📚 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 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。
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。
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。
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。
ach 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。
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)。
t 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。
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。
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。
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
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。
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。
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。
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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