洛必达法则求解不定式

AP 微积分 BC· AP Calculus BC CED — Contextual Applications of Differentiation· 14 分钟阅读

1. 针对0/0和∞/∞型的核心洛必达法则★★☆☆☆⏱ 4 min

洛必达法则是一种基于微分的技巧，用于计算不定式的极限，这类极限直接代入后无法得到确定的极限值。该法则仅能直接应用于 $\frac{0}{0}$ 和 $\frac{\infty}{\infty}$ 两种不定式；所有其他不定式必须先改写为这两种结构之一，才能使用洛必达法则。

📘 定义

洛必达法则

$lim_{x \to a} \frac{f ( x )}{g ( x )}$

If $lim_{x \to a} f (x) = 0$ and $lim_{x \to a} g (x) = 0$ , or $lim_{x \to a} f (x) = \pm \infty$ and $lim_{x \to a} g (x) = \pm \infty$ , $f$ and $g$ are differentiable near $a$ (except possibly at $a$ ), and $g^{'} (x) \neq = 0$ near $a$ (except possibly at $a$ ), then $lim_{x \to a} \frac{f ( x )}{g ( x )} = lim_{x \to a} \frac{f ^{'} ( x )}{g ^{'} ( x )}$ . This holds for one-sided limits and limits at infinity.

例:

Applies to $lim_{x \to 0} \frac{e ^{x} - 1}{x}$ but not to $lim_{x \to 0} \frac{x}{x + 1}$

The rule can be applied repeatedly: if after one application you still get an indeterminate $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , you can apply it again as long as all conditions hold.

📐 例题

Evaluate $lim_{x \to 0} \frac{e ^{2 x} - 1 - 2 x}{x ^{2}}$

1
Check indeterminacy by direct substitution: numerator $= e^{0} - 1 - 2 (0) = 0$ , denominator $= 0^{2} = 0$ . This is a valid $\frac{0}{0}$ indeterminate form, so we can apply L'Hopital's rule.
2
Differentiate numerator and denominator separately:
3
$f'(x) = 2e^{2x} - 2, \quad g'(x) = 2x$
4
The new limit is $lim_{x \to 0} \frac{2 e ^{2 x} - 2}{2 x} = \frac{0}{0}$ , which is still indeterminate, so apply L'Hopital's rule again.
5
Differentiate a second time:
6
$f''(x) = 4e^{2x}, \quad g''(x) = 2$
7
Evaluate the new limit:
8
$\lim_{x \to 0} \frac{4e^{2x}}{2} = \frac{4(1)}{2} = 2$
9
The original limit equals 2.

Exam tip:

Always confirm the indeterminate form explicitly on FRQ answers; AP graders require this step to award full points for using L'Hopital's rule.

2. Indeterminate Products: $0 \cdot \infty$★★★☆☆⏱ 3 min

An indeterminate product occurs when one term approaches 0 and the other approaches $\pm \infty$ , written $0 \cdot \infty$ . The product's limit is indeterminate because 0 pulls the product toward 0 while $\infty$ pulls it toward infinity, so the result can be any finite value, 0, or infinity. To apply L'Hopital's rule, rewrite the product as a fraction by moving one term to the denominator, resulting in either $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

Both conversions are mathematically valid, but one is almost always simpler to differentiate, so choose the conversion that minimizes extra work like the quotient rule.

📐 例题

Evaluate $lim_{x \to 0^{+}} x ln (4 x)$

1
Check the form: as $x \to 0^{+}$ , $x \to 0$ and $ln (4 x) \to - \infty$ , so we have an indeterminate product $0 \cdot (- \infty)$ .
2
Rewrite as a fraction: move $x$ to the denominator to get $lim_{x \to 0^{+}} \frac{l n ( 4 x )}{1/ x}$ , which becomes $- \infty/\infty$ , a valid form for L'Hopital's rule. (Moving $ln (4 x)$ to the denominator would result in a more complex differentiation, so this conversion is preferred.)
3
Differentiate numerator and denominator separately:
4
$\frac{d}{dx}[\ln(4x)] = \frac{1}{x}, \quad \frac{d}{dx}\left[\frac{1}{x}\right] = -\frac{1}{x^2}$
5
Simplify the ratio of derivatives:
6
$\frac{1/x}{-1/x^2} = -x$
7
Evaluate the limit: $lim_{x \to 0^{+}} (- x) = 0$ , so the original limit is 0.

Exam tip:

When converting a product, always leave the simpler term to differentiate in the numerator to avoid introducing extra chain rule or quotient rule errors.

