Limits Involving Factoring

Learning Objectives

In this section, we explore a fundamental algebraic technique for evaluating limits: limits involving factoring in calculus. By the end of this reading, we will be able to:

Why Factoring Appears in Limits

When we first study limits, we usually start with direct substitution. If ##f(x)## is continuous at ##x = a##, then

###\lim_{x \to a} f(x) = f(a).###

However, sometimes direct substitution produces an expression such as ##\frac{0}{0}##. This is not a real number but an indeterminate form. It signals that more algebraic work is needed before we can decide the limit.

A very common situation is when we deal with rational functions of the form

###\frac{P(x)}{Q(x)},###

where both ##P(x)## and ##Q(x)## are polynomials. If at a certain point ##x = a## we have ##P(a) = 0## and ##Q(a) = 0##, then both polynomials have ##(x – a)## as a factor according to the Factor Theorem. That common factor is exactly what creates the ##\frac{0}{0}## form. Once we factor and cancel it (for ##x \neq a##), the resulting simplified expression usually has a well-defined limit at ##x = a##.

The Strategy for Limits Involving Factoring

For rational expressions, the general strategy is:

Graphically, this procedure corresponds to “filling in a hole” in the graph of the function. The value of the limit at ##x = a## is the height at which that hole would be filled.

Worked Examples: Basic Polynomial Limits

Example 1

Evaluate ###\lim_{x \to 2} \frac{x^2 – 4}{x – 2}.###

Step 1: Check direct substitution.
Substituting ##x = 2## gives ###\frac{2^2 – 4}{2 – 2} = \frac{4 – 4}{0} = \frac{0}{0},### an indeterminate form.

Step 2: Factor.
###x^2 – 4 = (x – 2)(x + 2).###
So ###\frac{x^2 – 4}{x – 2} = \frac{(x – 2)(x + 2)}{x – 2}.###

Step 3: Cancel common factor.
For ##x \neq 2##, ###\frac{(x – 2)(x + 2)}{x – 2} = x + 2.###

Step 4: Evaluate the simplified limit.
###\lim_{x \to 2} (x + 2) = 2 + 2 = 4.###
So ###\lim_{x \to 2} \frac{x^2 – 4}{x – 2} = 4.###

Example 2

Evaluate ###\lim_{x \to 3} \frac{x^2 – 9}{x – 3}.###

Step 1: Direct substitution gives ###\frac{3^2 – 9}{3 – 3} = \frac{9 – 9}{0} = \frac{0}{0}.###

Step 2: Factor the numerator: ###x^2 – 9 = (x – 3)(x + 3).###

Step 3: Cancel the common factor ##(x – 3)## (for ##x \neq 3##): ###\frac{(x – 3)(x + 3)}{x – 3} = x + 3.###

Step 4: Evaluate the simplified limit: ###\lim_{x \to 3} (x + 3) = 6.###

Example 3

Evaluate ###\lim_{x \to 1} \frac{x^2 – 1}{x – 1}.###

Solution:

We have ###x^2 – 1 = (x – 1)(x + 1).### Thus

###\frac{x^2 – 1}{x – 1} = \frac{(x – 1)(x + 1)}{x – 1} = x + 1 \quad (x \neq 1).###

Therefore, ###\lim_{x \to 1} \frac{x^2 – 1}{x – 1} = \lim_{x \to 1} (x + 1) = 2.###

Example 4

Evaluate ###\lim_{x \to 5} \frac{x^2 – 10x + 25}{x – 5}.###

The numerator is a perfect square: ###x^2 – 10x + 25 = (x – 5)^2.### Hence

###\frac{x^2 – 10x + 25}{x – 5} = \frac{(x – 5)^2}{x – 5} = x – 5 \quad (x \neq 5).###

So ###\lim_{x \to 5} \frac{x^2 – 10x + 25}{x – 5} = \lim_{x \to 5} (x – 5) = 0.###

Example 5

Evaluate ###\lim_{x \to -2} \frac{x^2 + 3x + 2}{x + 2}.###

Factor the numerator: ###x^2 + 3x + 2 = (x + 1)(x + 2).### Then

###\frac{x^2 + 3x + 2}{x + 2} = \frac{(x + 1)(x + 2)}{x + 2} = x + 1 \quad (x \neq -2).###

