# 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：
$$
\displaystyle \forall \varepsilon >0,\ \exists \delta >0,\ s.t.\ 0<|x−c|< \del
ta \Longrightarrow |f(x)-L|< \varepsilon
$$
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。
$\mathnormal{Proof.}$
$$
\begin{aligned}
& |x-5| < \delta \Longrightarrow |2x-10|< \varepsilon \
& |2x-10| < 2\delta \
& 2\delta =\varepsilon \Rightarrow \delta =\frac{\varepsilon }{2}\
& |2x-10| < \varepsilon \
& \forall \varepsilon >0,\ \exists \delta >0 ,\ s.t.\ |x-5|< \delta \Longright
arrow |2x-10|< \varepsilon \ \quad Q.E.D.
\end{aligned}
$$
---
## 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: $\displaystyle \lim _{\theta \rightarrow 0}\frac{1-\cos \theta }{\the
ta } =0$
$\mathnormal{Proof.}$
$$
\begin{aligned}
\lim _{\theta \rightarrow 0}\frac{1-\cos \theta }{\theta } & =\lim _{\theta \rig
htarrow 0}\frac{( 1-\cos \theta )( 1+\cos \theta )}{\theta ( 1+\cos \theta )}\
& =\lim _{\theta \rightarrow 0}\frac{\sin^{2} \theta }{\theta ( 1+\cos \theta )
}\
& =\lim _{\theta \rightarrow 0}\frac{\sin \theta }{\theta } \cdot \lim _{\theta
\rightarrow 0}\frac{\sin \theta }{1+\cos \theta }\
& =1\cdot 0\
& =0
& Q.E.D.
\end{aligned}
$$
## 连续性的定义
函数在某一点处连续：$f$ is continuous at $x=c\Longleftrightarrow \displaystyle \
lim _{x\rightarrow c} f( x) =f( c)$
函数在开区间连续：$f$ is continuous over $( a,\ b) \Longleftrightarrow f$ is con
tinuous over every point in the interval
函数在闭区间连续：$f$ is continuous over $[ a,\ b] \Longleftrightarrow f$ is con
tinuous over $( a,\ b)$ and $\displaystyle \lim _{x\rightarrow a^{+}} f( x) =f(
a)$, $\displaystyle \lim _{x\rightarrow b^{-}} f( x) =f( b)$
## Intermediate Value Theorem
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：
1. **not continuous**
2. **vertical tangent**
3. **"sharp turn"**
## Proof: Differentiability implies continuity
$\mathnormal{Proof.}$
Assume: $f$ differentiability at $x=c$
$
\begin{array}{l}
\because f\ differentiability\ at\ x=c\
\therefore \displaystyle f^{\prime }( c) = \lim _{x\rightarrow c}\frac{f( x) -f(
c)}{x-c}
\end{array}
$
$$
\begin{aligned}
\lim _{x\rightarrow c}[ f( x) -f( c)] & =\lim _{x\rightarrow c}( x-c) \cdot \fra
c{f( x) -f( c)}{x-c}\
& =\lim _{x\rightarrow c}( x-c) \cdot \lim _{x\rightarrow c}\frac{f( x) -f( c)}
{x-c}\
& =0\cdot f^{\prime }( c)\
& =0\
& \
\lim _{x\rightarrow c}[ f( x) -f( c)] & =0\
\lim _{x\rightarrow c} f( x) -\lim _{x\rightarrow c} f( c) & =0\
\lim _{x\rightarrow c} f( x) -f( c) & =0\
\lim _{x\rightarrow c} f( x) & =f( c)
& Q.E.D.
