Using L'Hospital's Rule for Determining Limits of Indeterminate Forms · 使用洛必达法则确定不定式的极限
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| L'Hospital's Rule/ˈelhɒspɪtlz ruːl/ | 洛必达法则 | luò bì dá fǎ zé |
| indeterminate forms/ˌɪndɪˈtɜːmɪnət fɔːmz/ | 不定式 | bù dìng shì |
A derivative trick for stubborn limits
- Back in Unit 1, $\tfrac00$ meant "simplify by algebra." Sometimes no algebra helps.
- L'Hospital's Rule 洛必达法则 uses derivatives to crack those stubborn limits.
- It applies to the indeterminate forms $\tfrac00$ and $\tfrac{\infty}{\infty}$.
- The idea: replace the ratio of functions by the ratio of their derivatives.
对付顽固极限的求导技巧
- 回到第 1 单元,$\tfrac00$ 意味着"用代数化简"。有时代数帮不上忙。
- 洛必达法则用导数攻克那些顽固的极限。
- 它适用于不定式 $\tfrac00$ 和 $\tfrac{\infty}{\infty}$。
- 想法是:用两个函数导数之比代替它们本身之比。
Check the form first
- L'Hospital only applies to a genuine indeterminate form: $\dfrac00$ or $\dfrac{\infty}{\infty}$.
- Substitute first to confirm you have one — if direct substitution already gives a number, you are done, and the rule does not apply.
- Using it on a non-indeterminate limit (like $\tfrac{5}{2}$) gives a wrong answer.
- So always verify the form before reaching for the rule.
先检查形式
- 洛必达只适用于真正的不定式:$\dfrac00$ 或 $\dfrac{\infty}{\infty}$。
- 先代入确认你确实有一个——若直接代入已经给出一个数,那就完成了,该法则不适用。
- 把它用在非不定式极限(如 $\tfrac{5}{2}$)上会得到错误答案。
- 所以动用该法则前永远先核实形式。
L'Hospital's Rule applies to which indeterminate forms? · 洛必达法则适用于 哪些 不定式？
Only $\tfrac00$ and $\tfrac\infty\infty$. $\tfrac52$ is already determinate; $\tfrac30$ is an asymptote, not indeterminate. · 仅 $\tfrac00$ 和 $\tfrac\infty\infty$。$\tfrac52$ 已是定式；$\tfrac30$ 是渐近线，不是不定式。
Before applying the rule, confirm the limit is an ____ form by substituting. · 应用法则前，通过代入确认极限是否为 ____ 形式。
Only $\tfrac00$ or · 或 $\tfrac\infty\infty$ qualify. · 仅 $\tfrac00$ 或 $\tfrac\infty\infty$ 符合资格。
Differentiate top and bottom separately
- The rule: $$\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$$(when the form is indeterminate).
- Warning: this is not the quotient rule — you differentiate $f$ and $g$ independently, no minus, no $g^2$.
- Then re-evaluate the new, simpler limit.
- $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{\cos x}{1}=1$ — the famous limit, in one line.
分别对上下求导
- 法则:$$\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$$(当形式为不定式时)。
- 警告: 这不是商法则——你各自独立地对 $f$ 和 $g$ 求导,没有减号,没有 $g^2$。
- 然后重新求那个更简单的新极限。
- $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{\cos x}{1}=1$——那个著名极限,一行搞定。
A 0/0 ratio with a real limit · 具有实数极限的 0/0 型比值
y = sin(x)/x (near 0) · y = sin(x)/x (靠近 0)
$\tfrac{\sin x}{x}$ looks like $\tfrac00$ at $0$ but the curve heads smoothly to $1$ — L'Hospital finds that value fast. · $\tfrac{\sin x}{x}$ 看起来像 $\tfrac00$ 在 $0$ 但曲线平滑地趋向于 $1$ ——洛必达法则能快速求得该值。
L'Hospital's Rule uses the quotient rule $\dfrac{f'g-fg'}{g^2}$ on the numerator and denominator. · 洛必达法则对分子和分母分别使用商法则 $\dfrac{f'g-fg'}{g^2}$。
It differentiates $f$ and $g$ separately: $\tfrac{f'}{g'}$, not the quotient rule. · 它分别对 $f$ 和 $g$ 求导：得到 $\tfrac{f'}{g'}$，而非商法则。
Evaluate · 评价 $\displaystyle\lim_{x\to0}\dfrac{e^x-1}{x}$ using the rule. · 使用该法则计算 $\displaystyle\lim_{x\to0}\dfrac{e^x-1}{x}$。
$\tfrac00\Rightarrow\lim\tfrac{e^x}{1}=e^0=1$.
