Indeterminate Forms and L’Hôpital’s Rule | 未定式与洛必达法则

📚 Indeterminate Forms and L’Hôpital’s Rule | 未定式与洛必达法则

In IB Mathematics Analysis and Approaches HL, evaluating limits is a fundamental skill that leads directly into calculus. However, certain limits resist direct substitution and yield ambiguous results known as indeterminate forms. This is where L’Hôpital’s Rule becomes an indispensable tool in your mathematical arsenal.

在IB数学分析与方法（AA）HL课程中，求极限是一项直接衔接微积分的基础技能。然而，某些极限无法通过直接代入得到结果，会产生称为”未定式”的模糊结果。此时，洛必达法则便成为你数学工具箱中不可或缺的利器。

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1. What Are Indeterminate Forms? | 什么是未定式？

An indeterminate form is a mathematical expression obtained after direct substitution into a limit where the result does not determine the limit’s value. The most familiar example appears when evaluating expressions where both the numerator and denominator approach zero simultaneously.

未定式是指将数值直接代入极限表达式后，所得到的结果无法确定该极限值的数学形式。最常见的例子出现在分子和分母同时趋近于零的情形。

For instance, when we substitute x = 2 into the limit expression (x² – 4) / (x – 2), we obtain 0/0. Yet the limit itself clearly exists and equals 4, which we can verify through algebraic simplification.

例如，将 x = 2 代入极限表达式 (x² – 4) / (x – 2)，我们得到 0/0。然而该极限本身显然存在且等于 4，这可以通过代数化简来验证。

The key insight is that 0/0 does not mean “zero divided by zero is undefined, so the limit does not exist.” Rather, it signals that further analysis is required to determine the actual behaviour of the function near that point.

关键的理解是：0/0 并不意味着”零除以零无定义，所以极限不存在”。相反，它提示我们需要进一步分析才能确定函数在该点附近的实际行为。

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2. The Seven Classic Indeterminate Forms | 七种经典未定式

In IB HL, you are expected to recognise the following seven indeterminate forms. The first two are the most frequently tested, while the remaining five require algebraic transformation before L’Hôpital’s Rule can be applied.

在IB HL课程中，你需要识别以下七种未定式。前两种最为常见，其余五种需要先经过代数变换才能应用洛必达法则。

Type | 类型	Form | 形式	Direct Substitution Result | 直接代入结果
1	0 / 0	Zero divided by zero | 零除以零
2	∞ / ∞	Infinity divided by infinity | 无穷大除以无穷大
3	0 × ∞	Zero times infinity | 零乘以无穷大
4	∞ − ∞	Infinity minus infinity | 无穷大减无穷大
5	0⁰	Zero to the power of zero | 零的零次方
6	∞⁰	Infinity to the zero power | 无穷大的零次方
7	1^∞	One to the power of infinity | 一的无穷大次方

It is crucial to remember that only forms 1 and 2 can be handled directly by L’Hôpital’s Rule. The remaining five forms must first be converted into either 0/0 or ∞/∞.

务必牢记：只有形式1和形式2可以直接使用洛必达法则处理。其余五种形式必须先转化为 0/0 或 ∞/∞ 的形式。

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3. Statement of L’Hôpital’s Rule | 洛必达法则的陈述

L’Hôpital’s Rule provides a powerful method for evaluating limits of quotients that produce indeterminate forms. The formal statement is presented below.

洛必达法则为求商形式的未定式极限提供了一种强有力的方法。其正式陈述如下。

如果 lim[ f(x) / g(x) ] 在 x → a 时产生 0/0 或 ∞/∞，且 lim [ f′(x) / g′(x) ] 存在（或趋于 ±∞），则：

lim [ f(x) / g(x) ] = lim [ f′(x) / g′(x) ]

Here, x → a may also be replaced by x → a⁺, x → a⁻, x → +∞, or x → −∞, provided the corresponding conditions are satisfied.

其中，x → a 也可是 x → a⁺（右极限）、x → a⁻（左极限）、x → +∞ 或 x → −∞，只要相应条件满足即可。

The rule requires three essential conditions: first, both f(x) and g(x) must be differentiable in an open interval containing a (except possibly at a itself); second, g′(x) must not be zero near a (except possibly at a); and third, the limit of the quotient of derivatives must exist.

