📚 Review Set 26A - Non-Calculator | 复习题26A - 非计算器
Review Set 26A is a dedicated non-calculator assessment that consolidates the introductory calculus and limit techniques typically covered in the IB Mathematics: Analysis and Approaches course (SL/HL). Working through these problems without a calculator sharpens your algebraic manipulation, reinforces limit properties, and builds fluency in differentiation rules. This article guides you through the core question types found in the set, offering worked examples and strategic commentary to strengthen your mathematical reasoning.
复习题26A是一套完全非计算器的评估练习，旨在巩固IB数学分析与方法课程（SL/HL）中常见的微积分入门与极限技巧。在不使用计算器的前提下完成这些题目，能有效锻炼代数操作能力、加深对极限性质的理解，并熟练运用各类求导法则。本文带你梳理题集中的核心题型，通过典型例题和策略讲解，夯实你的数学推理功底。
1. Evaluating Limits by Factorisation | 因式分解求极限
When evaluating limits algebraically, always attempt direct substitution first. If the result is an indeterminate form such as 0/0, factor the numerator and denominator to cancel common factors. For instance, evaluate limx→2 (x2 e numerator as (x − 2)(x + 2). Cancel the (x − 2) term, then substitute x = 2 to obtain 4.
在代数法求极限时，始终先尝试直接代入。若得出 0/0 型不定式，就对分子和分母进行因式分解，约去公因式。例如求 limx→2 (x2 − 4)/(x − 2)。直接代入得 0/0，因此将分子分解为 (x − 2)(x + 2) ，约去 (x − 2) 后代入 x=2，得 4。
Another typical example: limx→−1 (x2 check that the simplified expression is defined at the limiting point before substituting.
另一个典型例子：limx→−1 (x2 + 3x + 2)/(x + 1)。将分子分解为 (x + 1)(x + 2) ，约去 (x + 1) 并代入 x = −1，得 1。务必在代入前检查化简后的表达式在极限点是否已定义。
2. Limits Involving Square Roots | 含根式的极限
A limit containing a square root that yields 0/0 usually requires rationalisation. Multiply numerator and denominator by the conjugate of the radical part. Evaluate limx→0 (√x + 4 − 2)/x. The conjugate is √x + 4 + 2. Multiplying produces ((x + 4) − 4) / [x(√x + 4 + 2)] = x / [x(√x + 4 + 2)] = 1/(√x + 4 + 2). Substituting x = 0 yields 1/4.
含根式的极限若出现 0/0 型，通常需要有理化。将分子分母同时乘以根式部分的共轭根式。求 limx→0 (√x + 4 − 2)/x，共轭为 √x + 4 + 2。相乘后得 ((x + 4) − 4) / [x(√x + 4 + 2)] = 1/(√x + 4 + 2)。代入 x = 0 得到 1/4。
You may also encounter limits of the form √x + a − √a over x. The same conjugate method applies. Keep the expression tidy by cancelling common x-factors immediately after expansion, then substitute.
你也会遇到形如 (√x + a − √a) / x 的极限。同样使用共轭法处理。展开后立即约去公因式 x，再代入，保持表达式整洁。
3. Fundamental Trigonometric Limits | 基本三角极限
One of the most important limits in calculus is limθ→0 ears routinely in derivative proofs and related limit questions. For example, evaluate limx→0 sin(3x) / (2x). Rewrite as (3/2) · limx→0 sin(3x)/(3x) = (3/2) · 1 = 3/2.
微积分中最重要的极限之一是 limθ→0 sinθ / θ = 1。这一结果在导数证明和相关极限题中反复出现。例如求 limx→0 sin(3x) / (2x)。可改写为 (3/2) · limx→0 sin(3x)/(3x) = 3/2。
A similar approach works for limx→0 tan(ax) / x and other combinations. Always try to match the argument of the sine function with the denominator. For limx→0 (1 − cos x) / x2, you can use 1 − cos x = 2 sin2(x/2) to transform it into a known limit, yielding 1/2.
类似的方法适用于 limx→0 tan(ax) / x 等组合。始终尝试让分母与正弦函数的参数相匹配。对于 limx→0 (1 − cos x) / x2，可利用 1 − cos x = 2 sin2(x/2) 将其转化为已知极限，结果为 1/2。
4. First Principles Differentiation | 第一原理求导
The definition of the derivative, f'(x) = limh→0 [f(x + h) − f(x)] / h, is fundamental in non-calculator calculus. For f(x) = x2, we compute f'(x) = limh→0 ((x + h)2 − x2)/h = limh→0 (2xh + h2)/h = limh→0 (2x + h) = 2x. Cancelling h is the critical step.
导数的定义 f'(x) = limh→0 [f(x + h) − f(x)] / h 是非计算器微积分的基础。对 f(x) = x2，计算 f'(x) = limh→0 ((x + h)2 − x2)/h = limh→0 (2x + h) = 2x。约去 h 是关键步骤。
he difference quotient: [1/(x + h) − 1/x] / h = ((x − (x + h)) / (x(x + h))) / h = −1/(x(x
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