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
- 周期
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