📚 Use of Series Expansions to Find Limits | 利用级数展开求极限
When evaluating limits, especially those of the form 0/0, we often need to describe how a function behaves near a particular point. A series expansion rewrites a function as an infinite polynomial, and taking the first few terms can reveal the exact leading order of cancellation that determines the limit.
求极限时，尤其是遇到 0/0 型的未定式，我们往往需要刻画函数在某一点附近的行为。级数展开将函数表示成无限多项式，取其中前几项就能揭示出极限中起到决定作用的“主项”，从而消去分子中的抵消部分。
1. Why Use Series Expansions? | 为什么用级数展开？
Direct substitution often fails for limits such as limx→0 (eˣ − 1)/x, because both numerator and denominator approach 0. Raising both to a common polynomial form lets us see exactly how quickly each part tends to 0.
直接代入法在处理像 limx→0 (eˣ − 1)/x 这样的极限时会失效，因为分子和分母都趋近于 0。把函数统一写成多项式形式，就能清楚看出每一项趋于 0 的速度。
Series expansions convert transcendental functions (exponential, logarithmic, trigonometric) into algebraic expressions. In many AQA exam questions, this is the only way to obtain a limit without advanced tools such as L’Hôpital’s rule.
级数展开把超越函数（指数、对数、三角函数）转化为代数表达式。在 AQA 考试中，这常常是免去洛必达法则等超纲工具就能求得极限的唯一方法。
2. Standard Maclaurin Expansions | 常用麦克劳林展开
The following expansions are valid for x near 0 and form the toolkit for solving limit problems. You should know them from the AQA formula book or memory.
下面这些展开式在 x 趋近于 0 时成立，是求解极限问题的基础工具。它们既在 AQA 公式册中给出，也需要熟练记忆。
- eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + ⋯
- ln(1+x) = x − x²/2 + x³/3 − x⁴/4 + ⋯ (｜x｜ < 1)
- sin x = x − x³/3! + x⁵/5! − x⁷/7! + ⋯
- cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + ⋯
- tan x = x + x³/3 + 2x⁵/15 + ⋯ (sometimes useful)
- (1+x)ⁿ = 1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + ⋯ (valid for ｜x｜ < 1 when n is not a positive integer)
For example, the first three terms of eˣ are 1 + x + x²/2. The symbol “⋯” means all higher-order terms, which we usually ignore after the required order.
例如，eˣ 的前三项是 1 + x + x²/2。符号“⋯”表示所有更高阶的项，在展开到所需阶数之后通常忽略。
3. The Key Idea: Substitute and Simplify | 核心思想：代入并化简
As x → 0, every function that is analytic at 0 can be written as a polynomial plus a remainder that vanishes faster than the last included term. For example, if we substitute sin x = x − x³/6 + ⋯ into limx→0 sin x / x , we get
当 x → 0 时，任何在 0 处解析的函数都可以写成多项式加一个余项，而余项比所保留的最后一项更快地趋于 0。例如，将 sin x = x − x³/6 + ⋯ 代入 limx→0 sin x / x，得到
sin x / x = (x − x³/6 + ⋯) / x = 1 − x²/6 + ⋯ → 1.
Thus the limit is found by ignoring terms that tend to 0 after division. The same idea applies to any combination of standard functions.
因此，忽略那些除以 x 的幂之后仍趋于 0 的项，就得到了极限。这个方法同样适用于任何标准函数的组合。
4. Worked Example 1: limx→0 (eˣ − 1 − x)/x² | 例 1：limx→0 (eˣ − 1 − x)/x²
Start with the expansion eˣ = 1 + x + x²/2 + x³/6 + ⋯. Subtract 1 and x:
首先写出展开式 eˣ = 1 + x + x²/2 + x³/6 + ⋯，然后减去 1 和 x：
eˣ − 1 − x = (1 + x + x²/2 + x³/6 + ⋯) − 1 − x = x²/2 + x³/6 + ⋯.
Divide by x²:
两边除以 x²：
(eˣ − 1 − x)/x² = 1/2 + x/6 + ⋯ → 1/2.
The best constant term is 1/2. All other terms contain positive powers of x and vanish as x → 0. Therefore the limit is 1/2.
常数项为 1/2，其余各项均含有 x 的正次幂，随着 x → 0 都趋于 0，所以极限为 1/2。
5. Worked Example 2: limx→0 (sin x − x)/x³ | 例 2：limx→0 (sin x − x)/x³
Use sin x = x − x³/3! + x⁵/5! − ⋯. The x terms cancel immediately:
使用 sin x = x − x³/3! + x⁵/5! − ⋯，x 项立即消去：
sin x − x = (x − x³/6 + x⁵/120 − ⋯) − x = −x³/6 + x⁵/120 − ⋯.
Divide by x³:
除以 x³：
(sin x − x)/x³ = −1/6 + x²/120 − ⋯ → −1/6.
Notice that we needed the x³ term of sin x, because the denominator is x³. If we had only used sin x ≈ x, we would have obtained 0/0 and could not distinguish the limit.
