📚 Trigonometric Limits | 三角函数的极限
Trigonometric limits are a cornerstone of IB Mathematics: Analysis and Approaches, particularly in the study of differentiation. The derivative of sin x is built on the fundamental limit limx→0 (sin x)/x = 1. This article will guide you through the standard results, proofs, and exam-style techniques needed to handle trigonometric limits with confidence.
三角函数的极限是 IB 数学（分析与方法）的核心内容，尤其是在导数的学习中。(sin x)/x 在 x 趋向 0 时的极限 1 是推导 sin x 导数的基石。本文将带领你系统掌握三角极限的标准结论、证明过程以及考试必备的解题技巧。
1. The Fundamental Limit: sin x / x | 基本极限：sin x / x
The starting point for all trigonometric limits is the fundamental result
所有三角函数的极限都始于一个基础结论
limx→0 (sin x) / x = 1
This result holds when x is measured in radians. In degree mode the limit is π/180, so always work in radians for calculus.
该结论仅在 x 以弧度制度量时成立。若使用角度制，极限会变成 π/180，因此在微积分中始终使用弧度。
The graph of y = (sin x)/x shows a removable discontinuity at x = 0; the function is not defined there, but the limit exists and equals 1.
函数 y = (sin x)/x 的图像在 x = 0 处有一个可去间断点：函数在该点无定义，但极限存在且等于 1。
2. Proof by Squeeze Theorem | 夹逼定理证明
To prove the fundamental limit, begin with a unit circle and an angle x > 0.
证明基本极限时，先取单位圆和一个正角 x > 0。
Area relations inside the unit circle give
单位圆内的面积关系给出
sin x < x < tan x for 0 < x < π/2
Dividing by sin x > 0 produces
除以 sin x > 0 可得
1 < x / sin x < 1 / cos x
Taking reciprocals reverses the inequalities:
取倒数将改变不等号方向：
cos x < (sin x) / x < 1
As x → 0, both cos x and 1 approach 1, so by the Squeeze Theorem the middle expression also approaches 1.
当 x → 0 时，cos x 和 1 都趋近于 1，根据夹逼定理，中间的表达式的极限也为 1。
3. Related Standard Limits | 相关标准极限
From the fundamental limit we get two standard companions.
由基本极限可以推出两个常用的伴随极限。
limx→0 (1 − cos x) / x = 0
This follows by multiplying numerator and denominator by (1 + cos x); the numerator becomes sin²x, so the limit is (lim sin x/x) · lim sin x/(1+cos x) = 1 · 0 = 0.
证明时将分子分母同乘 (1 + cos x)，分子变为 sin²x，于是极限为 (lim sin x/x) · lim sin x/(1+cos x) = 1 · 0 = 0。
Similarly
类似地
limx→0 (tan x) / x = 1
Because tan x / x = (sin x / x) · (1 / cos x), and the second factor tends to 1.
因为 tan x / x = (sin x / x) · (1 / cos x)，而第二个因子趋于 1。
4. Limits with Compound Arguments | 复合变元的极限
When the argument contains a constant, use a substitution or the standard rate adjustment.
当角度中含有常数时，可通过换元或比例微调来处理。