\end{aligned}
$$
## Justifying the power rule
### Proof: $\displaystyle \frac{d}{dx}\left( x^{n}\right) =nx^{n-1}$
$\mathnormal{Proof.}$
$\displaystyle \frac{d}{dx}\left( x^{n}\right) =\lim _{\Delta x\rightarrow 0}\fr
ac{( x+\Delta x)^{n} -x^{n}}{\Delta x}$
According to Binomial theorem:
$$
\begin{aligned}
\displaystyle \lim _{\Delta x\rightarrow 0}\frac{( x+\Delta x)^{n} -x^{n}}{\Delt
a x} & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\cancel{x^{n}} +\binom{
n}{1} x^{n-1} \Delta x+\binom{n}{2} x^{n-2} \Delta x^{2} +...+\binom{n}{n} x^{0}
\Delta x^{n}\cancel{-x^{n}}}{\Delta x}\
& =\displaystyle \lim _{\Delta x\rightarrow 0}\binom{n}{1} x^{n-1} +\cancel{\bi
nom{n}{2} x^{n-2} \Delta x} +...+\cancel{\binom{n}{n} \Delta x^{n-1}}\
& =\displaystyle \lim _{\Delta x\rightarrow 0}\binom{n}{1} x^{n-1}\
& =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{n!}{\cancel{1!}( n-1) !} x^
{n-1}\
& =\displaystyle \lim _{\Delta x\rightarrow 0} nx^{n-1}
& Q.E.D.
\end{aligned}
$$
### Proof: $\displaystyle \frac{d}{dx}\left(\sqrt{x}\right) =\frac{1}{2} x^{-\fr
ac{1}{2}}$
$\mathnormal{Proof.}$
$$
\begin{aligned}
\frac{d}{dx}\left(\sqrt{x}\right) & =\displaystyle \lim _{\Delta x\rightarrow 0}
\frac{\sqrt{x+\Delta x} -\sqrt{x}}{\Delta x}\
& =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\left(\sqrt{x+\Delta x} -\s
qrt{x}\right)\left(\sqrt{x+\Delta x} +\sqrt{x}\right)}{\Delta x\left(\sqrt{x+\De
lta x} +\sqrt{x}\right)}\
& =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{1}{\sqrt{x+\Delta x} +\sqrt
{x}}\
& =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{1}{2\sqrt{x}}\
& =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{1}{2} x^{-\frac{1}{2}}
& Q.E.D.
\end{aligned}
$$
## Justifying the basic derivative rules
### Proof: Constant rule ($\displaystyle \frac{d}{dx} k=0$)
$\mathnormal{Proof.}$
$$
\begin{array}{l}
\because k\ is\ constant\
\therefore y\ does\ not\ change\ as\ x\ changes\
\therefore f( x+h) -f( x) =0\
\therefore \displaystyle \frac{d}{dx} k= \lim _{h\rightarrow 0}\frac{f( x+h) -f(
x)}{h} =\lim _{h\rightarrow 0}\frac{0}{h} =0
\end{array}
$$
### Proof: Constant multiple and sum/difference rules
**Constant multiple rule:** $\displaystyle \dfrac{d}{dx}[k\cdot f(x)]=k\cdot\dfr
ac{d}{dx}f(x)$
**Sum rule:** $\displaystyle \dfrac{d}{dx}[f(x)+g(x)]=\dfrac{d}{dx}f(x)+\dfrac{d
}{dx}g(x)$
**Difference rule:** $\displaystyle \dfrac{d}{dx}[f(x)-g(x)]=\dfrac{d}{dx}f(x)-\
dfrac{d}{dx}g(x)$
$\mathnormal{Proof.}$
$\displaystyle 1.\ f( x) =kg( x) \Longrightarrow f^{\prime }( x) =kg^{\prime }(
x)$
$$
\begin{aligned}
f^{\prime }( x) & =\displaystyle \lim _{h\rightarrow 0}\frac{f( x+h) -f( x)}{h}\
& =\displaystyle \lim _{h\rightarrow 0}\frac{kg( x+h) -kg( x)}{h}\
& =\displaystyle \lim _{h\rightarrow 0} k\left(\frac{g( x+h) -g( x)}{h}\right)\
& =k\displaystyle \lim _{h\rightarrow 0}\frac{g( x+h) -g( x)}{h}\
& =kg^{\prime }( x)
& Q.E.D.
\end{aligned}
$$
$\displaystyle 2.\ f( x) =g( x) \pm j( x) \Longrightarrow f^{\prime }( x) =g^{\p
rime }( x) \pm j^{\prime }( x)$
$$
\begin{aligned}
f^{\prime }( x) & =\displaystyle \lim _{h\rightarrow 0}\frac{g( x+h) \pm j( x+h)
-( g( x) \pm j( x))}{h}\
& =\displaystyle \lim _{h\rightarrow 0}\left(\frac{g( x+h) -g( x)}{h} \pm \frac
{j( x+h) -j( x)}{h}\right)\
& =\displaystyle \lim _{h\rightarrow 0}\frac{g( x+h) -g( x)}{h} \pm \lim _{h\ri
ghtarrow 0}\frac{j( x+h) -j( x)}{h}\
& =g^{\prime }( x) \pm j^{\prime }( x)
& Q.E.D.
