Решебник по алгебре 11 класс Никольский Параграф 2. Предел функции и непрерывность Задание 15

\[\boxed{\mathbf{15.}}\]

\[\textbf{а)}\ \lim_{x \rightarrow \frac{\pi}{2}}\left( \sin x + \cos x \right) =\]

\[= \lim_{x \rightarrow \frac{\pi}{2}}{\sin x} + \lim_{x \rightarrow \frac{\pi}{2}}{\cos x} =\]

\[= 1 + 0 = 1.\]

\[\textbf{б)}\ \lim_{x \rightarrow 1}\left( x^{4} - 2x^{2} + x + 1 \right) =\]

\[= 1 - 2 + 1 + 1 = 0.\]

\[\textbf{в)}\ \lim_{x \rightarrow 1}\frac{x^{3} - 1}{x - 1}\]

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

\[= x^{2} + x + 1;\ \ x \neq 0.\]

\[\lim_{x \rightarrow 1}{(x^{2} + x + 1)} = 1 + 1 + 1 = 3.\]

\[\textbf{г)}\ \lim_{x \rightarrow - 2}\frac{\sin{(x + 2)}}{x + 2}\]

\[x + 2 = t:\]

\[\lim_{x \rightarrow - 2}\frac{\sin{(x + 2)}}{x + 2} = \lim_{x \rightarrow 0}\frac{\sin t}{t} = 1.\]

\[\textbf{д)}\ \lim_{x \rightarrow 0}\frac{1 - \cos{2x}}{x^{2}}\]

\[\text{si}n^{2}\frac{a}{2} = \frac{1 - \cos a}{2}\]

\[\frac{1 - \cos{2x}}{x^{2}} = \frac{2\sin^{2}x}{x^{2}}\]

\[\lim_{x \rightarrow 0}\frac{1 - \cos{2x}}{x^{2}} = \lim_{x \rightarrow 0}\frac{2\sin^{2}x}{x^{2}} =\]

\[= \lim_{x \rightarrow 0}{2\left( \frac{\sin x}{x} \right)^{2}} =\]

\[= 2\left( \lim_{x \rightarrow 0}\frac{\sin x}{x} \right)^{2} = 2 \cdot 1^{2} = 2.\]

\[\textbf{е)}\ \lim_{x \rightarrow 0}(1 + 3x)^{\frac{1}{x}}\]

\[\lim_{x \rightarrow 0}(1 + x)^{\frac{1}{x}} = e - второй\ \]

\[замечательный\ предел.\]

\[t = 3x:\]

\[(1 + 3x)^{\frac{1}{x}} = (1 + t)^{\frac{3}{t}}\]

\[\lim_{x \rightarrow 0}(1 + t)^{\frac{3}{t}} = \lim_{x \rightarrow 0}\left( (1 + t)^{\frac{1}{t}} \right)^{3} =\]

\[= \left( \lim_{x \rightarrow 0}(1 + t)^{\frac{1}{t}} \right)^{3} = e^{3}.\]

\[\textbf{ж)}\ \lim_{x \rightarrow 0}(1 + 3x)^{\frac{2}{x}}\]

\[\lim_{x \rightarrow 0}(1 + 3x)^{\frac{2}{x}} = \lim_{x \rightarrow 0}\left( 1 + \frac{3x}{2} \right)^{\frac{2}{3x} \cdot 3}\]

\[\lim_{x \rightarrow 0}(1 + x)^{\frac{1}{x}} = e - второй\ \]

\[замечательный\ предел.\]

\[t = \frac{3x}{2}:\]

\[t \rightarrow 0.\]

\[\lim_{x \rightarrow 0}\left( 1 + \frac{3x}{2} \right)^{\frac{2}{3x} \cdot 3} = \lim_{x \rightarrow 0}(1 + t)^{\frac{3}{t}} =\]

\[= \lim_{x \rightarrow 0}\left( (1 + t)^{\frac{1}{t}} \right)^{3} =\]

\[= \left( \lim_{x \rightarrow 0}(1 + t)^{\frac{1}{t}} \right)^{3} = e^{3}.\]

\[\textbf{з)}\ \lim_{x \rightarrow 0}\left( \frac{2 + x}{2} \right)^{\frac{4}{x}}\]

\[\lim_{x \rightarrow 0}(1 + x)^{\frac{1}{x}} = e - второй\ \]

\[замечательный\ предел.\]

\[f(x) = \left( \frac{2 + x}{2} \right)^{\frac{4}{x}} = \left( 1 + \frac{1}{2}x \right)^{\frac{4}{x}}\]

\[t = \frac{1}{2}x;\ \ t \rightarrow 0:\]

\[\lim_{x \rightarrow 0}(1 + t)^{\frac{2}{t}} = \lim_{x \rightarrow 0}\left( (1 + t)^{\frac{1}{t}} \right)^{2} =\]

\[= \left( \lim_{x \rightarrow 0}(1 + t)^{\frac{1}{t}} \right)^{2} = e^{2}.\]