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The World of Limits in Advanced Mathematics

Basic Operation Rules

The basic operation rules of limits are the fundamental tools for computing limits. Mastering these rules can greatly simplify the process of evaluating limits.

Prerequisites

Let lim⁡f(x)=A\lim f(x) = A and lim⁡g(x)=B\lim g(x) = B, where AA and BB are both finite numbers.

The Four Arithmetic Rules

Addition Rule

Addition rule

lim⁡[f(x)±g(x)]=A±B\lim [f(x) \pm g(x)] = A \pm B

That is: the limit of the sum (or difference) of two functions equals the sum (or difference) of their limits.

证明

  1. Since lim⁡f(x)=A\lim f(x) = A, for any ε>0\varepsilon > 0, there exists δ1>0\delta_1 > 0 such that when 0<∣x−x0∣<δ10 < \vert x - x_0 \vert < \delta_1, we have ∣f(x)−A∣<ε2\vert f(x) - A \vert < \frac{\varepsilon}{2}

  2. Since lim⁡g(x)=B\lim g(x) = B, for any ε>0\varepsilon > 0, there exists δ2>0\delta_2 > 0 such that when 0<∣x−x0∣<δ20 < \vert x - x_0 \vert < \delta_2, we have ∣g(x)−B∣<ε2\vert g(x) - B \vert < \frac{\varepsilon}{2}

  3. Let δ=min⁡{δ1,δ2}\delta = \min\{\delta_1, \delta_2\}; then when 0<∣x−x0∣<δ0 < \vert x - x_0 \vert < \delta: ∣f(x)+g(x)−(A+B)∣≤∣f(x)−A∣+∣g(x)−B∣<ε2+ε2=ε\vert f(x) + g(x) - (A + B) \vert \leq \vert f(x) - A \vert + \vert g(x) - B \vert < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon

  4. Therefore lim⁡[f(x)+g(x)]=A+B\lim [f(x) + g(x)] = A + B

符号说明
SymbolTypePronunciation/DescriptionMeaning in this context
ε\varepsilonGreek letterEpsilonAn arbitrarily small positive number
δ\deltaGreek letterDeltaA positive number related to ε\varepsilon

Multiplication Rule

Multiplication rule

lim⁡[f(x)⋅g(x)]=A⋅B\lim [f(x) \cdot g(x)] = A \cdot B

That is: the limit of the product of two functions equals the product of their limits.

证明

Idea of the proof:

  1. Use the identity: f(x)⋅g(x)−A⋅B=[f(x)−A]⋅g(x)+A⋅[g(x)−B]f(x) \cdot g(x) - A \cdot B = [f(x) - A] \cdot g(x) + A \cdot [g(x) - B]

  2. Since lim⁡g(x)=B\lim g(x) = B, g(x)g(x) is bounded in some neighborhood of x0x_0

  3. Use the property of infinitesimals: the product of a bounded function and an infinitesimal is still an infinitesimal

  4. Conclude lim⁡[f(x)⋅g(x)]=A⋅B\lim [f(x) \cdot g(x)] = A \cdot B

Generalization: lim⁡[c⋅f(x)]=c⋅A\lim [c \cdot f(x)] = c \cdot A (where cc is a constant)

Division Rule

Division rule

lim⁡f(x)g(x)=AB\lim \frac{f(x)}{g(x)} = \frac{A}{B} (where B≠0B \neq 0)

That is: the limit of the quotient of two functions equals the quotient of their limits (provided the limit of the denominator is nonzero).

证明

Idea of the proof:

  1. First prove lim⁡1g(x)=1B\lim \frac{1}{g(x)} = \frac{1}{B}

  2. Since B≠0B \neq 0, g(x)g(x) is nonzero in some neighborhood of x0x_0

  3. Use the multiplication rule: lim⁡f(x)g(x)=lim⁡f(x)⋅lim⁡1g(x)=A⋅1B=AB\lim \frac{f(x)}{g(x)} = \lim f(x) \cdot \lim \frac{1}{g(x)} = A \cdot \frac{1}{B} = \frac{A}{B}

Power Rule

Power rule

lim⁡[f(x)]n=An\lim [f(x)]^n = A^n (where nn is a positive integer)

That is: the limit of a power of a function equals the power of its limit.