\end{aligned}
$$
## Proof: The derivatives of sin(x) and cos(x)
Known $\displaystyle \lim _{x\rightarrow 0}\frac{\sin x}{x} =1$ and $\displaysty
le \lim _{x\rightarrow 0}\frac{1-\cos x}{x} =0$
$\mathnormal{Proof.}$
$\displaystyle 1.\ \frac{d}{dx}[\sin x] =\cos x$
$$
\begin{aligned}
\frac{d}{dx}[\sin x] & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\sin( x
+\Delta x) -\sin( x)}{\Delta x}\
& =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\cos x\sin \Delta x+\sin x\
cos \Delta x-\sin x}{\Delta x}\
& =\displaystyle \lim _{\Delta x\rightarrow 0}\left(\frac{\cos x\sin \Delta x}{
\Delta x} +\frac{\sin x\cos \Delta x-\sin x}{\Delta x}\right)\
& =\displaystyle \lim _{\Delta x\rightarrow 0}\cos x\left(\frac{\sin \Delta x}{
\Delta x}\right) +\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\sin x(\cos \
Delta x-1)}{\Delta x}\
& =\cos x\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\sin \Delta x}{\Delta
x} -\sin x\displaystyle \lim _{\Delta x\rightarrow 0}\frac{1-\cos \Delta x}{\De
lta x}\
& =\cos x\cdot 1-\sin x\cdot 0\
& =\cos x
& Q.E.D.
\end{aligned}
$$
$\displaystyle 2.\ \frac{d}{dx}[\cos x] =-\sin x$
## Proof: The derivative of $e^{x}$ is $e^{x}$
Know the limit definition of $\mathbb{e}$ is $e=\displaystyle \lim _{n\rightarro
w \infty }\left( 1+\frac{1}{n}\right)^{n} =\displaystyle \lim _{n\rightarrow 0}(
1+n)^{\frac{1}{n}}$
$\mathnormal{Proof.}$
$$
\begin{aligned}
\frac{d}{dx}\left( e^{x}\right) & =\displaystyle \lim _{\Delta x\rightarrow 0}\f
rac{e^{x+\Delta x} -e^{x}}{\Delta x}\
& =e^{x}\displaystyle \lim _{\Delta x\rightarrow 0}\frac{e^{\Delta x} -1}{\Delt
a x}
\end{aligned}
$$
$\displaystyle Let\ n=e^{\Delta x} -1,\ we\ can\ get\ n+1=e^{\Delta x} ,\ such\
that\ \Delta x=\ln( n+1) \ and\ as\ \Delta x\rightarrow 0=n\rightarrow 0$
$We\ can\ rewrite\ to:$
$$
\begin{aligned}
\frac{d}{dx}\left( e^{x}\right) & =e^{x}\displaystyle \lim _{n\rightarrow 0}\fra
c{n}{\ln( n+1)}\
& =e^{x}\displaystyle \lim _{n\rightarrow 0}\frac{\frac{1}{n} n}{\frac{1}{n}\ln
( n+1)}\
& =e^{x}\displaystyle \lim _{n\rightarrow 0}\frac{1}{\ln\left[( 1+n)^{\frac{1}{
n}}\right]}\
& =e^{x}\frac{1}{\ln\left[\displaystyle \lim _{n\rightarrow 0}( 1+n)^{\frac{1}{
n}}\right]}\
& =e^{x}
& Q.E.D.
