Chapter 1 Limits and Derivatives(1/2) - 微積分甲 (一) - 自修課程 (大學)

Limits of the Sequence
The Limits of Function
Evaluating Limits Analytically

# Limits of the Sequence

We hope to find $\lim_{n \to \infty}a_n = ?$

# Define

設 $\{a_n\},a_n \in \mathbb{R},n = N,L\in \mathbb{R}$ ，
若 $\forall \epsilon > 0 ,\exist N_0 \in \mathbb{N}$ s.t. $|a_n-L|<\epsilon$ if $n > N_0$ ，
則稱數列 $\{a_n\}$ 收斂於 $L$ ，記作 $\lim_{n \to \infty} a_n = L$ 或 $a_n \rightarrow L(n \rightarrow \infty)$

# Define

若 $\lim_{n \to \infty} a_n = L$ ，則稱數列 $\{a_n\}$ 為收斂 convergent ，反之則為發散 divergent

# Example

find the limit of the sequences:
Ex1. $\{ (-1)^n \}$ , Ex2. $\{ sin \; n \}$ , Ex3. $\{ 2n+5 \}$

$\lim_{n \to \infty}(-1)^n$ does not exist (震盪)
$\Rightarrow \{ (-1)^n \}$ is divergent.

$\lim_{n \to \infty}sin\; n$ does not exist
$\Rightarrow \{ sin \; n \}$ is divergent.

$\lim_{n \to \infty}2n+5 = \infty$ .The limit does not exist
$\Rightarrow \{ 2n+5 \}$ is divergent.

# The Limits of Functions

# 直觀的角度

$f(x)$ is defined $,\; \lim_{x\to a}f(x) = L ( = f(a))$
$f(x)$ is not defined but $\; \lim_{x\to a}f(x) = L$
$f(x)$ is defined but $\; f(a) \neq L , \; \lim_{x\to a}f(x) = L$

左右趨近值不相同
Infinite limits
震盪行為

$H(x) = \begin{cases} 1 & \text{ if } x > 0 \\ -1 & \text{ if } x \le 0 \end{cases}$
$\lim_{x \to 0}H(x)$ does not exist

$G(x) = \frac{1}{x^2}$
$\lim_{x \to 0}G(x) = \infty$ does not exist

$F(x) = sin(\frac{1}{x})$
$\lim_{x \to 0}F(x)$ does not exist

# 數列觀點下的函數極限

# Define

設 $A \in \mathbb{R}$ ， $A$ 的導集為 $\{ a | \exists \{ a_n \} \in A \setminus \{a \} \; s.t. \; \lim_{n \to \infty} a_n = a \}$
$A$ 的導集以 $A'$ 表示

# Define

$A \in \mathbb{R} \; f: A \rightarrow \mathbb{R}$ is a function

若所有取值在 $A \setminus \{ a \}$ 中並收斂到 $a$ 的數列 $\{ a \} ^\infty_{n=1}$ , $\{f(a_n)\}^\infty_{n=1}$ 也會收斂到同一個值，
則我們說當 $x$ 趨近於 $a$ 時， $f(x)$ 的極限存在。

# Proposition

若 f (x) 的極限存在，則極限必唯一。換句話說，
$\exists L \in \mathbb{R}$ 滿足對取值於 $A \setminus \{ a \}$ 並收斂到 $a$ 的數列 $\{ a_n \}$
都有 $\lim_{n \to \infty}=L$ ，我們稱「當 $x$ 趨近於 $a$ 時， $f(x)$ 的極限為 $L$ 」

# Example

Find the limit of $\lim_{x \to 0}f(x)$ ,where $f(x) = \begin{cases} sin(\frac{1}{x}) & \text{ if } x \neq 0 \\ 0 & \text{ if } x = 0 \end{cases}$

取 $\begin{matrix} \{x_n\} &= \{ \frac{1}{2n\pi} \} \in \mathbb{R}\setminus \{0\} & (\lim_{n \to \infty} \frac{1}{2n\pi} = 0) \\ \{y_n\} &= \{ \frac{1}{2n\pi+\frac{\pi}{2}} \} \in \mathbb{R}\setminus \{0\} & (\lim_{n \to \infty} \frac{1}{2n\pi+\frac{\pi}{2}} = 0) \end{matrix}$

But $\begin{matrix} \lim_{n\to \infty}f(x_n) =& \lim_{n\to \infty}sin(2n\pi)&=0 \\ \lim_{n\to \infty}f(y_n) =& \lim_{n\to \infty}sin(2n\pi+\frac{\pi}{2}&=1 \end{matrix}$
$\Rightarrow \lim_{x \to 0}f(x)$ does not exist $(0 \neq 1)$

# 數學語言

# define

$\lim_{x\to a}f(x) =L \leftrightarrow \forall \epsilon >0, \exists \delta>0 \; s.t. \;|f(x)-L|<\epsilon \; whenever \; 0<|x-a|< \delta$

