---
tags: Calculus, 2022-Fall-CalculusI
---
{%hackmd theme-dark %}
# 極限、連續、微分大雜燴
###### tags: `Calculus` `2022-Fall-CalculusI`

# 重要定理及其證明
## 連續必極限
:::success
**[Theorem]**
$$f:\text{continuous at }x=a\Longrightarrow \lim_{x \to a}f(x): \text{exists}$$

Proof:
$\because f : \text{cont. at } x = a, \ \therefore \text{by definition of continuity, the limit}\\
\text{exists and }f(a) \text{ exists, too.}\quad_\blacksquare$

或是你要大喊"trivial!!"，我也不能反駁你。
Warning: "$\Leftarrow$" 不能反推，也就極限存在不一定會連續。cex : $y=\frac{x^2-4}{x-2}$
:::

## 微分必連續
:::success
**[Theorem 2.8.4]**
$$f:\text{differentiable at }x=a\Longrightarrow f:\text{continuous at }x=a$$

Proof:
Note that $f : \text{diff. at } x = a$. Therefore, the following limit exists:
$$\lim_{x\to a}{\frac{f(x)-f(a)}{x-a}}$$, say(令成是/叫做) $C \in \mathbb{R}$.
Also, we have $\lim_{x\to a} (x-a)=0$.
Hence, we have:
$$\lim_{x\to a}f(x)=\lim_{x\to a}[(f(x)-f(a))+f(a)]=\lim_{x\to a}[(f(x)-f(a))]+\lim_{x\to a}f(a)$$$$=\lim_{x\to a}[\frac{f(x)-f(a)}{x-a}\cdot (x-a)]+f(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a} \cdot \lim_{x\to a} (x-a)+f(a)$$$$=C \cdot 0 + f(a)=f(a)\qquad_\blacksquare$$
Warning: "$\Leftarrow$" 不能反推，cex : $y=|x|$
:::