3. Indeterminate Differences: $\infty - \infty$★★★☆☆⏱ 4 min

Indeterminate differences occur when we have the difference of two terms, both approaching $+ \infty$ or both approaching $- \infty$ , written $\infty - \infty$ . This is indeterminate because the two infinite terms compete, and the result can be 0, any finite number, or $\pm \infty$ . To solve this type of limit, convert the difference into a single fraction, almost always by combining terms over a common denominator, factoring, or multiplying by a conjugate to eliminate radicals. The resulting fraction will almost always be a $\frac{0}{0}$ or $\frac{\infty}{\infty}$ indeterminate form suitable for L'Hopital's rule.

📐 例题

Evaluate $lim_{x \to 1} (\frac{1}{x - 1} - \frac{1}{l n x})$

1
Check the form: as $x \to 1$ , both terms approach $\pm \infty$ , so this is an indeterminate difference $\infty - \infty$ .
2
Combine the fractions over a common denominator to get a single fraction:
3
$\lim_{x \to 1} \frac{\ln x - (x-1)}{(x-1)\ln x}$
4
Check indeterminacy: numerator at $x = 1$ is $ln 1 - 0 = 0$ , denominator is $(0) (ln 1) = 0$ , so we have $\frac{0}{0}$ , valid for L'Hopital's rule.
5
Differentiate numerator and denominator:
6
$\text{Numerator derivative: } \frac{1}{x} - 1 = \frac{1-x}{x}, \quad \text{Denominator derivative: } \ln x + 1 - \frac{1}{x}$
7
Substitute $x = 1$ : $\frac{0}{0 + 1 - 1} = \frac{0}{0}$ , which is still indeterminate, so apply L'Hopital's rule again.
8
Differentiate a second time:
9
$\text{Numerator derivative: } -\frac{1}{x^2}, \quad \text{Denominator derivative: } \frac{1}{x} + \frac{1}{x^2}$
10
Evaluate the limit:
11
$\lim_{x \to 1} \frac{-1/x^2}{1/x + 1/x^2} = \frac{-1}{1 + 1} = -\frac{1}{2}$
12
The original limit is $- \frac{1}{2}$ .

Exam tip:

Never differentiate each term of the difference separately; you must always combine into a single fraction first before applying L'Hopital's rule.

4. Indeterminate Powers: $0^0, 1^\infty, \infty^0$★★★★☆⏱ 3 min

These three indeterminate forms are exponential expressions of the form $lim_{x \to a} f (x)^{g (x)}$ , where the limits of the base and exponent create one of the three indeterminate combinations. To solve these, use the natural logarithm to convert the exponential into a product, which you can then convert to $\frac{0}{0}$ or $\frac{\infty}{\infty}$ to apply L'Hopital's rule. The standard process is: (1) Let $L = lim f (x)^{g (x)}$ , (2) Take natural log of both sides: $ln L = lim g (x) ln f (x)$ , (3) Solve the resulting indeterminate product limit, (4) Exponentiate to get $L = e^{l n L}$ .

📐 例题

Evaluate $lim_{x \to 0^{+}} (3 x)^{x}$

1
Check the form: as $x \to 0^{+}$ , $3 x \to 0$ and $x \to 0$ , so we have $0^{0}$ , an indeterminate power.
2
Let $L = lim_{x \to 0^{+}} (3 x)^{x}$ , so take the natural log of both sides:
3
$\ln L = \lim_{x \to 0^+} x \ln(3x)$
4
This is an indeterminate product $0 \cdot (- \infty)$ . Rewrite as a fraction:
5
$\lim_{x \to 0^+} \frac{\ln(3x)}{1/x}, \text{ which is } -\infty/\infty$
6
Differentiate numerator and denominator, then simplify:
7
$\frac{1/x}{-1/x^2} = -x$
8
Evaluate the limit: $ln L = lim_{x \to 0^{+}} (- x) = 0$ .
9
Exponentiate to get the original limit: $L = e^{0} = 1$ .

Exam tip:

Don't forget to undo the natural logarithm at the end; forgetting the final exponentiation is one of the most common mistakes AP graders see on this question type.

5. AP-Style Concept Check★★★☆☆⏱ 2 min

✓ 快速检测

Test your understanding of L'Hopital's rule with this AP-style multiple choice question:

What is the value of $lim_{x \to \infty} \frac{5 x ^{3} + 2 x - 1}{3 x ^{3} - 4 x ^{2}}$ ?
- (0)
- (\frac{5}{3})
- (\infty)
- (\frac{3}{5})
显示答案
1 —
This is an indeterminate $\infty/\infty$ form. Applying L'Hopital's rule three times (or comparing leading terms of the polynomials) gives the result $\frac{5}{3}$ .

