Use of Series Expansions to Find Limits | 利用级数展开求极限

📚 Use of Series Expansions to Find Limits | 利用级数展开求极限

When evaluating limits, algebraic manipulation can be difficult, especially with expressions involving trigonometric, exponential, or logarithmic functions. Series expansions provide a powerful method by replacing such functions with polynomial approximations, often turning indeterminate forms into computable limits.

在求极限时，代数变形有时非常困难，特别是涉及三角函数、指数函数或对数函数时。级数展开通过将这些函数替换为多项式近似，往往能把不定式转化为可计算的极限，是一种非常有力的方法。

1. Maclaurin Series: The Foundation | 麦克劳林级数：基础

A Maclaurin series is a Taylor series centred at x = 0. For a function f(x) that is infinitely differentiable at 0, the series is:

f(x) = f(0) + f′(0)x + (f″(0)/2!)x² + (f‴(0)/3!)x³ + …

This expansion is valid within its radius of convergence, which depends on the function. Using the first few terms gives a polynomial approximation, which is often easier to manipulate when finding limits.

麦克劳林级数是在 x = 0 处的泰勒级数。若函数 f(x) 在 0 处可无穷次求导，则其展开式为：

f(x) = f(0) + f′(0)x + (f″(0)/2!)x² + (f‴(0)/3!)x³ + …

该展开在函数本身的收敛半径内有效。只取前几项就能得到多项式近似，在求极限时更容易处理。

2. Taylor Series: A Generalisation | 泰勒级数：推广

If we approximate a function around a point x = a instead of 0, we use the Taylor series. The formula is:

f(x) = f(a) + f′(a)(x−a) + (f″(a)/2!)(x−a)² + (f‴(a)/3!)(x−a)³ + …

This is useful when a limit involves x approaching a non-zero value. By changing the variable t = x − a, we can often convert it to a Maclaurin-type series in t.

如果我们在点 x = a 附近近似函数，则应使用泰勒级数。其公式为：

f(x) = f(a) + f′(a)(x−a) + (f″(a)/2!)(x−a)² + (f‴(a)/3!)(x−a)³ + …

当极限涉及 x 趋向非零值时，这一形式非常有用。通过变量替换 t = x − a，往往可以转化为关于 t 的麦克劳林型级数。

3. Standard Expansions You Must Know | 必须掌握的常用展开式

The AQA specification expects you to recall and apply the following Maclaurin series. These are essential for evaluating limits efficiently.

AQA 考纲要求你熟记并运用下列麦克劳林级数。它们是高效求极限的关键。

Function	Series expansion	Valid for
eˣ	1 + x + x²/2! + x³/3! + …	all x
sin x	x − x³/3! + x⁵/5! − …	all x
cos x	1 − x²/2! + x⁴/4! − …	all x
ln(1 + x)	x − x²/2 + x³/3 − x⁴/4 + …	−1 < x ≤ 1
(1 + x)ⁿ	1 + n x + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + …	\|x\| < 1 (for non-integer n)

Notice the alternating signs for sin and cos, and the limited validity of the logarithmic expansion. In an exam, copying these accurately is the first step toward the correct limit.

注意 sin 和 cos 的交错符号，以及对数展开的收敛区间限制。在考试中，准确写出这些级数是求对极限的第一步。

4. The Binomial Series | 二项式级数

The binomial expansion applies to (1 + x)ⁿ for any real n. When n is positive integer, the series is finite, but for rational or negative n it is an infinite series.

二项式展开适用于任意实数 n 的 (1 + x)ⁿ。当 n 为正整数时，级数有限；当 n 为分数或负数时，则为无穷级数。

Positive integer n: (1 + x)ⁿ = Σ from k=0 to n C(n,k) xᵏ — an exact polynomial.

正整数 n： (1 + x)ⁿ = Σ from k=0 to n C(n,k) xᵏ —— 是精确多项式。
Negative or fractional n: use the infinite series 1 + nx + n(n−1)/2! x² + …, valid for |x| < 1.

负数或分数 n： 使用无穷级数 1 + nx + n(n−1)/2! x² + …，有效范围为 |x| < 1。

The first two terms are often enough for limits, but sometimes the x² term is needed to distinguish between similar functions.

通常前两项就够了，但有时需要 x² 项才能区分不同函数。

5. Using Series to Evaluate 0/0 Limits | 利用级数求 0/0 型极限

A common limit problem is the indeterminate form 0/0. For example, lim(x→0) (sin x − x)/x³. Direct substitution gives 0/0. Replacing sin x by its series x − x³/6 + x⁵/120 − … , we get:

常见的极限问题是 0/0 型未定式。例如，求 lim(x→0) (sin x − x)/x³。直接代入得 0/0。将 sin x 替换为级数 x − x³/6 + x⁵/120 − …，得到：

lim(x→0) [x − x³/6 + … − x]/x³ = lim(x→0) (−x³/6 + …)/x³ = −1/6

Notice how the leading terms cancel, leaving only the first non-zero term. This is the essence of using series for 0/0 limits: expand each function, subtract common leading terms, then divide by the lowest power of x.

