Решебник по алгебре 11 класс Никольский Параграф 2. Предел функции и непрерывность Задание 17

\[\boxed{\mathbf{17.}}\]

\[\mathbf{а)\ }\lim_{x \rightarrow 0}\frac{\text{tgx}}{7x} = \frac{1}{7}\lim_{x \rightarrow 0}\frac{\text{tgx}}{x} = \frac{1}{7} \cdot 1 = \frac{1}{7}.\]

\[\mathbf{б)\ }\lim_{x \rightarrow 0}\frac{tg7x}{x}\]

\[7x = y:\]

\[\lim_{x \rightarrow 0}\frac{7 \cdot tg7x}{7x} = 7\lim_{x \rightarrow 0}\frac{tg7x}{7x} =\]

\[= 7\lim_{x \rightarrow 0}\frac{\text{tgy}}{y} = 7 \cdot 1 = 7.\]

\[\mathbf{в)\ }\lim_{x \rightarrow 0}\frac{tg5x}{10x}\]

\[5x = y:\]

\[\lim_{x \rightarrow 0}\frac{tg5x}{10x} = \lim_{x \rightarrow 0}\frac{\text{tgy}}{2y} =\]

\[= \frac{1}{2}\lim_{x \rightarrow 0}\frac{\text{tgy}}{y} = \frac{1}{2} \cdot 1 = \frac{1}{2}.\]

\[\textbf{г)}\ \lim_{x \rightarrow 0}\frac{\sin x}{5x} = \frac{1}{5}\lim_{x \rightarrow 0}\frac{\sin x}{x} =\]

\[= \frac{1}{5} \cdot 1 = \frac{1}{5}.\]

\[\textbf{д)}\ \lim_{x \rightarrow 0}\frac{5x}{\sin{5x}}\]

\[\frac{5x}{\sin{5x}} = \frac{5x\ :5x}{\sin{5x\ :5x}} = 1\ :\frac{\sin{5x}}{5x}\]

\[5x = y:\]

\[\frac{1}{\frac{\sin{5x}}{5x}} = \frac{1}{\frac{\sin y}{y}}\]

\[\lim_{x \rightarrow 0}\frac{5x}{\sin{5x}} = \lim_{x \rightarrow 0}\frac{1}{\frac{\sin y}{y}} = \frac{1}{1} = 1.\]

\[\textbf{е)}\ \lim_{x \rightarrow 0}\frac{\sin{3x}}{2x}\]

\[\frac{\sin{3x}}{2x} = \frac{3 \cdot \sin{3x}}{3 \cdot 2x} = \frac{3 \cdot \sin{3x}}{2 \cdot 3x}\]

\[3x = y:\]

\[\frac{3 \cdot \sin{3x}}{2 \cdot 3x} = \frac{3 \cdot \sin y}{2 \cdot y};\]

\[\lim_{x \rightarrow 0}\frac{\sin{3x}}{2x} = \lim_{x \rightarrow 0}\frac{3 \cdot \sin{3x}}{2 \cdot 3x} =\]

\[= \frac{3}{2}\lim_{x \rightarrow 0}\frac{\sin{3x}}{3x} = \frac{3}{2}\lim_{x \rightarrow 0}\frac{\sin y}{y} =\]

\[= \frac{3}{2} \cdot 1 = 1,5.\]

\[\textbf{ж)}\ \lim_{x \rightarrow 0}\frac{\text{tgx}}{\sin x}\]

\[\frac{\text{tgx}}{\sin x} = \frac{\sin x}{\cos x}\ :\sin x = \frac{1}{\cos x};\]

\[\lim_{x \rightarrow 0}\frac{\text{tgx}}{\sin x} = \lim_{x \rightarrow 0}\frac{1}{\cos x} = \frac{1}{\cos 0} =\]

\[= \frac{1}{1} = 1.\]

\[\textbf{з)}\ \lim_{x \rightarrow 0}\frac{tg5x}{\sin x}\]

\[\frac{tg5x}{\sin x} = \frac{5x \cdot \frac{tg5x}{5x}}{\sin x} = \frac{5x \cdot \frac{tg5x}{5x}}{\frac{x \cdot \sin x}{x}};\]

\[\lim_{x \rightarrow 0}\frac{tg5x}{\sin x} = \lim_{x \rightarrow 0}\frac{5x \cdot \frac{tg5x}{5x}}{\frac{x \cdot \sin x}{x}} =\]

\[= \frac{\lim_{x \rightarrow 0}{5x \cdot \frac{tg5x}{5x}}}{\lim_{x \rightarrow 0}\frac{x \cdot \sin x}{x}} =\]

\[= \frac{\lim_{x \rightarrow 0}{5x} \cdot \lim_{x \rightarrow 0}\frac{tg5x}{5x}}{\lim_{x \rightarrow 0}x \cdot \lim_{x \rightarrow 0}\frac{\sin x}{x}} =\]

\[= \lim_{x \rightarrow 0}\frac{5x}{x} \cdot \lim_{x \rightarrow 0}\frac{tg5x}{5x} = 5\lim_{x \rightarrow 0}\frac{tg5x}{5x};\]

\[5x = y:\]

\[\lim_{x \rightarrow 0}\frac{tg5x}{\sin x} = 5\lim_{x \rightarrow 0}\frac{tg5x}{5x} =\]

\[= 5\lim_{x \rightarrow 0}\frac{\text{tgy}}{y} = 5 \cdot 1 = 5.\]

\[\textbf{и)}\ \lim_{x \rightarrow 0}\frac{tg2x}{\sin{5x}}\]

\[\frac{tg2x}{\sin{5x}} = \frac{2x \cdot \frac{tg2x}{2x}}{\frac{5x \cdot \sin{5x}}{5x}};\]

\[\lim_{x \rightarrow 0}\frac{tg2x}{\sin{5x}} = \lim_{x \rightarrow 0}\frac{2x \cdot \frac{tg2x}{2x}}{\frac{5x \cdot \sin{5x}}{5x}} =\]

\[= \frac{\lim_{x \rightarrow 0}{2x} \cdot \lim_{x \rightarrow 0}\frac{tg2x}{2x}}{\lim_{x \rightarrow 0}{5x} \cdot \lim_{x \rightarrow 0}\frac{\sin{5x}}{5x}} =\]

\[= \lim_{x \rightarrow 0}\frac{2x}{5x} \cdot \frac{\lim_{x \rightarrow 0}\frac{tg2x}{2x}}{\lim_{x \rightarrow 0}\frac{\sin{5x}}{5x}} =\]

\[= \frac{2}{5} \cdot \frac{\lim_{x \rightarrow 0}\frac{tg2x}{2x}}{\lim_{x \rightarrow 0}\frac{\sin{5x}}{5x}};\]

\[2x = y;\ \ 5x = z:\]

\[\frac{2}{5} \cdot \frac{\lim_{x \rightarrow 0}\frac{\text{tgy}}{y}}{\lim_{x \rightarrow 0}\frac{\sin z}{z}} = \frac{2}{5} \cdot \frac{1}{1} = \frac{2}{5} = 0,4.\]