📚 Two Important Limits | 两个重要极限
In A-Level Mathematics, two trigonometric limits appear again and again when we study differentiation and small-angle approximations. They are limx→0 (sin x)/x = 1 and limx→0 (1 – cos x)/x = 0.
在 A-Level 数学中，有两个三角极限在学习微分和小角度近似时反复出现，它们是 limx→0 (sin x)/x = 1 和 limx→0 (1 – cos x)/x = 0。
1. The Two Limits at a Glance | 两个极限速览
The first limit tells us that near x = 0, the value of sin x is almost equal to x itself. This fact is the foundation for differentiating sine and cosine from first principles.
第一个极限告诉我们，在 x = 0 附近，sin x 的值几乎等于 x 本身。这一事实是使用第一性原理对正弦函数和余弦函数求导的基础。
The second limit tells us that the gap between 1 and cos x shrinks faster than x does, as x approaches 0. It is often used to simplify expressions containing 1 – cos x.
第二个极限说明，当 x 趋近于 0 时，1 与 cos x 之间的差距比 x 缩小得更快。它常用于化简含有 1 – cos x 的式子。
limx→0 (sin x)/x = 1, limx→0 (1 – cos x)/x = 0
2. The First Limit: Geometric Proof | 第一个极限的几何证明
Consider a unit circle and an angle x (in radians) measured from the positive x-axis. Let x be small and positive. Three regions can be compared inside the first quadrant.
考虑单位圆中从 x 轴正方向开始的一个角 x（以弧度为单位）。令 x 为很小的正角。我们可以在第一象限内比较三个区域的面积。
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The area of triangle OPQ, where O is the centre and P is on the circle, Q is the foot of the perpendicular from P to the x-axis, is (1/2) sin x.
三角形 OPQ 的面积为 (1/2) sin x，其中 O 为圆心，P 在圆上，Q 为 P 到 x 轴的垂足。
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The area of the circular sector from the x-axis to P is (1/2) x.
从 x 轴到 P 点的扇形面积为 (1/2) x。
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The area of the right triangle formed by x-axis, the vertical line through P, and the tangent at P is (1/2) tan x.
由 x 轴、过 P 的垂线与 P 点切线构成的直角三角形面积为 (1/2) tan x。
Thus we have (1/2) sin x < (1/2) x < (1/2) tan x. Dividing by sin x gives 1 < x / sin x < 1 / cos x. Taking reciprocals gives cos x < (sin x)/x < 1.
因此有 (1/2) sin x < (1/2) x < (1/2) tan x。除以 sin x 得到 1 < x / sin x < 1 / cos x，取倒数得到 cos x < (sin x)/x < 1。
Since cos x → 1 as x → 0, the squeeze theorem forces (sin x)/x to approach 1. For negative x, sin(-x)/(-x) = (sin x)/x, so the same result holds from both sides.
因为当 x → 0 时 cos x → 1，由夹逼定理可知 (sin x)/x 必然趋近于 1。当 x 为负数时，sin(-x)/(-x) = (sin x)/x，所以左右两侧结论一致。
3. The Second Limit: Algebraic Proof | 第二个极限的代数证明
The second limit can be proved by applying the difference of two squares and using the first limit.
第二个极限可以通过平方差公式并借助第一个极限来证明。
(1 – cos x)/x = [(1 – cos x)(1 + cos x)] / [x(1 + cos x)] = (1 – cos² x) / [x(1 + cos x)] = sin² x / [x(1 + cos x)]
Rewrite the last expression as (sin x)/x × (sin x)/(1 + cos x).
将最后的式子改写为 (sin x)/x × (sin x)/(1 + cos x)。
As x → 0, the first factor tends to 1 by the first important limit. The second factor tends to 0/(1 + 1) = 0. Therefore the whole product tends to 0.
当 x → 0 时，第一个因子由第一个重要极限可知趋近于 1；第二个因子趋近于 0/(1 + 1) = 0。因此整个乘积趋近于 0。
4. Why These Limits Are Important | 为什么这两个极限重要
The two limits are not just isolated facts. They are the bridge between geometric intuition and calculus of trigonometric functions.
这两个极限不只是孤立的结论，它们连接了几何直观与三角函数的微积分运算。
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The first limit is needed to show that the derivative of sin x is cos x.
第一个极限用于证明 sin x 的导数是 cos x。
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The second limit is needed to complete the differentiation of cos x and to prove the derivative of sin x in full.
第二个极限用于完整推导 cos x 的导数，并在证明 sin x 导数时不可缺少。
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Both limits appear in many exam problems where a rational expression contains trigonometric functions.
当题目中的分式含有三角函数时，这两个极限经常出现。
5. Differentiation of Trigonometric Functions | 三角函数的导数
Using the first principle definition f'(x) = limh→0 [f(x+h) – f(x)]/h, we can differentiate sin x.
