📚 Finding Limits in Simple Cases | 简单情形下求极限
In AQA A-Level Mathematics, limits appear most often when you meet differentiation from first principles, investigate asymptotic behaviour, and check continuity. This article shows you how to find limits in simple cases using substitution, algebraic simplification, and limit laws.
在 AQA A-Level 数学中，极限最常出现在导数第一性原理、渐近行为研究和连续性判断中。本文将展示如何通过代入、代数化简和极限运算法则求简单情形下的极限。
1. What Is a Limit? | 什么是极限？
A limit describes the value that a function f(x) approaches as the input x approaches a given number a. We write the limit as L if f(x) can be made as close as desired to L by taking x sufficiently close to a, without necessarily letting x equal a.
极限描述的是当自变量 x 趋近于某个数 a 时，函数 f(x) 趋近的值。如果通过让 x 足够接近 a，可以使 f(x) 任意接近 L，则记极限为 L。这里不要求 x 恰好等于 a。
lim x→a f(x) = L
It is important to understand that the limit asks what happens near a, not what happens at a.
需要理解的是，极限关注的是在 a 附近发生的情况，而不是在 a 点本身发生的情况。
2. Notation and Basic Language | 记号与基本用语
x→a⁺ f(x) for approaching from the right and lim x→a⁻ f(x) for approaching from the left.
对于双侧极限，我们写作 lim x→a f(x)。对于单侧极限，我们用 lim x→a⁺ f(x) 表示从右侧趋近，用 lim x→a⁻ f(x) 表示从左侧趋近。
∞ or lim x→a f(x) = −∞. If no unique value is approached, we say the limit does not exist.
如果当 x 趋近于 a 时函数无限增大，我们写作 lim x→a f(x) = ∞ 或 lim x→a f(x) = −∞。如果没有趋近于唯一的值，我们就说极限不存在。
3. Limit Laws for Simple Functions | 简单函数的极限运算法则
When limits exist, they can be combined using straightforward rules. These laws allow you to break a complicated limit into smaller parts.
当极限存在时，可以用简单的规则将它们组合起来。这些法则允许你把复杂的极限拆成较小的部分。
| If lim f(x) = L and lim g(x) = M, then lim [f(x) + g(x)] = L + M | 若 lim f(x) = L 且 lim g(x) = M，则 lim [f(x) + g(x)] = L + M |
| lim [f(x) − g(x)] = L − M | lim [f(x) − g(x)] = L − M |
| lim [k f(x)] = kL for any constant k | 对任意常数 k，lim [k f(x)] = kL |
| lim [f(x) g(x)] = LM | lim [f(x) g(x)] = LM |
| lim [f(x) ÷ g(x)] = L ÷ M provided M ≠ 0 | 若 M ≠ 0，则 lim [f(x) ÷ g(x)] = L ÷ M |
These rules are especially useful for polynomial, rational, and trigonometric limits after simplification.
这些法则在化简后的多项式、有理式和三角函数的极限中尤其有用。
4. Direct Substitution | 直接代入法
y into the expression. If the result is a finite number, that number is usually the limit.
对于许多简单函数，第一步就是直接将 x = a 代入表达式。如果得到的是一个有限数值，这个数值通常就是极限。
lim x→2 (3x² − 5x + 1) = 3(2)² − 5(2) + 1 = 12 − 10 + 1 = 3
Direct substitution works whenever the function is continuous at the point. Polynomials, rational functions at allowed inputs, and basic trigonometric functions often behave this way.
只要函数在该点连续，直接代入就有效。多项式、在允许输入处的有理函数以及基本三角函数通常都具有这种性质。
5. Indeterminate Form 0/0 | 0/0 未定式
Sometimes substitution gives 0/0. This does not mean the limit is 1 or undefined; it means the expression needs algebraic simplification before the true limit can be seen.
有时代入会得到 0/0。这并不意味着极限是 1 或者不存在；它意味着需要先进行代数化简，才能看出真正的极限。
nsider lim x→3 (x² − 9)/(x − 3). Substituting x = 3 gives 0/0, so we factor the numerator.
考虑 lim x→3 (x² − 9)/(x − 3)。代入 x = 3 会得到 0/0，因此我们对分子进行因式分解。
lim x→3 (x² − 9)/(x − 3) = lim x→3 [(x − 3)(x + 3)]/(x − 3)
After cancelling the common factor, we can substitute directly: lim x→3 (x + 3) = 6.
约去公因式后，我们可以直接代入：lim x→3 (x + 3) = 6。
6. Factorising and Cancelling | 因式分解与约分
When a rational expression gives 0/0, factorising the numerator or denominator often reveals a common factor that can be cancelled for all x except the limiting value.
当有理式出现 0/0 时，对分子或分母进行因式分解通常可以发现一个公因式，该公因式在除极限点外的所有 x 上都可以约去。
x + 2)/(x + 1). The numerator factorises as (x + 1)(x + 2), so the expression becomes x + 2 for x ≠ −1.
例如，求 lim x→−1 (x² + 3x + 2)/(x + 1)。分子可因式分解为 (x + 1)(x + 2)，因此当 x ≠ −1 时，表达式变为 x + 2。
lim x→−1 (x² + 3x + 2)/(x + 1) = lim x→−1 (x + 2) = 1
Always state that the cancellation is valid because x ≠ −1 while taking the limit.
