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\begin{document}

\title{\begin{CJK}{UTF8}{gbsn}淺談羅必達法則 \end{CJK}\selectlanguage{english}(L'H\^{o}pital's Rule)\selectlanguage{english}}


\author{Chih-Hung Chang}
\affil{Affiliation not available}



\date{\today}

\maketitle




\begin{CJK}{UTF8}{gbsn}相信大家都還記得剛開始學極限時\end{CJK}\selectlanguage{english},Qi Zhong [?] Ge Jing Dian De Ti Xing Jiu Shi \selectlanguage{english}$\displaystyle \lim_{x \to 1} \frac{x^2 - 1}{x - 1}$\begin{CJK}{UTF8}{gbsn}這種被稱為不定型\end{CJK}\selectlanguage{english}(Indeterminate Form)\begin{CJK}{UTF8}{gbsn}的極限問題。當時我們的做法很簡單\end{CJK}\selectlanguage{english},Jiu Shi Ba Gong Yin Shi Xiao Diao \selectlanguage{english},
\selectlanguage{english}$$
\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x + 1)(x - 1)}{x - 1} = \lim_{x \to 1} (x + 1) = 2
$$
\begin{CJK}{UTF8}{gbsn}但是\end{CJK}\selectlanguage{english},Ru Guo Jin Tian Wo Men Yu Dao Xiang Shi \selectlanguage{english}$\displaystyle \lim_{x \to 1} \frac{\ln x}{x-1}$\begin{CJK}{UTF8}{gbsn}這種問題時\end{CJK}\selectlanguage{english},Yi Qian De Zuo Fa Jiu Bu Guan Yong Liao . Ci Shi \selectlanguage{english},Ru Guo Wo Men Qiu Jiu Yu \selectlanguage{english}Google\begin{CJK}{UTF8}{gbsn}大神\end{CJK}\selectlanguage{english},Da Shen Jiu Hui Gao Su Wo Men \selectlanguage{english}:[Shi Yong \selectlanguage{english}L'\^{o}pital Rule\begin{CJK}{UTF8}{gbsn}吧\end{CJK}\selectlanguage{english}!] 

\selectlanguage{english}\begin{CJK}{UTF8}{gbsn}\section{不定型極限問題 (Indeterminate Form)}\end{CJK}\selectlanguage{english}
\begin{CJK}{UTF8}{gbsn}什麼叫做不定型極限問題\end{CJK}\selectlanguage{english}?Jian Dan Lai Shuo \selectlanguage{english},Jiu Shi Zhe Zhong Zhi Jie Ba \selectlanguage{english}$x$\begin{CJK}{UTF8}{gbsn}的極限值代入原始方程式之後\end{CJK}\selectlanguage{english},Hui Chan Sheng \selectlanguage{english}$\dfrac{0}{0}$\begin{CJK}{UTF8}{gbsn}或是\end{CJK}\selectlanguage{english}$\dfrac{\infty}{\infty}$\begin{CJK}{UTF8}{gbsn}的問題\end{CJK}\selectlanguage{english},[?] Lu Cheng Zhi Wei Bu Ding Xing De Ji Xian Wen Ti . 

Cong Xiao Xue San , Si Nian Ji Kai Shi \selectlanguage{english},Wo Men Jiu Zhi Dao \selectlanguage{english}$\dfrac{0}{0}$\begin{CJK}{UTF8}{gbsn}是沒有定義的。不定型問題之所以麻煩\end{CJK}\selectlanguage{english},Dao Bu Shi Yin Wei \selectlanguage{english}$\dfrac{0}{0}$\begin{CJK}{UTF8}{gbsn}沒有定義\end{CJK}\selectlanguage{english},Er Zai Yu Zhe Zhong Wen Ti Ge Zhong Jie Lun \selectlanguage{english}---Ji Xian Bu Cun Zai , Ji Xian Cun Zai Qie Wei Ding Zhi \selectlanguage{english},Yi Ji Ji Xian Shi Wu Qiong Da \selectlanguage{english}---Du You Ke Neng Fa Sheng . \selectlanguage{english}(\begin{CJK}{UTF8}{gbsn}這裡我們將無窮大這種不收斂的極限獨立出來討論。\end{CJK}\selectlanguage{english})\begin{CJK}{UTF8}{gbsn}底下我們來看一個簡單的例子。
\end{CJK}\selectlanguage{english}\begin{example}
Find (a) $\displaystyle \lim_{x \to 0} \frac{x^2}{x}$ \qquad (b) $\displaystyle \lim_{x \to 0} \frac{x}{x^2}$ \qquad (c) $\displaystyle \lim_{x \to 0^+} \frac{x}{x^2}$
\end{example}
\begin{CJK}{UTF8}{gbsn}顯然\end{CJK}\selectlanguage{english}(a)\begin{CJK}{UTF8}{gbsn}的極限是\end{CJK}\selectlanguage{english}0,(b)\begin{CJK}{UTF8}{gbsn}的極限不存在\end{CJK}\selectlanguage{english},Er \selectlanguage{english}(c)\begin{CJK}{UTF8}{gbsn}的極限為無窮大。