注意前几项相互抵消，只剩下第一个非零项。这就是用级数求 0/0 型极限的核心：展开每个函数，减去相同的首项，再除以 x 的最低次幂。

6. Limits at Infinity and Change of Variable | 无穷远处的极限与变量替换

For limits as x → ∞, direct series expansion in x is often invalid because the power terms grow. Instead, introduce t = 1/x and consider t → 0. Then use a series expansion in t.

当 x → ∞ 时，直接按 x 展开级数通常无效，因为幂次项趋于无穷。此时令 t = 1/x，研究 t → 0，然后按 t 作级数展开。

Example: lim(x→∞) x ln(1 + 1/x). Set t = 1/x, so t → 0. Then x ln(1 + 1/x) = (1/t) ln(1 + t). Using ln(1 + t) = t − t²/2 + t³/3 − …, we get (1/t)(t − t²/2 + …) = 1 − t/2 + … → 1.

例：求 lim(x→∞) x ln(1 + 1/x)。令 t = 1/x，则 t → 0。于是 x ln(1 + 1/x) = (1/t) ln(1 + t)。利用 ln(1 + t) = t − t²/2 + t³/3 − …，得 (1/t)(t − t²/2 + …) = 1 − t/2 + … → 1。

This technique extends to expressions involving e^(1/x) or arctan(1/x), making the behaviour near infinity transparent.

这一技巧同样适用于含 e^(1/x) 或 arctan(1/x) 的表达式，使无穷远处的行为一目了然。

7. Leading Order Analysis and Truncation | 主项分析与截断

When using series expansions, you must decide how many terms to keep. For a limit where the denominator is xᵏ, you generally need to expand the numerator up to the xᵏ term and ignore higher-order terms. Keeping too few terms leads to errors; keeping too many wastes time.

使用级数展开时，必须决定保留多少项。如果分母是 xᵏ，通常需要把分子展开到 xᵏ 项，并忽略高阶项。保留项数过少会出错，过多则浪费时间。

A systematic approach is to first expand all functions to the same order, then simplify. Use the “big-O” idea: write O(xᵐ) to denote terms of order xᵐ and higher, making it clear when terms can be discarded.

系统的方法是先一致展开到相同阶数，再化简。可用“大 O”记号：写 O(xᵐ) 表示 xᵐ 及更高阶的项，明确何时可舍弃。

For example, lim(x→0) (cos x − 1 + x²/2)/x⁴. Using cos x ≈ 1 − x²/2 + x⁴/24, we get (1 − x²/2 + x⁴/24 − 1 + x²/2)/x⁴ = 1/24. If we only kept cos x ≈ 1 − x²/2, the result would be 0, which is wrong.

例如，lim(x→0) (cos x − 1 + x²/2)/x⁴。使用 cos x ≈ 1 − x²/2 + x⁴/24，得 (1 − x²/2 + x⁴/24 − 1 + x²/2)/x⁴ = 1/24。如果只保留 cos x ≈ 1 − x²/2，结果为 0，这是错误的。

8. Worked Examples (AQA-Style) | 例题（AQA 风格）

Now let’s apply the techniques to typical AQA exam questions. Each example shows the full method.

下面我们将这些技巧应用到典型的 AQA 考试题中。每个例子都展示完整解法。

Example 1: Find lim(x→0) (eˣ − 1 − x)/x².

例 1： 求 lim(x→0) (eˣ − 1 − x)/x²。

Expand eˣ = 1 + x + x²/2 + x³/6 + …. Then eˣ − 1 − x = x²/2 + x³/6 + …. Dividing by x² gives 1/2 + x/6 + … → 1/2. So the limit is 1/2.

展开 eˣ = 1 + x + x²/2 + x³/6 + …。则 eˣ − 1 − x = x²/2 + x³/6 + …。除以 x² 得 1/2 + x/6 + … → 1/2。因此极限为 1/2。

Example 2: Find lim(x→0) (tan x − x)/x³, given tan x = x + x³/3 + 2x⁵/15 + ….

例 2： 求 lim(x→0) (tan x − x)/x³，已知 tan x = x + x³/3 + 2x⁵/15 + …。

Subtract x: tan x − x = x³/3 + 2x⁵/15 + …. Divide by x³ to get 1/3 + 2x²/15 + … → 1/3. Hence the limit is 1/3.

减去 x 得 tan x − x = x³/3 + 2x⁵/15 + …。除以 x³ 得 1/3 + 2x²/15 + … → 1/3。因此极限为 1/3。

Example 3: Find lim(x→0) (√(1+x) − 1 − x/2)/x².