该法则需要三个基本条件：第一，f(x) 和 g(x) 在包含 a 的开区间内可导（a 点本身可以例外）；第二，在 a 附近 g′(x) 不等于零（a 点本身可以例外）；第三，导数之比的极限必须存在。

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4. Applying the Rule to 0/0 Forms | 应用于 0/0 型未定式

To apply L’Hôpital’s Rule to a 0/0 indeterminate form, simply differentiate the numerator and the denominator separately, then re-evaluate the limit. Note that you do NOT use the quotient rule of differentiation in this step.

将洛必达法则应用于 0/0 型未定式时，只需分别对分子和分母求导，然后重新求极限。请注意，此步骤中不要使用商的求导法则。

Example 1: Evaluate lim (x → 0) [ sin(3x) / x ].

例1：求 lim (x → 0) [ sin(3x) / x ]。

Direct substitution gives 0/0. Applying L’Hôpital’s Rule, we differentiate the numerator to obtain 3cos(3x) and the denominator to obtain 1. The new limit is lim (x → 0) [ 3cos(3x) / 1 ] = 3cos(0) = 3.

直接代入得到 0/0。应用洛必达法则，对分子求导得到 3cos(3x)，对分母求导得到 1。新极限为 lim (x → 0) [ 3cos(3x) / 1 ] = 3cos(0) = 3。

Example 2: Evaluate lim (x → 0) [ (eˣ − 1 − x) / x² ].

例2：求 lim (x → 0) [ (eˣ − 1 − x) / x² ]。

Substituting x = 0 gives (1 − 1 − 0) / 0 = 0/0. Applying the rule once yields lim (x → 0) [ (eˣ − 1) / (2x) ], which is still 0/0. Apply the rule a second time to get lim (x → 0) [ eˣ / 2 ] = 1/2.

代入 x = 0 得到 (1 − 1 − 0) / 0 = 0/0。应用一次法则得到 lim (x → 0) [ (eˣ − 1) / (2x) ]，这仍是 0/0。再次应用法则得到 lim (x → 0) [ eˣ / 2 ] = 1/2。

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5. Applying the Rule to ∞/∞ Forms | 应用于 ∞/∞ 型未定式

The ∞/∞ form arises frequently when analysing rational functions or logarithmic expressions as x tends to infinity. L’Hôpital’s Rule applies identically to this case.

∞/∞ 型未定式经常出现在分析有理函数或对数表达式当 x 趋于无穷大时的情形。洛必达法则在此情况下同样适用。

Example 3: Evaluate lim (x → ∞) [ ln(x) / x ].

例3：求 lim (x → ∞) [ ln(x) / x ]。

As x → ∞, both ln(x) and x tend to ∞, giving ∞/∞. Differentiating the numerator and denominator separately, we obtain lim (x → ∞) [ (1/x) / 1 ] = lim (x → ∞) (1/x) = 0. This result shows that the natural logarithm grows more slowly than any positive power of x.

当 x → ∞ 时，ln(x) 和 x 都趋于 ∞，得到 ∞/∞。分别对分子和分母求导，得到 lim (x → ∞) [ (1/x) / 1 ] = lim (x → ∞) (1/x) = 0。该结果表明自然对数的增长速度慢于 x 的任何正次幂。

Example 4: Evaluate lim (x → ∞) [ x² / eˣ ].

例4：求 lim (x → ∞) [ x² / eˣ ]。

This yields ∞/∞. Differentiating twice, we obtain lim (x → ∞) [ 2x / eˣ ] = lim (x → ∞) [ 2 / eˣ ] = 0. Exponential functions dominate polynomial functions at infinity.

这产生 ∞/∞。求导两次后得到 lim (x → ∞) [ 2x / eˣ ] = lim (x → ∞) [ 2 / eˣ ] = 0。指数函数在无穷远处主导多项式函数。

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6. Transformable Form: 0 × ∞ | 可转化形式：0 × ∞

To handle a 0 × ∞ indeterminate form, we rearrange the product into a quotient so that L’Hôpital’s Rule becomes applicable. The general strategy is to divide one factor by the reciprocal of the other.

