Limits

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Definition of Limits

Limits of Sequence

$$\underset{n\to \infty }{\lim }{x}_n=A$$

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 $c$

$$\underset{x\to a}{\lim }f(x)=A$$

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

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

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 ${\lim }_{x\to a}f(x)=A>0$

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 ${\lim }_{x\to a}f(x)=A<0$

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 $a$

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

Order Property of Limits

Order Property of Limits

If $f(x)\le g(x)$, and ${\lim }_{x\to a}f(x)=A,{\lim }_{x\to a}g(x)=B$, 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 Boundness

Local Boundness

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

Rules with Real Number Operations

$$\underset{x\to a}{\lim }f(x)=A,\underset{x\to a}{\lim }g(x)=B$$

Constant Rule

swappable to scalar product

$$\underset{x\to a}{\lim }kf(x)=k\underset{x\to a}{\lim }f(x)$$

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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

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

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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

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

Product Rule

distributive to function product

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

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

$$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\frac{{\lim }_{x\to a}f(x)}{{\lim }_{x\to a}g(x)}$$

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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:

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

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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:

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

applying the product rule:

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

Composition Rule

composition rule

if $f$ is continuous, then

$$\underset{x\to x_0}{\lim }f(g(x))=f\left(\underset{x\to x_0}{\lim }g(x))\right)$$

$$\begin{array}{r}\forall x\in \dot{U}(x_0,{\delta }_x),u=g(x)\in U(u_0,{\delta }_u)\\ \forall u\in \dot{U}(u_0,{\delta }_u),y=f(x)\in U(y_0,ϵ)\\ f(u_0)=\underset{\text{ the limit}}{\underbrace{y_0}}\\ ⟹\forall x\in \dot{U}(x_0,{\delta }_x),y=f(x)\in U(y_0,ϵ)\end{array}$$