📚 Using Series Expansions to Find Limits | 用级数展开求极限
ll quantities. In AQA A-Level Mathematics and Further Mathematics, the Maclaurin expansion gives a systematic way to replace functions by polynomials near x = 0, making limits quick and exact.
当极限出现 0/0 等未定式时，级数展开通常能揭示隐藏在相互抵消的小量中的有限比值。在 AQA A-Level 数学与进阶数学中，麦克劳林展开提供了一种把函数在 x = 0 附近替换为多项式的方法，使极限计算快速且精确。
1. The Core Idea: Replace with Local Polynomials | 核心思想：用局部多项式替换
ten as a constant plus linear, quadratic and higher powers of x. When a limit gives 0/0, the constant terms vanish, so the leading non-zero power of x controls the value of the limit.
在 x = 0 附近，光滑函数可以写成常数项加 x 的一次、二次及更高次幂。当极限为 0/0 时，常数项相互抵消，因此第一个非零的 x 幂决定极限值。
Series expansion turns a limit of transcendental functions into a limit of polynomials, where cancellation is visible term by term and no repeated differentiation is needed.
级数展开把超越函数的极限转化为多项式的极限，使每一项的抵消都清晰可见，而且不需要反复求导。
f(x) = a₀ + a₁x + a₂x² + a₃x³ + ⋯ near x = 0
over, and so on. The limit is therefore decided by the lowest power of x remaining after cancellation.
若 f(0) = 0，则 a₀ = 0，当 x → 0 时第一个存活的项 a₁x 起主导作用。如果 a₁ 也为 0，则二次项接管，依此类推。因此极限由抵消之后剩余的最低次幂决定。
2. Standard Maclaurin Expansions | 标准麦克劳林展开
for limits as x → 0, and they must be quoted accurately with correct signs and factorials.
你应当熟记 AQA 公式手册中的下列展开式，尤其是前三到四项。它们是处理 x → 0 型极限的主要工具，引用时必须符号与阶乘都准确无误。
at another point, first shift the variable to bring it to zero, as explained in Section 8.
这些展开式只适用于 x = 0 附近。如果极限在别的点，应先作变量平移使其变为零，方法见第 8 节。
| Function | 函数 | Maclaurin expansion | 麦克劳林展开 | Valid for | 有效范围 |
|---|---|---|
| eˣ | 1 + x + x²/2! + x³/3! + x⁴/4! + ⋯ | all real x | 所有实数 |
| sin x | x − x³/3! + x⁵/5! − ⋯ | all real x | 所有实数 |
| cos x | 1 − x²/2! + x⁴/4! − ⋯ | all real x | 所有实数 |
| ln(1 + x) | x − x²/2 + x³/3 − x⁴/4 + ⋯ | −1 < x ≤ 1 |
| (1 + x)ⁿ | 1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + ⋯ | |x| < 1 |
| tan x | x + x³/3 + 2x⁵/15 + ⋯ | |x| < π/2 |
| arcsin x | x + x³/6 + 3x⁵/40 + ⋯ | |x| < 1 |
| arctan x | x − x³/3 + x⁵/5 − ⋯ | |x| ≤ 1 |
| 1/(1 − x) | 1 + x + x² + x³ + ⋯ | |x| < 1 |
| sinh x | x + x³/3! + x⁵/5! + ⋯ | all real x | 所有实数 |
| cosh x | 1 + x²/2! + x⁴/4! + ⋯ | all real x | 所有实数 |
Notice that sin x, tan x and arcsin x contain only odd powers, while cos x and cosh x contain only even powers. This parity often tells you immediately which terms can cancel.
注意 sin x、tan x、arcsin x 只含奇次幂，而 cos x、cosh x 只含偶次幂。这种奇偶性常常能让你立刻看出哪些项可能抵消。
3. Why Series Beat L'Hôpital for This Job | 为何此处级数优于洛必达法则
L'Hôpital' sy for products, roots or compositions. Worse, applying it carelessly to a form that is not genuinely 0/0 or ∞/∞ gives nonsense.
洛必达法则正确且常常很快，但它需要反复对分子分母求导，而面对乘积、根式或复合函数时，每一次求导都可能变得很乱。更糟的是，若对并非真正 0/0 或 ∞/∞ 的形式滥用它，就会得到荒谬结果。
Series expansion replaces all of that with one algebraic step: write both functions as polynomials, cancel, and read off the ratio of the lowest surviving terms. It also shows the order of the error term, which is useful when a question asks for the limit and the leading behaviour.
级数展开把这一切替换为一步代数操作：把两个函数写成多项式，约去公因子，读出最低存活项之比。它还能显示误差项的阶，这在题目同时要求极限与主导行为时十分有用。
lim (f(x)/g(x)) as x → 0 = (leading term of f) ÷ (leading term of g)
In AQA papers, marks are usually given for a correct expansion, a correct cancellation and a clearly stated final value, so the series method is also easier to score full marks on.
在 AQA 试卷中，得分点通常是：正确的展开、正确的约分、明确写出的最终值。因此级数法也更容易拿满分。
4. Deciding How Many Terms to Keep | 决定保留几项
The golden rule is: expand far enough that a non-zero term survives in both the numerator and the denominator. If you stop too early everything cancels and you get 0/0 again; if you go too far you waste time but you are still correct.