推论

Generalization: lim⁡[f(x)]g(x)=AB\lim [f(x)]^{g(x)} = A^B (subject to certain conditions)

Worked Examples

Example 1

Compute lim⁡x→2(x2+3x−1)\lim_{x \to 2} (x^2 + 3x - 1)

Solution: lim⁡x→2(x2+3x−1)=lim⁡x→2x2+lim⁡x→23x−lim⁡x→21=4+6−1=9\lim_{x \to 2} (x^2 + 3x - 1) = \lim_{x \to 2} x^2 + \lim_{x \to 2} 3x - \lim_{x \to 2} 1 = 4 + 6 - 1 = 9

Example 2

Compute lim⁡x→1x2+2xx+1\lim_{x \to 1} \frac{x^2 + 2x}{x + 1}

Solution: lim⁡x→1x2+2xx+1=lim⁡x→1(x2+2x)lim⁡x→1(x+1)=1+21+1=32\lim_{x \to 1} \frac{x^2 + 2x}{x + 1} = \frac{\lim_{x \to 1} (x^2 + 2x)}{\lim_{x \to 1} (x + 1)} = \frac{1 + 2}{1 + 1} = \frac{3}{2}

Practice Problems

Exercise 1

Compute the limit lim⁡x→3(2x2−5x+1)\lim_{x \to 3} (2x^2 - 5x + 1).

Reference Answer(1 个标签)
limit operation rules

Idea: Use the addition rule and the power rule.

Detailed steps:

  1. lim⁡x→3(2x2−5x+1)=2lim⁡x→3x2−5lim⁡x→3x+lim⁡x→31\lim_{x \to 3} (2x^2 - 5x + 1) = 2 \lim_{x \to 3} x^2 - 5 \lim_{x \to 3} x + \lim_{x \to 3} 1

  2. =2⋅9−5⋅3+1=18−15+1=4= 2 \cdot 9 - 5 \cdot 3 + 1 = 18 - 15 + 1 = 4

Answer: The limit is 4.

Exercise 2

Compute the limit lim⁡x→2x3+1x2−1\lim_{x \to 2} \frac{x^3 + 1}{x^2 - 1}.

Reference Answer(1 个标签)
limit operation rules

Idea: Use the division rule.

Detailed steps:

  1. lim⁡x→2x3+1x2−1=lim⁡x→2(x3+1)lim⁡x→2(x2−1)\lim_{x \to 2} \frac{x^3 + 1}{x^2 - 1} = \frac{\lim_{x \to 2} (x^3 + 1)}{\lim_{x \to 2} (x^2 - 1)}

  2. =8+14−1=93=3= \frac{8 + 1}{4 - 1} = \frac{9}{3} = 3

Answer: The limit is 3.

Exercise 3

Compute the limit lim⁡x→0sin⁡x+xx\lim_{x \to 0} \frac{\sin x + x}{x}.

Reference Answer(1 个标签)
limit operation rules

Idea: Use the addition rule and the first important limit.

Detailed steps:

  1. lim⁡x→0sin⁡x+xx=lim⁡x→0(sin⁡xx+xx)\lim_{x \to 0} \frac{\sin x + x}{x} = \lim_{x \to 0} \left(\frac{\sin x}{x} + \frac{x}{x}\right)

  2. =lim⁡x→0sin⁡xx+lim⁡x→01= \lim_{x \to 0} \frac{\sin x}{x} + \lim_{x \to 0} 1

  3. =1+1=2= 1 + 1 = 2

Answer: The limit is 2.

Summary

Symbols Used in This Article

符号类型读音/说明在本文中的含义
ε\varepsilon希腊字母Epsilon（伊普西隆）An arbitrarily small positive number
δ\delta希腊字母Delta（德尔塔）A positive number related to ε\varepsilon
nn数学符号positive integerA positive integer
lim⁡\lim数学符号limitThe limit of a function or sequence
→\to数学符号tends toA variable tending to some value
±\pm数学符号plus or minusAddition or subtraction
⋅\cdot数学符号multiplicationMultiplication

中英对照

中文术语英文术语音标说明
加法法则addition rule/əˈdɪʃən ruːl/The limit of a sum equals the sum of limits
乘法法则multiplication rule/ˌmʌltɪplɪˈkeɪʃən ruːl/The limit of a product equals the product of limits
除法法则division rule/dɪˈvɪʒən ruːl/The limit of a quotient equals the quotient of limits
幂运算法则power rule/ˈpaʊə ruːl/The limit of a power equals the power of the limit
有界函数bounded function/ˈbaʊndɪd ˈfʌŋkʃən/A function whose values stay within some range
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Operation Rules of Limits
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