IB Mathematics: Three Important Limits | IB 数学：三个重要极限

📚 IB Mathematics: Three Important Limits | IB 数学：三个重要极限

In IB Mathematics: Analysis and Approaches (AA) and Applications and Interpretation (AI), limits are the foundation of calculus. Three limits appear again and again: the sine limit, the cosine limit, and the exponential limit. They explain why the derivatives of sin x, cos x, and eˣ have the forms they do, and they are frequently tested in Paper 1 and Paper 2.

在 IB 数学（AA 与 AI）中，极限是微积分的基石。其中三个极限反复出现：正弦极限、余弦极限和指数极限。它们解释了 sin x、cos x 和 eˣ 的导数为何具有特定形式，也是 Paper 1 和 Paper 2 的高频考点。

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1. Why These Three Limits Matter | 为什么这三个极限如此重要

A limit describes the value a function approaches as the input approaches a point. In IB, you are not usually asked to prove limits from first principles, but you must recognise and apply them. The three limits below are called “important” because they unlock the derivatives of trigonometric and exponential functions, and because they let you evaluate many otherwise difficult limits quickly.

极限描述的是当自变量趋近某一点时，函数值趋近的值。在 IB 中，通常不要求从第一原理证明极限，但必须识别并熟练运用它们。下面三个极限之所以”重要”，是因为它们能推出三角函数与指数函数的导数，并能快速求解许多原本复杂的极限。

They also connect to the definition of the number e and to small-angle approximations used in physics and engineering. Mastering them is not about memorising three isolated facts; it is about recognising their structure inside more complicated expressions.

它们还与数 e 的定义以及物理、工程中的小角近似密切相关。掌握它们不是死记三个孤立结论，而是要在更复杂的表达式中识别出它们的结构。

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2. Limit 1 — The Sine Limit | 极限一：正弦极限

lim (x → 0) (sin x) / x = 1

This is the most fundamental trigonometric limit. It says that for very small angles measured in radians, sin x is approximately equal to x. The limit is only true when x is in radians; if x were in degrees, the limit would be π/180, not 1. That is one reason IB calculus always uses radians.

这是最基本的三角极限。它表明，当角度用弧度表示且 x 很小时，sin x 近似等于 x。该极限仅在弧度制下成立；若用角度制，极限将是 π/180 而不是 1。这也是 IB 微积分始终使用弧度的原因之一。

x	0.5	0.1	0.01	0.001	−0.001
sin x / x	0.9589	0.9983	0.99998	0.9999998	0.9999998

The table gives numerical evidence: as x gets closer to 0 from either side, sin x / x gets closer to 1. This agrees with the small-angle approximation sin x ≈ x for |x| small.

上表给出了数值证据：当 x 从两侧趋近 0 时，sin x / x 趋近 1。这与小角近似 sin x ≈ x（|x| 很小时）一致。

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3. Geometric Proof Sketch of the Sine Limit | 正弦极限的几何证明思路

For 0 < x < π/2, compare three areas: the triangle OAP, the circular sector OAP, and the triangle OAT, where O is the origin, A = (1, 0), P = (cos x, sin x), and T = (1, tan x). The area inequalities are:

当 0 < x < π/2 时，比较三个面积：三角形 OAP、扇形 OAP 和三角形 OAT。其中 O 为原点，A = (1, 0)，P = (cos x, sin x)，T = (1, tan x)。面积不等式为：

½ sin x < ½ x < ½ tan x

Multiplying by 2 and dividing by sin x (positive for 0 < x < π/2) gives 1 < x / sin x < 1 / cos x. Taking reciprocals reverses the inequalities: cos x < sin x / x < 1. Since cos x → 1 as x → 0, the Squeeze Theorem gives sin x / x → 1. The same result holds for negative x because sin x / x is an even function.

两边乘以 2 并除以 sin x（在 0 < x < π/2 时为正），得到 1 < x / sin x < 1 / cos x。取倒数会反转不等号：cos x < sin x / x < 1。由于 x → 0 时 cos x → 1，由夹逼定理得 sin x / x → 1。因为 sin x / x 是偶函数，负 x 的情形同样成立。

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4. Limit 2 — The Cosine Limit | 极限二：余弦极限

lim (x → 0) (1 − cos x) / x = 0

This limit is often written as lim (x → 0) (cos x − 1) / x = 0. It shows that 1 − cos x approaches 0 faster than x does. In fact, 1 − cos x behaves like x²/2 for small x, which is why the limit with denominator x is 0, but the limit with denominator x² is 1/2.

这个极限也常写成 lim (x → 0) (cos x − 1) / x = 0。它表明 1 − cos x 趋于 0 的速度比 x 更快。事实上，当 x 很小时，1 − cos x 的行为像 x²/2，所以分母为 x 时极限是 0，而分母为 x² 时极限是 1/2。

The distinction between denominator x and denominator x² is a classic IB trap. Always check what power of x appears in the denominator before evaluating.

分母是 x 还是 x²，是 IB 的经典陷阱。求极限前务必检查分母中 x 的幂次。

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5. Deriving the Cosine Limit from the Sine Limit | 由正弦极限推导余弦极限

Use the identity 1 − cos x = sin²x / (1 + cos x). Then rewrite the quotient:

利用恒等式 1 − cos x = sin²x / (1 + cos x)，将商改写为：

(1 − cos x) / x = (sin x / x) × (sin x / (1 + cos x))

As x → 0, sin x / x → 1 and sin x / (1 + cos x) → 0 / (1 + 1) = 0. Therefore the product tends to 1 × 0 = 0.

当 x → 0 时，sin x / x → 1，且 sin x / (1 + cos x) → 0 / (1 + 1) = 0。因此乘积趋于 1 × 0 = 0。

This derivation shows that the cosine limit is not independent: it follows directly from the sine limit. In an exam, if you forget the cosine limit, you can rebuild it in a few lines.

这个推导说明余弦极限并非独立结论：它直接由正弦极限推出。考试中若忘记余弦极限，可以用几行重新推出。

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6. Limit 3 — The Exponential Limit | 极限三：指数极限

lim (x → 0) (eˣ − 1) / x = 1

This limit is the derivative of eˣ at x = 0. It states that for small x, eˣ ≈ 1 + x. It is the exponential counterpart of the sine limit and is essential for differentiating exponential functions from first principles.

这个极限就是 eˣ 在 x = 0 处的导数。它表明当 x 很小时，eˣ ≈ 1 + x。它是正弦极限的指数对应物，也是从第一原理推导指数函数导数的关键。

Equivalently, lim (x → 0) (aˣ − 1) / x = ln a for any a > 0. In particular, when a = e, the limit is 1. This general form is useful when the base is not e.

等价地，对任意 a > 0，有 lim (x → 0) (aˣ − 1) / x = ln a。特别地，当 a = e 时极限为 1。当底数不是 e 时，这个一般形式很有用。

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7. The Number e and the Compound-Interest Limit | 数 e 与复利极限

lim (n → ∞) (1 + 1/n)ⁿ = e

The number e is defined by this limit. It arises from continuous compounding: if you invest 1 unit at 100% annual interest compounded n times per year, the balance approaches e. Setting x = 1/n, the limit becomes lim (x → 0) (1 + x)^(1/x) = e.

数 e 由此极限定义。它来自

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