高数 01.[08-10]习题课

函数的连续与间断点
$1.f(x)在点x_0连续的等价形式 \\ \lim_{x \rightarrow x_0}{f(x)} = f(x_0) \\ \Longleftrightarrow \lim_{\Delta{x} \rightarrow 0}{[f(x_0 + \Delta{x}) -f(x_0)]} = 0 \\ \Longleftrightarrow f(x_0^-) = f(x_0) = f(x_0^+) \\ \qquad \qquad \color{blue}{\overline{左连续}} \qquad \color{blue}{\overline{右连续}} \\ $
$2.f(x)在点x_0剪断的类型 \\ 第一类间断点 \left \{ \begin{array}{l}可去间断点 \\ 跳跃间断点 \end{array} \right \}左右极限存在 \\ 第二类间断点 \left \{ \begin{array}{l}无穷间断点 \\ 振荡间断点 \end{array} \right \}左右极限至少有一个不存在 \\ $

$\left. \begin{array}{l}基本初等函数在定义域内连续 \\ 连续函数的四则运算的结果连续 \\ 连续函数的反函数连续 \\ 连续函数的复合函数连续 \end{array} \right \} 初等函数在定义区间内连续$
$说明：分段函数在界点处是否连续需讨论其左、右极限连续性$

$\color{blue}{连续函数的性质}$
$设f(x) \in C[a, b],则 \\ 1.f(x)在[a, b]上有界; \\ 2.f(x)在[a, b]上达到最大值与最小值; \\ 3.f(x)在[a, b]上可取最大值和最小值之间的任何值; \\ 4.当f(a)f(b) < 0时,必存在\xi \in (a, b),使f(\xi) = 0.$

例题分析
一、选择题
$例1.设f(x) = \left \{ \begin{array}{l} \dfrac{\sin{3x}}{x}, x \neq 0 \\ a, \qquad x = 0 \end{array} \right. 在=0处连续，则a = (\ \ {\color{blue}{D}}\ \ )$
$A.-1; \quad B.1; \quad C.2; \quad D.3$
$\color{blue}{ 解：\lim_{x \rightarrow 0}{\dfrac{\sin{3x}}{x}} = \lim_{x \rightarrow 0}{\dfrac{3x}{x}} = 3 \\ \because f(0^-) = f(0) = f(0^+) \\ \therefore a = 3 }$

$例2.设f(x) = \dfrac{1 - \cos^2{x}}{x^2},当x \neq 0时F(x) = f(x),若F(x)在点x = 0处连续，则F(0)等于(\ \ {\color{blue}{C}}\ \ )$
$A.-1; \quad B.0; \quad C.1; \quad D.2$
$\color{blue}{ 解：F(0) = \lim_{x \rightarrow 0}{F(x)} = \lim_{x \rightarrow 0}{f(x)} \\ = \lim_{x \rightarrow 0}{\dfrac{\sin^2{x}}{x^2} } \\ = \lim_{x \rightarrow 0}{(\dfrac{x}{x})^2}\\ = 1 }$

$例3.函数f(x) = \dfrac{x - 4}{x^2 - 3x - 4}的间断点的个数是(\ \ {\color{blue}{C}}\ \ )$
$A.0; \quad B.1; \quad C.2; \quad D.3$
$\color{blue}{ f(x)的间断点是连续性的三个条件之一，满足的点，本例中f(x)为有理函数，\\ 只需考察分母为零的点.由于x^2 - 3x - 4 = (x + 1)(x - 4).可知x_1 = -1, \\ x_2 = 4为f(x)的间断点，故选C }$

$例4.函数f(x) 在 x = a 连续，是在x = a 处有极限的(\ \ {\color{blue}{B}}\ \ )$
$A.既充分又必要条件; \quad B.充分非必要条件; \\\ C.必要非充分条件; \quad D.无关条件$
$\color{blue}{ 由连续定义可知选B }$

$例5.要使函数f(x) = \dfrac{\sqrt{1 + x} - \sqrt{1 - x}}{x}在x = 0处连续，应给f(0)补充定义的数值是(\ \ {\color{blue}{C}}\ \ )$
$A.1/2; \quad B.2; \quad C.1; \quad D.0$
$\color{blue}{ \lim_{x \rightarrow 0}{f(x)} = \lim_{x \rightarrow 0}{\dfrac{\sqrt{1 + x} - \sqrt{1 - x}}{x}} \\ = \lim_{x \rightarrow 0}{\dfrac{(1 + x) - (1 - x)}{x(\sqrt{1 + x} + \sqrt{1 - x})}} \\ = \lim_{x \rightarrow 0}{\dfrac{2}{\sqrt{1 + x} + \sqrt{1 - x}}} \\ = 1 }$

