函数概念与极限

函数

对数运算法则

商变差，奇变和，幂次变倍数

$$\left\{ \begin{array}{} \ln ab = \ln a + \ln b, & \\ \ln \frac{a}{b} = \ln a - \ln b, & \\ \ln a^b = b \ln a, & \end{array} \right.$$

反双曲正弦函数

$$\operatorname{arsinh} = \ln \left( x + \sqrt{x^2 + 1} \right)$$

双曲余弦函数（悬链线）偶函数

$$\cosh x = \frac{e^x + e^{-x}}{2}$$

定义域

$$\left\{ \begin{array}{} \frac{1}{x}, & x \neq 0 \\ \sqrt[2n]{x}, & x \geq 0 \\ \log_{n} x, & x > 0 \\ \arcsin x, \arccos x, & -1 \leq x \leq 1 \\ \tan x, & x \neq k\pi + \frac{\pi}{2} \\ \cot x, & x \neq k\pi \end{array} \right.$$

性质

1. 奇偶性

   • 内偶则偶，内奇同外
   • 可导函数求导奇偶互换

2. 周期

   • 三角函数：奇变偶不变，符号看象限

极限

$$\lim_{x \to -\infty} e^x = 0, \lim_{x \to +\infty} e^x = +\infty$$ $$e^{x_a} - 1 = e^{x_a} - e^0$$ $$e^{\tan x} - e^{\sin x} = e^{\sin x} \left( e^{\tan x - \sin x} - 1 \right)$$

特殊函数值

$$\ln 1 = 0, \ln e = 1$$

判断反函数

$$a = f(f^{-1}(a)) \quad 存在反函数 \Rightarrow f^{-1}(f(a))$$

常考对数运算（幂次变倍数）

$$\begin{gather} \ln \sqrt{x} = \frac{1}{2} \ln x, \\ \ln \frac{1}{x} = -\ln x \end{gather}$$

拉格朗日中值定理

下式先通分再商变差：

$$\ln \left( 1 + \frac{1}{x} \right) = \ln \frac{x + 1}{x} = \ln (x + 1) - \ln x$$

自然常数

$$e = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n$$

三角函数

正余弦

$$\left\{ \begin{array}{} \sin x, & T = 2\pi, & x \in (-\infty, \infty), & y \in [-1, 1], & x = 0, y = 0 \\ \cos x, & T = 2\pi, & x \in (-\infty, \infty), & y \in [-1, 1], & x = 0, y = 1 \end{array} \right.$$ $$\sin^2 \alpha + \cos^2 \alpha = 1$$

正余切

$$\left\{ \begin{array}{} \tan x = \frac{\sin x}{\cos x}, \\ \cot x = \frac{\cos x}{\sin x} = \frac{1}{\tan x} \end{array} \right.$$ $$\left\{ \begin{array}{} \tan x, & T = \pi, & x \neq k\pi + \frac{\pi}{2} (k \in \mathbb{Z}), & y \in (-\infty, +\infty), & x = 0, y = 0 \\ \cot x, & T = \pi, & x \neq k\pi (k \in \mathbb{Z}), & y \in (-\infty, +\infty), & x = \frac{\pi}{2}, y = 0 \end{array} \right.$$

正余割

$$\left\{ \begin{array}{} 偶函数 \ \sec x = \frac{1}{\cos x}, \\ 奇函数 \ \csc x = \frac{1}{\sin x} \end{array} \right.$$ $$1 + \tan^2 \alpha = \sec^2 \alpha, \quad 1 + \cot^2 \alpha = \csc^2 \alpha$$

反三角函数

恒等式

$$\sin(\arcsin x) = x, \quad \cos(\arccos x) = x$$ $$\sin(\arccos x) = \sqrt{1 - x^2} = \cos(\arcsin x)$$

推导

$$t = \arccos x, \quad \sin t = \sqrt{1 - \cos^2 t}$$ $$\arcsin(\sin y) = y, \quad y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$ $$\arccos(\cos y) = y, \quad y \in [0, \pi]$$ $$\arcsin x + \arccos x = \frac{\pi}{2} \quad (-1 \leq x \leq 1)$$ $$\arctan x + \operatorname{arccot} x = \frac{\pi}{2} \quad (-\infty < x < +\infty)$$

函数极限

$$常数a是数列\{x_{n}\}的极限，或者数列x_n收敛于a,记为$$ $$\lim_{n \to \infty} x_{n} = a$$

邻域

$$U(x_{0}, \delta), \delta \text{ 邻域的半径}, x_{0} \text{ 邻域的中心}$$ $$\mathring{U}(x_{0}, \delta) \text{ 去心邻域}$$ $$右 \delta \text{ 邻域} U^+(x_{0}, \delta)，左 \delta \text{ 邻域} U^-(x_{0}, \delta)$$ $$\forall \text{ Arbitrary}, \exists \text{ Exist}$$ $$\forall M > 0, \exists \delta > 0, \text{ 当 } 0 < |x - x_{0}| < \delta, \text{ 有 } |f(x)| > M$$

