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Resolving Indeterminate Forms Algebraically

Reference: Stewart 2.3  •  Chapter: 1  •  Section: 3

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When Direct Substitution Fails

You try to compute $\lim_{x \to 2}\frac{x^2 - 4}{x - 2}$ by substituting $x = 2$, and you get $\frac{0}{0}$.

This is not "undefined" in the usual sense. It is an indeterminate form. The limit might exist; we just cannot see it yet. The expression $\frac{0}{0}$ is like a locked door: behind it could be any number, or no number at all. The key is algebra.

Prerequisite Map

PrerequisitesBasic Limit LawsFactoring Techniques
This skillResolving Indeterminate Forms Algebraically
Leads tono further branch yet

Quick Reference

Property Value
Concept Limits
Chapter 1.6
Difficulty Intermediate
Time ~25 minutes

The Core Idea

When substitution gives $\frac{0}{0}$, the numerator and denominator share a common factor that's causing both to vanish. Find and cancel that factor.

The Three-Step Strategy

Step 1: Try direct substitution
         │
         ▼
    Get 0/0?
     /      \
   No        Yes
    │          │
    ▼          ▼
  Done!    Step 2: Simplify
           (factor/rationalize)
                │
                ▼
           Step 3: Try substitution
           on simplified expression

Technique 1: Factoring and Cancellation

When to Use

Why It Works

If $x = a$ makes both numerator and denominator zero, then $(x - a)$ is a factor of both.

Example

Evaluate $\lim_{x \to 3}\frac{x^2 - 9}{x - 3}$.

Step 1: Substitution gives $\frac{9-9}{3-3} = \frac{0}{0}$. ✗

Step 2: Factor the numerator: $$\frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3}$$

Step 3: Cancel (valid for $x \neq 3$): $$= x + 3$$

Step 4: Now substitute: $$\lim_{x \to 3}(x + 3) = 6$$

Common Factoring Patterns

Pattern Factored Form
$x^2 - a^2$ $(x-a)(x+a)$
$x^2 + bx + c$ $(x - r_1)(x - r_2)$ where $r_1 + r_2 = -b$, $r_1 r_2 = c$
$x^3 - a^3$ $(x-a)(x^2 + ax + a^2)$
$x^3 + a^3$ $(x+a)(x^2 - ax + a^2)$

Technique 2: Rationalizing

When to Use

Why It Works

Multiplying by the conjugate eliminates the radical, often revealing a cancelable factor.

The Conjugate

Expression Conjugate
$\sqrt{x} - a$ $\sqrt{x} + a$
$\sqrt{x} + a$ $\sqrt{x} - a$
$\sqrt{x} - \sqrt{y}$ $\sqrt{x} + \sqrt{y}$

Example

Evaluate $\lim_{x \to 4}\frac{\sqrt{x} - 2}{x - 4}$.

Step 1: Substitution gives $\frac{2-2}{4-4} = \frac{0}{0}$. ✗

Step 2: Multiply by conjugate (top and bottom): $$\frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \frac{x - 4}{(x-4)(\sqrt{x}+2)}$$

Step 3: Cancel $(x - 4)$: $$= \frac{1}{\sqrt{x} + 2}$$

Step 4: Substitute: $$\lim_{x \to 4}\frac{1}{\sqrt{x} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$$

Technique 3: Expanding and Simplifying

When to Use

Example

Evaluate $\lim_{h \to 0}\frac{(2+h)^2 - 4}{h}$.

Step 1: Substitution gives $\frac{4-4}{0} = \frac{0}{0}$. ✗

Step 2: Expand $(2+h)^2 = 4 + 4h + h^2$: $$\frac{4 + 4h + h^2 - 4}{h} = \frac{4h + h^2}{h}$$

Step 3: Factor and cancel: $$= \frac{h(4 + h)}{h} = 4 + h$$

Step 4: Substitute: $$\lim_{h \to 0}(4 + h) = 4$$

Practice Problems

Level 1 Basic Factoring

Evaluate $\lim_{x \to 5}\frac{x^2 - 25}{x - 5}$.

Thought Process

Direct substitution gives $\frac{0}{0}$, so we need to factor.

The numerator $x^2 - 25$ is a difference of squares: $x^2 - 5^2 = (x-5)(x+5)$.

After canceling $(x-5)$, try substitution again.

Show Answer

$$\lim_{x \to 5}\frac{x^2 - 25}{x - 5} = \lim_{x \to 5}\frac{(x-5)(x+5)}{x-5} = \lim_{x \to 5}(x+5) = 10$$

Level 2 Quadratic over Linear

Evaluate $\lim_{x \to -2}\frac{x^2 + 5x + 6}{x + 2}$.

Thought Process

Check: at $x = -2$, numerator is $4 - 10 + 6 = 0$ and denominator is $0$. Indeterminate form confirmed.

Factor the quadratic: we need two numbers that multiply to 6 and add to 5. That's 2 and 3.

So $x^2 + 5x + 6 = (x+2)(x+3)$.