二 填空题
$例6.函数f(x) = \dfrac{x}{\sqrt[3]{3 - x}}的间断点是 \underline{\ \ {\color{blue}{3}}\ \ }$
$\color{blue}{ \sqrt[3]{3 - x} = 0 \Longrightarrow x = 3 }$

$例7.设f(x) = \left \{ \begin{array}{l}\dfrac{\sin{x}}{x}, x \neq 0 \\ 0, x = 0 \end{array} \right. 则x = 0为f(x)的第\underline{\ \ {\color{blue}{一}}\ \ }类间断点.$
$\color{blue}{ \lim_{x \rightarrow 0^-}{f(x)} = \lim_{x \rightarrow 0^+}{f(x)} \\ = \lim_{x \rightarrow 0}{\dfrac{\sin{x}}{x}} = \lim_{x \rightarrow 0}{\dfrac{x}{x}} = 1 \\ f(0) = 0 \neq 1 }$

$例8.设函数f(x) = \left \{ \begin{array}{l} 1 - e^{-x}, x < 0 \\ A + x, x geq 0 \end{array} \right. 在x = 0处连续，则常数A = \underline{\ \ {\color{blue}{0}}\ \ }.$
$\color{blue}{ \because f(0^-) = f(0^+) = f(0) \\ \therefore \lim_{x \rightarrow 0^-}{(1 - e^{-x})} = 0 = f(0) = A + 0 \\ A = 0 }$

三、解答题
$例9.设函数f(x) = \left \{ \begin{array}{l} \dfrac{x^4 + ax + b}{(x - 1)(x + 2)}, x \neq 1, x \neq -2 \\ 2, \qquad x = 1 \end{array} \right. 在点 x = 1 处连续，试确定常数a,b的值.$
$\color{blue}{ \because 当x = 1时，(x - 1) \rightarrow 0,且f(x) 在1处连续，故 \\ 1^4 + a \cdot 1 + b = 0 \\ 即 b = -a - 1 \\ 2 = \lim_{x \rightarrow 1}{\dfrac{x^4 + ax + b}{(x - 1)(x + 2)}}\\ = \lim_{x \rightarrow 1}{\dfrac{x^4 + ax -a - 1}{(x - 1)(x + 2)}}\\ = \lim_{x \rightarrow 1}{\dfrac{a(x - 1) + (x^2 + 1)(x + 1)(x -1)}{(x - 1)(x + 2)}}\\ = \lim_{x \rightarrow 1}{\dfrac{a + (x^2 + 1)(x + 1)}{(x + 2)}}\\ = \lim_{x \rightarrow 1}{\dfrac{a + (1^2 + 1)(1 + 1)}{(1 + 2)}}\\ = \lim_{x \rightarrow 1}{\dfrac{a + 4}{3}}\\ \therefore \\ a = 2 \\ b = -3 }$

$例10.确定A的值，使函数f(x) = \left \{ \begin{array}{l} 5e^x - \cos{x}, x \leq 0 \\ \dfrac{\sin{3x}}{\tan{Ax}}, \quad x > 0 \end{array} \right. 在点 x = 0 处连续.$
$\color{blue}{ 要使f(x) 在 x = 0 处连续，必须满足条件：\\ \lim_{x \rightarrow 0^-}{f(x)} = \lim_{x \rightarrow 0^+}{f(x)} = f(0). \\ \lim_{x \rightarrow 0^-}{(5e^x - \cos{x})} = 5 - 1 =4 = f(0)\\ \lim_{x \rightarrow 0^+}{\dfrac{\sin{3x}}{\tan{Ax}}} = \dfrac{3x}{Ax} = \dfrac{3}{A} = f(0) = 4 \\ A = \dfrac{3}{4} }$

