Kuzee
The World of Limits in Advanced Mathematics

Factorization Method

Factorization is an important method for solving limits of the 00\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 00\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 00\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)P(x) = (x - a)Q(x)
  • Rational function: P(x)Q(x)\frac{P(x)}{Q(x)}, where P(a)=Q(a)=0P(a) = Q(a) = 0

Solving Steps

  1. Identify the type of indeterminate form
  2. Factor both the numerator and the denominator
  3. Cancel the common zero factor
  4. Substitute directly to evaluate

Common Factorization Formulas

Difference of Squares

Difference of squares

a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b)

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

Applications:

  • x2−4=(x+2)(x−2)x^2 - 4 = (x + 2)(x - 2)
  • x2−9=(x+3)(x−3)x^2 - 9 = (x + 3)(x - 3)

Perfect Square

Perfect square

a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2

a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a - b)^2

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

Applications:

  • x2+4x+4=(x+2)2x^2 + 4x + 4 = (x + 2)^2
  • x2−6x+9=(x−3)2x^2 - 6x + 9 = (x - 3)^2

Sum of Cubes

Sum of cubes

a3+b3=(a+b)(a2−ab+b2)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:

  • x3+8=(x+2)(x2−2x+4)x^3 + 8 = (x + 2)(x^2 - 2x + 4)
  • x3+27=(x+3)(x2−3x+9)x^3 + 27 = (x + 3)(x^2 - 3x + 9)

Difference of Cubes

Difference of cubes

a3−b3=(a−b)(a2+ab+b2)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:

  • x3−8=(x−2)(x2+2x+4)x^3 - 8 = (x - 2)(x^2 + 2x + 4)
  • x3−27=(x−3)(x2+3x+9)x^3 - 27 = (x - 3)(x^2 + 3x + 9)

Factoring Polynomials

Factoring a difference of powers

xn−an=(x−a)(xn−1+axn−2+a2xn−3+⋯+an−1)x^n - a^n = (x - a)(x^{n-1} + ax^{n-2} + a^2x^{n-3} + \cdots + a^{n-1})

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

Applications:

  • x4−1=(x−1)(x3+x2+x+1)x^4 - 1 = (x - 1)(x^3 + x^2 + x + 1)
  • x5−32=(x−2)(x4+2x3+4x2+8x+16)x^5 - 32 = (x - 2)(x^4 + 2x^3 + 4x^2 + 8x + 16)

Worked Examples

Example 1

Find the limit lim⁡x→3x2−9x−3\lim_{x \to 3} \frac{x^2 - 9}{x - 3}

factorization

Approach: This is a 00\frac{0}{0} indeterminate form, which can be handled by factoring using the difference of squares.

Steps:

  1. Check the type of indeterminate form:

    • When x=3x = 3, the numerator x2−9=9−9=0x^2 - 9 = 9 - 9 = 0
    • When x=3x = 3, the denominator x−3=3−3=0x - 3 = 3 - 3 = 0
    • Hence it is a 00\frac{0}{0} indeterminate form

  2. Factor the numerator: x2−9=x2−32=(x+3)(x−3)x^2 - 9 = x^2 - 3^2 = (x + 3)(x - 3)

  3. Cancel the zero factor: lim⁡x→3x2−9x−3=lim⁡x→3(x+3)(x−3)x−3=lim⁡x→3(x+3)\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)

  4. Substitute directly: lim⁡x→3(x+3)=3+3=6\lim_{x \to 3} (x + 3) = 3 + 3 = 6

Answer: lim⁡x→3x2−9x−3=6\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6

Example 2

Find the limit lim⁡x→2x3−8x2−4\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}

factorization

Approach: This is a 00\frac{0}{0} indeterminate form; we need to factor both the numerator and the denominator.