黄金法则是：展开到分子和分母中都有非零项存活为止。若停得太早，一切都抵消，你又会得到 0/0；若展开过多，只是浪费时间，但结果仍然正确。
A practical procedure for AQA questions is the following.
对 AQA 题目，可采用如下实用步骤。
- Check that the limit is of the form 0/0 or a related indeterminate form, and that x → 0. | 先确认极限是 0/0 或相关未定式，并且 x → 0。
- Expand the numerator and the denominator to the same power of x, normally up to x³ or x⁴. | 把分子与分母展开到相同的 x 次幂，通常到 x³ 或 x⁴。
- Collect like powers and cancel any common factor of x. | 合并同类项，约去公因子 x。
- Let x → 0 in what remains; the constant term is the limit. | 对剩余部分令 x → 0，常数项即为极限。
If after cancelling you still have a factor of x left, the limit is 0; if the denominator vanishes first, the limit is infinite or does not exist as a finite number.
若约分之后仍剩下因子 x，则极限为 0；若分母先消失，则极限为无穷大，或不存在有限极限值。
5. Worked Examples: Basic Trigonometric Limits | 例题：基础三角极限
Start with the limit everyone meets first. Using sin x = x − x³/3! + ⋯ and dividing by x gives 1 − x²/6 + ⋯, and the limit is 1.
从最常见的极限开始。用 sin x = x − x³/3! + ⋯ 除以 x 得 1 − x²/6 + ⋯，极限为 1。
lim (sin x)/x = 1 as x → 0
Now use the higher-order terms, which is exactly where the series method earns its keep. Subtracting x from sin x removes the linear term, leaving the cubic term to govern the limit.
现在利用更高阶项，这正是级数法真正发挥优势之处。从 sin x 减去 x 消去一次项，剩下的三次项就决定了极限值。
lim (sin x − x)/x³ = −1/6 as x → 0
The same trick works for cosine, where the first non-constant term is quadratic, and for tangent, where the cubic coefficient is 1/3 rather than −1/6.
同样的技巧对余弦也成立，其第一个非常数项是二次项；对正切，其三次项系数是 1/3 而不是 −1/6。
lim (1 − cos x)/x² = 1/2, lim (tan x − x)/x³ = 1/3
ghtly richer example combines two expansions in one expression. Writing tan x = x + x³/3 + ⋯ and sin x = x − x³/6 + ⋯ gives tan x − sin x = x³(1/3 + 1/6) + ⋯ = x³/2 + ⋯, so the limit is 1/2.
一个稍丰富的例子在同一式中结合两个展开。写成 tan x = x + x³/3 + ⋯ 与 sin x = x − x³/6 + ⋯，得 tan x − sin x = x³(1/3 + 1/6) + ⋯ = x³/2 + ⋯，故极限为 1/2。
6. Worked Examples: Exponential and Logarithmic Limits | 例题：指数与对数极限
Exponential limits behave in the same way. Since eˣ = 1 + x + x²/2! + ⋯, subtracting 1 leaves a linear term, and subtracting 1 and x leaves a quadratic term.
指数型极限遵循同样的规律。由于 eˣ = 1 + x + x²/2! + ⋯，减去 1 后剩一次项，减去 1 和 x 后剩二次项。
lim (eˣ − 1)/x = 1, lim (eˣ − 1 − x)/x² = 1/2, lim (eˣ − 1 − x − x²/2)/x³ = 1/6
Logarithms follow the alternating pattern of ln(1 + x) = x − x²/2 + x³/3 − ⋯ . Watch that minus sign on the quadratic term; it is the most common source of lost marks.
对数遵循 ln(1 + x) = x − x²/2 + x³/3 − ⋯ 的交错模式，因此符号至关重要。请注意二次项前的负号，它是失分最常见的原因。
lim ln(1 + x)/x = 1, lim (ln(1 + x) − x)/x² = −1/2, lim (ln(1 + x) − x + x²/2)/x³ = 1/3
When the argument is scaled, substitute carefully. For e2x replace every x in the expansion by 2x, obtaining 1 + 2x + 2x² + (4/3)x³ + ⋯, so that (e2x − 1 − 2x)/x² → 2.
当自变量被缩放时，替换要谨慎。对 e2x，把展开式中每个 x 换成 2x，得 1 + 2x + 2x² + (4/3)x³ + ⋯，于是 (e2x − 1 − 2x)/x² → 2。
A mixed example is (ln(1 + x) − sin x)/x². The linear terms cancel, and the quadratic terms give −x²/2 from the logarithm while sine contributes nothing at order x², so the limit is −1/2.
混合例子 (ln(1 + x) − sin x)/x² 中，一次项抵消，二次项由对数给出 −x²/2，而正弦在 x² 阶无贡献，故极限为 −1/2。
7. Worked Examples: Cancellation of Leading Terms | 例题：主导项的抵消
Harder exam questions deliberately cancel the first two terms of an expansion. Then the limit depends on the cubic or even the quartic coefficient, so you must write enough terms.
较难的考题会故意让展开式的前两项都抵消。此时极限取决于三次甚至四次项系数，因此你必须写足够多的项。
Consider (cos x − 1 + x²/2)/x⁴. The expansion cos x = 1 − x²/2 + x⁴/24 − ⋯ that the constant and quadratic terms vanish exactly, leaving x⁴/24, so the limit is 1/24.
考虑 (cos x − 1 + x²/2)/x⁴。展开 cos x = 1 − x²/2 +
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