Trigonometric Limits | 三角函数的极限

📚 Trigonometric Limits | 三角函数的极限

Trigonometric limits form a cornerstone of calculus and are heavily tested in the IB Mathematics: Analysis and Approaches course, both at SL and HL. Understanding the behaviour of sine, cosine, tangent, and their related functions as the variable approaches a particular value is essential for evaluating derivatives of trig functions, solving complex limit problems, and grasping the underpinnings of series expansions. This article systematically covers the most important trigonometric limits, their geometric proofs, key techniques for evaluation, and common pitfalls, providing bilingual explanations to support English and Chinese learners.

三角函数的极限是微积分的基石，也是 IB 数学分析与方法课程（SL 和 HL）中的重点考查内容。理解正弦、余弦、正切及其相关函数在变量趋近某个值时的行为，对于计算三角函数的导数、解决复杂的极限问题以及理解级数展开都至关重要。本文系统地涵盖了最重要的三角函数极限、它们的几何证明、关键的求值技巧以及常见误区，提供中英双语解释以帮助英语和中文学习者。

---

1. Introduction to Trigonometric Limits | 三角函数极限简介

In calculus, we often need to find the limit of expressions involving trigonometric functions as the variable approaches a specific value, most commonly 0. Unlike polynomial limits, direct substitution can lead to indeterminate forms such as 0/0. The results of these limits are not arbitrary; they rely on the properties of the unit circle and radian measure. The most fundamental of these is the limit of sin x / x as x → 0, which equals 1. This single result unlocks nearly all other trigonometric limits.

在微积分中，我们经常需要求包含三角函数的表达式在变量趋近某个特定值（最常见的是 0）时的极限。与多项式极限不同，直接代入可能会导致 0/0 型的不定式。这些极限的结果并非任意，它们依赖于单位圆的性质和弧度制。其中最根本的是当 x → 0 时 sin x / x 的极限，它等于 1。这个单一结果几乎可以解锁所有其他的三角函数极限。

---

2. The Fundamental Limit: lim(x→0) sin x / x = 1 | 基本极限：lim(x→0) sin x / x = 1

The limit lim (x → 0) (sin x) / x = 1 is the single most important trigonometric limit. It is valid only when x is measured in radians. This limit describes the behaviour of the sine function near the origin, showing that sin x behaves almost identically to its argument x for very small values. Any attempt to use degrees will give a different constant, which is why radian measure is standard in calculus.

极限 lim (x → 0) (sin x) / x = 1 是最重要的三角函数极限。它仅在 x 以弧度为单位时成立。这个极限描述了正弦函数在原点附近的行为，表明对于非常小的值，sin x 几乎与其自变量 x 相同。任何使用度数的尝试都会得到一个不同的常数，这就是为什么弧度制是微积分中的标准。

---

3. Geometric Proof of lim(x→0) sin x / x = 1 | lim(x→0) sin x / x = 1 的几何证明

A classic geometric proof uses the unit circle and the squeeze theorem. Consider a small positive angle x (in radians) that subtends an arc of length x. The area of sector OAB is x/2, the area of triangle OAB is (sin x)/2, and the area of triangle OAC is (tan x)/2. By comparing these areas for 0 < x < π/2, we obtain the inequality sin x < x < tan x. Dividing by sin x (which is positive) gives 1 < x / sin x < 1 / cos x. Taking reciprocals reverses the inequalities, yielding cos x < (sin x) / x < 1.

一个经典的几何证明利用单位圆和夹逼定理。考虑一个小的正角 x（弧度），其所对的弧长为 x。扇形 OAB 的面积为 x/2，三角形 OAB 的面积为 (sin x)/2，三角形 OAC 的面积为 (tan x)/2。比较 0 < x < π/2 时的这些面积，我们得到不等式 sin x < x < tan x。除以 sin x（为正）得到 1 < x / sin x < 1 / cos x。取倒数将不等号反向，得出 cos x < (sin x) / x < 1。

Now, as x → 0⁺, both cos x and 1 approach 1. By the squeeze theorem, the middle expression (sin x)/x is forced to approach 1. The limit as x → 0⁻ can be established similarly using the even-even property, because sin(-x)/(-x) = sin x / x. Therefore, the two-sided limit exists and equals 1.

