limit of functions

Prerequisites

• converge

Definition of convergence of functions at a point

Convergence of functions at a point: Let $X$ be a subset of ${\mathbb{R}}^n$, let $f:X\to {\mathbb{R}}^m$ be a function, let $E$ be a subset of $X$, $x_0$ be an adherent point of $E$, and let $L\in {\mathbb{R}}^m$. We say that $f$ $\text{converges to}$ $L$ $\text{at}$ $x_0$ $\text{in}$ $E$ and write

$$\underset{x\to x_0,x\in E}{\lim }f(x)=L,$$

iff $\forall ϵ>0,\exists \delta >0$, s.t. (($\forall x\in E\wedge d(x_0,x)<\delta$) $⟶$ $d(f(x),L)<ϵ$) is true.

Equivalent definition

The following statements are equivalent:

1. $f$ converges to $L$ at $x_0$ in $E$.
2. For every sequence $(a_n{)}_{n=0}^{\infty }$ which consists entirely of elements of $E$ and converges to $x_0$, the sequence $(f(a_n){)}_{n=0}^{\infty }$ converges to $L$.

proof:
(1) $⟶$ (2):
(1) $⟺$ $\forall ϵ>0,\exists \delta >0$, s.t. (($\forall x\in E\wedge d(x_0,x)<\delta$) $⟶$ $d(f(x),L)<ϵ$) is true.
$(a_n\to x_0)⟺\forall \delta >0,\exists N\in {\mathbb{Z}}_{\ge 0}$, s.t. $\forall n\ge N,d(a_n,x_0)<\delta$
Then, $\forall ϵ>0,\exists N\in {\mathbb{Z}}_{\ge 0}$, s.t. $\forall n\ge N,d(f(a_n),L)<ϵ$
so, $(f(a_n){)}_{n=0}^{\infty }$ converges to $L$.
(2) $⟶$ (1):
Suppose $f$ does not converge to $L$ at $x_0$ in $E$.
It means that $\exists ϵ>0,\forall \delta >0,\exists x\in E\wedge d(x_0,x)<\delta \text{ s.t. }d(f(x),L)\ge ϵ$
Choose $\delta =\frac{1}{n+1},n\in {\mathbb{Z}}_{\ge 0}$
Then, $\exists x_n\in E,d(x_0,x_n)<\frac{1}{n+1},d(f(x_n),L)\ge ϵ$
so, $(x_n{)}_{n=0}^{\infty }$ is a sequence which consists entirely of elements of $E$ and converges to $x_0$, but $(f(x_n){)}_{n=0}^{\infty }$ does not converge to $L$.
This is a contradiction.
Therefore, $f$ converges to $L$ at $x_0$ in $E$.

For n=1

We can get the law of limit of functions in one variable from the law of limit of sequences.

$$\begin{array}{rl}\underset{x\to x_0}{\lim }(f\pm g)(x)& =\underset{x\to x_0}{\lim }f(x)\pm \underset{x\to x_0}{\lim }g(x)\\ \underset{x\to x_0}{\lim }\max (f,g)(x)& =\max \left(\underset{x\to x_0}{\lim }f(x),\underset{x\to x_0}{\lim }g(x)\right)\\ \underset{x\to x_0}{\lim }\min (f,g)(x)& =\min \left(\underset{x\to x_0}{\lim }f(x),\underset{x\to x_0}{\lim }g(x)\right)\\ \underset{x\to x_0}{\lim }(fg)(x)& =\underset{x\to x_0}{\lim }f(x)\underset{x\to x_0}{\lim }g(x)\\ \underset{x\to x_0}{\lim }(f/g)(x)& =\frac{{\lim }_{x\to x_0}f(x)}{{\lim }_{x\to x_0}g(x)}\end{array}$$

where we have dropped the restriction $x\in E$ for brevity

For n>1

$\mathbf{x}=(x_1,x_2,\cdots ,x_n{)}^T\in {\mathbb{R}}^n$

${\mathbf{x}}^n\to \mathbf{y}$ means that $\forall ϵ>0,\exists N\in {\mathbb{Z}}_{\ge 0}$, s.t. $\forall n\ge N,d({\mathbf{x}}^n,\mathbf{y})<ϵ$

