微积分笔记

极限和连续性

何为极限？

假设对于函数 $f$ 有：

$\displaystyle \lim _{x\to c} f(x)=L$

即：只要 $x$ 无限接近于 $c$, 则 $f(x)$ 必然无限接近于 $L$。

用 $\varepsilon -\delta$ 语言来描述就是：

说白了就是：无论给定任何一个数字 $\varepsilon (\varepsilon >0)$，总能找到一个数 $\delta ( \delta >0)$。使当 $x$ 在 $c$ 的 $\delta$ 范围内时，$f(x)$ 在极限 $L$ 的 $\varepsilon$ 范围内。

例：已知 $f( x) =\begin{cases} 2x & x\neq 5\\ x & x=5 \end{cases}，$ 证明 $\displaystyle \lim _{x\rightarrow 5} f( x) =10$

根据定义，给定任意 $\varepsilon (\varepsilon >0)$，有 $\delta ( \delta >0)$。因此，我们本质上是要找到一个 $\delta =function\ of\ \varepsilon$ 的函数。

$\mathnormal{Proof.}$

夹逼定理

设 $I$ 为包含某点 $c$ 的区间，$f, g, h$ 为定义在 $I$ 上的函数。若对于所有属于 $I$ 而不等于 $c$ 的 $x$，有：

• $g( x) \leqslant f( x) \leqslant h( x)$
• $\displaystyle \lim _{x\rightarrow c} g( x) =\lim _{x\rightarrow c} h( x) =L$

则，$\displaystyle \lim _{x\rightarrow c} f( x) =L$。

$g(x)$ 和 $h(x)$ 分别被称为 $f(x)$ 的下界和上界。

Proof: $\displaystyle \lim _{\theta \rightarrow 0}\frac{\sin \theta }{\theta } =1$

Proof: $\displaystyle \lim _{\theta \rightarrow 0}\frac{1-\cos \theta }{\theta } =0$

$\mathnormal{Proof.}$

连续性的定义

函数在某一点处连续：$\displaystyle f\ is\ continuous\ at\ x=c\Longleftrightarrow \lim _{x\rightarrow c} f( x) =f( c)$

函数在开区间连续：$f\ is\ continuous\ over\ ( a,\ b) \Longleftrightarrow f\ is\ continuous\ over\ every\ point\ in\ the\ interval$

函数在闭区间连续：$\displaystyle f\ is\ continuous\ over\ [ a,\ b] \Longleftrightarrow f\ is\ continuous\ over\ ( a,\ b) \ and\ \lim _{x\rightarrow a^{+}} f( x) =f( a) ,\ \lim _{x\rightarrow b^{-}} f( x) =f( b)$

中值定理

$Suppose\ f\ is\ a\ continuous\ function\ at\ every\ point\ of\ the\ interval\ [ a,\ b]$

• $f\ will\ take\ on\ every\ value\ between\ f( a) \ and\ f( b) \ over\ the\ interval$
• $For\ any\ L\ between\ the\ values\ f( a) \ and\ f( b) ,\ there\ exists\ a\ number\ c\ in\ [ a,\ b] \ for\ which\ f( c) =L$

怎么会有这么简单的定理…

导数

导数的符号表示

Lagrange's notation: $f^{\prime }$

Leibniz's notation: $\frac{dy}{dx}$

Newton's notation: $\dot{y}$

导数的两种定义形式

$\displaystyle f^{\prime }( x) =\lim _{h\rightarrow 0}\frac{f( x+h) -f( x)}{h}$

$\displaystyle f^{\prime }( c) =\lim _{x\rightarrow c}\frac{f( x) -f( c)}{x-c}$

可微性

$f\ is\ differentiability\ at\ x=c\Longrightarrow f\ is\ continuous\ at\ x=c \\ f\ is\ not\ continuous\ at\ x=c\Longrightarrow f\ is\ not\ differentiability\ at\ x=c$

不可微的三种情况：$\begin{cases} 1.\ not\ continuous\\ 2.\ vertical\ tanget\\ 3.\ sharp\ turn \end{cases}$

Proof: Differentiability implies continuity

$\mathnormal{Proof.}$

$Assume:\ f\ differentiability\ at\ x=c$

$\begin{array}{l} \because f\ differentiability\ at\ x=c\\ \therefore f^{\prime }( c) =\displaystyle \lim _{x\rightarrow c}\frac{f( x) -f( c)}{x-c} \end{array}$

