Limits

Essential Concepts

Open Ball

The set of points $B(\mathbf{a},r)$ such that their distance to a given point $\mathbf{a}$ is strictly less than the given distance $r$ , i.e.

B(\mathbf{a},r)=\{\mathbf{x}\in\mathbb{R}^n:\Vert\mathbf{x}-\mathbf{a}\Vert\textcolor{red}{\lt}r\}

Closed Ball

The set of points $B(\mathbf{a},r)$ such that their distance to a given point $\mathbf{a}$ is less than or equal to the given distance $r$ , i.e.

\bar{B}(\mathbf{a},r)=\{\mathbf{x}\in\mathbb{R}^n:\Vert\mathbf{x}-\mathbf{a}\Vert\textcolor{red}{\le}r\}

Boundary Points

Suppose set $X$ is a subset of $\mathbb{R}^n$ , and some point $\mathbf{a}\in X$ . If any open ball centered at $\mathbf{a}$ simultaneously contains points that belongs to $X$ and points that do not belong to $X$ , then $\mathbf{a}$ is called a boundary point of $X$ .

The boundary of $X$ is the set of all boundary points of $X$ , denoted as $\partial X$

Interior Points

Suppose $X$ is a subset of $\mathbb{R}^n$ and some point $\mathbf{a}\in X$ . If there exists some $r\gt 0$ such that $B(\mathbf{a},r)\subset X$ , then $\mathbf{a}$ is called an interior point of $X$ .

The interior of $X$ is the set of all interior points of $X$ .

Open Set

Suppose $X\subset \mathbb{R}^n$ , if any $x \in X$ is an interior point of $X$ , then $X$ is an open set

Closed Set

Suppose $X\subset \mathbb{R}^n$ , if its complement $\complement_U X$ is an open set, then $X$ is a closed set

Proposition

$X$ is a closed set $\iff$ $X$ includes all its boundary points

$\varnothing,\mathbb{R}^2$ are both open sets and closed sets.

Limit Points, Isolated Points

Suppose $X\subset \mathbb{R}^n$ , if some point $\mathbf{a}\in \mathbb{R}^n$ satisfies that for all $\delta \gt 0$ , there always exists some point $\mathbf{x}\in X \cap B(\mathbf{a},\delta)$ and $\mathbf{x}\ne\mathbf{a}$ , then $\mathbf{a}$ is a limit point of $X$ , or an accumulation point of $X$ .

On the other hand, if $\mathbf{a}\in X$ is not a limit point, then it’s an isolated point.

The Limit of a Vector-Valued Function

With the concept of limit points, we can now define the limit of a vector-valued function with $\epsilon-\delta$ language.

Suppose function $\mathbf{f}: X\mapsto \mathbb{R}^m$ , where $X\subset \mathbb{R}^n$ and $\mathbf{a}$ is a limit point of $X$ . We say $\mathbf{L}$ to be the limit of $\mathbf{f}$ at $\mathbf{a}$ , if and only if for any $\epsilon$ , there exists some $\delta\gt 0$ such that when $\mathbf{x}\in X$ and $0\lt \Vert\mathbf{x-a}\Vert\lt \delta$ , we have $\Vert \mathbf{f(x)-L}\Vert\lt \epsilon$ . So we denote

\lim_{\mathbf{x}\to\mathbf{a}} \mathbf{f}(\mathbf{x})=\mathbf{L}

Proposition: Unique-ness of Limit

If $\mathbf{f}$ has a limit $\mathbf{L}$ at $\mathbf a$ , then this limit is unique.

极限运算法则

多元函数极限不存在的判定方法

有两种常用方法判断函数在某个点的极限不存在：

考虑穿过极限点的一条曲线，如果这条曲线在该点处没有极限，那么原函数在该点处也没有极限
考虑穿过极限点的两条曲线，如果这两条曲线在该点处的极限不同，那么也没有极限。

【例一】证明 $\lim_{(x,y)\to(0,0)}\frac{x+y^3}{x^3+y^2}$ 的极限不存在。

考虑 $y=0$ 这条直线穿过 $(0,0)$ 这个点，此时化简为 $\lim_{x\to 0}\frac{1}{x^2}$ ，而其极限不存在。所以原极限不存在。

【例二】证明 $\lim_{(x,y,z)\to (0,0,0)}\frac{2x^2+y^2-3z^2}{x^2+y^2+z^2}$ 的极限不存在

考虑沿直线 $y=z=0$ 去逼近，原极限等于

\lim_{x\to 0} \frac{2x^2}{x^2}=2

而考虑沿直线 $x=z=0$ 取逼近，原极限等于

\lim_{y\to 0} \frac{y^2}{y^2}=1

这两个极限不等，所以原极限不存在。

Sandwich Theorem

Sandwich Theorem

Let $f,g,h:X\mapsto \mathbb{R}$ be 3 functions, such that $X\in\mathbb{R}^n$ .

Suppose $\mathbf{a}\in X$ be a limit point of $X$ , and $f(\mathbf{x})\le g(\mathbf{x})\le h(\mathbf{x})$ holds for any $\mathbf{x}\in X$ . Then, if $\lim_{\mathbf{x}\to \mathbf{a}} f(\mathbf{x})=\lim_{\mathbf{x}\to \mathbf{a}} h(\mathbf{x})=L$ , we have $\lim_{\mathbf{x}\to\mathbf{a}} g(\mathbf{x})=L$ ．

[Example 3] Find $\lim_{(x,y)\to (0,0)}\frac{x^3+y^3}{x^2+y^2}$

Convert to polar coordinates by letting $x=r\cos\theta,y=r\sin\theta$ , so $(x,y)\to(0,0)\iff r\to 0^+$ , the expression is equivalent to

\lim_{r\to 0^+}r(\cos^3 \theta+\sin^3\theta)

And we have $-2r\le r(\cos^3\theta+\sin^3\theta)\le 2r$ and $\plusmn 2r\to 0$ , thus by sandwich theorem, we have

\lim_{r\to 0} r(\cos^3\theta+\sin^3\theta)=0