Limits

If by selecting a small punctuated neighborhood of $x$, a variable $y$ determined by $x$ can close enough to a fix value in any level, then that variable $y$ is converge to that fix value.

Definition of Limits

Limits of Sequence

$$\lim \_n\to \infty x\_n=A$$

is saying:

$$\forall ϵ>0,\exists N>0,\forall n>N,|x\_n-A|\le ϵ$$

Limits of function

the limit of $f$ of $x$ as $x$ approaches $c$

$$\lim \_x\to af(x)=A$$

it means that the value of the function $f$ can be made arbitrary close to $L$ by choosing x close to $c$

\forall \epsilon > 0, \exists \delta > 0, \forall x \in \left{ x : 0 < |x - a| < \delta \right}, |f(x) - A| < \epsilon $$\lim \_x\to \infty f(x)=A$$

is saying:

$$\forall ϵ>0,\exists x\_0>0,\forall x>x\_0,|f(x)-A|<ϵ$$

Heine’s Theorem

人话：所有路径都收敛于一个值，那么就在该区域无死角收敛了。

证明：

• 如果函数收敛，自然每一条路径都收敛
• 如果函数不收敛，那么能够构造不收敛的坏数列

Sign-Preserving Property

Sign-Preserving Property

If a function $f$ has a limit $A$ on $a$ , then there’s a punctured neighborhood $I\_a$ of $a$, one that set, $f$ keeps the same sign with the limit $A$

Suppose $\lim \_x\to af(x)=A>0$

Set $ϵ=\frac{A}{2}$, then $\exists \delta >0$, $\forall x\in (a-\delta ,a)\cup (a,a+\delta )$, $\left|f(x)-A\right|<ϵ=\frac{A}{2}⟹f(x)>\frac{A}{2}>0$

So as when $\lim \_x\to af(x)=A<0$

Squeeze Theorem

Squeeze Theorem

if $g(x)\le f(x)\le h(x)$, and $g(x)$ and $h(x)$ has same limitation $A$ on $a$, then $f(x)$ also has limitation $A$ on $a$

$$\begin{array}{rl}& \lim \_x\to ag(x)=\lim \_x\to ah(x)=A\\ ⟹& \forall ϵ>0,\exists \delta \_1,\delta \_2>0,\delta =\min (\delta \_1,\delta \_2),\forall x\in (a-\delta ,a)\cup (a,a+\delta ),|g(x)-A|<ϵ,|h(x)-A|<ϵ\\ ⟹& A-ϵ<g(x)<A+ϵ,A-ϵ<h(x)<A+ϵ\\ ⟹& A-ϵ<g(x)\le f(x)\le h(x)<A+ϵ\\ ⟹& |f(x)-A|<ϵ\\ ⟹& \lim \_x\to af(x)=A\end{array}$$

Order Property of Limits

Order Property of Limits

If $f(x)\le g(x)$, and $\lim \_x\to af(x)=A,\lim \_x\to ag(x)=B$, then $A\le B$

if $A>B$, then for $ϵ=\frac{A-B}{2}$, $\exists \delta \_1,\delta \_2>0,\delta =\min (\delta \_1,\delta \_2),\forall x\in (a-\delta ,a)\cup (a,a+\delta )$:

$$\begin{array}{rl}& |f(x)-A|<\frac{A-B}{2},|g(x)-B|<\frac{A-B}{2}\\ ⟹& f(x)>\frac{A+B}{2},g(x)<\frac{A+B}{2}\\ ⟹& g(x)<f(x)\end{array}$$

which contradicts with $f(x)\le g(x)$. So $A\le B$

Local Boundedness

Local Boundness

If $\lim \_x\to af(x)=A$, $f(x)$ is bounded in some punctured neighborhood of $a$

Rules with Real Number Operations

$$\lim \_x\to af(x)=A,\lim \_x\to ag(x)=B$$

Constant Rule

swappable to scalar product

$$\lim \_x\to akf(x)=k\lim \_x\to af(x)$$

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Proof

with $\delta$, $|f(x)-L|\le \frac{ϵ}{|k|}$, thus:

$$|kf(x)-kL|=|k||f(x)-L|\le ϵ$$

Sum Rule

distributive to function addition

\lim\_{ x \to a } \left\[ f(x) + g(x) \right] = \lim\_{ x \to a } f(x) + \lim\_{ x \to a } g(x)