Steps:

  1. Check the type of indeterminate form:

    • When x=2x = 2, the numerator x3−8=8−8=0x^3 - 8 = 8 - 8 = 0
    • When x=2x = 2, the denominator x2−4=4−4=0x^2 - 4 = 4 - 4 = 0
    • Hence it is a 00\frac{0}{0} indeterminate form

  2. Factor the numerator: x3−8=x3−23=(x−2)(x2+2x+4)x^3 - 8 = x^3 - 2^3 = (x - 2)(x^2 + 2x + 4)

  3. Factor the denominator: x2−4=x2−22=(x+2)(x−2)x^2 - 4 = x^2 - 2^2 = (x + 2)(x - 2)

  4. Cancel the zero factor: lim⁡x→2x3−8x2−4=lim⁡x→2(x−2)(x2+2x+4)(x+2)(x−2)=lim⁡x→2x2+2x+4x+2\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}

  5. Substitute directly: lim⁡x→2x2+2x+4x+2=4+4+44=124=3\lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 2} = \frac{4 + 4 + 4}{4} = \frac{12}{4} = 3

Answer: lim⁡x→2x3−8x2−4=3\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4} = 3

Practice Problems

Exercise 1

Find the limit lim⁡x→1x4−1x2−1\lim_{x \to 1} \frac{x^4 - 1}{x^2 - 1}

factorization

Approach: This is a 00\frac{0}{0} indeterminate form; we need to factor both the numerator and the denominator.

Steps:

  1. Check the type of indeterminate form:

    • When x=1x = 1, the numerator x4−1=1−1=0x^4 - 1 = 1 - 1 = 0
    • When x=1x = 1, the denominator x2−1=1−1=0x^2 - 1 = 1 - 1 = 0
    • Hence it is a 00\frac{0}{0} indeterminate form

  2. Factor the numerator: x4−1=(x2)2−12=(x2+1)(x2−1)=(x2+1)(x+1)(x−1)x^4 - 1 = (x^2)^2 - 1^2 = (x^2 + 1)(x^2 - 1) = (x^2 + 1)(x + 1)(x - 1)

  3. Factor the denominator: x2−1=(x+1)(x−1)x^2 - 1 = (x + 1)(x - 1)

  4. Cancel the zero factor: lim⁡x→1x4−1x2−1=lim⁡x→1(x2+1)(x+1)(x−1)(x+1)(x−1)=lim⁡x→1(x2+1)\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)

  5. Substitute directly: lim⁡x→1(x2+1)=1+1=2\lim_{x \to 1} (x^2 + 1) = 1 + 1 = 2

Answer: lim⁡x→1x4−1x2−1=2\lim_{x \to 1} \frac{x^4 - 1}{x^2 - 1} = 2

Exercise 2

Find the limit lim⁡x→−2x3+8x2+4x+4\lim_{x \to -2} \frac{x^3 + 8}{x^2 + 4x + 4}

factorization

Approach: This is a 00\frac{0}{0} indeterminate form; we need to factor both the numerator and the denominator.

Steps:

  1. Check the type of indeterminate form:

    • When x=−2x = -2, the numerator x3+8=−8+8=0x^3 + 8 = -8 + 8 = 0
    • When x=−2x = -2, the denominator x2+4x+4=4−8+4=0x^2 + 4x + 4 = 4 - 8 + 4 = 0
    • Hence it is a 00\frac{0}{0} indeterminate form

  2. Factor the numerator: x3+8=x3+23=(x+2)(x2−2x+4)x^3 + 8 = x^3 + 2^3 = (x + 2)(x^2 - 2x + 4)

  3. Factor the denominator: x2+4x+4=(x+2)2x^2 + 4x + 4 = (x + 2)^2

  4. Cancel the zero factor: lim⁡x→−2x3+8x2+4x+4=lim⁡x→−2(x+2)(x2−2x+4)(x+2)2=lim⁡x→−2x2−2x+4x+2\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}

  5. Substitute directly: lim⁡x→−2x2−2x+4x+2=4+4+40\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.

  6. 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

Symbol notation
SymbolTypePronunciation/ExplanationMeaning in This Article
lim⁡x→a\lim_{x \to a}Mathematical symbollimit as x approaches aThe limit as x→ax \to a
P(x),Q(x)P(x), Q(x)Mathematical symbolP/Q of xPolynomials in numerator and denominator
aaMathematical symbolaThe limit point
a2−b2a^2 - b^2Mathematical symbola squared minus b squaredDifference of squares
a3+b3a^3 + b^3Mathematical symbola cubed plus b cubedSum of cubes
a3−b3a^3 - b^3Mathematical symbola cubed minus b cubedDifference of cubes
xn−anx^n - a^nMathematical symbolx to the n minus a to the nDifference of powers
00\frac{0}{0}Mathematical symbolzero over zeroIndeterminate form
Previous
Direct Substitution
Next
Rationalization Method

Found an issue with this content?

Tell us what could be improved.