现在，当 x → 0⁺ 时，cos x 和 1 都趋近于 1。根据夹逼定理，中间表达式 (sin x)/x 被迫趋近于 1。当 x → 0⁻ 时的极限可以利用奇偶性类似地建立，因为 sin(-x)/(-x) = sin x / x。因此，双侧极限存在且等于 1。

---

4. Applications: lim(x→0) sin(kx) / x and Variations | 应用：lim(x→0) sin(kx) / x 及其变形

Once the basic limit is known, we can evaluate limits of the form sin(kx) / x by a simple algebraic manipulation. Write sin(kx) / x = k · sin(kx) / (kx). As x → 0, kx → 0, so sin(kx)/(kx) → 1. Hence, the limit is k. For example, lim (x→0) (sin 5x) / x = 5.

一旦知道了基本极限，我们就可以通过简单的代数变换来求形如 sin(kx) / x 的极限。将 sin(kx) / x 写成 k · sin(kx) / (kx)。当 x → 0 时，kx → 0，所以 sin(kx)/(kx) → 1。因此，极限为 k。例如，lim (x→0) (sin 5x) / x = 5。

A common extension is lim (x→0) (sin(ax)) / (sin(bx)). This can be split as (a/b) · [sin(ax)/(ax)] · [bx/sin(bx)]. Both bracketed limits are 1, giving the result a/b. Similarly, limits involving sin² x can be reduced: lim (x→0) (sin² 3x) / x² = (lim (x→0) (sin 3x)/x)² = 9.

一个常见的推广是 lim (x→0) (sin(ax)) / (sin(bx))。它可以拆分为 (a/b) · [sin(ax)/(ax)] · [bx/sin(bx)]。两个中括号内的极限都是 1，结果为 a/b。类似地，涉及 sin² x 的极限可以化简：lim (x→0) (sin² 3x) / x² = (lim (x→0) (sin 3x)/x)² = 9。

---

5. Limit of (1 – cos x) / x as x→0 | 极限 lim(x→0) (1 – cos x) / x

This limit appears frequently when differentiating cosine. Direct substitution gives 0/0. By multiplying the numerator and denominator by (1 + cos x) and using the Pythagorean identity, we get (1 – cos x)/x = (1 – cos² x) / [x (1 + cos x)] = sin² x / [x (1 + cos x)] = (sin x / x) · [sin x / (1 + cos x)].

这个极限在求余弦的导数时频繁出现。直接代入得 0/0。通过将分子分母同时乘以 (1 + cos x) 并利用勾股恒等式，我们得到 (1 – cos x)/x = (1 – cos² x) / [x (1 + cos x)] = sin² x / [x (1 + cos x)] = (sin x / x) · [sin x / (1 + cos x)]。

As x → 0, the first factor → 1, and the second factor → 0/(1+1) = 0. Therefore, lim (x→0) (1 – cos x) / x = 0. A closely related limit, which is also essential, is lim (x→0) (1 – cos x) / x² = 1/2, obtained by using the double-angle identity 1 – cos x = 2 sin²(x/2).

当 x → 0 时，第一个因子 → 1，第二个因子 → 0/(1+1) = 0。因此，lim (x→0) (1 – cos x) / x = 0。一个紧密相关且同样重要的极限是 lim (x→0) (1 – cos x) / x² = 1/2，可利用倍角公式 1 – cos x = 2 sin²(x/2) 求得。

---

6. Limit of tan x / x as x→0 | 极限 lim(x→0) tan x / x

Using the definition tan x = sin x / cos x, we have tan x / x = (sin x / x) · (1 / cos x). Since both sin x / x and 1 / cos x approach 1 as x → 0, their product also approaches 1. Thus, lim (x→0) (tan x) / x = 1. This result is vital for differentiating the tangent function and for evaluating limits like lim (x→0) (tan 2x) / (3x) = 2/3.

利用定义 tan x = sin x / cos x，我们有 tan x / x = (sin x / x) · (1 / cos x)。由于当 x → 0 时，sin x / x 和 1 / cos x 都趋近于 1，它们的乘积也趋近于 1。因此，lim (x→0) (tan x) / x = 1。这个结果对于求正切函数的导数以及计算类似 lim (x→0) (tan 2x) / (3x) = 2/3 的极限至关重要。

---

7. Using Trigonometric Identities to Evaluate Limits | 使用三角恒等式求极限

Many limits require rewriting the expression using identities before the basic limits can be applied. Key identities include the double-angle formulas (sin 2θ = 2 sin θ cos θ, cos 2θ = 1 – 2 sin² θ), the sum-to-product formulas, and the Pythagorean identity. For instance, to evaluate lim (x→0) (sin 2x – 2 sin x) / x³, we expand sin 2x and use the small-angle approximations derived from the fundamental limits.