Justifying the power rule

Proof: $\frac{d}{dx}\left( x^{n}\right) =nx^{n-1}$

$\mathnormal{Proof.}$

$\displaystyle \frac{d}{dx}\left( x^{n}\right) =\lim _{\Delta x\rightarrow 0}\frac{( x+\Delta x)^{n} -x^{n}}{\Delta x}$

$According\ to\ Binomial\ theorem:$

Proof: $\frac{d}{dx}\left(\sqrt{x}\right) =\frac{1}{2} x^{-\frac{1}{2}}$

$\mathnormal{Proof.}$

Justifying the basic derivative rules

Proof: Constant rule ($\frac{d}{dx} k=0$)

$\mathnormal{Proof.}$

Proof: Constant multiple and sum/difference rules

Constant multiple rule: $\dfrac{d}{dx}[k\cdot f(x)]=k\cdot\dfrac{d}{dx}f(x)$

Sum rule: $\dfrac{d}{dx}[f(x)+g(x)]=\dfrac{d}{dx}f(x)+\dfrac{d}{dx}g(x)$

Difference rule: $\dfrac{d}{dx}[f(x)-g(x)]=\dfrac{d}{dx}f(x)-\dfrac{d}{dx}g(x)$

$\mathnormal{Proof.}$

$1.\ f( x) =kg( x) \Longrightarrow f^{\prime }( x) =kg^{\prime }( x)$

$2.\ f( x) =g( x) \pm j( x) \Longrightarrow f^{\prime }( x) =g^{\prime }( x) \pm j^{\prime }( x)$

Proof: The derivatives of sin(x) and cos(x)

Known $\displaystyle \lim _{x\rightarrow 0}\frac{\sin x}{x} =1$ and $\displaystyle \lim _{x\rightarrow 0}\frac{1-\cos x}{x} =0$

$\mathnormal{Proof.}$

$1.\ \frac{d}{dx}[\sin x] =\cos x$

$2.\ \frac{d}{dx}[\cos x] =-\sin x$

Proof: The derivative of $e^{x}$ is $e^{x}$

Know the limit definition of $\mathbb{e}$ is $e=\displaystyle \lim _{n\rightarrow \infty }\left( 1+\frac{1}{n}\right)^{n} =\displaystyle \lim _{n\rightarrow 0}( 1+n)^{\frac{1}{n}}$

$\mathnormal{Proof.}$

$Let\ n=e^{\Delta x} -1,\ we\ can\ get\ n+1=e^{\Delta x} ,\ such\ that\ \Delta x=\ln( n+1) \ and\ as\ \Delta x\rightarrow 0=n\rightarrow 0$

$We\ can\ rewrite\ to:$

Proof: The derivative of $\ln( x)$ is $\frac{1}{x}$

Method 1 (Directly from the definition of the derivative as a limit)

$\mathnormal{Proof.}$

$Let\ n=\frac{\Delta x}{x} ,\ \Delta x=nx,\ \frac{1}{\Delta x} =\frac{1}{n} \cdot \frac{1}{x} \ and\ as\ \Delta x\rightarrow 0=n\rightarrow 0$

$We\ can\ rewrite\ to:$

Method 2 (Using the fact that $\frac{d}{dx}\left( e^{x}\right) =e^{x}$ and applying implicit differentiation)

$\mathnormal{Proof.}$

$Known\ \frac{d}{dx}\left( e^{x}\right) =e^{x}$

$Let\ y=\ln( x) ,\ we\ can\ get:$

Proof: The product rule

$\mathnormal{Proof.}$

Proof: The derivatives of $\tan( x)$、$\cos( x)$、$\sec( x)$ and $\csc( x)$

$\mathnormal{Proof.}$

Proof: The derivatives of $a^{x}$ (For any positive base a)

$\mathnormal{Proof.}$

$Known\ \frac{d}{dx}\left( e^{x}\right) =e^{x}$

$Let\ a=e^{\ln a}$

Proof: The derivatives of $\log_{a} x$ (For any positive base $a\neq 1$)

$\mathnormal{Proof.}$

$Known\ \frac{d}{dx}(\ln x) =\frac{1}{x}$

Proof: Chain Rule and Quotient Rule

$\mathnormal{Chain\ Rule\ Proof.}$

$For\ why\ if\ function\ u\ is\ continuous\ at\ x,\ then\ \Delta u\rightarrow 0\ as\ \Delta x\rightarrow 0$:

$The\ chain\ rule\ tell\ us:\ \frac{d}{dx}[ y( u( x))] =\frac{dy}{dx} =\frac{dy}{du} \cdot \frac{du}{dx}$

$Assuming\ y,\ u\ differentiable\ at\ x.\ We\ can\ get:$

$\mathnormal{Quotient\ Rule\ Proof.}$