\end{aligned}
$$
## Proof: The derivative of $\ln( x)$ is $\frac{1}{x}$
### Method 1 (Directly from the definition of the derivative as a limit)
$\mathnormal{Proof.}$
$$
\begin{aligned}
\frac{d}{dx}(\ln x) & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\ln( x+\
Delta x) -\ln( x)}{\Delta x}\
& =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\ln\left(\frac{x+\Delta x}{
x}\right)}{\Delta x}\
& =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\ln\left( 1+\frac{\Delta x}
{x}\right)}{\Delta x}\
& =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{1}{\Delta x}\ln\left( 1+\fr
ac{\Delta x}{x}\right)\
& =\displaystyle \lim _{\Delta x\rightarrow 0}\ln\left[\left( 1+\frac{\Delta x}
{x}\right)^{\frac{1}{\Delta x}}\right]
\end{aligned}
$$
$\displaystyle Let\ n=\frac{\Delta x}{x} ,\ \Delta x=nx,\ \frac{1}{\Delta x} =\f
rac{1}{n} \cdot \frac{1}{x} \ and\ as\ \Delta x\rightarrow 0=n\rightarrow 0$
$We\ can\ rewrite\ to:$
$$
\begin{aligned}
\displaystyle \lim _{\Delta x\rightarrow 0}\ln\left[\left( 1+\frac{\Delta x}{x}\
right)^{\frac{1}{\Delta x}}\right] & =\frac{1}{x}\displaystyle \lim _{n\rightarr
ow 0}\ln\left[( 1+n)^{\frac{1}{n}}\right]\
& =\frac{1}{x}\ln\left[\displaystyle \lim _{n\rightarrow 0}( 1+n)^{\frac{1}{n}}
\right]\
& =\frac{1}{x}
& Q.E.D.
\end{aligned}
$$
### Method 2 (Using the fact that $\displaystyle \frac{d}{dx}\left( e^{x}\right)
=e^{x}$ and applying implicit differentiation)
$\mathnormal{Proof.}$
$\displaystyle Known\ \frac{d}{dx}\left( e^{x}\right) =e^{x}$
$\displaystyle Let\ y=\ln( x) ,\ we\ can\ get:$
$$
\begin{aligned}
\frac{d}{dx}\left( e^{y}\right) & =\frac{d}{dx}( x)\
e^{y} \cdot \frac{dy}{dx} & =1\
\frac{dy}{dx} & =\frac{1}{e^{y}}\
& =\frac{1}{e^{\ln x}}\
& =\frac{1}{x}
& Q.E.D.
\end{aligned}
$$
## Proof: The product rule
$\mathnormal{Proof.}$
$$
\begin{aligned}
\frac{d}{dx}[ f( x) g( x)] & =\displaystyle \lim _{h\rightarrow 0}\frac{f( x+h)
g( x+h) -f( x+h) g( x) +f( x+h) g( x) -f( x) g( x)}{h}\
& =\displaystyle \lim _{h\rightarrow 0}\left[ f( x+h)\frac{g( x+h) -g( x)}{h} +
g( x)\frac{f( x+h) -f( x)}{h}\right]\
& =\left[\displaystyle \lim _{h\rightarrow 0} f( x+h)\right]\left[\displaystyle
\lim _{h\rightarrow 0}\frac{g( x+h) -g( x)}{h}\right] +\left[\displaystyle \lim
_{h\rightarrow 0} g( x)\right]\left[\displaystyle \lim _{h\rightarrow 0}\frac{f
( x+h) -f( x)}{h}\right]\
& =f( x) g^{\prime }( x) +g( x) f^{\prime }( x)
& Q.E.D.
\end{aligned}
$$
## Proof: The derivatives of $\tan( x)$、$\cos( x)$、$\sec( x)$ and $\csc( x)$
$\mathnormal{Proof.}$
$$
\begin{aligned}
\frac{d}{dx}(\tan x) & =\frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) & \frac{d
}{dx}(\cot x) & =\frac{d}{dx}\left(\frac{\cos x}{\sin x}\right)\
& =\frac{\cos^{2} x+\sin^{2} x}{\cos^{2} x} & & =\frac{-\left(\sin^{2} x+\cos^
{2} x\right)}{\sin^{2} x}\
& =\frac{1}{\cos^{2} x} & & =-\frac{1}{\sin^{2} x}\
& =\sec^{2} x & & =-\csc^{2} x\
\frac{d}{dx}(\sec x) & =\frac{d}{dx}\left(\frac{1}{\cos x}\right) & \frac{d}{dx}
(\csc x) & =\frac{d}{dx}\left(\frac{1}{\sin x}\right)\
& =\frac{0\cdot \cos x+1\cdot \sin x}{\cos^{2} x} & & =\frac{0\cdot \sin x-1\c
dot \cos x}{\sin^{2} x}\
& =\frac{\sin x}{\cos^{2} x} & & =-\frac{\cos x}{\sin^{2} x}\
& =\tan x\cdot \sec x & & =-\cot x\cdot \csc x
& Q.E.D.