许多极限需要先用三角恒等式改写表达式，然后才能运用基本极限。关键的恒等式包括倍角公式（sin 2θ = 2 sin θ cos θ，cos 2θ = 1 – 2 sin² θ）、和差化积公式以及勾股恒等式。例如，要计算 lim (x→0) (sin 2x – 2 sin x) / x³，我们展开 sin 2x 并利用由基本极限导出的微小角度近似。

Another common technique is to substitute a new variable to transform the limit into the standard form. If the variable approaches a non-zero constant, such as lim (x→π) sin x / (x – π), let t = x – π. Then x = t + π, sin(t+π) = – sin t, and the limit becomes lim (t→0) (- sin t) / t = -1. IB exam questions often test this type of substitution.

另一种常见技巧是引入新变量，将极限转化为标准形式。如果变量趋近于非零常数，例如 lim (x→π) sin x / (x – π)，令 t = x – π。那么 x = t + π，sin(t+π) = – sin t，极限变为 lim (t→0) (- sin t) / t = -1。IB 考试题常测试这种代换方法。

---

8. Limits at Infinity for Trigonometric Functions | 三角函数在无穷远处的极限

Trigonometric functions like sin x and cos x oscillate indefinitely and do not approach a limit as x → ∞. However, when they are divided by a power of x, the limit can be determined using the squeeze theorem. For example, since -1 ≤ sin x ≤ 1 for all real x, we have -1/x ≤ (sin x)/x ≤ 1/x for x > 0. As x → ∞, both -1/x and 1/x tend to 0, forcing lim (x→∞) (sin x)/x = 0.

像 sin x 和 cos x 这样的三角函数会无限振荡，当 x → ∞ 时并不趋近于一个极限。然而，当它们除以 x 的某次幂时，可以利用夹逼定理确定极限。例如，由于对所有实数 x 有 -1 ≤ sin x ≤ 1，对于 x > 0 有 -1/x ≤ (sin x)/x ≤ 1/x。当 x → ∞ 时，-1/x 和 1/x 都趋于 0，迫使 lim (x→∞) (sin x)/x = 0。

Similarly, limits like lim (x→∞) (cos x + 2) / x² = 0 and lim (x→∞) x sin(1/x) can be tackled. For the latter, let t = 1/x, so as x → ∞, t → 0⁺. Then x sin(1/x) = (sin t)/t → 1. This shows that oscillatory behaviour combined with algebraic manipulation often yields finite answers.

类似地，lim (x→∞) (cos x + 2) / x² = 0 以及 lim (x→∞) x sin(1/x) 这样的极限也可以求解。对于后者，令 t = 1/x，当 x → ∞ 时 t → 0⁺。那么 x sin(1/x) = (sin t)/t → 1。这表明振荡行为结合代数变换常常会得到有限的答案。

---

9. The Squeeze Theorem with Trig Functions | 三角函数的夹逼定理

The squeeze (or sandwich) theorem is indispensable for proving the fundamental trigonometric limits and for evaluating limits where direct algebraic simplification is messy. A typical problem is lim (x→0) x² cos(1/x). Since -1 ≤ cos(1/x) ≤ 1, multiplying by x² gives -x² ≤ x² cos(1/x) ≤ x². As x → 0, both outer bounds approach 0, so the limit is 0.

夹逼定理（或称三明治定理）对于证明基本三角函数极限以及评估那些直接代数化简很麻烦的极限是不可或缺的。一个典型问题是 lim (x→0) x² cos(1/x)。由于 -1 ≤ cos(1/x) ≤ 1，乘以 x² 得到 -x² ≤ x² cos(1/x) ≤ x²。当 x → 0 时，两边的界都趋近于 0，所以极限为 0。

In IB HL papers, you may also encounter limits like lim (x→0) (x sin x) / (1 – cos x). By applying identities and the squeeze theorem, you can find the value is 2. Mastering the setup of inequalities is crucial for rigorous limit evaluation.

在 IB HL 试卷中，你可能还会遇到像 lim (x→0) (x sin x) / (1 – cos x) 这样的极限。通过应用恒等式和夹逼定理，可以求出其值为 2。掌握不等式的构造对于严格的极限求值至关重要。

---

10. Limits Involving Inverse Trigonometric Functions | 涉及反三角函数的极限

Inverse trigonometric limits often reduce to the same fundamental forms. For example, lim (x→0) (arcsin x) / x = 1 and lim (x→0) (arctan x) / x = 1. These can be proved by substituting x = sin θ or x = tan θ. If x = sin θ, then as x → 0, θ → 0, and (arcsin x)/x = θ / (sin θ) → 1. Similarly, for arctan x, the limit also equals 1.