By L'Hospital, $\displaystyle\lim_{x\to0}\dfrac{\sin x}{x}=\lim_{x\to0}\dfrac{\cos x}{1}=$ · 根据洛必达法则，$\displaystyle\lim_{x\to0}\dfrac{\sin x}{x}=\lim_{x\to0}\dfrac{\cos x}{1}=$
$\cos 0=1$.
Repeat if it's still indeterminate
- If the new ratio $\tfrac{f'}{g'}$ is also $\tfrac00$ or $\tfrac\infty\infty$, apply the rule again.
- Keep differentiating top and bottom until you get a determinate value.
- $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}=\lim_{x\to0}\frac{\sin x}{2x}=\lim_{x\to0}\frac{\cos x}{2}=\tfrac12$ (two applications).
- Each round must still be indeterminate to justify another application.
若仍是不定式就重复
- 若新的比 $\tfrac{f'}{g'}$ 仍然是 $\tfrac00$ 或 $\tfrac\infty\infty$,就再次套用该法则。
- 不断对上下求导,直到得到一个确定的值。
- $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}=\lim_{x\to0}\frac{\sin x}{2x}=\lim_{x\to0}\frac{\cos x}{2}=\tfrac12$(用了两次)。
- 每一轮都必须仍然是不定式,才有理由再用一次。
Applying the rule twice, $\displaystyle\lim_{x\to0}\dfrac{1-\cos x}{x^2}=$ ? (a decimal) · 连续应用两次法则后，$\displaystyle\lim_{x\to0}\dfrac{1-\cos x}{x^2}=$ ? (一个小数)
$\to\tfrac{\sin x}{2x}\to\tfrac{\cos x}{2}=\tfrac12$.
$\dfrac{f'}{g'}$ is not the quotient rule. L'Hospital differentiates numerator and denominator separately: $\frac{d}{dx}[f]$ over $\frac{d}{dx}[g]$ — no $\frac{f'g-fg'}{g^2}$. And only use it on $\tfrac00$ or $\tfrac{\infty}{\infty}$; applying it when the limit is already determinate produces nonsense.
$\dfrac{f'}{g'}$ 不是商法则。洛必达是分别对分子和分母求导:$\frac{d}{dx}[f]$ 除以 $\frac{d}{dx}[g]$——没有 $\frac{f'g-fg'}{g^2}$。而且只对 $\tfrac00$ 或 $\tfrac{\infty}{\infty}$ 使用它;在极限已经确定时套用会得出胡话。
Evaluate $\displaystyle\lim_{x\to0}\frac{e^x-1}{x}$.
- Substitute: $\frac{e^0-1}{0}=\frac{0}{0}$ — indeterminate, so L'Hospital applies.
- Differentiate top and bottom: $\dfrac{e^x}{1}$.
- $\displaystyle\lim_{x\to0}\frac{e^x}{1}=e^0=1$.
求 $\displaystyle\lim_{x\to0}\frac{e^x-1}{x}$。
- 代入:$\frac{e^0-1}{0}=\frac{0}{0}$——不定式,所以洛必达适用。
- 分别对上下求导:$\dfrac{e^x}{1}$。
- $\displaystyle\lim_{x\to0}\frac{e^x}{1}=e^0=1$。
L'Hospital's Rule: for the indeterminate forms $\tfrac00$ or $\tfrac{\infty}{\infty}$, $\lim\frac{f}{g}=\lim\frac{f'}{g'}$ — differentiate top and bottom separately (not the quotient rule) and re-evaluate. Confirm the form first, and repeat the rule while the result stays indeterminate.
洛必达法则:对不定式 $\tfrac00$ 或 $\tfrac{\infty}{\infty}$,$\lim\frac{f}{g}=\lim\frac{f'}{g'}$——分别对上下求导(不是商法则)再重新求值。先确认形式,并在结果仍为不定式时重复该法则。