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Proof: (triangular Inequality)

with $\delta \_1$, $|f(x)-L|\le \frac{ϵ}{2}$ with $\delta \_2$, $|g(x)-M|\le \frac{ϵ}{2}$ Thus

$$\begin{array}{rl}|f(x)+g(x)-(L+M)|\le & |f(x)-L|+|g(x)-M|\\ \le & \frac{ϵ}{2}+\frac{ϵ}{2}=ϵ\end{array}$$ \lim\_{ x \to a }\left\[ f(x) + g(x) \right] = \lim\_{ x \to a } f(x) + \lim\_{ x \to a } g(x)

Product Rule

distributive to function product

\lim\_{ x \to a } \left\[ f(x)g(x) \right] = \lim\_{ x \to a } f(x) \lim\_{ x \to a } g(x)

Proof:

1. involve a trick of constructing $|f(x)-L|$ and $|g(x)-M|$
2. be aware that $f(x)$ is locally bounded due to the Local Boundness

$\forall ϵ>0$:

• with $\delta \_1$, $|f(x)-L|<\frac{ϵ}{2(|M|+1)}$, thus $|f(x)|<|f(x)-L|+|L|<\frac{ϵ}{2(|M|+1)}+|L|=K$
• with $\delta \_2,|g(x)-M|<\frac{ϵ}{2K}$,

so, with $\delta =\min (\delta \_1,\delta \_2)$,

$$\begin{array}{rl}|f(x)g(x)-ML|& =|f(x)g(x)-Mf(x)+Mf(x)-ML|\\ & \le |f(x)||g(x)-M|+|M||f(x)-L|\\ & \le K\frac{ϵ}{2K}+|M|\frac{ϵ}{2(|M|+1)}\\ & \le ϵ\end{array}$$

Quotient Rule

distributive to function devision

$$\lim \_x\to a\frac{f(x)}{g(x)}=\frac{\lim \_x\to af(x)}{\lim \_x\to ag(x)}$$

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Proof

key steps:

step1: Proof $\lim \_x\to a\frac{1}{g(x)}=\frac{1}{B}$

$$\begin{array}{rl}\left|\frac{1}{g(x)}-\frac{1}{B}\right|& =\left|\frac{B-g(x)}{Bg(x)}\right|\\ & =\frac{|B-g(x)|}{|B||g(x)|}\end{array}$$

step2: use distributive to function product:

$$\lim \_x\to a\frac{f(x)}{g(x)}=\lim \_x\to af(x)\frac{1}{g(x)}=\frac{A}{B}$$

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Detail Proof

This proof is divided into two steps:

1. proof $\lim \_x\to a\frac{1}{g(x)}=\frac{1}{M}$ and reverse triangular inequality
2. product rule

with $\delta \_1$, $|g(x)-M|\le \frac{|M|}{2}$, which saids:

$$\begin{array}{r}||g(x)|-|M||\le |g(x)-M|\le \frac{|M|}{2}\\ -\frac{|M|}{2}\le |g(x)|-|M|\le \frac{|M|}{2}\\ \frac{|M|}{2}\le |g(x)|\le \frac{3}{2}|M|\end{array}$$

with $\delta \_2$, $|g(x)-M|\le \frac{|M{|}^2ϵ}{2}$

Thus:

$$\left|\frac{1}{g(x)}-\frac{1}{M}\right|=\left|\frac{M-g(x)}{Mg(x)}\right|=\frac{|g(x)-M|}{|M||g(x)|}\le ϵ$$

Thus:

$$\lim \_x\to c\frac{1}{g(x)}=\frac{1}{M}$$

applying the product rule:

$$\lim \_x\to c\frac{f(x)}{g(x)}=\lim \_x\to cf(x)\lim \_x\to c\frac{1}{g(x)}=\frac{L}{M}$$

Composition Rule

composition rule

if $f$ is continuous, then

$$\lim \_x\to x\_0f(g(x))=f\left(\lim \_x\to x\_0g(x))\right)$$

$$\begin{array}{r}\forall x\in \dot{U}(x\_0,\delta \_x),u=g(x)\in U(u\_0,\delta \_u)\\ \forall u\in \dot{U}(u\_0,\delta \_u),y=f(x)\in U(y\_0,ϵ)\\ f(u\_0)=\underset{\text{ the limit}}{\underbrace{y\_0}}\\ ⟹\forall x\in \dot{U}(x_0,\delta \_x),y=f(x)\in U(y\_0,ϵ)\end{array}$$