# Evaluating Limits Analytically

# `Limit Laws`

Suppose that c is a constant and $\lim_{x\to a}f(x), \lim_{x\to a}f(x)$ exist. Then

$\lim_{x\to a}(f(x) \pm g(x)) = \lim_{x\to a}f(x) \pm \lim_{x\to a}g(x)$
$\lim_{x\to a}cf(x) = c \lim_{x\to a}f(x)$
$\lim_{x\to a}(f(x)g(x)) = \lim_{x\to a}f(x) \times \lim_{x\to a}g(x)$
$\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}\; if \; \lim_{x\to a}g(x) \neq 0$
$\lim_{x \to a}c = c$
$\lim_{x \to a}x = a$
$\lim_{x \to a}(f(x))^n = (\lim_{x \to a}f(x))^n ,\; n \in \mathbb{N}$ (from 3.)
$\lim_{x \to a}(x^n) = (\lim_{x \to a}x)^n = a^n, \; n \in \mathbb{N}$ (from 6. 7.)
若 $p(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_0$ is a polynomial function (多項式函數)，
則 $\lim_{x \to a}p(x) = p(a)$
若 $r(x) = \frac{p(x)}{q(x)}$ is a rational function (有理函數).
This means $p(x)$ and $q(x)$ are polynomial functions and
$c \in \mathbb{R} \; s.t. \; g(c) \neq 0$ then $\lim_{x \to c}r(x) = \frac{\lim_{x \to c}p(x)}{\lim_{x \to c}q(x)} = \frac{p(c)}{q(c)}$

Remark

$f(x) = x^2sin(\frac{1}{x})$ ，find $\lim_{x \to 0}f(x)$
$\lim_{x \to 0}x^2sin(\frac{1}{x}) \neq \lim_{x \to 0}x^2 \times \lim_{x \to 0}sin(\frac{1}{x})$
Because $\lim_{x \to 0}sin(\frac{1}{x})$ does not exist

# Thm

運用 左右極限 來判斷 極限是否存在
$\lim_{x \to a}f(x)=L \Leftrightarrow \lim_{x \to a^+}f(x) = \lim_{x \to a^-}f(x) = L$

Prove that $\lim_{x\to 0 }\frac{|x|}{x}$ does not exist

$|x| = \begin{cases} x &\text{ if } x \ge 0 \\ -x &\text{ if } x < 0 \end{cases} \; \Rightarrow \frac{|x|}{x} = \begin{cases} 1 &\text{ if } x \ge 0 \\ -1 &\text{ if } x < 0 \end{cases}$
$\because \lim_{x \to 0^+} \frac{|x|}{x}=1$ and $\lim_{x \to 0^-} \frac{|x|}{x}=-1$
$\therefore \lim_{x \to 0} \frac{|x|}{x}$ does not exist

$f(x) = \begin{cases} \sqrt{x-3} &\text{ if } x \ge 3 \\ 6-2x &\text{ if } x < 3 \end{cases}$
Find $\lim_{x\to 3}f(x)$

$\lim_{x \to 3^+}f(x) = \lim_{x \to 3^+}\sqrt{x-3} = 0$
$\lim_{x \to 3^-}f(x) = \lim_{x \to 3^-}6-2x = 0$
$\Rightarrow \lim_{x\to 3}f(x) = 0$

# Thm

If $\lim_{x\to a}|f(x)| = 0$ then $\lim_{x\to a}f(x) = 0$

Remark

If $L \neq 0$ and $\lim_{x\to a}|f(x)| = L$ then this Thm fails.

# Thm

If $f$ and $g$ are two function such that $\lim_{x\to a}g(x) = L$ and $\lim_{x\to L}f(L)$ ,
then $\lim_{x\to a}(f \circ g)(x)= \lim_{x\to a}f(g(x)) = f(L)$

Example

Let $f(x) = sin x , g(x) = x^2$

$(f \circ g)(x) = f(g(x)) = sin x^2$
$(g \circ f)(x) = g(f(x)) = sin^2 x$

Remark

$sin^{-1} x \neq \frac{1}{sin x} \Rightarrow f(x) = sin^{-1}x$ is the inverse function of $f(x) = sin x$

# Thm

If $f$ and $g$ are two functions defined on $I$ and $a \in I$ (except possibly at $a$ )
and $f(x) \le g(x)$ on $I \setminus \{a\},\; \lim_{x\to a}f(x),\; \lim_{x\to a}g(x)$ exist
then $\lim_{x\to a}f(x)\le \lim_{x\to a}g(x)$

# `The Squeeze Thm`

If $f(x) \le h(x) \le g(x)$ is near a (except possibly at $a$ )
and $\lim_{x\to a}f(x) = L = \lim_{x\to a}g(x)$ then $\lim_{x\to a}h(x) = L$

