L’Hôpital’s Rule | 洛必达法则

📚 L’Hôpital’s Rule | 洛必达法则

In calculus, evaluating limits of functions often leads to expressions like 0/0 or ∞/∞, which are undefined but can be resolved. L’Hôpital’s rule provides a systematic method to compute such limits by differentiating the numerator and denominator separately, under certain conditions. Named after the 17th-century French mathematician Guillaume de l’Hôpital, the rule was actually discovered by Johann Bernoulli, but it remains a cornerstone of limit evaluation.

在微积分中，计算函数的极限时常会遇到形如 0/0 或 ∞/∞ 的未定式，虽然它们无定义，但可以求解。洛必达法则提供了一种系统的方法，在一定条件下通过对分子和分母分别求导来计算这些极限。该法则以 17 世纪法国数学家纪尧姆·德·洛必达命名，实际上是由约翰·伯努利发现的，但它始终是极限计算的基石。

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1. Introduction to L’Hôpital’s Rule | 洛必达法则简介

When directly substituting x = a into a quotient f(x)/g(x) yields an indeterminate form such as 0/0 or ∞/∞, the limit cannot be determined without further analysis. L’Hôpital’s rule transforms the original limit into lim f'(x)/g'(x), which is often easier to evaluate. However, the rule is only valid when f and g are differentiable near a and the new limit exists or is ±∞. Misuse of the rule can lead to wrong answers, so understanding the underlying conditions is essential for any IB Mathematics student.

当直接将 x = a 代入分式 f(x)/g(x) 得到 0/0 或 ∞/∞ 等不定式时，极限无法直接确定。洛必达法则将原极限转化为 lim f'(x)/g'(x)，后者通常更容易计算。但该法则只有在 f 和 g 在 a 附近可导且新极限存在或为 ±∞ 时才成立。滥用法则会得出错误答案，因此理解其使用条件对每一位 IB 数学学生都至关重要。

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

Assume that f(x) and g(x) are differentiable on an open interval around a, except possibly at a itself, with g'(x) ≠ 0 near a. If lim (x→a) f(x) = 0 and lim (x→a) g(x) = 0, or lim (x→a) f(x) = ±∞ and lim (x→a) g(x) = ±∞, then

lim (x→a) f(x)/g(x) = lim (x→a) f'(x)/g'(x),

provided the limit on the right-hand side exists (or is ±∞). The rule also applies when x→a⁻, x→a⁺, or as x→∞ or x→-∞, by adjusting the conditions accordingly.

设 f(x) 和 g(x) 在 a 点附近的开区间上可导（a 点本身可能除外），且在 a 附近 g'(x) ≠ 0。若 lim (x→a) f(x) = 0 且 lim (x→a) g(x) = 0，或者 lim (x→a) f(x) = ±∞ 且 lim (x→a) g(x) = ±∞，则

lim (x→a) f(x)/g(x) = lim (x→a) f'(x)/g'(x)，

前提是右端的极限存在（或为 ±∞）。该法则同样适用于 x→a⁻、x→a⁺ 以及 x→∞ 或 x→-∞ 的情形，只需相应调整条件即可。

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3. Simple Applications (0/0 Form) | 简单应用（0/0 型）

A classic example is lim (x→0) sin x / x. Direct substitution gives 0/0. Since sin x and x are differentiable and their derivatives cos x and 1 satisfy the condition, L’Hôpital’s rule yields

lim (x→0) sin x / x = lim (x→0) cos x / 1 = 1.

一个经典例子是 lim (x→0) sin x / x。直接代入得到 0/0。由于 sin x 和 x 可导，且它们的导数 cos x 和 1 满足条件，应用洛必达法则得

lim (x→0) sin x / x = lim (x→0) cos x / 1 = 1。

Similarly, consider lim (x→0) (1 – cos x) / x². Substitution gives 0/0. Differentiating numerator and denominator gives sin x / (2x). This is again 0/0, so we apply the rule once more to obtain cos x / 2 → 1/2. Hence the original limit is 1/2.

