Two Important Limits | 两个重要极限

📚 Two Important Limits | 两个重要极限

The two limits known as “the two important limits” are fundamental pillars of calculus. They appear repeatedly when differentiating trigonometric functions, analysing exponential growth, and evaluating more complex limits. Mastering them is essential for any A-Level mathematician.

所谓「两个重要极限」，是微积分大厦的两块基石。无论是在求三角函数的导数、分析指数增长，还是在计算更复杂的极限时，它们都会反复出现。掌握它们是每一位 A-Level 数学学习者的必修课。

1. The First Important Limit | 第一重要极限

The first important limit states that as \(x\) approaches 0, the ratio of \(\sin x\) to \(x\) tends to 1. It is written as:

第一重要极限告诉我们：当 \(x\) 趋近于 0 时，\(\sin x\) 与 \(x\) 之比趋于 1。记为：

lim (x → 0) (sin x) / x = 1

This result is remarkable because both the numerator and denominator tend to 0, yet the ratio has a finite, non‑zero limit. It is often the first “0/0” limit students encounter.

这个结果之所以引人注目，是因为分子分母都趋于 0，但比值却存在一个非零的有限极限。这也是同学们遇到的第一个「0/0」型极限。

2. Geometric Intuition | 几何直观

Consider a unit circle and an angle x (in radians) measured from the positive x‑axis. The length of the arc subtended by x is exactly x. The vertical line from the point on the circle to the axis has length sin x. As x shrinks, the arc and the vertical line become nearly identical.

考虑单位圆，从 x 轴正方向起量一个角 x（以弧度为单位）。x 所对的圆弧长正好是 x。圆上点到 x 轴的垂线长度是 sin x。当 x 很小的时候，圆弧和垂线几乎重合。

Therefore, the ratio (sin x)/x approaches 1. This is not a proof, but it explains why the limit is intuitively obvious.

因此比值 (sin x)/x 趋近于 1。这虽然不是证明，但能直观地解释为什么这个极限如此自然。

3. Proof Using the Squeeze Theorem | 用夹逼定理证明

A rigorous proof uses the Squeeze Theorem. For 0 < x < π/2, consider the areas of a small triangle, a circular sector, and a larger triangle. Comparing areas gives:

严格的证明通常借助夹逼定理。当 0 < x < π/2 时，比较一个小三角形、一个扇形和一个大三角形的面积，可以得到：

sin x < x < tan x

Dividing by sin x (positive) and taking reciprocals yields:

除以正的 sin x 并取倒数，得到：

cos x < (sin x) / x < 1

As x → 0⁺, cos x → 1. By the Squeeze Theorem, (sin x)/x → 1. A similar argument for negative x completes the proof.

当 x → 0⁺ 时，cos x → 1。由夹逼定理，(sin x)/x → 1。对负 x 作类似讨论，即可完成证明。

4. Direct Applications | 直接应用

The first important limit is used to evaluate many other limits. For example:

第一重要极限常被用来计算许多其他极限。例如：

lim (x → 0) (tan x)/x = 1 because tan x = sin x / cos x.
lim (x → 0) (sin 2x)/x = 2 using substitution t = 2x.
lim (x → 0) (1 – cos x)/x² = 1/2 via the identity 1 – cos x = 2 sin²(x/2).

lim (x → 0) (tan x)/x = 1，因为 tan x = sin x / cos x。
lim (x → 0) (sin 2x)/x = 2，令 t = 2x 即可。
lim (x → 0) (1 – cos x)/x² = 1/2，利用 1 – cos x = 2 sin²(x/2)。

Each of these can be derived quickly by rewriting the expression in terms of (sin u)/u and taking the limit.

这些结果都可以通过将表达式改写为 (sin u)/u 的形式，并取极限而快速得到。

5. The Second Important Limit | 第二重要极限

The second important limit defines the number e. For real x,

第二重要极限定义了自然常数 e。对实数 x，有

lim (x → ∞) (1 + 1/x)ˣ = e

Equivalently, by substituting t = 1/x, we obtain the form used near zero:

等价地，令 t = 1/x，可得到在零点附近使用的形式：

lim (t → 0) (1 + t)^(1/t) = e

The number e ≈ 2.71828… is irrational and transcendental, and it forms the base of natural logarithms.

常数 e ≈ 2.71828… 是无理数且是超越数，它是自然对数的底。

6. Understanding the Limit Numerically | 数值理解

Consider the sequence aₙ = (1 + 1/n)ⁿ for positive integers n. The first few values show convergence:

考虑正整数序列 aₙ = (1 + 1/n)ⁿ。前几项已经显示收敛：

n	1	10	100	1000
aₙ	2	2.59374	2.70481	2.71692

The values creep upward and approach e. This limit also models compound interest: if £1 is invested at 100% annual interest compounded n times per year, the balance after one year is exactly (1 + 1/n)ⁿ. As compounding becomes continuous, the balance tends to e.

