Limit of mapping and function

1.4 Limit of mapping and function

Definition 1.4.1 Given $f : E \to R^{p}$ , $x_{0}$ is the cluster/accumulation point of $E \subseteq R^{m}$ (meaning for any $δ > 0$ , there exists $x \in E$ subjecting to $0 < ∥ x - x_{0} ∥ < δ$ ). If for any $ϵ > 0$ , there exists $δ_{ϵ} > 0$ such that for any $x \in E$ subjecting to $0 < ∥ x - x_{0} ∥ < δ_{ϵ}$ , $∥ f (x) - A ∥ < ϵ$ , then $lim_{x \to x_{0}} f (x)$ exists and $lim_{x \to x_{0}} f (x) = A$ .

Definition 1.4.2 Given $f : E \to R^{p}$ , $x_{0}$ is the cluster point of $E \subseteq R^{m}$ . If for any $M > 0$ , there exists $δ_{M} > 0$ such that for any $x \in E$ subjecting to $0 < ∥ x - x_{0} ∥ < δ_{M}$ , $∥ f (x) ∥ > M$ , then it’s marked as $lim_{x \to x_{0}} f (x) = + \infty$ . Similar for $- \infty$ and $\infty$ .

Definition 1.4.3 $lim_{x \to a, y \to + \infty} f (x, y) = A \Leftrightarrow$ For any $ϵ > 0$ , there exists $δ_{ϵ} > 0$ and $N_{ϵ} > 0$ such that $| x - a | < δ_{ϵ} \land y > N_{ϵ} \Rightarrow | f (x, y) - A | < ϵ$ . Similar for $lim_{x \to + \infty, y \to \infty}$ , etc.

Definition 1.4.4 $lim_{(x, y) \to \infty} f (x, y) = A \Leftrightarrow$ For any $ϵ > 0$ , there exists $N_{ϵ} > 0$ such that $∥ (x, y) ∥ > N_{ϵ} \Rightarrow | f (x, y) - A | < ϵ$ .

Theorem 1.4.5 If $x_{0}$ is the cluster point of $E$ , then $lim_{x \to x_{0}} f (x) = A$ is equivalent to that $\tilde{f} (x) = {\begin{cases} f (x) & x \in E ∖ {x_{0}} \\ A & x = x_{0} \end{cases}$ is continuous.

Theorem 1.4.6 If $x_{0}$ is the cluster point of $E$ , then $lim_{x \to x_{0}} f (x) = A$ is equivalent to that for any ${x_{n}} \in E$ where $lim_{n \to + \infty} x_{n} = x_{0}$ and $x_{n}$ , $lim_{n \to + \infty} f (x_{n}) = A$ .

Theorem 1.4.7 Composition. If $lim_{x \to x_{0}} f (x) = y_{0}$ , $lim_{y \to y_{0}} g (y) = A$ , and satisfies one of the conditions below

There exists $δ > 0$ such that for any $x \in E$ subjecting to $0 < ∥ x - x_{0} ∥ < δ$ , $f (x) \neq y_{0}$ .
$g (y_{0}) = A$ , or $g$ is continuous at $y_{0}$ .

then $lim_{x \to x_{0}} g (f (x)) = A$ .

Note Several notes for the limit.

1.: The limit has nothing to do with the selection of the norm.
2.: The computation of continuous mappings is true for the limit.
3.: Theorem 1.4.6 is commonly used to prove that the limit doesn’t exist.
4.: Theorem 1.4.7 could decompose the mapping into several simpler mappings. The following is to explain that $lim_{y \to y_{0}} g (y)$ doesn’t exist. Construct a continuous curve $y (t)$ where $lim_{t \to 0} y (t) = y_{0}$ yet $lim_{t \to 0} g (y (t))$ doesn’t exist.
5.: $lim = lim = lim x→a,y→b (x,y)→ (a,b)$

Example 1.4.1

1.

Seek $L (a, b) = lim_{(x, y) \to (a, b)} f (x, y)$ where $E = R^{2} ∖ {(0, 0)}$ and

3 3 f(x,y ) = exp(x--+-y-) −-1- x2 + y2

$(a, b) \neq (0, 0)$ . When $(x, y) \to (a, b)$ , $x^{2} + y^{2} \to a^{2} + b^{2} > 0$ , so $f (x, y)$ is continuous at $(a, b)$ , i.e. $L (a, b) = f (a, b)$ .
$(a, b) = (0, 0)$ . Here
$exp(x3 + y3) − 1 0 lim ------2---2------→ -- (x,y)→ (a,b) x + y 0$
So we could expand the numerator with Taylor series. Let $t = x^{3} + y^{3}$ , then when $t \to 0$ , $e^{t} - 1 = t + o (t)$ meaning there exists $δ_{0} > 0$ such that for any $| t | < δ_{0}$ , $| e^{t} - 1 | \leq 2 t$ . therefore $| x^{3} + y^{3} | \leq 2 ∥ x ∥_{\infty}^{3} < δ_{0}$ . Notice that $\begin{aligned} | \frac{\exp (x^{3} + y^{3}) - 1}{x^{2} + y^{2}} - 0 | & \leq \frac{2 | x^{3} + y^{3} |}{x^{2} + y^{2}} \leq 2 \frac{| x |^{3} + | y |^{3}}{x^{2} + y^{2}} \\ \leq \frac{4 ∥ x ∥_{\infty}^{3}}{∥ x ∥_{\infty}^{2}} = 4 ∥ x ∥_{\infty} < ϵ \end{aligned}$