使用导数定义 f'(x) = limh→0 [f(x+h) – f(x)]/h，我们可以对 sin x 求导。
d/dx (sin x) = limh→0 [sin(x+h) – sin x]/h = limh→0 [sin x cos h + cos x sin h – sin x]/h
Separating the terms gives
将各项分开得到
cos x × limh→0 (sin h)/h – sin x × limh→0 (1 – cos h)/h = cos x × 1 – sin x × 0 = cos x
Similarly, differentiating cos x from first principles gives -sin x. The two important limits are exactly the quantities needed to evaluate the two small parts.
类似地，用第一性原理对 cos x 求导得到 -sin x。这两个重要极限恰恰是我们求这两小部分所需的数值。
6. Small-Angle Approximations | 小角度近似
When x is measured in radians and is very small, the first limit implies sin x ≈ x.
当 x 以弧度为单位且非常小时，第一个极限给出了 sin x ≈ x。
The second limit implies that 1 – cos x is even smaller than x. In fact, using the identity 1 – cos x = 2 sin²(x/2), we obtain 1 – cos x ≈ x²/2 for small x.
第二个极限说明 1 – cos x 比 x 小得多。事实上，利用恒等式 1 – cos x = 2 sin²(x/2)，可得在 x 很小时 1 – cos x ≈ x²/2。
These approximations are frequently used in mechanics, such as the motion of a simple pendulum, and in binominal expansions involving small angles.
这些近似常用于力学问题中，例如单摆运动的研究，也用于涉及小角度的二项式展开。
7. Worked Examples | 例题讲解
Example 1: Evaluate limx→0 (sin 3x)/x.
例 1：求 limx→0 (sin 3x)/x。
Rewrite as 3 × limx→0 (sin 3x)/(3x). Since 3x → 0 as x → 0, the inner limit is 1. Therefore the answer is 3.
将其改写为 3 × limx→0 (sin 3x)/(3x)。因为当 x → 0 时 3x → 0，所以内层极限为 1，答案就是 3。
Example 2: Evaluate limx→0 (tan x)/x.
例 2：求 limx→0 (tan x)/x。
(tan x)/x = (sin x)/x × 1/cos x → 1 × 1 = 1
Example 3: Evaluate limx→0 (1 – cos 2x)/x.
例 3：求 limx→0 (1 – cos 2x)/x。
Using 1 – cos 2x = 2 sin² x, the expression becomes 2 sin² x / x = 2 × (sin x)/x × sin x → 2 × 1 × 0 = 0.
利用 1 – cos 2x = 2 sin² x，原式化为 2 sin² x / x = 2 × (sin x)/x × sin x → 2 × 1 × 0 = 0。
8. Common Mistakes and Pitfalls | 常见错误与陷阱
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Using degrees instead of radians. The limits only hold when x is measured in radians.
使用角度制而不是弧度制。这两个极限仅在 x 以弧度为单位时才成立。
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Writing sin x/x → 0 by treating sin x as zero. The correct limit is 1, not 0.
错误地认为 sin x/x 趋近于 0。正确的极限是 1，而不是 0。
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Ignoring the second limit. Some students only remember the first limit and then cannot finish the first-principles proof for sin x.
忽略第二个极限。有些学生只记住第一个极限，结果无法完成 sin x 的第一性原理证明。
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Applying the limits to the variable 3x without changing the denominator. Always rescale with the same factor.
在分母没有相应改变时，将极限直接用在不等于 x 的变量比如 3x 上。总是要用相同的因子进行缩放。
9. Exam-Style Questions | 考试题型
Question: Given that f(x) = sin x, use the first principles definition to prove f'(x) = cos x.
例题：已知 f(x) = sin x，请用导数的第一性原理定义证明 f'(x) = cos x。
f'(x) = limh→0 [sin(x+h) – sin x]/h
Using the addition formula and the two important limits, the proof follows exactly as in Section 5. This style of question appears regularly on AQA papers.
借助和角公式以及两个重要极限，可按第 5 节的步骤完成证明。这类题型在 AQA 试卷中经常出现。
Question: Find limx→0 (1 – cos x)/(x²).
例题：求 limx→0 (1 – cos x)/(x²)。
Using 1 – cos x = 2 sin²(x/2), we get
利用 1 – cos x = 2 sin²(x/2)，得到
limx→0 2 sin²(x/2) / x² = (1/2) × [limx→0 sin(x/2)/(x/2)]² = (1/2) × 1² = 1/2
10. Summary | 总结
The two important limits are limx→0 (sin x)/x = 1 and limx→0 (1 – cos x)/x = 0. They are proved using geometry and algebra, and they underpin the differentiation of sine and cosine.
两个重要极限是 limx→0 (sin x)/x = 1 和 limx→0 (1 – cos x)/x = 0。它们分别通过几何法和代数法证明，并且是正弦与余弦求导的基础。
Master these two results, always work in radians, and remember the small-angle approximations sin x ≈ x and 1 – cos x ≈ x²/2. They will serve you well in both pure mathematics and applied topics.
掌握这两个结果，始终使用弧度制，并牢记小角度近似 sin x ≈ x 与 1 – cos x ≈ x²/2。它们在纯数学和应用数学中都会让你受益匪浅。
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