Show Answer

Factor the numerator: $$x^2 + 5x + 6 = (x + 2)(x + 3)$$

Cancel and substitute: $$\lim_{x \to -2}\frac{(x+2)(x+3)}{x+2} = \lim_{x \to -2}(x+3) = 1$$

Level 3 Rationalizing the Numerator

Evaluate $\lim_{x \to 9}\frac{\sqrt{x} - 3}{x - 9}$.

Thought Process

At $x = 9$: numerator is $3 - 3 = 0$, denominator is $0$. We have $\frac{0}{0}$.

The numerator has a square root, so rationalize by multiplying by the conjugate $\frac{\sqrt{x}+3}{\sqrt{x}+3}$.

Note that $(\sqrt{x}-3)(\sqrt{x}+3) = x - 9$, which cancels with the denominator.

Show Answer

Multiply by conjugate: $$\frac{\sqrt{x} - 3}{x - 9} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \frac{x - 9}{(x-9)(\sqrt{x}+3)}$$

Cancel $(x - 9)$: $$= \frac{1}{\sqrt{x} + 3}$$

Substitute: $$\lim_{x \to 9}\frac{1}{\sqrt{x} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$$

Level 4 Difference Quotient Preview

Evaluate $\lim_{h \to 0}\frac{(3+h)^3 - 27}{h}$.

Thought Process

At $h = 0$: $(3)^3 - 27 = 0$ and denominator is $0$. Indeterminate.

Expand $(3+h)^3$ using the binomial formula or by hand: $(3+h)^3 = 27 + 27h + 9h^2 + h^3$

Then simplify and factor out $h$ from the numerator.

(This type of limit appears constantly when computing derivatives!)

Show Answer

Expand $(3+h)^3$: $$(3+h)^3 = 27 + 3(9)h + 3(3)h^2 + h^3 = 27 + 27h + 9h^2 + h^3$$

Substitute into the limit: $$\frac{27 + 27h + 9h^2 + h^3 - 27}{h} = \frac{27h + 9h^2 + h^3}{h}$$

Factor out $h$: $$= \frac{h(27 + 9h + h^2)}{h} = 27 + 9h + h^2$$

Now substitute $h = 0$: $$\lim_{h \to 0}(27 + 9h + h^2) = 27$$

Level 5 Double Rationalization

Evaluate $\lim_{x \to 0}\frac{\sqrt{1+x} - \sqrt{1-x}}{x}$.

Thought Process

At $x = 0$: numerator is $1 - 1 = 0$, denominator is $0$. Indeterminate.

The numerator has a difference of two square roots. Multiply by its conjugate: $$\frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}$$

This will give $(1+x) - (1-x) = 2x$ in the numerator, which cancels with $x$ in the denominator.

Show Answer

Multiply by conjugate: $$\frac{\sqrt{1+x} - \sqrt{1-x}}{x} \cdot \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}$$

Numerator becomes: $$(\sqrt{1+x})^2 - (\sqrt{1-x})^2 = (1+x) - (1-x) = 2x$$

Full expression: $$= \frac{2x}{x(\sqrt{1+x} + \sqrt{1-x})} = \frac{2}{\sqrt{1+x} + \sqrt{1-x}}$$

Substitute $x = 0$: $$\lim_{x \to 0}\frac{2}{\sqrt{1+x} + \sqrt{1-x}} = \frac{2}{1 + 1} = 1$$

Conceptual Check (CCI-Style)

Conceptual Why Cancellation Works

When evaluating $\lim_{x \to 2}\frac{x^2 - 4}{x - 2}$, we cancel $(x-2)$ from numerator and denominator to get $\lim_{x \to 2}(x+2) = 4$.

But wait: is $\frac{x^2-4}{x-2}$ not undefined at $x = 2$, while $x+2$ is defined there? How can they have the same limit?

Thought Process

Think about what a limit actually measures. Does it care about the value AT the point?

The two expressions $\frac{x^2-4}{x-2}$ and $x+2$ disagree only at $x = 2$. Everywhere else, they're identical.

Show Answer

The limit only cares about values NEAR the point, not AT the point.

The expressions $\frac{x^2-4}{x-2}$ and $x+2$ are equal for all $x \neq 2$.

Since limits examine behavior as $x$ approaches 2 (but never equals 2), both expressions give the same limit.

This is precisely why we can cancel: we're only looking at values where $x \neq 2$, so the cancellation $(x-2)/(x-2) = 1$ is valid.

Mastery Checklist

Mental Model

The $\frac{0}{0}$ form is a disguise:

The numerator and denominator both going to zero means they share a common factor, like two fractions that can be reduced. The expression $\frac{6}{9}$ looks different from $\frac{2}{3}$, but they're the same number. Similarly, $\frac{x^2-4}{x-2}$ looks different from $x+2$, but near $x = 2$, they behave identically.

Your job is to simplify the fraction to reveal the limit hiding underneath.

Connections

Looking back:

Looking ahead:

Why this matters:

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Last updated: 2026-01-22