\end{aligned}
$$
## Proof: The derivatives of $a^{x}$ (For any positive base a)
$\mathnormal{Proof.}$
$\displaystyle Known\ \frac{d}{dx}\left( e^{x}\right) =e^{x}$
$\displaystyle Let\ a=e^{\ln a}$
$$
\begin{aligned}
\frac{d}{dx}\left( a^{x}\right) & =\frac{d}{dx}\left[\left( e^{\ln a}\right)^{x}
\right]\
& =\frac{d}{dx}\left[ e^{(\ln a) x}\right]\
& =e^{(\ln a) x} \cdot \ln a\
& =a^{x} \cdot \ln a
& Q.E.D.
\end{aligned}
$$
## Proof: The derivatives of $\log_{a} x$ (For any positive base $a\neq 1$)
$\mathnormal{Proof.}$
$\displaystyle Known\ \frac{d}{dx}(\ln x) =\frac{1}{x}$
$$
\begin{aligned}
\frac{d}{dx}(\log_{a} x) & =\frac{d}{dx}\left(\frac{1}{\ln a} \cdot \ln x\right)
\
& =\frac{1}{x\ln a}
& Q.E.D.
\end{aligned}
$$
## Proof: Chain Rule and Quotient Rule
$\mathnormal{Chain\ Rule\ Proof.}$
$$
\begin{aligned}
Known:\ & 1.\ If\ a\ function\ is\ differentiable,\ then\ it\ is\ also\ continu
ous.\
& 2.\ If\ function\ u\ is\ continuous\ at\ x,\ then\ \Delta u\rightarrow 0\ as\
\Delta x\rightarrow 0
\end{aligned}
$$
For why if function $u$ is continuous at $x$, then $\Delta u\rightarrow 0$ as $\
Delta x\rightarrow 0$:
$\displaystyle The\ chain\ rule\ tell\ us:\ \frac{d}{dx}[ y( u( x))] =\frac{dy}{
dx} =\frac{dy}{du} \cdot \frac{du}{dx}$
Assuming $y$, $u$ differentiable at $x$. We can get:
$$
\begin{aligned}
\frac{dy}{dx} & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\Delta y}{\Del
ta x}\
& =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta u} \cdot \
frac{\Delta u}{\Delta x}\
& =\left(\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta u}\r
ight)\left(\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\Delta u}{\Delta x}\
right)\
& =\left(\displaystyle \lim _{\Delta u\rightarrow 0}\frac{\Delta y}{\Delta u}\r
ight)\left(\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\Delta u}{\Delta x}\
right)\
& =\frac{dy}{du} \cdot \frac{du}{dx}
& Q.E.D.
\end{aligned}
$$
$\mathnormal{Quotient\ Rule\ Proof.}$
$$
\begin{aligned}
\frac{d}{dx}\left[\frac{f( x)}{g( x)}\right] & =\frac{d}{dx}\left[ f( x) \cdot [
g( x)]^{-1}\right]\
& =f^{\prime }[ x]( g( x))^{-1} -f[ x]( g( x))^{-2} g^{\prime }( x)\
& =\frac{f^{\prime }( x)}{g( x)} -\frac{f( x) g^{\prime }( x)}{[ g( x)]^{2}}\
& =\frac{f^{\prime }( x) g( x) -f( x) g^{\prime }( x)}{[ g( x)]^{2}}
& Q.E.D.