總的來說\end{CJK}\selectlanguage{english},Ju Fan \selectlanguage{english}$\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)}$\begin{CJK}{UTF8}{gbsn}滿足\end{CJK}\selectlanguage{english}(i)$f(x) \to 0$\begin{CJK}{UTF8}{gbsn}以及\end{CJK}\selectlanguage{english}$g(x) \to 0$\begin{CJK}{UTF8}{gbsn}或是\end{CJK}\selectlanguage{english}(ii)$f(x) \to \pm \infty$\begin{CJK}{UTF8}{gbsn}以及\end{CJK}\selectlanguage{english}$g(x) \to \pm \infty$,\begin{CJK}{UTF8}{gbsn}我們就將其稱為不定型的極限問題。

\end{CJK}\selectlanguage{english}\section{L'H\^{o}pital's Rule}
L'H\^{o}pital's Rule\begin{CJK}{UTF8}{gbsn}的定理敘述如下。
\end{CJK}\selectlanguage{english}\begin{theorem}
Suppose $f$ and $g$ are differentiable and $g'(x) \neq 0$ on an open interval $I$ that contains $a$ (except possibly at $a$). Suppose that
$$
\lim_{x \to a} f(x) = 0 \quad \text{and} \quad \lim_{x \to a} g(x) = 0
$$
or that
$$
\lim_{x \to a} f(x) = \pm \infty \quad \text{and} \quad \lim_{x \to a} g(x) = \pm \infty
$$
Then
$$
\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}
$$
if the limit on the right side exists (or is $\pm \infty$).
\end{theorem}
\begin{CJK}{UTF8}{gbsn}以微積分教科書上常見的例子而言\end{CJK}\selectlanguage{english},Wo Men Ke Yi Jian Dan Li Jie Ru Xia \selectlanguage{english}:
\selectlanguage{english}\begin{quote}
\begin{CJK}{UTF8}{gbsn}若將\end{CJK}\selectlanguage{english}$x$\begin{CJK}{UTF8}{gbsn}代入原方程式會產生\end{CJK}\selectlanguage{english}$\dfrac{0}{0}$\begin{CJK}{UTF8}{gbsn}或是\end{CJK}\selectlanguage{english}$\dfrac{\infty}{\infty}$\begin{CJK}{UTF8}{gbsn}的話\end{CJK}\selectlanguage{english},Yuan Shi Wen Ti Jiu Ke Yi Zhuan Cheng Fen Zi Fen Mu Ge Zi Wei Fen Hou De Ji Xian Wen Ti . Dan Zhe Li You [?] Dian Wu Bi Te Bie Liu Xin \selectlanguage{english}:\selectlanguage{english}\begin{CJK}{UTF8}{gbsn}\textbf{轉換後的問題,Qi Ji Xian Bi Xu Cun Zai . (這裡我們視無窮大為極限存在。)}\end{CJK}\selectlanguage{english}\selectlanguage{english}\begin{CJK}{UTF8}{gbsn}\end{CJK}\selectlanguage{english}
\end{quote}

\begin{CJK}{UTF8}{gbsn}回到前面\end{CJK}\selectlanguage{english}$\displaystyle \lim_{x \to 1} \frac{\ln x}{x-1}$\begin{CJK}{UTF8}{gbsn}的例子\end{CJK}\selectlanguage{english},Sui Ran Fen Zi Fen Mu Mei You Gong Yin Shi Ke Yi Dui Xiao \selectlanguage{english},Dan Tou Guo Luo Bi Da Fa Ze \selectlanguage{english},Wo Men Huan Shi Ke Yi De Dao 
\selectlanguage{english}$$
\lim_{x \to 1} \frac{\ln x}{x-1} = \lim_{x \to 1} \frac{1/x}{1} = 1
$$
\begin{CJK}{UTF8}{gbsn}的結論。