Terminally, for any $0 < ∥ x ∥_{\infty} < δ_{ϵ} = min {\frac{ϵ}{3}, \sqrt[3]{\frac{δ_{0}}{2}}}$ , $| f (x, y) | < ϵ$ , i.e. $L (0, 0) = 0$ .

2.

Seek

x+y+1 x-+-y-+-1- L = (x,ly)im→ (1,0)(x + y)x+y−1 = (x,yli)→m(1,0)exp ∖lef t[x + y − 1 ln(x + y)∖right]

Let $t = x + y - 1 = t (x, y) \to 0$ , here $L = lim_{(x, y) \to (1, 0)} \exp f (t)$ where $f (t) = \frac{t + 2}{t} \ln (t + 1)$ and $lim_{t \to 0} f (t) = 2$ , therefore $L = e^{2}$ .

3.

Seek $L (a, b) = lim_{(x, y) \to (a, b)} f (x, y)$ where $E = {(x, y) | y > 0}$ and

x2 f(x, y) = exp∖lef t(− √---∖right) y

$(a, b) \in E$ , $f$ is continuous at $(a, b)$ , i.e. $L (a, b) = f (a, b)$ .
$a \neq 0, b = 0$ , here $x^{2} \to a^{2} > 0$ . Let $| x - a | < \frac{| a |}{2} \Rightarrow | x | > \frac{| a |}{2}$ , so
$x2-- --a2- ? --a4-- √y--> 4 √y- > M ⇔ 0 < y < 16M 2$
For any $ϵ > 0$ , take $M > max {1, - \ln ϵ}$ and $δ_{M} = min {\frac{| a |}{2}, \frac{a^{4}}{16 M^{2}}}$ , then for any $(x, y) \in E$ subjecting to $0 < ∥ (x, y) - (a, 0) ∥_{\infty} < δ_{M}$ , $| f (x, y) | < ϵ$ , i.e. $L (a, 0) = 0$ .
$(a, b) = (0, 0)$ , let $\frac{x^{2}}{\sqrt{y}} = α$ , here $(t, \frac{t^{4}}{α^{2}}) \to (0, 0)$ , $f (t, \frac{t^{4}}{α^{2}}) = e^{- α}$ , it’s related to $α$ . So $L (0, 0)$ doesn’t exist.

Multiple limit

L(a,b) = lim f(x,y) (x,y)→ (a,b)

and repeated limit

Lxy (a, b) = xli→ma lyi→mb f(x,y), Lyx(a,b) = liym→bxli→ma f(x, y)

Take $f (x, y) = \exp (- \frac{x^{2}}{\sqrt{y}})$ as an example.

Multiple limit.
$L(a,b) =$
When $b > 0$ ,
$Lxy(a,b) = Lyx(a,b) = f(a,b)$
.
When $b = 0$ , $\begin{aligned} g (x) & = lim_{y \to 0^{+}} \exp (- \frac{x^{2}}{\sqrt{y}}) = {\begin{cases} 1 & x = 0 \\ 0 & x \neq 0 \end{cases} \\ lim_{x \to a} g (x) & = {\begin{cases} 0 & a \neq 0 \\ 0 & a = 0 \end{cases} = 0 \\ h (y) & = lim_{x \to a} \exp (- \frac{x^{2}}{\sqrt{y}}) = \exp (- \frac{a^{2}}{\sqrt{y}}) \\ lim_{y \to 0^{+}} h (y) & = {\begin{cases} 0 & a \neq 0 \\ 0 & a = 0 \end{cases} = 0 \end{aligned}$

Therefore $L_{x y} (a, 0) = L_{y x} (a, 0) = 0$ for any $a$ .

Theorem 1.4.8 Generally, assuming $U \subseteq R^{2}$ is an open set, $(a, b) \in U$ , $f$ is defined on $U {∖ (a, b)}$ satisfying that

L (a,b) = (x,ly)i→m(a,b)f (x,y), Lxy (a, b) = xli→ma lyim→b f(x,y), Lyx(a,b) = liym→b lxi→ma f(x, y)

all exist, then $L (a, b) = L_{x y} (a, b) = L_{y x} (a, b)$ . In another word, if $L_{x y} (a, b) \neq L_{y x} (a, b)$ , then $L (a, b)$ doesn’t exist.