\end{aligned}
$$
# Proof: L'Hôpital's rule
> [!NOTE]
> This isn't full proof of L'Hôpital's rule, just a special case. But it should
give some intuition for why it works.
$$
f( a) =0,\ g( a) =0;\ f^{\prime} ( a) \ exists,\ g^{\prime} ( a) \ exists\ \Long
leftrightarrow \ \displaystyle \lim _{x\rightarrow a}\frac{f( x)}{g( x)} =\frac{
f^{\prime} ( a)}{g^{\prime} ( a)}
$$
$$
\begin{aligned}
\frac{f^{\prime }( a)}{g^{\prime }( a)} & =\frac{\displaystyle \lim _{x\rightarr
ow a}\frac{f( x) -f( a)}{x-a}}{\displaystyle \lim _{x\rightarrow a}\frac{g( x) -
g( a)}{x-a}} & \
& =\displaystyle \lim _{x\rightarrow a}\frac{f( x) -f( a)}{g( x) -g( a)} & \
& =\displaystyle \lim _{x\rightarrow a}\frac{f( x)}{g( x)} & We\ know\ f( a) \
and\ g( a) \ both\ equal\ to\ zero \
& & Q.E.D
\end{aligned}
$$
# Mean Value Theorem
If $f$ is continuous over $[ a,\ b]$ and every point over $( a,\ b)$ is differen
tiable. Then there exists some $c\in ( a,\ b)$ where $\displaystyle \frac{\Delta
y}{\Delta x} =\frac{f( b) -f( a)}{b-a} =f^{\prime }( c)$
# Extreme Value Theorem
$f$ continuous over $[ a,\ b] \Longrightarrow \exists \ c,\ d\in [ a,\ b] :f( c)
\leqslant f( x) \leqslant f( d)$ for all $x\in [ a,\ b]$
critical points exists when non endpoint point at $$x=a\ \begin{cases} f^{\prime
}( a) =0\ f^{\prime }( a) \ undefined \end{cases}$$
# Definite Integral & Riemann Sum
The definite integral of a continuous function $f$ over the interval $[ a,\ b]$,
denoted by $\displaystyle \int _{a}^{b} f( x) dx$, is the limit of a Riemann su
m as the number of subdivisions approaches infinity.
$$
\displaystyle \int _{a}^{b} f( x) dx=\lim _{n\rightarrow \infty }\sum _{i=1}^{n}
f( x_{i}) \Delta x
$$
Where $\displaystyle \Delta x=\frac{b-a}{n}$ and $x_{i} =a+\Delta x\cdot i$
# Definite integrals properties
**Sum/Difference:**
$$\displaystyle \int _{a}^{b}[ f( x) \pm g( x)] dx=\int _{a}^{b} f( x) dx\pm \in
t _{a}^{b} g( x) dx$$
**Constant multiple:**
$$\displaystyle \int _{a}^{b} k\cdot f( x) dx=k\int _{a}^{b} f( x) dx$$
**Reverse interval:**
$$\displaystyle \int _{a}^{b} f( x) dx=-\int _{b}^{a} f( x) dx$$
**Zero-length interval:**
$$\displaystyle \int _{a}^{a} f( x) dx=0$$
**Adding intervals:**
$$\displaystyle \int _{a}^{b} f( x) dx+\int _{b}^{c} f( x) dx=\int _{a}^{c} f( x
) dx$$
这么简单的东西相信你一定也知道怎么证。~~那证明就略略略了吧～~~
# First fundamental theorem of calculus
Let $f$ be a continuous real−valued function defined on $[ a,\ b]$. And $F$ be t
he function defined, for all $x$ in $[ a,\ b]$, by $\displaystyle F( x) =\int _{
a}^{x} f( t) dt$
Then $F$ is uniformly continuous on $[ a,\ b]$ and differentiable on the open in
terval $( a,\ b)$, and $\displaystyle F^{\prime }( x) =f( x)$ for all $x$ in $(
a,\ b)$ so $F$ is an antiderivative of $f$.
# Second fundamental theorem of calculus / Newton–Leibniz theorem
Let $f$ be a continuous real−valued function defined on $[ a,\ b]$ and $F$ is a
continuous function on $[ a,\ b]$ which is an antiderivative of $f$ in $( a,\ b)
$: $\displaystyle F^{\prime }( x) =f( x)$
If $f$ is Riemann integrable on $[ a,\ b]$ then $\displaystyle \int _{a}^{b} f(
x) dx=F( b) -F( a)$
# Reverse power rule
$$\displaystyle \int x^{n} dx=\frac{x^{n+1}}{n+1} +C,\ n\neq -1$$
Yes that just simple!
