Examining sin(x)/x near x=0 | 考察 x 趋近 0 时 sin(x)/x 的行为

📚 Examining sin(x)/x near x=0 | 考察 x 趋近 0 时 sin(x)/x 的行为

In calculus, few limits rival the importance of sin x / x as x approaches 0. It appears in the derivative of sine, the behaviour of oscillatory systems, and the foundation of trigonometric limits. An investigation into this function reveals a subtle cancellation between the almost‑linear growth of sin x and the linear denominator near zero, producing a finite, non‑trivial limit of exactly 1. This article explores numerical, graphical, and analytical approaches to that limit, building a rigorous proof using the squeeze theorem.

在微积分中，极少有极限能与 x → 0 时 sin x / x 的重要性相提并论。它出现在正弦函数的导数、振荡系统的行为以及三角极限的基础中。对此函数的探究揭示了 sin x 近乎线性的增长与分母 x 的线性趋势在零点附近的精巧抵消，从而产生一个有限且非平凡的极限，精确值为 1。本文将从数值、图形以及分析三种途径探索这一极限，并用夹逼定理建立严格证明。

---

1. Introduction to the Limit | 极限的引入

The expression sin x / x is undefined at x = 0, because division by zero is not allowed. However, the limit as x → 0 examines the behaviour of the ratio for values increasingly close to 0, and it turns out that the ratio approaches a definite number. Intuitively, for small angles measured in radians, the length of a chord and the corresponding arc are nearly equal, making sin x ≈ x. The ratio therefore tends to 1.

表达式 sin x / x 在 x = 0 处无定义，因为零不能作为除数。但 x → 0 时的极限考察了当 x 无限接近 0 时该比值的趋势，结果发现比值趋近于一个确定的数。直观地说，对于以弧度衡量的微小角度，弦长与对应的弧长几乎相等，因此 sin x ≈ x，比值便趋向于 1。

The central question of this investigation is: how can we be certain that the limit equals exactly 1, and how can that fact be proved without circular reasoning? We begin by gathering numerical evidence.

本次探究的核心问题是：我们何以确信该极限恰好等于 1，又如何在不陷入循环论证的前提下证明这一点？我们首先从数值证据开始。

---

2. Numerical Exploration | 数值探究

Using a GDC or spreadsheet, we can evaluate sin x / x for a sequence of x‑values approaching 0 from both sides. The table below displays values rounded to six decimal places. Notice that as |x| shrinks, the ratio converges steadily towards 1.

利用图形计算器或电子表格，我们可以对从两侧趋近 0 的一系列 x 值计算 sin x / x。下表展示了四舍五入至六位小数的结果。可以看到，随着 |x| 逐渐减小，比值稳定地向 1 收敛。

x (rad)	sin x / x	|Difference from 1|
–0.5	0.958851	0.041149
–0.1	0.998334	0.001666
–0.01	0.999983	0.000017
–0.001	0.99999983	0.00000017
0.001	0.99999983	0.00000017
0.01	0.999983	0.000017
0.1	0.998334	0.001666
0.5	0.958851	0.041149

These data strongly suggest that the limit is 1, but numerical evidence alone cannot constitute a proof. We must also consider the behaviour of the function from both the left and right, ensuring that the two‑sided limit exists.

这些数据强烈表明极限是 1，但单纯的数值证据并不能构成证明。我们还必须考察函数从左、右两侧的行为，确保双侧极限确实存在。

---

3. Graphical Behaviour | 图形行为

Plotting y = sin x / x on a graphing application reveals a smooth curve with a hole at (0, 1). The hole signals a removable discontinuity: the function is not defined at x = 0, yet the graph approaches a height of 1 from either side. Away from zero, the curve oscillates with decreasing amplitude as |x| grows, but the local behaviour near the origin is dominated by the limit.

在绘图软件上绘制 y = sin x / x，会看到一条平滑曲线在 (0, 1) 处有一个“洞”。这个洞表明函数在 x = 0 处有一个可去间断点：函数在 x = 0 无定义，但图像从两侧都接近高度 1。远离零点时，曲线会随着 |x| 增大而衰减振荡，但在原点附近的局部行为完全由该极限主导。

Zooming in repeatedly on the origin shows the curve flattening into what appears to be a horizontal line at y = 1. This visual evidence aligns with the numerical table and motivates a more formal analytic approach.

