This is a beta course, so its structure, chapters, and examples may continue to change.

Factorization Method

Factorization is an important method for solving limits of the $\frac{0}{0}$ indeterminate form. By factoring, we can cancel the common zero factor in the numerator and denominator and thus find the limit.

Basic Principle

Basic principle of factorization

When a limit takes the $\frac{0}{0}$ indeterminate form, both the numerator and the denominator vanish at the limit point. Through factorization, we can find the common zero factor and cancel it.

Applicable Conditions

1. The $\frac{0}{0}$ Indeterminate Form

The numerator vanishes at the limit point
The denominator vanishes at the limit point
The numerator and denominator share a common zero factor

2. Common Forms

Polynomial: $P(x) = (x - a)Q(x)$
Rational function: $\frac{P(x)}{Q(x)}$ , where $P(a) = Q(a) = 0$

Solving Steps

Identify the type of indeterminate form
Factor both the numerator and the denominator
Cancel the common zero factor
Substitute directly to evaluate

Common Factorization Formulas

Difference of Squares

Difference of squares

$a^2 - b^2 = (a + b)(a - b)$

Note: The difference of two squares equals the product of their sum and difference.

Applications:

$x^2 - 4 = (x + 2)(x - 2)$
$x^2 - 9 = (x + 3)(x - 3)$

Perfect Square

Perfect square

$a^2 + 2ab + b^2 = (a + b)^2$

$a^2 - 2ab + b^2 = (a - b)^2$

Note: A perfect square trinomial can be factored into the product of two identical factors.

Applications:

$x^2 + 4x + 4 = (x + 2)^2$
$x^2 - 6x + 9 = (x - 3)^2$

Sum of Cubes

Sum of cubes

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Note: The sum of two cubes can be factored as the product of their sum and a quadratic trinomial.

Applications:

$x^3 + 8 = (x + 2)(x^2 - 2x + 4)$
$x^3 + 27 = (x + 3)(x^2 - 3x + 9)$

Difference of Cubes

Difference of cubes

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Note: The difference of two cubes can be factored as the product of their difference and a quadratic trinomial.

Applications:

$x^3 - 8 = (x - 2)(x^2 + 2x + 4)$
$x^3 - 27 = (x - 3)(x^2 + 3x + 9)$

Factoring Polynomials

Factoring a difference of powers

$x^n - a^n = (x - a)(x^{n-1} + ax^{n-2} + a^2x^{n-3} + \cdots + a^{n-1})$

Note: The $n$ -th power difference can be factored into a linear factor times a polynomial of degree $n-1$ .

Applications:

$x^4 - 1 = (x - 1)(x^3 + x^2 + x + 1)$
$x^5 - 32 = (x - 2)(x^4 + 2x^3 + 4x^2 + 8x + 16)$

Worked Examples

Example 1

Find the limit $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$

Reference Answer(1 个标签)

factorization

Idea: This is a $\frac{0}{0}$ indeterminate form, which can be handled by factoring using the difference of squares.

Detailed steps:

Check the type of indeterminate form:
- When $x = 3$ , the numerator $x^2 - 9 = 9 - 9 = 0$
- When $x = 3$ , the denominator $x - 3 = 3 - 3 = 0$
- Hence it is a $\frac{0}{0}$ indeterminate form
Factor the numerator: $x^2 - 9 = x^2 - 3^2 = (x + 3)(x - 3)$
Cancel the zero factor: $\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x + 3)(x - 3)}{x - 3} = \lim_{x \to 3} (x + 3)$
Substitute directly: $\lim_{x \to 3} (x + 3) = 3 + 3 = 6$

Answer: $\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6$

Example 2

Find the limit $\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}$

Reference Answer(1 个标签)

factorization

Idea: This is a $\frac{0}{0}$ indeterminate form; we need to factor both the numerator and the denominator.

Detailed steps:

Check the type of indeterminate form:
- When $x = 2$ , the numerator $x^3 - 8 = 8 - 8 = 0$
- When $x = 2$ , the denominator $x^2 - 4 = 4 - 4 = 0$
- Hence it is a $\frac{0}{0}$ indeterminate form
Factor the numerator: $x^3 - 8 = x^3 - 2^3 = (x - 2)(x^2 + 2x + 4)$
Factor the denominator: $x^2 - 4 = x^2 - 2^2 = (x + 2)(x - 2)$
Cancel the zero factor: $\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4} = \lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{(x + 2)(x - 2)} = \lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 2}$
Substitute directly: $\lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 2} = \frac{4 + 4 + 4}{4} = \frac{12}{4} = 3$

Answer: $\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4} = 3$

Practice Problems

Exercise 1

Find the limit $\lim_{x \to 1} \frac{x^4 - 1}{x^2 - 1}$

Reference Answer(1 个标签)

factorization

Idea: This is a $\frac{0}{0}$ indeterminate form; we need to factor both the numerator and the denominator.