$例11.设函数f(x) = \left \{ \begin{array}{l} \dfrac{\cos{x}}{x + 2}, x \geq 0 \\ \dfrac{\sqrt{a} - \sqrt{a - x}}{x}, \quad x < 0 \end{array} \right. \\ 试问：(1)点a为何值时，x = 0是f(x)的连续点. \\ (2)点a为何值时，x = 0是f(x)的间断点.$
$\color{blue}{ 要使f(x) 在 x = 0 处连续，必须满足条件：\\ \lim_{x \rightarrow 0^-}{f(x)} = \lim_{x \rightarrow 0^+}{f(x)} = f(0). \\ \lim_{x \rightarrow 0^+}{\dfrac{\cos{x}}{x + 2}} = \dfrac{1}{2}\\ \lim_{x \rightarrow 0^-}{\dfrac{\sqrt{a} - \sqrt{a - x}}{x}} \\ = \lim_{x \rightarrow 0^-}{\dfrac{(a) - (a - x)}{x(\sqrt{a} + \sqrt{a - x})}} \\ = \lim_{x \rightarrow 0^-}{\dfrac{1}{\sqrt{a} + \sqrt{a - x}}} \\ = \lim_{x \rightarrow 0^-}{\dfrac{1}{2\sqrt{a}}} \\ = \dfrac{1}{2\sqrt{a}} \\ a = 1 \\ 当a \neq 1时，x = 0是f(x)的间断点. }$

练习1.3
一、选择题
$1.设f(x) = \left \{ \begin{array}{l} \dfrac{\ln{(1 + x)}}{x}, x \neq 0 \\ k, \quad x = 0 \end{array} \right. \\ 在x = 0处连续,则k等于(\ \ { \color{blue}{D} }\ \ ).$
$A.0; \quad B.e; \quad C.-1; \quad D.1.$
$\color{blue}{ \lim_{x \rightarrow 0}{\dfrac{\ln{(1 + x)}}{x}} = \lim_{x \rightarrow 0}{\dfrac{x}{x}} = 1 = k }$

$2.函数f(x) = \dfrac{\sin{\sqrt{x}}}{(x + 1)(x^2 - 1)}的间断点的个数为 (\ \ { \color{blue}{B} }\ \ ).$
$A.0; \quad B.1; \quad C.2; \quad D.3.$
$\color{blue}{ 分母(x+1)(x + 1)(x - 1) \neq 0 \Rightarrow x \neq -1, x \neq 1 \\ 分子\sin{\sqrt{x}} \Rightarrow x \geq 0 \\ 定义域内只有x \neq 1是间断点 }$

$3.设函数f(x) = \left \{ \begin{array}{l} 2x, \quad 0 \leq x < 1 \\ 3 - x, 1 \leq x \leq 2 \end{array} \right. 则f(x)的连续区间为(\ \ { \color{blue}{D} }\ \ ).$
$A.[-1, 1),(1, 3]; \quad B.[1, 3]; \quad C.[-1, 3); \quad D.[0, 2].$
$\color{blue}{ \lim_{x \rightarrow 1^+}{f(x)} = \lim_{x \rightarrow 1^+}{(3 - x)} = 2 \\ \lim_{x \rightarrow 1^-}{f(x)} = \lim_{x \rightarrow 1^-}{(2x)} = 2 \\ \lim_{x \rightarrow 1^-}{f(x)} =\lim_{x \rightarrow 1^+}{f(x)} = 2 = f(1) \\ 可知f(x)在[0, 2]上连续 }$

二、填空题
$1.f(x) = \cos{\dfrac{1}{x}}的间断点为\underline{\ \ {\color{blue}{x = 0}} \ \ }$

$2.f(x) = \left \{ \begin{array}{l} \dfrac{1 - x^2}{1 + x}, x \neq -1 \\ 0, x = -1 \end{array} \right. 则f(x)的间断点为\underline{\ \ {\color{blue}{x = -1}} \ \ }$
$解：\lim_{x \rightarrow -1}{\dfrac{1 - x^2}{1 + x}} = \lim_{x \rightarrow -1}{(1-x)} = 2 \neq f(-1)$

$3.设f(x) = \left \{ \begin{array}{l} ke^{2x}, x < 0 \\ 1 + cos{x}, x \geq 0 \end{array} \right. 在点x = 0处连续，则常数k = \underline{\ \ {\color{blue}{2}} \ \ }$
$解：\lim_{x \rightarrow 0^-}{ke^{2x}} = k = \lim_{x \rightarrow 0^+}{(1+ \cos{x})} = 2$

$4.设f(x) = \left \{ \begin{array}{l} e^{x}, x \leq 0 \\ a + x, x > 0 \end{array} \right. 在点x = 0处连续，则常数a = \underline{\ \ {\color{blue}{1}} \ \ }$
$解：\lim_{x \rightarrow 0^-}{e^{x}} = f(0) = e^0 = 1 = \lim_{x \rightarrow 0^+}{(a+ x)} = a + 0 = a$