# Indefinite integration rules
**Polynomials**
$$\displaystyle \int x^{n} dx=\frac{x^{n+1}}{n+1} +C$$
**Radicals**
$$\displaystyle \int \sqrt[m]{x^{n}} dx=\frac{x^{\frac{n}{m} +1}}{\frac{n}{m} +1
} +C$$
**Trigonometric functions**
$$\displaystyle \int \sin( x) dx=-\cos( x) +C$$
$$\displaystyle \int \cos( x) dx=\sin( x) +C$$
$$\displaystyle \int \sec^{2}( x) dx=\tan( x) +C$$
$$\displaystyle \int \csc^{2}( x) dx=-\cot( x) +C$$
$$\displaystyle \int \sec( x)\tan( x) dx=\sec( x) +C$$
$$\displaystyle \int \csc( x)\cot( x) dx=-\csc( x) +C$$
$$\displaystyle \int \sec xdx=\ln| \sec x+\tan x| +C$$（分子分母同乘 $\sec x+\ta
n x$）
$$\displaystyle \int \csc xdx=\ln| \csc x-\cot x| +C$$（分子分母同乘 $\csc x-\co
t x$）
下面是另一种方法求这两个不定积分：
$$
\begin{array}{ l l l }
\displaystyle \int \sec xdx & =\displaystyle \int \frac{1}{\cos x} dx & \
& =\displaystyle \int \frac{\cos x}{\cos^{2} x} dx & \
& =\displaystyle \int \frac{1}{1-\sin^{2} x}\cos xdx & Let\ u=\sin x\
& =\displaystyle \int \frac{1}{( 1+u)( 1-u)} du & \
& =\displaystyle \frac{1}{2}\int \left(\frac{1}{1+u} +\frac{1}{1-u}\right) du &
Partial\ fractions\
& =\displaystyle \frac{1}{2}\ln\left| \frac{1+u}{1-u}\right| +C & \
& =\displaystyle \frac{1}{2}\ln\left| \frac{1+\sin x}{1-\sin x}\right| +C &
\end{array}
$$
$$
\begin{array}{ l l l }
\displaystyle \int \csc xdx & =\displaystyle \int \frac{1}{\sin x} dx & \
& =\displaystyle \int \frac{\sin x}{\sin^{2} x} dx & \
& =\displaystyle \int \frac{1}{1-\cos^{2} x}\sin xdx & Let\ u=\cos x\
& =\displaystyle -\int \frac{1}{( 1+u)( 1-u)} du & \
& =\displaystyle -\frac{1}{2}\int \left(\frac{1}{1+u} +\frac{1}{1-u}\right) du
& Partial\ fractions\
& =\displaystyle -\frac{1}{2}\ln\left| \frac{1+u}{1-u}\right| +C & \
& =\displaystyle -\frac{1}{2}\ln\left| \frac{1+\cos x}{1-\cos x}\right| +C &
\end{array}
$$
**Exponential functions**
$$\displaystyle \int e^{x} dx=e^{x} +C$$
$$\displaystyle \int a^{x} dx=\frac{a^{x}}{\ln( a)} +C$$
**Logarithmic functions**
$$\displaystyle \int \frac{1}{x} dx=\ln |x|+C$$
**Inverse trigonometric functions**
$$\displaystyle \int \frac{1}{\sqrt{a^{2} -x^{2}}} dx=\arcsin\left(\frac{x}{a}\r
ight) +C$$
$$\displaystyle \int \frac{1}{a^{2} +x^{2}} dx=\frac{1}{a}\arctan\left(\frac{x}{
a}\right) +C$$
# Integration by parts
$$\displaystyle \int uvdx=u\int vdx-\int \left( u^{\prime }\int vdx\right) dx$$
# Integration by reduction formulae
$$\displaystyle \int \sin^{n} xdx=-\frac{1}{n}\sin^{n-1} x\cos x+\frac{n-1}{n}\i
nt \sin^{n-2} xdx$$
$$\displaystyle \int \cos^{n} xdx=\frac{1}{n}\cos^{n-1} x\sin x+\frac{n-1}{n}\in
t \cos^{n-2} xdx$$
$$\displaystyle \int \tan^{n} xdx=\frac{1}{n-1}\tan^{n-1} x-\int \tan^{n-2} xdx$
$
$$\displaystyle \int (\ln x)^{n} dx=x(\ln x)^{n} -n\int (\ln x)^{n-1} dx$$