Finding Limits in Simple Cases | 简单情形下的极限求法

📚 Finding Limits in Simple Cases | 简单情形下的极限求法

In A-Level Mathematics, a limit describes the value that a function approaches as its input approaches a given point. Finding limits in simple cases is a core skill for calculus, forming the basis for differentiation and integration.

在A-Level数学中，极限描述的是当自变量趋近某个点时函数所趋近的值。在简单情形下求极限是微积分的一项核心技能，也是微分和积分的基础。

1. What is a Limit? | 什么是极限？

A limit is written as limₓ→ₐ f(x) = L. This means that as x gets closer and closer to a, the value of f(x) gets arbitrarily close to L. It is important to notice that x does not have to be equal to a; we are only interested in the behaviour near a.

极限写作 limₓ→ₐ f(x) = L。这意味着当x越来越接近a时，函数值f(x)会任意接近L。需要注意的是，x不必等于a；我们只关心x在a附近时的行为。

For example, consider f(x) = 2x + 1 as x approaches 3. Even if we never let x equal 3, the values f(2.9), f(2.99), f(3.1) all tend to 7. Therefore limₓ→₃ (2x + 1) = 7.

例如，考虑 f(x) = 2x + 1 在x趋近3时的情况。即使我们不让x等于3，f(2.9)、f(2.99)、f(3.1)的值都趋向7。因此 limₓ→₃ (2x + 1) = 7。

2. Direct Substitution | 直接代入法

For many functions, the limit can be found simply by substituting the value of a directly into the expression. This works when the function is continuous at that point, meaning the limit equals the function value.

对于许多函数，可以直接将a的值代入表达式来求极限。当函数在该点连续时，这种方法有效，此时极限等于函数值。

limₓ→₃ (2x² − 5) = 2(3)² − 5 = 18 − 5 = 13

Polynomials are continuous everywhere, so direct substitution is always valid for a polynomial. Rational functions are also continuous wherever the denominator is not zero.

多项式处处连续，因此对于多项式，直接代入法总是有效。有理函数在其分母不为零的地方也是连续的。

3. Indeterminate Forms 0/0 | 不定式 0/0

Sometimes direct substitution gives 0/0, which is called an indeterminate form. This does not mean the limit does not exist; it means we need to simplify the expression before evaluating the limit.

有时直接代入会得到0/0，这称为不定式。这并不表示极限不存在，而是意味着在求极限之前我们需要先化简表达式。

if limₓ→ₐ f(x) = 0 and limₓ→ₐ g(x) = 0, then limₓ→ₐ f(x)/g(x) requires further work

Common techniques to remove the 0/0 form include factorisation, rationalisation, and using known standard limits.

消除0/0形式的常用技巧包括因式分解、有理化以及使用已知的标准极限。

4. Factorisation Method | 因式分解法

When f(x) and g(x) are polynomials and direct substitution gives 0/0, factorise both the numerator and denominator. A common factor that causes the zero may be cancelled.

当f(x)和g(x)都是多项式且直接代入得到0/0时，可以对分子分母进行因式分解。导致零的公因子可以被约去。

limₓ→₂ (x² − 4)/(x − 2) = limₓ→₂ (x + 2)(x − 2)/(x − 2) = limₓ→₂ (x + 2) = 4

Here the factor (x − 2) is cancelled because x ≠ 2 during the limiting process. The resulting expression x + 2 is continuous at x = 2, so we substitute directly.

这里约去因子(x − 2)，是因为在极限过程中x ≠ 2。得到的表达式x + 2在x = 2处连续，因此可以直接代入。

5. Rationalisation Method | 有理化方法

If the expression contains square roots, rationalisation is often effective. Multiply the numerator and denominator by the conjugate of the term involving the root, then simplify.

如果表达式中含有平方根，有理化通常很有效。将分子分母同乘以含有根号项的共轭式，然后化简。

limₓ→₀ (√(1 + x) − 1)/x = limₓ→₀ ((√(1 + x) − 1)(√(1 + x) + 1))/(x(√(1 + x) + 1))

The numerator simplifies to (1 + x) − 1 = x. Cancelling the x gives limₓ→₀ 1/(√(1 + x) + 1) = 1/(1 + 1) = 1/2.

分子化简为(1 + x) − 1 = x。约去x后得到 limₓ→₀ 1/(√(1 + x) + 1) = 1/(1 + 1) = 1/2。

6. Limits at Infinity | 无穷大处的极限

For rational functions as x approaches infinity, the behaviour is determined by the highest powers of x. Divide every term by the highest power appearing in the denominator.