反三角函数的极限通常可归结为相同的基本形式。例如，lim (x→0) (arcsin x) / x = 1 以及 lim (x→0) (arctan x) / x = 1。可以通过令 x = sin θ 或 x = tan θ 来证明。如果 x = sin θ，那么当 x → 0 时 θ → 0，且 (arcsin x)/x = θ / (sin θ) → 1。类似地，arctan x 的极限也等于 1。

Limits such as lim (x→0) (arctan 2x) / (3x) = 2/3 are common in assessments. Additionally, limits combining inverse trig and algebraic functions, like lim (x→∞) x arctan(1/x), can be solved by substitution t = 1/x, giving lim (t→0⁺) (arctan t)/t = 1.

诸如 lim (x→0) (arctan 2x) / (3x) = 2/3 的极限在考试中很常见。此外，结合反三角函数和代数函数的极限，如 lim (x→∞) x arctan(1/x)，可以通过代换 t = 1/x 求解，得到 lim (t→0⁺) (arctan t)/t = 1。

---

11. L’Hopital’s Rule and Trigonometric Limits | 洛必达法则与三角极限

When a trigonometric limit yields an indeterminate form 0/0 or ∞/∞, L’Hopital’s rule can be a powerful alternative to algebraic manipulation, provided the derivatives of the numerator and denominator exist. For instance, lim (x→0) (x – sin x) / x³ gives 0/0. Applying L’Hopital’s rule three times (or using series) yields 1/6. However, be careful: applying L’Hopital to lim (x→0) (sin x)/x directly is circular because the derivative of sin x relies on this very limit.

当一个三角函数极限出现 0/0 或 ∞/∞ 的不定式，并且分子分母的导数存在时，洛必达法则可以作为代数变换之外的有力替代。例如，lim (x→0) (x – sin x) / x³ 是 0/0 型。三次应用洛必达法则（或使用级数展开）可得 1/6。但要小心：直接对 lim (x→0) (sin x)/x 使用洛必达法则会造成循环论证，因为 sin x 的导数正依赖于这个极限本身。

IB HL students are expected to know L’Hopital’s rule but also to recognise when it is appropriate. Often, a combination of basic limits, identities, and L’Hopital’s rule provides the most efficient solution. Always simplify the expression using trig identities before taking derivatives, as this can drastically reduce the workload.

IB HL 学生应掌握洛必达法则，但也应知道何时适用。通常，将基本极限、恒等式和洛必达法则结合使用能提供最有效的解法。在求导之前，务必先用三角恒等式化简表达式，这样可以大幅减少计算量。

---

12. Common Mistakes and Exam Tips | 常见错误与考试技巧

• Forgetting radian measure: The limits sin x / x → 1 and tan x / x → 1 only hold for radians. If the variable is in degrees, convert to radians first.

  忘记弧度制：sin x / x → 1 和 tan x / x → 1 的极限仅在弧度下成立。若变量以度为单位，需先转换为弧度。

• Misusing the squeeze theorem: Ensure the lower and upper bounds genuinely converge to the same value. Inequalities like cos x ≤ sin x / x ≤ 1 are only valid for specific intervals.

  误用夹逼定理：确保下界和上界确实收敛到同一个值。像 cos x ≤ sin x / x ≤ 1 这样的不等式仅在特定区间内有效。

• Ignoring double-sided limits: Always check both sides of 0. For example, lim (x→0) |sin x|/x does not exist because the right-hand limit is 1 and the left-hand limit is -1.

  忽略双侧极限：务必检查 0 的两侧。例如，lim (x→0) |sin x|/x 不存在，因为右极限为 1 而左极限为 -1。

• Circular reasoning: Do not use L’Hopital’s rule to evaluate lim (x→0) sin x / x in a derivation where you have not yet proved the derivative of sin x.

  循环论证：在尚未证明 sin x 的导数时，不要用洛必达法则来求 lim (x→0) sin x / x。

• Not recognising standard forms: Practice rewriting expressions like (1 – cos ax) / x² as (a²/2) · [sin(ax/2)/(ax/2)]² to apply fundamental limits quickly.

  未能识别标准形式：练习将如 (1 – cos ax) / x² 的表达式改写为 (a²/2) · [sin(ax/2)/(ax/2)]²，以快速应用基本极限。

In the exam, always show clear steps, state the key limit used, and justify the application of the squeeze theorem or L’Hopital’s rule. A well-structured solution that references sin x / x → 1 will earn method marks even if the arithmetic is flawed.

在考试中，务必展示清晰的步骤，声明所用到的关键极限，并说明使用夹逼定理或洛必达法则的理由。一个结构良好、引用了 sin x / x → 1 的解法即使数字计算有误也能得到方法分。

---

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）