Detailed steps:

Check the type of indeterminate form:
- When $x = 1$ , the numerator $x^4 - 1 = 1 - 1 = 0$
- When $x = 1$ , the denominator $x^2 - 1 = 1 - 1 = 0$
- Hence it is a $\frac{0}{0}$ indeterminate form
Factor the numerator: $x^4 - 1 = (x^2)^2 - 1^2 = (x^2 + 1)(x^2 - 1) = (x^2 + 1)(x + 1)(x - 1)$
Factor the denominator: $x^2 - 1 = (x + 1)(x - 1)$
Cancel the zero factor: $\lim_{x \to 1} \frac{x^4 - 1}{x^2 - 1} = \lim_{x \to 1} \frac{(x^2 + 1)(x + 1)(x - 1)}{(x + 1)(x - 1)} = \lim_{x \to 1} (x^2 + 1)$
Substitute directly: $\lim_{x \to 1} (x^2 + 1) = 1 + 1 = 2$

Answer: $\lim_{x \to 1} \frac{x^4 - 1}{x^2 - 1} = 2$

Exercise 2

Find the limit $\lim_{x \to -2} \frac{x^3 + 8}{x^2 + 4x + 4}$

Reference Answer(1 个标签)

factorization

Idea: This is a $\frac{0}{0}$ indeterminate form; we need to factor both the numerator and the denominator.

Detailed steps:

Check the type of indeterminate form:
- When $x = -2$ , the numerator $x^3 + 8 = -8 + 8 = 0$
- When $x = -2$ , the denominator $x^2 + 4x + 4 = 4 - 8 + 4 = 0$
- Hence it is a $\frac{0}{0}$ indeterminate form
Factor the numerator: $x^3 + 8 = x^3 + 2^3 = (x + 2)(x^2 - 2x + 4)$
Factor the denominator: $x^2 + 4x + 4 = (x + 2)^2$
Cancel the zero factor: $\lim_{x \to -2} \frac{x^3 + 8}{x^2 + 4x + 4} = \lim_{x \to -2} \frac{(x + 2)(x^2 - 2x + 4)}{(x + 2)^2} = \lim_{x \to -2} \frac{x^2 - 2x + 4}{x + 2}$
Substitute directly: $\lim_{x \to -2} \frac{x^2 - 2x + 4}{x + 2} = \frac{4 + 4 + 4}{0}$ — the denominator is zero, so we need further handling.
Use L’Hôpital’s rule or re-factor: since the denominator is zero, this limit does not exist.

Answer: The limit does not exist.

Summary

Symbols Used in This Article

符号	类型	读音/说明	在本文中的含义
$\lim_{x \to a}$	数学符号	limit as x approaches a	The limit as $x \to a$
$P(x), Q(x)$	数学符号	P/Q of x	Polynomials in numerator and denominator
$a$	数学符号	a	The limit point
$a^2 - b^2$	数学符号	a squared minus b squared	Difference of squares
$a^3 + b^3$	数学符号	a cubed plus b cubed	Sum of cubes
$a^3 - b^3$	数学符号	a cubed minus b cubed	Difference of cubes
$x^n - a^n$	数学符号	x to the n minus a to the n	Difference of powers
$\frac{0}{0}$	数学符号	zero over zero	Indeterminate form

中英对照

中文术语	英文术语	音标	说明
因式分解法	factorization	/ˌfæktəraɪˈzeɪʃən/	Writing an expression as a product of factors
零因子	zero factor	/ˈzɪərəʊ ˈfæktə/	A factor that makes the expression zero
平方差公式	difference of squares	/ˈdɪfrəns əv skweəz/	$a^2 - b^2 = (a+b)(a-b)$
完全平方	perfect square	/ˈpɜːfɪkt skweə/	The product of two identical factors
立方和	sum of cubes	/sʌm əv kjuːbz/	Factorization of $a^3 + b^3$
立方差	difference of cubes	/ˈdɪfrəns əv kjuːbz/	Factorization of $a^3 - b^3$
不定式	indeterminate form	/ˌɪndɪˈtɜːmɪnət fɔːm/	A limit form that cannot be determined directly

Direct Substitution

Rationalization Method

Factorization Method

Basic Principle

Applicable Conditions

1. The 00\frac{0}{0}00​ Indeterminate Form

2. Common Forms

Solving Steps

Common Factorization Formulas

Difference of Squares

Perfect Square

Sum of Cubes

Difference of Cubes

Factoring Polynomials

Worked Examples

Example 1

Example 2

Practice Problems

Exercise 1

Exercise 2

Summary

Symbols Used in This Article

中英对照

1. The $\frac{0}{0}$ Indeterminate Form