处理 0 × ∞ 型未定式时，我们需要将乘积重新排列为商的形式以便应用洛必达法则。一般策略是将其中一个因子除以另一个因子的倒数。

Example 5: Evaluate lim (x → 0⁺) [ x ln(x) ].

例5：求 lim (x → 0⁺) [ x ln(x) ]。

Here x → 0 and ln(x) → −∞, so we have 0 × ∞. Rewrite the expression as ln(x) / (1/x), which now gives ∞/∞. Applying L’Hôpital’s Rule:

这里 x → 0 且 ln(x) → −∞，所以是 0 × ∞。将表达式改写为 ln(x) / (1/x)，现在变为 ∞/∞。应用洛必达法则：

lim (x → 0⁺) [ ln(x) / (1/x) ] = lim (x → 0⁺) [ (1/x) / (−1/x²) ] = lim (x → 0⁺) (−x) = 0

A useful trick is to move the more complicated factor to the numerator, since its derivative may be simpler. In practice, converting x ln(x) to ln(x) / (1/x) is usually more effective than converting it to x / (1/ln(x)).

一个有用的技巧是将较复杂的因子移到分子位置，因为其导数可能更简单。在实际操作中，将 x ln(x) 转化为 ln(x) / (1/x) 通常比转化为 x / (1/ln(x)) 更有效。

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7. Transformable Form: ∞ − ∞ | 可转化形式：∞ − ∞

When a limit produces ∞ − ∞, the goal is to combine the terms into a single fraction using a common denominator, thereby transforming it into 0/0 or ∞/∞.

当极限产生 ∞ − ∞ 时，目标是通过通分将各项合并为一个分式，从而将其转化为 0/0 或 ∞/∞。

Example 6: Evaluate lim (x → 0) [ 1/x − 1/sin(x) ].

例6：求 lim (x → 0) [ 1/x − 1/sin(x) ]。

Both terms tend to infinity, giving ∞ − ∞. Using the common denominator x·sin(x), we rewrite the expression as:

两项都趋于无穷大，得到 ∞ − ∞。利用公分母 x·sin(x)，将表达式改写为：

[ sin(x) − x ] / [ x·sin(x) ]

This now produces 0/0. Applying L’Hôpital’s Rule twice:

现在这产生 0/0。应用洛必达法则两次：

lim (x → 0) [ (cos(x) − 1) / (sin(x) + x·cos(x)) ] = lim (x → 0) [ (−sin(x)) / (2cos(x) − x·sin(x)) ] = 0/2 = 0

When the expression involves radicals or rational functions, multiplying by the conjugate is another effective technique before applying the rule.

当表达式涉及根式或有理函数时，在应用法则前乘以共轭式也是一种有效的技巧。

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8. Exponential Forms: 0⁰, ∞⁰, and 1^∞ | 指数型未定式：0⁰、∞⁰ 与 1^∞

Limits that produce 0⁰, ∞⁰, or 1^∞ require a logarithmic transformation. This is one of the most elegant uses of L’Hôpital’s Rule and frequently appears on IB Paper 2 (calculator) and Paper 3 (HL only).

产生 0⁰、∞⁰ 或 1^∞ 的极限需要对数变换。这是洛必达法则最优雅的应用之一，经常出现在IB Paper 2（可使用计算器）和 Paper 3（仅HL）中。

The general procedure is to set y = f(x)^g(x), then take the natural logarithm of both sides:

通用步骤是令 y = f(x)^g(x)，然后对两边取自然对数：

ln(y) = g(x) · ln[ f(x) ]

The product g(x) · ln[ f(x) ] now typically yields 0 × ∞, which can be handled as discussed in Section 6. Once lim ln(y) = L is found, the original limit is e^L.

乘积 g(x) · ln[ f(x) ] 现在通常产生 0 × ∞，可按第6节讨论的方法处理。一旦求得 lim ln(y) = L，原极限即为 e^L。

Example 7: Evaluate lim (x → ∞) (1 + 1/x)ˣ.

