Resolving Indeterminate Forms Algebraically

5 min read

Reference: Stewart §2.3

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

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

The expression is a ratio of polynomials
Direct substitution gives $\frac{0}{0}$

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

The expression contains square roots
Direct substitution gives $\frac{0}{0}$

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

The expression involves binomials raised to powers
Common in limits that preview the derivative

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.

Common Misconceptions

Common misconception

$0/0$ is always zero because zero divided by anything is zero. Zero divided by a nonzero number is zero. But $0/0$ is not “zero divided by something nonzero.” Both the numerator and denominator are approaching zero simultaneously, and the race between them determines the limit. For $\frac{x^2}{x}$, the numerator shrinks as $x^2$ while the denominator shrinks as $x$, so the ratio approaches 0. For $\frac{x}{x^2}$, the ratio approaches $\infty$. The form itself carries no information about the outcome.

Common misconception

rationalizing always means multiplying by a conjugate. Students learn to rationalize denominators by multiplying by a conjugate, and they often apply this mechanically to any square root. Rationalizing in limit problems can mean multiplying the numerator (not the denominator) by a conjugate, or it can mean a different substitution entirely. The goal is always the same: eliminate the indeterminate form, not follow a memorized pattern.

Mastery Checklist

I can recognize when direct substitution gives $\frac{0}{0}$
I can factor quadratics and difference of squares to cancel common terms
I can rationalize expressions with square roots using the conjugate
I can expand binomials to find cancelable factors
I understand why cancellation doesn’t affect the limit value

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:

Basic Limit Laws tells us what to do after simplification

Looking ahead:

Derivative Definition is ENTIRELY about this type of limit
L’Hôpital’s Rule (Calc II) provides an alternative approach

Why this matters:

Every derivative computation involves evaluating a $\frac{0}{0}$ limit
This is the single most important algebraic skill for calculus success

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Basic Limit Laws	Skills Index	Squeeze Theorem

Last updated: 2026-01-22