函数概念与极限
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
\[ \sinh x = \frac{e^x + e^{-x}}{2} \]
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
- 奇偶性
- 内偶则偶，内奇同外
- 可导函数求导奇偶互换
- 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
\[ e = \sum^{\infty}_{n=0} \left( 1 + \frac{1}{n} \right)^n \]
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
\[ \lim_{x \to x_{0}} f(x) = 0 \]
- 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：
\[ \lim_{n \to \infty} \sum^{n}_{i=1} \frac{1}{n + i} = \ln 2 \]
- 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：
\[ \lim_{x \to 0} \frac{1}{x} \cdot x = 1 \]
- 有限个无穷小的乘积是无穷小