再比如 lim (x→0) (1 – cos x) / x²。代入后为 0/0。对分子分母求导得 sin x / (2x)，这又是 0/0，于是再次使用法则得到 cos x / 2 → 1/2。因此原极限为 1/2。

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4. Applications with ∞/∞ Form | ∞/∞ 型的应用

For limits at infinity, the ∞/∞ form often appears. For instance, lim (x→∞) x / e^x. As x grows, both numerator and denominator tend to ∞. Applying L’Hôpital’s rule we differentiate to get 1 / e^x, which tends to 0. Thus

lim (x→∞) x / e^x = 0.

在无穷远处的极限常出现 ∞/∞ 型。例如 lim (x→∞) x / e^x。随着 x 增大，分子分母都趋于 ∞。应用洛必达法则，求导得 1 / e^x，其极限为 0。因此

lim (x→∞) x / e^x = 0。

Another common example is lim (x→∞) ln x / x, which is ∞/∞. After differentiation we get (1/x) / 1 = 1/x → 0. So logarithmic growth is slower than linear growth.

另一个常见例子是 lim (x→∞) ln x / x，为 ∞/∞。求导后得到 (1/x) / 1 = 1/x → 0。可见对数增长慢于线性增长。

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5. Repeated Use of L’Hôpital’s Rule | 洛必达法则的重复使用

Some limits require applying L’Hôpital’s rule more than once. For example, evaluate lim (x→0) (x – sin x) / x³. Both numerator and denominator approach 0. A first differentiation gives (1 – cos x) / (3x²), which is still 0/0. Differentiating again yields sin x / (6x), a standard 0/0 limit. A third application gives cos x / 6 → 1/6. Therefore

lim (x→0) (x – sin x)/x³ = 1/6.

每次应用前，都必须再次检查是否仍为不定式且导数存在。

有些极限需要多次使用洛必达法则。例如，计算 lim (x→0) (x – sin x) / x³。分子分母均趋向 0。第一次求导得 (1 – cos x) / (3x²)，仍是 0/0。再次求导得 sin x / (6x)，这是标准的 0/0 极限。第三次应用得 cos x / 6 → 1/6。因此

lim (x→0) (x – sin x)/x³ = 1/6。

Before each application, you must check again that the indeterminate form persists and the derivatives satisfy the conditions.

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6. Indeterminate Form 0 × ∞ | 不定式 0 × ∞

Products of the form 0 × ∞ can be transformed into a quotient so that L’Hôpital’s rule applies. For instance, lim (x→0⁺) x ln x. Here x → 0 and ln x → -∞, forming 0 × (-∞). Rewrite the product as

ln x / (1/x).

Now this is -∞/∞ as x→0⁺. Differentiating numerator and denominator gives (1/x) / (-1/x²) = -x, which tends to 0. Hence lim (x→0⁺) x ln x = 0.

形如 0 × ∞ 的乘积可以通过变形转化为分式，从而使用洛必达法则。例如 lim (x→0⁺) x ln x。这里 x → 0，ln x → -∞，构成 0 × (-∞)。将乘积改写为

ln x / (1/x)。

此时为 x→0⁺ 时的 -∞/∞ 型。对分子分母求导得 (1/x) / (-1/x²) = -x，它趋向于 0。因此 lim (x→0⁺) x ln x = 0。

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7. Indeterminate Form ∞ − ∞ | 不定式 ∞ − ∞

Differences like ∞ − ∞ can often be combined into a single fraction. Consider lim (x→0) (1/x – 1/sin x). As x→0, both terms tend to ∞. Combine the fractions:

(sin x – x) / (x sin x).

Now this is 0/0. Differentiating the numerator gives cos x – 1, and the denominator gives sin x + x cos x. The result is again 0/0. Differentiating once more yields -sin x in the numerator, and cos x + cos x – x sin x = 2 cos x – x sin x in the denominator. Substituting x = 0 gives 0/2 = 0. Thus the original limit is 0.

像 ∞ − ∞ 这样的差的极限通常可以通分化为一个分式。考虑 lim (x→0) (1/x – 1/sin x)。当 x→0 时，两项都趋向 ∞。通分得

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