对于有理函数，当x趋近无穷大时，其行为由x的最高次幂决定。将每一项除以分母中出现的最高次幂。

limₓ→∞ (3x² + 2x)/(x² − 5) = limₓ→∞ (3 + 2/x)/(1 − 5/x²) = 3

Since 2/x and 5/x² both tend to 0, the limit is 3. If the numerator has a lower degree than the denominator, the limit is 0; if the numerator has a higher degree, the limit is infinite.

因为2/x和5/x²都趋向0，所以极限为3。如果分子的次数低于分母，极限为0；如果分子的次数高于分母，极限为无穷大。

7. One-Sided Limits | 单侧极限

The left-hand limit limₓ→a⁻ f(x) describes the behaviour as x approaches a from values less than a. The right-hand limit limₓ→a⁺ f(x) describes behaviour from values greater than a.

左极限 limₓ→a⁻ f(x) 描述的是x从小于a的值趋近a时函数的行为。右极限 limₓ→a⁺ f(x) 描述的是从大于a的值趋近a时函数的行为。

limₓ→₀⁻ |x|/x = −1 and limₓ→₀⁺ |x|/x = 1

Since the one-sided limits are different, limₓ→₀ |x|/x does not exist. A two-sided limit exists only when the left-hand and right-hand limits are equal.

由于单侧极限不同，limₓ→₀ |x|/x 不存在。只有当左极限和右极限相等时，双侧极限才存在。

8. Continuity and Limits | 连续性与极限

A function f is continuous at x = a if and only if limₓ→ₐ f(x) = f(a). This condition includes three parts: f(a) is defined, the limit exists, and the two values are equal.

函数f在x = a处连续当且仅当 limₓ→ₐ f(x) = f(a)。这个条件包含三部分：f(a)有定义，极限存在，且两者相等。

For example, f(x) = x² is continuous at x = 2 because limₓ→₂ x² = 4 and f(2) = 4. If a function is continuous at a point, direct substitution is always valid there.

例如，f(x) = x²在x = 2处连续，因为 limₓ→₂ x² = 4 且 f(2) = 4。如果函数在某点连续，那么直接代入在该点总是有效。

9. Common Mistakes | 常见错误

Assuming that 0/0 means the limit does not exist. In fact, it usually means the expression must be simplified.

认为0/0表示极限不存在。事实上，它通常意味着需要对表达式进行化简。
Cancelling a factor without noting that it is zero at the exact point, although this is allowed in the limit because x approaches the point but never reaches it.

约去一个因子时没有注意到它在精确点处为零，然而在极限中这是允许的，因为x趋近该点但从不等于它。
Confusing the value of the function with the value of the limit. They are equal only when the function is continuous.

混淆函数值与极限值。只有当函数连续时它们才相等。

10. Worked Examples | 例题讲解

Example 1: Find limₓ→₁ (x² + x − 2)/(x − 1).

例1：求 limₓ→₁ (x² + x − 2)/(x − 1)。

x² + x − 2 = (x − 1)(x + 2), so limₓ→₁ (x + 2) = 3

Factorising the numerator cancels the problematic factor and the limit is 3.

对分子因式分解后约去问题因子，得到极限为3。

Example 2: Find limₓ→∞ (5x − 3)/(2x + 1).

例2：求 limₓ→∞ (5x − 3)/(2x + 1)。

Divide by x: limₓ→∞ (5 − 3/x)/(2 + 1/x) = 5/2

Since 3/x and 1/x tend to 0, the limit is 5/2.

因为3/x和1/x都趋向0，所以极限为5/2。

11. Practice Questions | 练习题

Try the following limits on your own before reading the answers.

请先自己尝试求解下列极限，再查看答案。

1. limₓ→₄ (x² − 16)/(x − 4)

Answer: 8. Since x² − 16 = (x − 4)(x + 4), the limit is 4 + 4 = 8.

答案：8。因为x² − 16 = (x − 4)(x + 4)，所以极限为4 + 4 = 8。

2. limₓ→₀ (√(4 + x) − 2)/x

Answer: 1/4. Rationalise the numerator to get 1/(√(4 + x) + 2), then substitute x = 0.

答案：1/4。将分子有理化得1/(√(4 + x) + 2)，然后代入x = 0。

3. limₓ→∞ (7x³ + 2)/(4x³ − 1)

Answer: 7/4. Divide every term by x³.

答案：7/4。将每一项除以x³。

12. Summary | 总结

In simple cases, first try direct substitution. If it produces 0/0, use factor

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Finding Limits in Simple Cases | 简单情形下的极限求法