一定要说明约去是有效的，因为在取极限时 x ≠ −1。
7. Rationalising the Numerator or Denominator | 有理化分子或分母
If an expression contains a square root and direct substitution gives 0/0, rationalising can help eliminate the root and reveal the limit.
如果表达式中含有平方根，并且直接代入得到 0/0，那么有理化可以帮助消去根号并求出极限。
Evaluate lim x→0 (√x + 1 1)/x. Substitute x = 0 to get 0/0. Multiply the numerator and denominator by the conjugate √x + 1 + 1.
求 lim x→0 (√x + 1 − 1)/x。代入 x = 0 会得到 0/0。将分子和分母同时乘以共轭式 √x + 1 + 1。
lim x→0 (√x + 1 − 1)/x = lim x→0 [(x + 1) − 1]/[x(√x + 1 + 1)]
This simplifies to 1/(√x + 1 + 1), and substituting x = 0 gives 1/2.
这可以化简为 1/(√x + 1 + 1)，代入 x = 0 得到 1/2。
8. Limits at Infinity | 无穷远处的极限
To find the limit of a rational function as x approaches infinity, divide every term by the highest power of x in the denominator. This often shows which terms dominate.
要求有理函数在 x 趋近于无穷大时的极限，可以将每一项都除以分母中 x 的最高次幂。这通常能显示出哪些项起主导作用。
For lim x→∞ (3x² + x)/(2x² − 5), divide numerator and denominator by x².
对于 lim x→∞ (3x² + x)/(2x² − 5)，将分子和分母都除以 x²。
lim x→∞ (3 + 1/x)/(2 − 5/x²) = 3/2
terms like 1/x and 5/x² tend to 0, so the limit is the ratio of the leading coefficients.
当 x→∞ 时，1/x 和 5/x² 这样的项都趋于 0，因此极限就是首项系数之比。
9. One-Sided Limits | 单侧极限
One-sided limits are useful when a function behaves differently on each side of a point, for example near a vertical asymptote.
当函数在某点的两侧行为不同时，例如在垂直渐近线附近，单侧极限非常有用。
proaches 2 from the right, the denominator is a small positive number, so f(x) tends to ∞.
对于 f(x) = 1/(x − 2)，当 x 从右侧趋近于 2 时，分母是一个很小的正数，因此 f(x) 趋于 ∞。
lim x→2⁺ 1/(x − 2) = ∞ and lim x→2⁻ 1/(x − 2) = −∞
two one-sided limits are different, the two-sided limit lim x→2 1/(x − 2) does not exist.
由于两个单侧极限不同，所以双侧极限 lim x→2 1/(x − 2) 不存在。
10. Continuity and Limits | 连续性与极限
three conditions hold: f(a) is defined, lim x→a f(x) exists, and lim x→a f(x) equals f(a).
函数 f 在 x = a 处连续需要满足三个条件：f(a) 有定义，lim x→a f(x) 存在，并且 lim x→a f(x) 等于 f(a)。
ntinuous there. However, the limit exists: lim x→1 (x² − 1)/(x − 1) = lim x→1 (x + 1) = 2.
考虑 f(x) = (x² − 1)/(x − 1)。在 x = 1 处，f(1) 没有定义，因此函数在该点不连续。但极限存在：lim x→1 (x² − 1)/(x − 1) = lim x→1 (x + 1) = 2。
of a removable discontinuity, where defining f(1) = 2 would make the function continuous.
这是一个可去间断点的例子，若定义 f(1) = 2，函数就会变得连续。
11. Limits and Differentiation from First Principles | 极限与导数第一性原理
defined as a limit. For y = f(x), the derivative is given by the first-principles formula:
函数的导数被定义为一个极限。对于 y = f(x)，导数由第一性原理公式给出：
f'(x) = lim h→0 [f(x + h) − f(x)]/h
For f(x) = x², we expand and simplify before taking the limit.
对于 f(x) = x²，我们先展开并化简，再取极限。
f'(x) = lim h→0 [(x + h)² − x²]/h = lim h→0 (2xh + h²)/h = lim h→0 (2x + h) = 2x
This shows why simplifying a limit is essential: it turns an indeterminate form into one that can be evaluated by substitution.
这说明了为什么化简极限至关重要：它能把未定式转化为可以通过代入求值的形式。
12. Exam-Style Tips | 考试技巧
In AQA exam questions on limits, clear method marks are awarded for showing algebraic simplification and correct limit notation.
在 AQA 有关极限的考题中，清晰的方法分来自展示代数化简和正确的极限记号。
- Always try direct substitution first, but do not stop if you get 0/0.
- 如果代入后得到 0/0，不要就此停下，要继续化简。
- Factorise, expand, rationalise, or divide by the highest power when needed.
- 根据需要，进行因式分解、展开、有理化或除以最高次幂。
- State the condition x ≠ a when cancelling common factors.
- 约去公因式时，要说明 x ≠ a。
- For piecewise functions, check the left-hand and right-hand limits separately.
- 对于分段函数，要分别检查左极限和右极限。
- Keep the limit sign until you actually substitute the value after simplification.
- 在化简后真正代入数值之前，要一直保留极限符号。
These habits will help you avoid common errors and communicate your reasoning clearly.
这些习惯能帮助你避免常见错误，并清晰地表达推理过程。
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