反复对原点进行放大观察，会发现曲线逐渐趋于平坦，近似一条 y = 1 的水平线。这种视觉证据与数值表格一致，也促使我们寻求更严格的分析方法。

---

4. Formal Definition of the Limit | 极限的正式定义

The statement lim (x → 0) sin x / x = 1 means that for every positive number ε, there exists a corresponding δ > 0 such that whenever 0 < |x| < δ, we have |sin x / x – 1| < ε. This ε‑δ definition captures the idea that the ratio can be made arbitrarily close to 1 by taking x sufficiently close to 0.

命题 lim (x → 0) sin x / x = 1 意味着对于任意正数 ε，总存在一个对应的 δ > 0，使得当 0 < |x| < δ 时，都有 |sin x / x – 1| < ε。这一 ε‑δ 定义捕捉到的思想是：只要 x 足够接近 0，比值就可以任意地接近 1。

While we will not produce the full ε‑δ proof here, this definition reminds us that the limit must be approached from both negative and positive directions. Because sin x / x is an even function (as we shall verify), it suffices to prove the right‑hand limit, and the left‑hand limit follows automatically.

虽然此处不给出完整的 ε‑δ 证明，但这一定义提醒我们，极限必须从负方向和正方向同时趋近。由于 sin x / x 是偶函数（稍后将验证），我们只需证明右极限，左极限便自动成立。

---

5. Geometric Setup for the Squeeze Theorem | 夹逼定理的几何架构

The standard proof uses a unit circle centred at O. Let x be a positive angle (in radians) less than π/2. On the circle, let A be the point (1, 0), B the point (cos x, sin x), and C the intersection of the tangent at A with the line OB extended. The area of triangle OAB, the area of sector OAB, and the area of triangle OAC then satisfy a chain of inequalities that bound sin x / x.

标准的证明借助一个以 O 为圆心的单位圆。设 x 是一个小于 π/2 的正角（以弧度计）。在圆上，令 A 为点 (1, 0)，B 为点 (cos x, sin x)，C 为 A 处切线与 OB 延长线的交点。那么三角形 OAB 的面积、扇形 OAB 的面积以及三角形 OAC 的面积满足一连串不等式，从而对 sin x / x 给出上下界。

Area(△OAB) = ½ · 1 · sin x = ½ sin x. Area(sector OAB) = ½ · 1² · x = ½ x. Area(△OAC) = ½ · 1 · tan x = ½ tan x. Because these three shapes are nested, we obtain ½ sin x ≤ ½ x ≤ ½ tan x for 0 < x < π/2.

面积(△OAB) = ½ · 1 · sin x = ½ sin x。面积(扇形 OAB) = ½ · 1² · x = ½ x。面积(△OAC) = ½ · 1 · tan x = ½ tan x。由于这三个图形互相嵌套，对 0 < x < π/2 可得 ½ sin x ≤ ½ x ≤ ½ tan x。

Dividing through by ½ sin x (which is positive for 0 < x < π/2) reverses one inequality and yields 1 ≤ x / sin x ≤ 1 / cos x. Taking reciprocals then gives cos x ≤ sin x / x ≤ 1.

两边同除以 ½ sin x（在 0 < x < π/2 时为正），会使得一个不等号反向，得到 1 ≤ x / sin x ≤ 1 / cos x。再取倒数即得 cos x ≤ sin x / x ≤ 1。

---

6. Applying the Squeeze Theorem | 夹逼定理的应用

We have established that for 0 < x < π/2, cos x ≤ sin x / x ≤ 1. As x → 0⁺, cos x → 1. Since the lower bound and the upper bound both tend to 1, the squeeze theorem forces sin x / x → 1 from the right. The same inequality can be shown to hold for negative x by symmetry, or by noting that both sides are even functions.

我们已经确立，对于 0 < x < π/2，有 cos x ≤ sin x / x ≤ 1。当 x → 0⁺ 时，cos x → 1。由于下界与上界都趋向于 1，夹逼定理迫使 sin x / x 从右侧趋于 1。通过对称性或注意到两边都是偶函数，同样的不等式对负 x 也成立。

Thus the two‑sided limit exists and equals exactly 1. This proof does not rely on L’Hôpital’s rule (which would require knowing the derivative of sin x, whose computation itself depends on this limit) and is therefore free of circular reasoning.

因此，双侧极限存在并精确等于 1。这一证明不依赖洛必达法则（洛必达法则需要知道 sin x 的导数，而该导数的计算本身又依赖于这个极限），从而避免了循环论证。

---

7. Even‑Function Symmetry | 偶函数对称性

The function f(x) = sin x / x is even, because sin(–x) = –sin x and dividing by –x returns precisely the original ratio: f(–x) = sin(–x)/(–x) = (–sin x)/(–x) = sin x / x = f(x). This symmetry means the graph is mirrored across the y‑axis, and the left‑hand limit mirrors the right‑hand limit.