例 3： 求 lim(x→0) (√(1+x) − 1 − x/2)/x²。

Use the binomial series (1+x)^(1/2) = 1 + (1/2)x + (1/2)(−1/2)/2! x² + … = 1 + x/2 − x²/8 + …. Then √(1+x) − 1 − x/2 = −x²/8 + …. Dividing by x² gives −1/8. So the limit is −1/8.

用二项式级数 (1+x)^(1/2) = 1 + (1/2)x + (1/2)(−1/2)/2! x² + … = 1 + x/2 − x²/8 + …。则 √(1+x) − 1 − x/2 = −x²/8 + …。除以 x² 得 −1/8。因此极限为 −1/8。

9. Common Mistakes and How to Avoid Them | 常见错误与避坑

Even when you know the series, pitfalls can lead to wrong limits. Here are the most frequent issues and tips to avoid them.

即使记住了级数，陷阱仍会导致错误结果。以下是最常见的问题及避坑建议。

Wrong truncation: Truncating the series too early may cancel the term you need. Always expand to at least the order of the denominator.

截断错误：过早截断可能抵消掉你需要的项。至少要展开到分母的阶数。
Mixing variables: When using t = 1/x, remember all occurrences of x must be replaced, and the limit in t must be as t → 0.

变量混淆：使用 t = 1/x 时，务必替换所有 x，并注意 t 应趋向 0。
Ignoring validity: The series for ln(1+x) is only valid for |x| < 1. If your limit involves x → ∞, you must use the substitution method.

忽略收敛域： ln(1+x) 的级数仅在 |x| < 1 时有效。若极限涉及 x → ∞，必须使用变量替换法。
Forgetting the factorial: A common sign of carelessness is writing cos x = 1 + x²/2 instead of 1 − x²/2. Double-check signs and factorials.

忽略阶乘或符号：常见笔误是写 cos x = 1 + x²/2，而应为 1 − x²/2。务必检查符号和阶乘。
Infinite limit with finite series: If the leading term after division still contains a variable that goes to infinity, the limit may be infinite. In that case, state the limit as ∞ or −∞, if appropriate.

有限级数与无穷极限：若除以后首项仍含有趋于无穷的变量，则极限可能就是无穷。此时应写为 ∞ 或 −∞（如适用）。

10. Practice Problems and Exam Tips | 练习与考试技巧

Try these on your own before checking the answers. They reflect common AQA exam patterns.

请先独立尝试下列题目，再核对答案。它们反映了常见的 AQA 考题模式。

Practice 1: lim(x→0) (sin x − x + x³/6)/x⁵

练习 1： lim(x→0) (sin x − x + x³/6)/x⁵
Practice 2: lim(x→∞) x²(1 − cos(1/x))

练习 2： lim(x→∞) x²(1 − cos(1/x))
Practice 3: lim(x→0) ((1+x)ⁿ − 1 − n x)/x² for n = 3

练习 3： lim(x→0) ((1+x)ⁿ − 1 − n x)/x²，其中 n = 3

Answers: Practice 1: use sin x = x − x³/6 + x⁵/120 − …, so after cancellation the limit is 1/120. Practice 2: set t = 1/x, then x²(1 − cos t)/t²? Wait: x²(1 − cos(1/x)) = (1 − cos t)/t², and cos t ≈ 1 − t²/2, so limit is 1/2. Practice 3: (1+x)³ = 1 + 3x + 3x² + x³, so ((1+x)³ − 1 − 3x)/x² = (3x² + x³)/x² → 3.

答案：练习 1：用 sin x = x − x³/6 + x⁵/120 − …，抵消后极限为 1/120。练习 2：令 t = 1/x，则 x²(1 − cos(1/x)) = (1 − cos t)/t²，而 cos t ≈ 1 − t²/2，因此极限为 1/2。练习 3：(1+x)³ = 1 + 3x + 3x² + x³，所以 ((1+x)³ − 1 − 3x)/x² = (3x² + x³)/x² → 3。

In the exam, always write the series expansion clearly before simplifying. Even if your final limit is wrong, part marks are given for the correct series and the substitution method. Practice identifying the required order from the denominator, and always check that your series are valid for the x-values involved.

考试中，务必先写出级数展开式再化简。即使最终极限算错，写出正确的级数和替换方法也能得到步骤分。练习时，要从分母判断所需阶数，并检查所用级数在 x 取值范围内是否有效。

By mastering series expansions, you turn tricky limits into routine polynomial handling. This skill is essential for AQA A-level Mathematics and serves as a foundation for further study in pure mathematics.

掌握了级数展开，你就能将棘手的极限转化为常规的多项式运算。这一技能对 AQA A-level 数学至关重要，也为后续高等数学学习打下坚实基础。

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Use of Series Expansions to Find Limits | 利用级数展开求极限