$f(x) = \begin{cases} x^2sin\frac{1}{x} &\text{ if } x \neq 0 \\ 0 &\text{ if } x = 0 \end{cases}$
Find $\lim_{x\to 0}f(x)$

$\because -1\le sin\frac{1}{x} \le 1 \; \forall x \in \mathbb{R} \setminus \{ 0 \}$
$\Rightarrow -x^2 \le x^2sin\frac{1}{x} \le x^2 \; \forall x \in \mathbb{R} \setminus \{ 0 \}$
$\because \lim_{x\to 0}(-x^2) = 0 = \lim_{x\to 0}x^2 \; \therefore \; \lim_{x\to 0}x^2sin\frac{1}{x} = 0$ (by the Squeeze Thm)

Prove that $\lim_{x \to 0}\frac{sin x}{x} = 1$

If $0<x<\frac{\pi}{2}$ then $sin \theta < \theta < tan \theta$ , we have

By $sin \theta < \theta \Rightarrow \frac{sin \theta}{\theta} < 1$
By $\theta < tan\theta = \frac{sin \theta}{cos \theta} \Rightarrow cos\theta < \frac{sin\theta}{\theta}$

由 1. 2. $\Rightarrow cos\theta < \frac{sin\theta}{\theta}<1$
$\because \lim_{x\to 0^+}cosx = 1 = \lim_{x\to 0^+} 1$
$\therefore$ By the Squeeze Thm. $\lim_{x\to 0^+}\frac{sin x}{x} = 1$

If $\frac{-\pi}{2}<x<0$ let $y = -x > 0$
$\Rightarrow 0 < y < \frac{\pi}{2} \Rightarrow \lim_{y\to 0^+}\frac{sin y}{y} = 1$
$\because y \rightarrow 0^+\Rightarrow -x \rightarrow 0^+ \Rightarrow x \rightarrow 0^-$

$\therefore \lim_{y\to 0^+}\frac{sin y}{y} = \lim_{x\to 0^-}\frac{sin (-x)}{-x} = \lim_{x\to 0^-}\frac{-sin x}{-x} = \lim_{x\to 0^-}\frac{sin x}{x} = 1$

$\Rightarrow \lim_{x\to 0}\frac{sin x}{x} = 1$

$\lim_{x\to 0}\frac{1-cosx}{x}$
$\lim_{x\to 0}\frac{tanx}{x}$
$\lim_{x\to 0}\frac{sin(ax)}{x}, a\in \mathbb{R}$
$\lim_{x\to 0}\frac{sinx^2}{x}$
$\lim_{x\to 0}\frac{xsinx}{1-cosx}$

$\lim_{x\to 0}\left ( \frac{1-cosx}{x}\cdot \frac{1+cosx}{1+cosx} \right ) = \lim_{x\to 0}\frac{1-cos^2x}{x(1+cosx)} = \lim_{x\to 0}\left ( \frac{sin x}{x} \right )\left ( \frac{sin x}{1+cosx} \right ) = 1 \cdot 0 = 0$

$\lim_{x\to 0}\frac{tan x}{x} = \lim_{x\to 0}\left ( \frac{sin x}{x} \cdot \frac{1}{cosx} \right ) = 1 \cdot 1 = 1$

$\lim_{x\to 0}\frac{sin ax}{x} = \lim_{x\to 0}\left ( \frac{sin ax}{ax} \cdot a \right ) = 1 \cdot a = a$

$\lim_{x\to 0}\frac{sin^2 x}{x} = \lim_{x\to 0}\left ( \frac{sin^2 x}{x^2} \cdot x \right ) = 1 \cdot 0 = 0$

$\lim_{x\to 0}\left ( \frac{xsinx}{1-cosx} \cdot \frac{1+cosx}{1+cosx} \right ) = \lim_{x\to 0}\frac{xsinx(1+cosx)}{sin^2x} = \lim_{x\to 0}\frac{x(1+cosx)}{sinx}$
$= \lim_{x\to 0}\left ( \frac{x}{sinx} \cdot (1+cosx) \right ) = 1 \cdot 2 = 2$

# Reference

蘇承芳老師 - 微積分甲（一）109 學年度 - Calculus (I) Academic Year 109

數學微積分大學自修

# Limits of the Sequence

# Define

# Define

# Example

# The Limits of Functions

# 直觀的角度

# 數列觀點下的函數極限

# Define

# Define

# Proposition

# Example

# 數學語言

# define

# Evaluating Limits Analytically

# Limit Laws

# Thm

# Thm

# Thm

# Thm

# The Squeeze Thm

# Reference

第05篇、生成函數

Chapter 1 Limits and Derivatives(2/2)

# `Limit Laws`

# `The Squeeze Thm`