函数 f(x) = sin x / x 是偶函数，因为 sin(–x) = –sin x，再除以 –x 恰好回到原有比值：f(–x) = sin(–x)/(–x) = (–sin x)/(–x) = sin x / x = f(x)。这种对称性意味着图像关于 y 轴对称，左极限与右极限互为镜像。

Consequently, once the right‑hand limit is proved, the left‑hand limit follows at no extra cost. The removable discontinuity at zero can be “filled” by defining f(0) = 1, making a continuous extension of the function.

因此，一旦证明了右极限，左极限便毫无额外代价地随之成立。零点处的可去间断点可以通过定义 f(0) = 1 来“填补”，从而得到该函数的一个连续延拓。

---

8. Extension: Limits Involving sin(ax)/x | 扩展：涉及 sin(ax)/x 的极限

Once the basic limit is established, related limits can be evaluated by simple substitutions. For any non‑zero constant a, we can write sin(ax)/x = a · sin(ax)/(ax). If we let u = ax, then as x → 0, u → 0 as well. Therefore, lim (x → 0) sin(ax)/x = a · lim (u → 0) sin u / u = a · 1 = a.

一旦基本极限得以确立，相关的极限便可通过简单的代换求值。对于任意非零常数 a，我们可写出 sin(ax)/x = a · sin(ax)/(ax)。若令 u = ax，则当 x → 0 时亦有 u → 0。因此，lim (x → 0) sin(ax)/x = a · lim (u → 0) sin u / u = a · 1 = a。

This trick also works for limits like sin(3x)/(2x), where the prefactor becomes 3/2. Recognizing the core structure sin(□)/□ → 1 is a key skill in IB calculus.

这一技巧同样适用于诸如 sin(3x)/(2x) 的极限，此时前置因子变为 3/2。识别出核心结构 sin(□)/□ → 1 是 IB 微积分中的关键技能。

---

9. Common Pitfalls and Misconceptions | 常见误区与误解

A frequent mistake is to substitute x = 0 directly into sin x / x, obtaining the undefined expression 0/0 and concluding the limit does not exist. The indeterminate form 0/0 signals that further analysis is required, not that the limit fails to exist.

一个常见错误是直接将 x = 0 代入 sin x / x，得到 0/0 的未定义表达式，并由此断定极限不存在。0/0 型不定式只表明需要进一步分析，而非极限不存在。

Another misconception is to invoke L’Hôpital’s rule without checking its prerequisites or acknowledging the circularity. While L’Hôpital’s rule does yield cos x / 1 → 1, its valid use here presupposes the derivative of sin x, whose derivation already uses this very limit. In an IB investigation, the squeeze‑theorem approach is preferred for its rigour.

另一个误解是未经检查前提或未意识到循环性就调用洛必达法则。尽管洛必达法则确实给出 cos x / 1 → 1，但在此处有效使用它，需要假设 sin x 的导数已知，而该导数的推导本身已经使用了这个极限。在 IB 探究中，夹逼定理的方法因其严格性而更受青睐。

Students also sometimes forget that x must be in radians. If degrees were used, the limit would become π/180 instead of 1, altering every subsequent result in calculus.

学生有时还会忘记 x 必须以弧度为单位。如果使用度数，极限将变为 π/180 而不是 1，这将改变微积分中所有后继结果。

---

10. Conclusion and Significance | 结论与意义

The limit lim (x → 0) sin x / x = 1 is a cornerstone of differential calculus. It enables the differentiation of trigonometric functions, underpins small‑angle approximations in physics, and serves as a model for handling indeterminate forms. Investigating it numerically, graphically, and analytically reveals the harmony between geometry and algebra.

极限 lim (x → 0) sin x / x = 1 是微分学的基石。它使三角函数的微分成为可能，支撑着物理学中的小角近似，并成为处理不定式的典范。从数值、图形和分析三种途径对其加以探究，揭示出几何与代数之间的和谐。

By carefully constructing the squeeze theorem proof, we avoid logical circularity and gain deeper insight into why the limit must equal 1. This result will reappear throughout calculus, linking circular motion, wave functions, and infinite series, and it rewards every effort taken to understand it fully.

通过精心构造夹逼定理的证明，我们避免了逻辑循环，并更深刻地理解了该极限为何必然等于 1。这一结果将贯穿整个微积分，将圆周运动、波函数与无穷级数联系起来，每一次努力去完全理解它，都将带来丰厚的回报。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）