Thus ###\lim_{x \to -2} \frac{x^2 + 3x + 2}{x + 2} = \lim_{x \to -2} (x + 1) = -1.###

Worked Examples with Squares and Cubes

Example 6

Evaluate ###\lim_{x \to 1} \frac{x^3 – 1}{x – 1}.###

Use the identity ###x^3 – 1 = (x – 1)(x^2 + x + 1).### Then

###\frac{x^3 – 1}{x – 1} = x^2 + x + 1 \quad (x \neq 1).###

Therefore, ###\lim_{x \to 1} \frac{x^3 – 1}{x – 1} = 1^2 + 1 + 1 = 3.###

Example 7

Evaluate ###\lim_{x \to 2} \frac{x^3 – 8}{x – 2}.###

Use the difference of cubes: ###x^3 – 8 = (x – 2)(x^2 + 2x + 4).### Hence

###\frac{x^3 – 8}{x – 2} = x^2 + 2x + 4 \quad (x \neq 2).###

Substituting ##x = 2## gives ###2^2 + 2 \cdot 2 + 4 = 4 + 4 + 4 = 12.###
So ###\lim_{x \to 2} \frac{x^3 – 8}{x – 2} = 12.###

Example 8

Evaluate ###\lim_{x \to -1} \frac{x^3 + 1}{x + 1}.###

Use the sum of cubes: ###x^3 + 1 = (x + 1)(x^2 – x + 1).### Then

###\frac{x^3 + 1}{x + 1} = x^2 – x + 1 \quad (x \neq -1).###

Substitute ##x = -1##: ###(-1)^2 – (-1) + 1 = 1 + 1 + 1 = 3.### Hence

###\lim_{x \to -1} \frac{x^3 + 1}{x + 1} = 3.###

Example 9

Evaluate ###\lim_{x \to 3} \frac{x^3 – 27}{x^2 – 9}.###

Factor numerator and denominator:

Thus

###\frac{x^3 – 27}{x^2 – 9} = \frac{(x – 3)(x^2 + 3x + 9)}{(x – 3)(x + 3)} = \frac{x^2 + 3x + 9}{x + 3} \quad (x \neq 3).###

Now substitute ##x = 3##:

###\frac{3^2 + 3 \cdot 3 + 9}{3 + 3} = \frac{9 + 9 + 9}{6} = \frac{27}{6} = \frac{9}{2}.###

Therefore, ###\lim_{x \to 3} \frac{x^3 – 27}{x^2 – 9} = \frac{9}{2}.###

Example 10

Evaluate ###\lim_{x \to 3} \frac{x^2 – 9}{x^2 – 5x + 6}.###

Factor both numerator and denominator:

So

###\frac{x^2 – 9}{x^2 – 5x + 6} = \frac{(x – 3)(x + 3)}{(x – 3)(x – 2)} = \frac{x + 3}{x – 2} \quad (x \neq 3).###

Now substitute ##x = 3##:

###\frac{3 + 3}{3 – 2} = \frac{6}{1} = 6.###

Hence ###\lim_{x \to 3} \frac{x^2 – 9}{x^2 – 5x + 6} = 6.###

Examples with More Algebraic Steps

Example 11

Evaluate ###\lim_{x \to 2} \frac{x^2 + 3x – 10}{x^2 – 4}.###

Factor numerator and denominator:

Then

###\frac{x^2 + 3x – 10}{x^2 – 4} = \frac{(x + 5)(x – 2)}{(x – 2)(x + 2)} = \frac{x + 5}{x + 2} \quad (x \neq 2).###

Substitute ##x = 2##:

###\frac{2 + 5}{2 + 2} = \frac{7}{4}.###

Therefore, ###\lim_{x \to 2} \frac{x^2 + 3x – 10}{x^2 – 4} = \frac{7}{4}.###

Example 12

Evaluate ###\lim_{x \to 1} \frac{x^4 – 1}{x – 1}.###

Factor step by step. First, treat the numerator as a difference of squares:

###x^4 – 1 = (x^2 – 1)(x^2 + 1).###

Now factor ##x^2 – 1## again:

###x^2 – 1 = (x – 1)(x + 1).###

So

###x^4 – 1 = (x – 1)(x + 1)(x^2 + 1).###

Then

###\frac{x^4 – 1}{x – 1} = (x + 1)(x^2 + 1) \quad (x \neq 1).###

Substituting ##x = 1## gives ###(1 + 1)(1^2 + 1) = 2 \cdot 2 = 4.###
So ###\lim_{x \to 1} \frac{x^4 – 1}{x – 1} = 4.###

Example 13

Evaluate ###\lim_{x \to 2} \frac{x^4 – 16}{x^2 – 4}.###

Again use difference of squares:

Then

###\frac{x^4 – 16}{x^2 – 4} = \frac{(x^2 – 4)(x^2 + 4)}{x^2 – 4} = x^2 + 4 \quad (x \neq 2).###

Now take the limit:

###\lim_{x \to 2} (x^2 + 4) = 2^2 + 4 = 8.###

Example 14

Evaluate ###\lim_{x \to 0} \frac{x^2 – 5x}{x}.###

Factor the numerator:

###x^2 – 5x = x(x – 5).###

So

###\frac{x^2 – 5x}{x} = \frac{x(x – 5)}{x} = x – 5 \quad (x \neq 0).###

Therefore, ###\lim_{x \to 0} \frac{x^2 – 5x}{x} = \lim_{x \to 0} (x – 5) = -5.###

Factoring in Difference Quotients

The technique of factoring is also central when we later define the derivative of a function. The following examples foreshadow that idea.

Example 15

Evaluate ###\lim_{h \to 0} \frac{(x + h)^2 – x^2}{h}.###

Step 1: Expand the numerator.

###(x + h)^2 – x^2 = x^2 + 2xh + h^2 – x^2 = 2xh + h^2.###

Step 2: Factor the numerator.

###2xh + h^2 = h(2x + h).###

So

###\frac{(x + h)^2 – x^2}{h} = \frac{h(2x + h)}{h} = 2x + h \quad (h \neq 0).###

Now

###\lim_{h \to 0} (2x + h) = 2x.###

Thus ###\lim_{h \to 0} \frac{(x + h)^2 – x^2}{h} = 2x.### This is exactly the derivative of ##x^2## with respect to ##x##.

Example 16

Evaluate ###\lim_{h \to 0} \frac{(x + h)^3 – x^3}{h}.###

Step 1: Expand the numerator.

###(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3.###

Therefore

###(x + h)^3 – x^3 = 3x^2h + 3xh^2 + h^3.###

Step 2: Factor out ##h##.

###3x^2h + 3xh^2 + h^3 = h(3x^2 + 3xh + h^2).###

So

###\frac{(x + h)^3 – x^3}{h} = 3x^2 + 3xh + h^2 \quad (h \neq 0).###

Now let ##h \to 0##:

###\lim_{h \to 0} (3x^2 + 3xh + h^2) = 3x^2.###

Thus ###\lim_{h \to 0} \frac{(x + h)^3 – x^3}{h} = 3x^2.###

Example 17

Evaluate ###\lim_{x \to 3} \frac{(x – 3)(x^2 + 2x + 1)}{x^2 – 9}.###

Factor the denominator:

###x^2 – 9 = (x – 3)(x + 3).###

The numerator can be noticed as

###(x – 3)(x^2 + 2x + 1) = (x – 3)(x + 1)^2.###

Thus

###\frac{(x – 3)(x^2 + 2x + 1)}{x^2 – 9} = \frac{(x – 3)(x + 1)^2}{(x – 3)(x + 3)} = \frac{(x + 1)^2}{x + 3} \quad (x \neq 3).###

Now substitute ##x = 3##:

###\frac{(3 + 1)^2}{3 + 3} = \frac{16}{6} = \frac{8}{3}.###

Therefore, ###\lim_{x \to 3} \frac{(x – 3)(x^2 + 2x + 1)}{x^2 – 9} = \frac{8}{3}.###

Key Takeaways

Suggested Further Reading

For a broader theoretical background, standard references on real analysis and calculus discuss limits and continuity in detail. Introductory explanations can also be found on reliable sources such as the Wikipedia article on limits and related pages on polynomial factorisation and the Factor Theorem.

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