注意：因为分母是 x³，所以要保留 sin x 的 x³ 项。若只用 sin x ≈ x，就会得到 0/0，无法判断极限。
6. The Binomial Series for (1+x)ⁿ | (1+x)ⁿ 的二项式级数
When n is a positive integer, the binomial theorem terminates. When n is negative or a fraction, the series is infinite and converges for ｜x｜ < 1. This is extremely useful for limits involving square roots, reciprocal powers, or cube roots.
当 n 是正整数时，二项式定理给出有限和；当 n 是负数或分数时，二项式级数是无限级数，且当 ｜x｜ < 1 时收敛。这在涉及平方根、倒数幂或立方根的极限问题中非常有用。
(1+x)ⁿ = 1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + ⋯, ｜x｜ < 1.
For example, with n = 1/2:
例如，取 n = 1/2：
(1+x)^(1/2) = 1 + x/2 − x²/8 + x³/16 − ⋯.
This expansion lets us replace a root by a polynomial, avoiding tricky rationalisation.
这个展开式能把根号替换成多项式，从而避免繁琐的有理化操作。
7. Worked Example 3: limx→0 ((1+x)^(1/2) − 1)/x | 例 3：limx→0 ((1+x)^(1/2) − 1)/x
Substitute the binomial expansion:
代入二项式展开：
(1+x)^(1/2) = 1 + x/2 − x²/8 + ⋯.
Then:
于是：
((1+x)^(1/2) − 1)/x = (x/2 − x²/8 + ⋯)/x = 1/2 − x/8 + ⋯ → 1/2.
Without series, one might multiply numerator and denominator by (1+x)^(1/2) + 1. Both methods agree, but the binomial series is more systematic for more complex powers.
若不使用级数，我们也可以将分子分母同乘 (1+x)^(1/2) + 1 来有理化。两种方法结果一致，但二项式级数对于更复杂的幂次更为系统。
8. Limits at Infinity Using Series | 利用级数求无穷远处的极限
For x → ∞, we can write expressions like √(x²+1) by factoring out the dominant term and then using a binomial expansion with a small variable such as 1/x. Consider
对于 x → ∞，可以将 √(x²+1) 这类式子先提出主项，再对 1/x 这样的“小量”使用二项式展开。考虑以下极限
limx→∞ x ( √(x²+1) − x ).
Rewrite:
改写为：
√(x²+1) = x √(1 + 1/x²) = x [1 + 1/(2x²) − 1/(8x⁴) + ⋯].
Subtract x:
再减去 x：
√(x²+1) − x = x[1 + 1/(2x²) − 1/(8x⁴) + ⋯] − x = 1/(2x) − 1/(8x³) + ⋯.
Multiply by x:
乘以 x：
x (√(x²+1) − x) = 1/2 − 1/(8x²) + ⋯ → 1/2.
Again the higher terms vanish, leaving a finite limit even though the original expression is of the form ∞ × 0.
同样地，高次项趋于 0，留下有限极限，尽管原式是 ∞ × 0 型的未定式。
9. How Many Terms Should You Keep? | 应该保留多少项？
A simple rule: if the denominator is xᵏ, expand the numerator until you have the xᵏ term, and ignore all higher powers. When leading terms cancel, you must go one order further until a nonzero constant appears.
一个简单的规则：如果分母是 xᵏ，就把分子展开到 xᵏ 项，忽略更高次幂。当首项相互抵消时，需要继续展开下一阶，直到出现非零常数项。
For products of series, multiply out but drop terms above the required order. For example, eˣ sin x = (1 + x + x²/2 + ⋯)(x − x³/6 + ⋯) = x + x² + (1/2 − 1/6)x³ + ⋯.
对于级数相乘，展开时只需保留所需阶数以内的项。例如 eˣ sin x = (1 + x + x²/2 + ⋯)(x − x³/6 + ⋯) = x + x² + (1/2 − 1/6)x³ + ⋯。
Always check the series is used within its radius of convergence. For x → 0, this is usually automatic, but for x → ∞ we must introduce a new variable like t = 1/x and then let t → 0.
始终检查级数是否在其收敛半径内使用。当 x → 0 时，这通常自动满足；但当 x → ∞ 时，我们需要引入新变量 t = 1/x，再让 t → 0。
10. Summary and Exam Tips | 小结与考试建议
The series expansion method follows a reliable five-step pattern: (1) identify the variable that tends to 0; (2) choose the appropriate standard expansion; (3) keep enough terms to match the denominator order; (4) simplify algebraically; (5) take the limit by ignoring all terms that vanish.
级数展开求极限遵循一个固定的五步模式：(1) 找出趋于 0 的变量；(2) 选择合适的标准展开式；(3) 保留足够项数以匹配分母阶数；(4) 进行代数化简；(5) 忽略所有趋于 0 的项，写出极限。
Common mistakes include stopping too early, using an expansion outside its radius of convergence, or forgetting to multiply by extra factors when combining series. With practice, this method becomes both fast and reliable on AQA exam questions.
常见错误包括展开项数不足、在收敛半径之外使用展开式，或者在组合级数时忘记乘以额外因子。多加练习后，这个方法在 AQA 考试中会变得既快速又可靠。
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