📚 Two Important Limits | 两个重要极限
In A-Level calculus, many results depend on two special limits. One controls the behaviour of the sine function near zero; the other defines Euler’s number e and underpins exponential growth. Mastering these limits allows you to evaluate a wide range of limit problems and to understand differentiation from first principles.
在 A-Level 微积分中，许多结论都依赖于两个特殊极限。一个控制正弦函数在零附近的行为，另一个定义欧拉数 e 并支撑指数增长。掌握这两个极限能帮助你求解大量极限问题，并理解从第一原理进行微分的本质。
1. Why Limits Matter in A-Level Calculus | 为何极限在 A-Level 微积分中重要
A limit describes the value a function approaches as the input approaches a given point, even if the function is not defined at that point. In the AQA A-Level specification, limits are used to define derivatives and to justify the behaviour of trigonometric, exponential and logarithmic functions. Without limits, the slope of a curve or the notion of continuous growth cannot be made precise.
极限描述的是当自变量趋近于某一点时函数所趋近的值，即使函数在该点没有定义。在 AQA A-Level 考纲中，极限被用来定义导数，并解释三角函数、指数函数与对数函数的行为。没有极限，曲线的斜率或连续增长的概念就无法精确表达。
f'(x) = lim h→0 [f(x+h) − f(x)] / h
The derivative itself is the limit of a difference quotient. Whenever we differentiate sin x, cos x, ln x or e^x from first principles, one of the two important limits appears naturally.
导数本身就是差商的极限。每当我们从第一原理求 sin x、cos x、ln x 或 e^x 的导数时，两个重要极限之一就会自然出现。
2. The First Important Limit: sin x over x | 第一个重要极限：sin x 除以 x
The first important limit is lim x→0 sin x / x = 1. This result only holds when x is measured in radians. It says that for very small angles, sin x and x are almost equal, so their ratio tends to 1 as x approaches 0.
第一个重要极限是 lim x→0 sin x / x = 1。这个结论只有在 x 以弧度为单位时才成立。它说明对于非常小的角度，sin x 与 x 几乎相等，因此当 x 趋近于 0 时，它们的比值趋近于 1。
lim x→0 sin x / x = 1
It is essential to remember that radian measure is not just a convention. If x were in degrees, the limit would equal π/180, which would make calculus results far less elegant.
必须牢记，弧度制不仅仅是一种习惯。如果 x 以角度为单位，这个极限将等于 π/180，这会让微积分的结果变得非常不简洁。
3. Geometric Intuition for sin x over x | sin x 除以 x 的几何直观
On the unit circle, consider an angle x with 0 < x < π/2. The arc length x lies between sin x and tan x, so sin x < x < tan x. Dividing by sin x gives 1 < x / sin x < 1 / cos x. Taking reciprocals reverses the inequalities: cos x < sin x / x < 1. Since cos x → 1 and 1 → 1, the squeeze theorem gives sin x / x → 1.
在单位圆上，考虑 0 < x < π/2 的角。弧长 x 介于 sin x 与 tan x 之间，因此 sin x < x < tan x。除以 sin x 得到 1 < x / sin x < 1 / cos x。取倒数会反转不等式：cos x < sin x / x < 1。由于 cos x → 1 且 1 → 1，夹逼定理给出 sin x / x → 1。
This geometric argument also shows why the limit does not depend on the direction of approach. For negative x, the same inequalities hold with reversed signs, so the two-sided limit still equals 1.
这个几何论证也说明了为什么该极限与趋近方向无关。对于负的 x，同样的不等式在符号反转后仍然成立，因此双侧极限仍等于 1。
4. Using the First Limit in Differentiation | 用第一个重要极限求导
To differentiate sin x from first principles, expand sin(x + h) using the compound-angle formula: sin(x + h) = sin x cos h + cos x sin h. Then the difference quotient becomes sin x (cos h − 1)/h + cos x (sin h / h). As h → 0, (cos h − 1)/h → 0 and sin h / h → 1, so the derivative is cos x.