例7：求 lim (x → ∞) (1 + 1/x)ˣ。

Direct substitution gives 1^∞. Let y = (1 + 1/x)ˣ. Then ln(y) = x · ln(1 + 1/x). As x → ∞, this becomes ∞ × 0. Rewriting as ln(1 + 1/x) / (1/x) produces 0/0. Applying L’Hôpital’s Rule:

直接代入得到 1^∞。令 y = (1 + 1/x)ˣ。则 ln(y) = x · ln(1 + 1/x)。当 x → ∞ 时，这变为 ∞ × 0。改写为 ln(1 + 1/x) / (1/x) 产生 0/0。应用洛必达法则：

lim (x → ∞) [ (−1/x²) / (1 + 1/x) ] / (−1/x²) = lim (x → ∞) 1 / (1 + 1/x) = 1

Therefore, ln(y) → 1, and y → e¹ = e. This is the celebrated definition of Euler’s number.

因此，ln(y) → 1，则 y → e¹ = e。这就是著名的欧拉数定义。

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9. Repeated Application and Simplification | 反复应用与化简

L’Hôpital’s Rule may be applied multiple times in succession, as long as each application results in an indeterminate form that justifies its continued use.

洛必达法则可以连续多次应用，只要每次应用后仍然得到未定式，就可以继续使用。

Example 8: Evaluate lim (x → 0) [ (x − sin(x)) / x³ ].

例8：求 lim (x → 0) [ (x − sin(x)) / x³ ]。

Substitution gives 0/0. First application: lim (x → 0) [ (1 − cos(x)) / (3x²) ], still 0/0. Second application: lim (x → 0) [ sin(x) / (6x) ], still 0/0. Third application: lim (x → 0) [ cos(x) / 6 ] = 1/6.

代入得到 0/0。第一次应用：lim (x → 0) [ (1 − cos(x)) / (3x²) ]，仍是 0/0。第二次应用：lim (x → 0) [ sin(x) / (6x) ]，仍是 0/0。第三次应用：lim (x → 0) [ cos(x) / 6 ] = 1/6。

An important piece of advice: always simplify algebraically before re-applying the rule. In the example above, noting that lim (x → 0) sin(x)/x = 1 could have shortened the solution; using known special limits in combination with L’Hôpital’s Rule is both permitted and encouraged in IB examinations.

一条重要建议：在重新应用法则之前，务必进行代数化简。在上面的例子中，利用 lim (x → 0) sin(x)/x = 1 可以缩短解答过程；在IB考试中，将已知的特殊极限与洛必达法则结合使用是被允许且值得鼓励的。

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10. Common Pitfalls and Important Cautions | 常见误区与重要警示

Several errors repeatedly cost students marks in IB examinations. The most serious is applying L’Hôpital’s Rule when the limit is not an indeterminate form. For instance, if direct substitution yields 2/0 rather than 0/0 or ∞/∞, the rule cannot be applied; the limit is either infinite or does not exist.

有几个错误在IB考试中反复导致学生失分。最严重的是在极限不是未定式时仍然使用洛必达法则。例如，如果直接代入得到 2/0 而不是 0/0 或 ∞/∞，则不能应用该法则；此时极限要么为无穷大，要么不存在。

• Never forget to verify the indeterminate form before applying the rule. Always write down “0/0” or “∞/∞” explicitly in your working.

• 切勿忘记在应用法则前验证未定式的类型。务必在答题过程中明确写下”0/0″或”∞/∞”。

• Differentiate the numerator and denominator separately — the quotient rule is not used when applying L’Hôpital’s Rule.

• 对分子和分母分别求导——应用洛必达法则时不要使用商的求导法则。

• Do not confuse the derivative of the quotient with the quotient of the derivatives. They are generally not equal.

• 不要将商的导数与导数的商混淆。二者通常并不相等。

• L’Hôpital’s Rule does not apply to forms like 1/0, 2/0, or 0/1, as these are not indeterminate forms. Such limits are evaluated by examining one-sided behaviour.

• 洛必达法则不适用于 1/0、2/0 或 0/1 等类型，因为它们不是未定式。这类极限应通过考察单侧行为来求解。

• When working with exponential forms, remember that the limit of ln(y) must be computed first; only then can you recover the original limit as e^L.

• 处理指数型未定式时，必须首先计算 ln(y) 的极限；只有这样才能通过 e^L 恢复原极限。

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11. IB Examination Worked Examples | IB考试典型例题

The following examples reflect the style and difficulty of L’Hôpital’s Rule questions encountered in IB Mathematics AA HL examinations.

