}} =\]

\[= \lim_{x \rightarrow \infty}\frac{3 + \frac{7}{x}}{2 + \frac{1}{x}} = \frac{3}{2} = 1,5.\]

\[\mathbf{б)\ }\lim_{x \rightarrow \infty}\frac{2x - 2}{5x + 6} = \lim_{x \rightarrow \infty}\frac{\frac{2x}{x} - \frac{2}{x}}{\frac{5x}{x} + \frac{6}{x}} =\]

\[= \lim_{x \rightarrow \infty}\frac{2 - \frac{2}{x}}{5 + \frac{6}{x}} = \frac{2}{5} = 0,4.\]

\[\mathbf{в)\ }\lim_{x \rightarrow + \infty}\frac{3x^{3} - 5x^{2} + 7}{5x^{2} + 7x - 5} =\]

\[= \lim_{x \rightarrow + \infty}\frac{\frac{3x^{3}}{x^{3}} - \frac{5x^{2}}{x^{3}} + \frac{7}{x^{3}}}{\frac{5x^{2}}{x^{3}} + \frac{7x}{x^{3}} - \frac{5}{x^{3}}} =\]

\[= \lim_{x \rightarrow + \infty}\frac{3 - \frac{5}{x} + \frac{7}{x^{3}}}{\frac{5}{x} + \frac{7}{x^{2}} - \frac{5}{x^{3}}} =\]

\[\mathbf{=}\lim_{x \rightarrow + \infty}\frac{3}{0} = + \infty.\]

\[\mathbf{г)\ }\lim_{x \rightarrow - \infty}\frac{3x^{4} + 5x - 1}{2x^{3} + 3x^{2} + 9x + 1} =\]

\[= \lim_{x \rightarrow - \infty}\frac{\frac{3x^{4}}{x^{4}} + \frac{5x}{x^{4}} - \frac{1}{x^{4}}}{\frac{2x^{3}}{x^{4}} + \frac{3x^{2}}{x^{4}} + \frac{9x}{x^{4}} + \frac{1}{x^{4}}} =\]

\[\mathbf{=}\lim_{x \rightarrow - \infty}\frac{3 + \frac{5}{x^{3}} - \frac{1}{x^{4}}}{\frac{2}{x} + \frac{3}{x^{2}} + \frac{9}{x^{3}} + \frac{1}{x^{4}}} =\]

\[= \lim_{x \rightarrow - \infty}\frac{3}{0} = - \infty.\]

\[\mathbf{д)\ }\lim_{x \rightarrow 2}\frac{x^{3} - 8}{x - 2} =\]

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

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

\[= 4 + 4 + 4 = 12.\]

\[\textbf{е)}\ \lim_{x \rightarrow - 1}\frac{x^{4} - 1}{x^{3} + 1} =\]

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

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

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

\[= \lim_{x \rightarrow - 1}\frac{( - 1 - 1)(1 + 1)}{(1 + 1 + 1)} =\]

\[= \frac{- 2 \cdot 2}{3} = - \frac{4}{3}.\]