當然了\end{CJK}\selectlanguage{english},Luo Bi Da Fa Ze Ye Bu Shi Wan Ling Dan \selectlanguage{english},Huan Shi You Xie Wen Ti \selectlanguage{english},Li Yong Yi Wang De Fang Fa Ke Yi Jie Jue \selectlanguage{english},Shi Yong Luo Bi Da Fa Ze Fan Er Zuo Bu Chu Lai . Li Ru Xiang Shi 
\selectlanguage{english}$\displaystyle \lim_{x \to 1^+} \frac{x - 1}{\sqrt{x^2 - 1}}$
\begin{CJK}{UTF8}{gbsn}這樣的題目。如果我們使用羅必達法則的話\end{CJK}\selectlanguage{english},Hui Xian Ru Wu Qiong Hui Quan De Jiong Jing . 
\selectlanguage{english}\begin{align*}
\lim_{x \to \infty} \frac{x + 2}{\sqrt{x^2 + 4x}} &\overset{L'H}{=} \lim_{x \to \infty} \frac{1}{(x + 2)/\sqrt{x^2 + 4x}} = \lim_{x \to \infty} \frac{\sqrt{x^2 + 4x}}{x + 2} \\
&\overset{L'H}{=} \lim_{x \to \infty} \frac{x + 2}{\sqrt{x^2 + 4x}} = \cdots
\end{align*}
\begin{CJK}{UTF8}{gbsn}那麼\end{CJK}\selectlanguage{english},Zheng Que De Zuo Fa Ying Gai Shi Shi Mo Ni \selectlanguage{english}?Zhe Ge Jiu Dang Zuo Da Jia De Fu Xi Gong Ke Luo \selectlanguage{english}!\selectlanguage{english}XDD

\begin{CJK}{UTF8}{gbsn}\section{L'H\^{o}pital's Rule簡易版證明}
這裡我們附上一個簡易版的證明。完整版的證明需要用到柯西均值定理\end{CJK}\selectlanguage{english}(Cauchy's Mean Value Theorem),\begin{CJK}{UTF8}{gbsn}由於比較抽象\end{CJK}\selectlanguage{english},You Xing Qu De Tong Xue Qing Can Kao Ge Da Wei Ji Fen Ke Ben . 

Jia She \selectlanguage{english}$f(a) = g(a) = 0$\begin{CJK}{UTF8}{gbsn}、\end{CJK}\selectlanguage{english}$f'$\begin{CJK}{UTF8}{gbsn}和\end{CJK}\selectlanguage{english}$g'$\begin{CJK}{UTF8}{gbsn}都是連續函數\end{CJK}\selectlanguage{english},Bing Qie \selectlanguage{english}$g'(a) \neq 0$;\begin{CJK}{UTF8}{gbsn}則
\end{CJK}\selectlanguage{english}\begin{align*}
\lim_{x \to a} \frac{f'(x)}{g'(x)} &= \frac{f'(a)}{g'(a)} 
= \frac{\displaystyle\lim_{x \to a} \dfrac{f(x) - f(a)}{x - a}}{\displaystyle\lim_{x \to a} \dfrac{g(x) - g(a)}{x - a}} \\
&= \lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} \\
&= \lim_{x \to a} \frac{f(x)}{g(x)}
\end{align*}
\begin{CJK}{UTF8}{gbsn}是不是輕輕鬆鬆就得到羅必達法則了呢\end{CJK}\selectlanguage{english}!

\selectlanguage{english}\begin{CJK}{UTF8}{gbsn}\section{後記}
關於羅必達法則\end{CJK}\selectlanguage{english},Wo Men Jiu Jian Dan Jie Shao Dao Zhe Er \selectlanguage{english};Qi Ta De Bian Xing Wen Ti \selectlanguage{english}(\begin{CJK}{UTF8}{gbsn}如\end{CJK}\selectlanguage{english}:\selectlanguage{english}$\infty - \infty$\begin{CJK}{UTF8}{gbsn}、\end{CJK}\selectlanguage{english}$\infty \cdot 0$\begin{CJK}{UTF8}{gbsn}、\end{CJK}\selectlanguage{english}$0^0$...)\begin{CJK}{UTF8}{gbsn}及應用\end{CJK}\selectlanguage{english},Qing Da Jia Zi Xing Can Kao Shu Ben Li De Xiang Guan Zhang Jie . 

\selectlanguage{english}\selectlanguage{english}
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