The following statements are equivalent:

1. ${\mathbf{x}}^{\mathbf{n}}\to \mathbf{y}$
2. $\forall i\in \{1,2,\cdots ,n\},(x_i^n{)}_{n=0}^{\infty }\to y_i$

proof:
(1) $⟶$ (2):
${\mathbf{x}}^n\to \mathbf{y}$ means that $\forall ϵ>0,\exists N\in {\mathbb{Z}}_{\ge 0}$, s.t. $\forall n\ge N,\sqrt{{\sum }_{i=1}^n(x_i^n-y_i{)}^2}<ϵ$
so, $\forall i\in \{1,2,\cdots ,n\},\exists N_i\in {\mathbb{Z}}_{\ge 0}$, s.t. $\forall n\ge N_i,|x_i^n-y_i|\le \sqrt{{\sum }_{i=1}^n(x_i^n-y_i{)}^2}<ϵ$
(2) $⟶$ (1):
$(x_i^n{)}_{n=0}^{\infty }\to y_i$ means that $\forall ϵ>0,\exists N_i\in {\mathbb{Z}}_{\ge 0}$, s.t. $\forall n\ge N_i,|x_i^n-y_i|<ϵ/n$
Let $N=\max \{N_1,N_2,\cdots ,N_n\}$
Then, $\forall n\ge N,\sqrt{{\sum }_{i=1}^n(x_i^n-y_i{)}^2}\le {\sum }_{i=1}^n|x_i^n-y_i|<n(ϵ/n)=ϵ$
so, ${\mathbf{x}}^n\to \mathbf{y}$

Let $f:X\to {\mathbb{R}}^m,X\subseteq {\mathbb{R}}^n,f(\mathbf{x})=(f_1(\mathbf{x}),f_2(\mathbf{x}),\cdots ,f_m(\mathbf{x}){)}^T$

From above, we can easily get that

${\lim }_{\mathbf{x}\to {\mathbf{x}}_0}f(\mathbf{x})=({\lim }_{\mathbf{x}\to {\mathbf{x}}_0}{f}_1(\mathbf{x}),{\lim }_{\mathbf{x}\to {\mathbf{x}}_0}{f}_2(\mathbf{x}),\cdots ,{\lim }_{\mathbf{x}\to {\mathbf{x}}_0}{f}_m(\mathbf{x}){)}^T$

Let $X\subseteq {\mathbb{R}}^n$, $f,g:X\to {\mathbb{R}}^m$

We can get the law of limit of functions in multi-variables from the law of limit of one variable.

$$\begin{array}{rl}\underset{\mathbf{x}\to {\mathbf{x}}_0}{\lim }(f\pm g)(\mathbf{x})& =\underset{\mathbf{x}\to {\mathbf{x}}_0}{\lim }f(\mathbf{x})\pm \underset{\mathbf{x}\to {\mathbf{x}}_0}{\lim }g(\mathbf{x})\\ \underset{\mathbf{x}\to {\mathbf{x}}_0}{\lim }(fg)(\mathbf{x})& =\underset{\mathbf{x}\to {\mathbf{x}}_0}{\lim }f(\mathbf{x})\underset{\mathbf{x}\to {\mathbf{x}}_0}{\lim }g(\mathbf{x})\\ \underset{\mathbf{x}\to {\mathbf{x}}_0}{\lim }(f/g)(\mathbf{x})& =\frac{{\lim }_{\mathbf{x}\to {\mathbf{x}}_0}f(\mathbf{x})}{{\lim }_{\mathbf{x}\to {\mathbf{x}}_0}g(\mathbf{x})}\end{array}$$

where multiplication and division are component-wise.