数值逐渐增大并趋近 e。这个极限也描述了复利：如果 £1 以 100% 的年利率一年复利 n 次，一年后的余额正是 (1 + 1/n)ⁿ。当复利连续化时，余额趋向 e。

7. Related Exponential Limits | 相关的指数极限

From the second important limit, several useful consequences follow. For example:

由第二重要极限，可以推出几个有用的结论。例如：

lim (x → 0) (ln(1 + x))/x = 1
lim (x → 0) (eˣ – 1)/x = 1
lim (x → ∞) (1 + a/x)ˣ = eᵃ

lim (x → 0) (ln(1 + x))/x = 1
lim (x → 0) (eˣ – 1)/x = 1
lim (x → ∞) (1 + a/x)ˣ = eᵃ

These forms are central when differentiating exponential and logarithmic functions from first principles.

这些形式在从定义出发求指数函数和对数函数的导数时至关重要。

8. Role in Differentiation | 在微分中的作用

The first important limit enables the derivative of sin x. Using the definition of the derivative:

第一重要极限使我们能求出 sin x 的导数。利用导数的定义：

d/dx (sin x) = lim (h → 0) [sin(x+h) – sin x] / h

Applying the identity sin(x+h) – sin x = 2 cos(x + h/2) sin(h/2) and letting h → 0, we obtain cos x. The second important limit similarly helps prove that d/dx (eˣ) = eˣ.

运用恒等式 sin(x+h) – sin x = 2 cos(x + h/2) sin(h/2)，并令 h → 0，可得 cos x。第二重要极限同样帮助证明了 d/dx (eˣ) = eˣ。

9. Common Mistakes and How to Avoid Them | 常见错误与规避

Students often misuse these limits. Common pitfalls include:

同学们经常误用这两个极限。常见的陷阱包括：

Assuming (sin x)/x → 1 when x → ∞. Actually, the limit is 0 because |sin x| ≤ 1 while x grows.
Writing lim (x → 0) (sin 2x)/(2x) = 1 without noticing the argument matches the denominator. Always match the variable inside sin with the denominator.
Confusing lim (x → ∞) (1 + 1/x)ˣ = e with lim (x → 0) (1 + x)^(1/x) = e. They are the same with t = 1/x.

误以为 x → ∞ 时 (sin x)/x → 1。实际上极限是 0，因为 |sin x| ≤ 1 而 x 趋于无穷。
写下 lim (x → 0) (sin 2x)/(2x) = 1 时没有检查角度与分母是否匹配。必须确保 sin 内的变量与分母一致。
混淆 lim (x → ∞) (1 + 1/x)ˣ = e 和 lim (x → 0) (1 + x)^(1/x) = e。两者在 t = 1/x 下是同一极限。

Always check the variable inside the trigonometric or exponential function and ensure it matches the expression in the denominator or exponent.

务必检查三角函数或指数函数内部的变量，并确保它与分母或指数中的表达式完全匹配。

10. Worked Examples | 例题精讲

Example 1: Evaluate lim (x → 0) (sin 3x)/(5x).

例 1：求 lim (x → 0) (sin 3x)/(5x)。

Rewrite as (3/5) · (sin 3x)/(3x). Since 3x → 0 as x → 0, the fraction tends to 1, so the limit is 3/5.

改写为 (3/5) · (sin 3x)/(3x)。因为当 x → 0 时 3x → 0，该分式趋于 1，所以极限为 3/5。

Example 2: Evaluate lim (x → 0) (1 + x)^(2/x).

例 2：求 lim (x → 0) (1 + x)^(2/x)。

Write (1 + x)^(2/x) = [(1 + x)^(1/x)]². The inner limit is e, so the answer is e².

将 (1 + x)^(2/x) 写成 [(1 + x)^(1/x)]²。内层极限为 e，因此答案为 e²。

11. Practice Problems | 巩固练习

Try the following exercises on your own:

请独立完成以下练习：

lim (x → 0) (sin 4x)/(sin 2x)
lim (x → 0) (tan x)/(3x)
lim (x → ∞) (1 + 2/x)ˣ
lim (x → 0) (e^(3x) – 1)/x

lim (x → 0) (sin 4x)/(sin 2x)
lim (x → 0) (tan x)/(3x)
lim (x → ∞) (1 + 2/x)ˣ
lim (x → 0) (e^(3x) – 1)/x

Answers: ① 2 ② 1/3 ③ e² ④ 3.

答案：① 2 ② 1/3 ③ e² ④ 3。

12. Summary | 总结

The two important limits are not isolated formulas; they are the bridge between algebra and calculus. The first one controls the behaviour of trigonometric ratios near zero, while the second one defines the exponential constant e. Mastering their proofs, variations, and applications will serve you well throughout the A-Level course and beyond.

两个重要极限不是孤立的公式，它们是连接代数和微积分的桥梁。第一个极限决定了三角函数比值在零点附近的行为，第二个极限则定义了指数常数 e。掌握它们的证明、变形和应用，将在整个 A-Level 课程以及更远的未来中助你一臂之力。

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Two Important Limits | 两个重要极限