为了从第一原理求 sin x 的导数，使用复合角公式展开：sin(x + h) = sin x cos h + cos x sin h。于是差商变为 sin x (cos h − 1)/h + cos x (sin h / h)。当 h → 0 时，(cos h − 1)/h → 0 且 sin h / h → 1，因此导数为 cos x。
d/dx sin x = cos x
The small companion limit lim h→0 (cos h − 1)/h = 0 is itself a consequence of the first important limit, because 1 − cos h = 2 sin²(h/2). This pair of limits makes trigonometric differentiation fully rigorous.
辅助极限 lim h→0 (cos h − 1)/h = 0 本身也是第一个重要极限的推论，因为 1 − cos h = 2 sin²(h/2)。这对极限使三角函数的微分变得完全严谨。
5. Common Variations of the Sine Limit | 正弦极限的常见变形
The standard sine limit can be scaled and combined. For constants a and b, lim x→0 sin(ax) / bx = a/b. Also lim x→0 tan x / x = 1 because tan x / x = (sin x / x) × (1 / cos x). The related cosine limit is lim x→0 (1 − cos x) / x² = 1/2, which follows from 1 − cos x = 2 sin²(x/2).
标准正弦极限可以进行缩放与组合。对于常数 a 和 b，lim x→0 sin(ax) / bx = a/b。此外 lim x→0 tan x / x = 1，因为 tan x / x = (sin x / x) × (1 / cos x)。相关的余弦极限是 lim x→0 (1 − cos x) / x² = 1/2，它由 1 − cos x = 2 sin²(x/2) 推出。
- lim x→0 sin(ax) / bx = a/b
- lim x→0 tan x / x = 1
- lim x→0 (1 − cos x) / x² = 1/2
These forms are not separate results to memorise blindly; they all come from algebraic rearrangement of the single standard limit and from the identity cos x → 1 as x → 0.
这些形式并不是需要盲目记忆的独立结论；它们都来自对唯一标准极限的代数变形，以及当 x → 0 时 cos x → 1 这一事实。
6. The Second Important Limit: (1 + x)^(1/x) | 第二个重要极限：(1+x)^(1/x)
The second important limit is lim x→0 (1 + x)^(1/x) = e. Equivalently, if n tends to infinity, lim n→∞ (1 + 1/n)^n = e. Here e ≈ 2.71828, the base of natural logarithms. This limit is an indeterminate form of type 1^∞, so it cannot be evaluated by simply substituting x = 0.
第二个重要极限是 lim x→0 (1 + x)^(1/x) = e。等价地，当 n 趋于无穷大时，lim n→∞ (1 + 1/n)^n = e。这里的 e ≈ 2.71828，是自然对数的底数。这个极限属于 1^∞ 型未定式，因此不能简单代入 x = 0 来求值。
lim x→0 (1 + x)^(1/x) = e
The limit is subtle because the base 1 + x tends to 1 while the exponent 1/x tends to infinity. The two effects compete, and their balance produces the finite number e.
这个极限很微妙，因为底数 1 + x 趋近于 1，而指数 1/x 趋近于无穷大。这两种效应相互竞争，它们的平衡产生了有限数 e。
7. Linking the Second Limit to e | 第二个极限与 e 的联系
The number e is often defined precisely by this limit. It appears in the derivative of ln x: d/dx ln x = lim h→0 [ln(x+h) − ln x]/h = lim h→0 ln(1 + h/x)/h. Let u = h/x, so u → 0, giving (1/x) lim u→0 ln(1+u)^(1/u) = 1/x. This is why natural logarithms are natural in calculus.
数 e 常常正是由这个极限精确定义的。它出现在 ln x 的导数中：d/dx ln x = lim h→0 [ln(x+h) − ln x]/h = lim h→0 ln(1 + h/x)/h。令 u = h/x，则 u → 0，得到 (1/x) lim u→0 ln(1+u)^(1/u) = 1/x。这就是为什么自然对数在微积分中如此自然。
d/dx ln x = 1/x
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