Trigonometric Limits | 三角函数的极限

📚 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 lim_x→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。

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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

lim_x→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。

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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。

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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。

lim_x→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

lim_x→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。

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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。

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