Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 155

\[1)\ f(x) = \left\{ \begin{matrix} \log_{2}(x + 1)\ при\ x \leq 1 \\ \frac{b}{x^{2}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }при\ x > 1 \\ \end{matrix} \right.\ \]

\[a = 1;\]

\[\lim_{\begin{matrix} x \rightarrow 1 \\ x < 1 \\ \end{matrix}}{\log_{2}(x + 1)} = \log_{2}2 = 1;\]

\[\lim_{\begin{matrix} x \rightarrow 1 \\ x > 1 \\ \end{matrix}}\frac{b}{x^{2}} = \frac{b}{1^{2}} = b = 1.\]

\[Ответ:\ \ 1.\]

\[2)\ f(x) = \left\{ \begin{matrix} \sin\frac{x}{2}\ при\ x < \pi \\ \text{bx\ \ \ \ \ \ }при\ x \geq \pi \\ \end{matrix} \right.\ \]

\[a = \pi;\]

\[\lim_{\begin{matrix} x \rightarrow \pi \\ x < \pi \\ \end{matrix}}{\sin\frac{x}{2}} = \sin\frac{\pi}{2} = 1;\]

\[\lim_{\begin{matrix} x \rightarrow \pi \\ x > \pi \\ \end{matrix}}\text{bx} = b\pi = 1\]

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

\[Ответ:\ \ \frac{1}{\pi}.\]

\[3)\ f(x) = \left\{ \begin{matrix} \frac{1 + x}{1 + x^{3}}\ при\ x \neq - 1 \\ \text{b\ \ \ \ \ \ \ \ \ \ \ }при\ x = - 1 \\ \end{matrix} \right.\ \]

\[a = - 1;\]

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

\[\lim_{x \rightarrow - 1}b = b = \frac{1}{3}.\]

\[Ответ:\ \ \frac{1}{3}.\]

\[4)\ f(x) = \left\{ \begin{matrix} \cos x\text{\ \ \ \ \ \ \ }при\ x \leq 0 \\ b(x - 1)\ при\ x > 0 \\ \end{matrix} \right.\ \]

\[a = 0;\]

\[\lim_{\begin{matrix} x \rightarrow 0 \\ x < 0 \\ \end{matrix}}{\cos x} = \cos 0 = 1;\]

\[\lim_{\begin{matrix} x \rightarrow 0 \\ x > 0 \\ \end{matrix}}{b(x - 1)} = - b = 1\]

\[b = - 1.\]

\[Ответ:\ - 1.\]