6. 常见陷阱

错误做法:

Applying L'Hopital's rule to a determinate form, e.g., evaluating $lim_{x \to 0} \frac{x}{x + 1}$ by differentiating to get $1/1 = 1$ when direct substitution gives $0/1 = 0$ .

原因:

Students get in the habit of using L'Hopital's for every limit and forget to check the indeterminacy condition first.

正确做法:

Always plug in the limit value first to confirm you have an indeterminate form before applying L'Hopital's.

错误做法:

Differentiating the entire fraction using the quotient rule, instead of differentiating numerator and denominator separately, e.g., differentiating $lim_{x \to 0} \frac{e ^{x} - 1}{x}$ as $\frac{x e ^{x} - ( e ^{x} - 1 )}{x ^{2}}$ .

原因:

Confusion between L'Hopital's rule and the derivative quotient rule, since we work with a ratio of functions.

正确做法:

Explicitly label $f (x)$ (numerator) and $g (x)$ (denominator) on scratch paper before differentiating to avoid mixing up rules.

错误做法:

Stopping after one application of L'Hopital's when the result is still indeterminate, leaving the answer as $0/0$ .

原因:

Students assume one differentiation is enough and don't check the new limit for indeterminacy.

正确做法:

After each differentiation step, substitute the limit value to check for indeterminacy; apply L'Hopital's again if the result is still indeterminate.

错误做法:

Forgetting to exponentiate after using the logarithm for an indeterminate power, leaving the answer as $ln L = 2$ instead of $L = e^{2}$ .

原因:

Students get focused on applying L'Hopital's to the product after taking the log and forget the original limit is for the power, not the log of the power.

正确做法:

Write "Original limit $L = e^{l n L}$ " explicitly before solving for $ln L$ to remind yourself of the final step.

错误做法:

Applying L'Hopital's rule directly to discrete sequences, e.g., evaluating $lim_{n \to \infty} \frac{l n n}{n}$ on the discrete sequence without extending to a continuous function.

原因:

L'Hopital's rule only applies to differentiable functions, which sequences are not.

正确做法:

When evaluating a sequence limit, restate it as the limit of the corresponding continuous function as $x \to \infty$ , apply L'Hopital's to the continuous version, then conclude the sequence limit matches.

7. 速查表

Category	Rule/Conversion	Key Notes
Core L'Hopital's Rule	If $lim \frac{f}{g} = \frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$ , then $lim \frac{f ( x )}{g ( x )} = lim \frac{f ^{'} ( x )}{g ^{'} ( x )}$	Holds for $x \to a, x \to a^{\pm}, x \to \pm \infty$ ; requires $g^{'} (x) \neq = 0$ near $a$
Indeterminate Product $0 \cdot \infty$	$lim f g = lim \frac{f}{1/ g}$ or $lim \frac{g}{1/ f}$	Convert to 0/0 or ∞/∞; choose conversion to simplify differentiation
Indeterminate Difference $\infty - \infty$	Combine to single fraction over common denominator	Use conjugate for radical differences; confirm indeterminate form before applying rule
Indeterminate Power $0^{0}, 1^{\infty}, \infty^{0}$	Let $L = lim f^{g}$ , $ln L = lim g ln f$ , $L = e^{l n L}$	Always undo the natural logarithm for the final answer
Repeated L'Hopital's	Apply rule multiple times if new limit is still indeterminate	Stop only when you get a determinate finite or infinite limit
Determinate Non-Indeterminate Forms	$\frac{0}{\infty} = 0$ , $\frac{\infty}{0} = \pm \infty$ , $\infty + \infty = \infty$ , $0^{\infty} = 0$	Never apply L'Hopital's to these; they resolve directly
Discrete Sequence Limits	$lim_{n \to \infty} a_{n} = lim_{x \to \infty} f (x)$ where $f (n) = a_{n}$	Extend discrete sequences to continuous functions before applying the rule

真题中的出现

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下一步

L'Hopital's rule is a foundational tool for evaluating indeterminate limits that you will use across multiple remaining topics in AP Calculus BC. The most immediate application is evaluating limits for improper integrals, where you will regularly need to resolve indeterminate forms to determine if an integral converges or diverges. You will also use it to compare growth rates of transcendental functions and evaluate limits of infinite sequences for series topics. Mastering the conversion steps for all indeterminate forms now will save you time and avoid common errors on later topics.