例题 1.14

$$\exists \lim_{x \to 0} \frac{f(x)}{x^2}, f(x) = \frac{x - \sin x}{x} + x^2 \lim_{x \to 0} \frac{f(x)}{1 - \cos x}$$ $$\lim_{x \to 0} \frac{f(x)}{x^2} =$$

解：

$$1 - \cos x \sim \frac{1}{2} x^2 \Rightarrow \frac{f(x)}{x^2} = \frac{x - \sin x}{x^3} + \lim_{x \to 0} \frac{2 f(x)}{x^2}$$ $$\Rightarrow \lim_{x \to 0} \frac{f(x)}{x^2} = \lim_{x \to 0} \frac{x - \sin x}{x^3} + \lim_{x \to 0} \frac{2 f(x)}{x^2} \Rightarrow \lim_{x \to 0} \frac{f(x)}{x^2} = -\lim_{x \to 0} \frac{x - \sin x}{x^3} = -\frac{1}{6}$$

注意：

$$1 - \cos x \sim \frac{1}{2} x^2, \frac{x - \sin x}{x^3} \sim \frac{1}{6}, \frac{\sin x}{x} \sim 1$$

性质

唯一性：

$$如果 \exists \lim_{x \to x_{0}} f(x), \quad 极限唯一$$ $$\lim_{x \to x_{0}} f(x) = A \iff f(x) = A + \alpha(x), \quad \lim_{x \to x_{0}} \alpha(x) = 0$$

例题 1.10

$$\lim_{x \to -1} f(x) = \frac{|x|^x - 1}{x(x + 1) \ln |x|} =$$

解：

$$\lim_{x \to -1} f(x) = \frac{e^{x \ln |x|} - 1}{x(x + 1) \ln |x|} = \frac{x \ln |x|}{x(x + 1) \ln |x|} = \frac{1}{x + 1}$$ $$\lim_{x \to -1^+} f(x) = +\infty$$ $$\lim_{x \to -1^-} f(x) = -\infty$$ $$因此: \lim_{x \to -1} f(x) = \infty$$

注意：

$$e^x - 1 \sim x, \quad |x|^x = e^{x \ln |x|}$$

例题 1.11

$$\exists \ f(x) = 2x + \sqrt{x^2 + 2x + 1}, \quad g(x) = \left\{ \begin{array}{} x + 2, & x \geq 0 \\ x - 1, & x < 0 \end{array} \right.$$ $$\lim_{x \to -\frac{1}{3}} g[f(x)] =$$

解：

$$\begin{gather} \lim_{x \to -\frac{1}{3}} g[f(x)] = \left\{ \begin{array}{} 3x + 3, & x \geq -\frac{1}{3} \\ 3x, & -1 \leq x < -\frac{1}{3} \\ x - 2, & x < -1 \end{array} \right. \end{gather}$$ $$\begin{gather} \Rightarrow \lim_{x \to -\frac{1}{3}^+} g[f(x)] = 2 \neq \lim_{x \to -\frac{1}{3}^-} g[f(x)] = -1 \end{gather}$$ $$因此 \nexists \lim_{x \to -\frac{1}{3}} g[f(x)]$$

注意：

$$\sqrt{x^2 + 2x + 1} = |x + 1|$$

局部有界性

$$|f(x)| = |f(x) - A + A| \leq |f(x) - A| + |A|$$ $$\exists \lim_{x \to a^+} f(x), \quad \exists \lim_{x \to b^-} f(x), \quad f(x) \in (a, b) \text{ 有界}$$

局部保号性

脱帽严格不等：

$$\lim_{x \to x_{0}} f(x) > 0 \Rightarrow f > 0, \quad \lim_{x \to x_{0}} f(x) < 0 \Rightarrow f < 0$$

戴帽非严格不等：

$$f \geq 0 \Rightarrow \lim_{x \to x_{0}} f(x) \geq 0, \quad f \leq 0 \Rightarrow \lim_{x \to x_{0}} f(x) \leq 0$$

无穷小

$$\lim_{x \to x_{0}} f(x) = 0$$

1. 有限个无穷小的和是无穷小，无穷个无穷小的和不一定是无穷小。

注意：

$$\lim_{n \to \infty} \sum^{n}_{i=1} \frac{1}{n + i} = \ln 2$$

2. 有界函数和无穷小的乘积是无穷小

$$\exists |\alpha_{1}| < M, \quad \exists \alpha_{2} \to 0$$ $$\Rightarrow |\alpha_{1} \cdot \alpha_{2}| = |\alpha_{1}| \cdot |\alpha_{2}| \leq M |\alpha_{2}| \to 0$$

注意：

$$\lim_{x \to 0} \frac{1}{x} \cdot x = 1$$

1. 有限个无穷小的乘积是无穷小