以下例题反映了IB数学AA HL考试中洛必达法则相关问题的典型风格和难度。

Example 9 (Paper 2 style): Find the value of lim (x → 0) [ (eˣ − 1 − x − x²/2) / x³ ].

例9（Paper 2风格）：求 lim (x → 0) [ (eˣ − 1 − x − x²/2) / x³ ]。

Substitution gives 0/0. Applying the rule three times:

代入得到 0/0。应用法则三次：

lim (x → 0) [ (eˣ − 1 − x) / (3x²) ] → lim (x → 0) [ (eˣ − 1) / (6x) ] → lim (x → 0) [ eˣ / 6 ] = 1/6

This limit is closely connected to the Taylor series expansion of eˣ, a topic that appears in the HL calculus option.

该极限与 eˣ 的泰勒级数展开密切相关，这是HL微积分选学模块中的内容。

Example 10 (Paper 3 style): The curve y = [ f(x) ]^(g(x)) passes through a point where behaviour appears indeterminate. Given f(x) = x + 1 and g(x) = 3/x, determine lim (x → ∞) (x + 1)^(3/x).

例10（Paper 3风格）：曲线 y = [ f(x) ]^(g(x)) 在某个点表现出未定式的特征。已知 f(x) = x + 1，g(x) = 3/x，求 lim (x → ∞) (x + 1)^(3/x)。

As x → ∞, the base x + 1 → ∞ and the exponent 3/x → 0, giving ∞⁰. Let y = (x + 1)^(3/x). Then ln(y) = (3/x)·ln(x + 1), which is 0 × ∞. Rewrite as ln(y) = 3·ln(x + 1)/x. This is ∞/∞, so L’Hôpital’s Rule applies:

当 x → ∞ 时，底数 x + 1 → ∞ 且指数 3/x → 0，得到 ∞⁰。令 y = (x + 1)^(3/x)。则 ln(y) = (3/x)·ln(x + 1)，这是 0 × ∞。改写为 ln(y) = 3·ln(x + 1)/x。这是 ∞/∞，因此可应用洛必达法则：

lim (x → ∞) [ 3 / (x + 1) ] / 1 = 0

Therefore, ln(y) → 0 and y → e⁰ = 1. The limit evaluates to 1.

因此，ln(y) → 0，y → e⁰ = 1。极限值为 1。

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12. Summary and Examination Strategy | 总结与应考策略

L’Hôpital’s Rule is a high-yield topic in IB Mathematics AA HL. Mastery requires three things: recognising indeterminate forms quickly, applying the rule correctly with separate differentiation of numerator and denominator, and knowing how to transform non-standard forms into quotients.

洛必达法则是IB数学AA HL中的高回报考点。掌握它需要三件事：快速识别未定式、正确地分别对分子分母求导、以及知道如何将非标准形式转化为商的形式。

For 0/0 and ∞/∞, apply the rule directly. For 0 × ∞ and ∞ − ∞, reorganise into a quotient first. For 0⁰, ∞⁰, and 1^∞, use logarithmic transformation and then apply the rule.

对于 0/0 和 ∞/∞，直接应用法则。对于 0 × ∞ 和 ∞ − ∞，先整理为商的形式。对于 0⁰、∞⁰ 和 1^∞，采用对数变换后再应用法则。

In the examination, always show the indeterminate form explicitly in your working, state that L’Hôpital’s Rule is being applied, and simplify between applications. These steps not only structure your solution but also ensure that you meet the GDC (Graphical Display Calculator) and method mark requirements of IB mark schemes.

在考试中，务必在解答中明确写出未定式的形式，说明正在应用洛必达法则，并在每次应用之间进行化简。这些步骤不仅使你的解答结构清晰，还能确保你满足IB评分标准中对方法分（M marks）和计算器使用（GDC）的要求。

Finally, remember that L’Hôpital’s Rule is not always the fastest approach. Algebraic simplification, special limits, and asymptotic reasoning can often produce results more efficiently. Use the rule when it genuinely simplifies your work, and always verify the conditions before applying it.

最后，请记住洛必达法则并不总是最快的解题路径。代数化简、特殊极限和渐近推理往往能更高效地得出结果。在确实能简化计算时使用该法则，并务必在使用前验证适用条件。

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