Cancelling Factors: Why the Missing Input Stays Missing

Nablio · 7 minPublished September 11, 2026

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Shared calculus foundations for Concordia MATH 203 and McGill MATH 140. Course sequences and classroom timing may differ.

You simplify a fraction, substitute a number, and obtain a perfectly ordinary answer. But was that number allowed in the original expression? This small algebra question becomes essential when calculus asks you to distinguish what a function does at an input from what it approaches near that input.

Cancellation uses division. Dividing the numerator and denominator by the same factor preserves the fraction only where that factor is nonzero. The habit to build is simple: record denominator restrictions before simplifying, and carry them beside the simplified expression.

The graph follows y = x + 3 with a hole at (2, 5). The limit as x approaches 2 is 5, although f(2) is undefined. — An open circle marks an excluded point, not the absence of a limit. The full reasoning appears below.

1. Keep the original domain

Consider $f(x)=\frac{x^2+x-6}{x-2}$ . The denominator is zero at $x=2$ , so the original function is defined for every real input except 2. Two numbers with product $-6$ and sum 1 are 3 and $-2$ . Factoring the numerator therefore gives $(x+3)(x-2)$ . We can cancel the common factor only when $x-2$ is nonzero. For every permitted input, the fraction and $x+3$ have the same output. The restriction is part of the simplified result, even though the new formula no longer visibly contains a denominator.

Worked example

$\begin{aligned}f(x)&=\frac{x^2+x-6}{x-2}\\&=\frac{(x+3)(x-2)}{x-2}\\&=x+3,\qquad x\ne2.\end{aligned}$ $f(3)=6$ . $f(2)$ is undefined: the original denominator is zero.

Common mistake. Cancel factors, not terms in a sum. In $\frac{x+3}{x}$ , there is no common factor $x$ to cancel. For $x\ne0$ , this expression equals $1+\frac{3}{x}$ , not 3.

Try it yourself

Let $p(x)=\frac{x^2-9}{x+3}$ . Simplify with its restriction, then find $p(-3)$ .

Show answer and reasoning

Factor $x^2-9$ as $(x-3)(x+3)$ . Cancelling gives $p(x)=x-3$ , with $x\ne-3$ . Thus $p(-3)$ remains undefined. The simplified formula does not restore an input excluded by the original denominator.

2. A limit asks about nearby inputs

Return to $f(x)=\frac{x^2+x-6}{x-2}$ . Although $f(2)$ is undefined, $f(x)$ follows $x+3$ whenever $x\ne2$ . As $x$ approaches 2 from either side, those outputs approach 5. Therefore its limit as $x$ approaches 2 is 5. Sample values illustrate the approach; the algebraic identity with $x+3$ at every nearby permitted input explains it. A limit can exist even when the function has no value at the point being approached.

Worked example

For $x=1.9$ and $x=1.99$ , the outputs are $4.9$ and $4.99$ . For $x=2.01$ and $x=2.1$ , the outputs are $5.01$ and $5.1$ . At $x=2$ , the function is undefined. $\lim_{x\to2}f(x)=5.$ This does not say that $f(2)=5$ .

Common mistake. Direct substitution giving $\frac{0}{0}$ does not determine a limit. It is not the number zero. Here, factoring reveals the behaviour near the excluded input.

Try it yourself

Let $q(x)=\frac{x^2-4}{x^2-x-2}$ . What restrictions remain after simplification, and what is its limit as $x$ approaches 2?

Show answer and reasoning

The denominator is $(x-2)(x+1)$ , so exclude both 2 and $-1$ . Factoring the numerator and cancelling gives $\frac{x+2}{x+1}$ , still with both restrictions. Near 2, this expression approaches $\frac{4}{3}$ . Cancellation has not restored the excluded input 2.

3. Filling the hole defines an extension

The graph of $f$ is the line $y=x+3$ with a missing point at $(2,5)$ . This is a removable discontinuity: we could define a new function that fills the hole. That extension would agree with $f$ at every input where $f$ is defined, while also assigning an output at 2. To make the extension continuous there, its value must match the limit. Assigning some different value leaves a mismatch between the point and the nearby graph.

Worked example

Original function: $f(x)=x+3$ for $x\ne2$ ; no value at 2. Continuous extension: $r(x)=x+3$ for every real $x$ , including $r(2)=5$ .

Try it yourself

Define $r(x)=x+3$ for $x\ne2$ and $r(2)=c$ . If $c=0$ , what are $r(2)$ and its limit at 2? Which value of $c$ makes $r$ continuous there?

Show answer and reasoning

With $c=0$ , the function value is 0 but the limit is 5. Changing one output does not change the nearby values. Continuity requires the value to equal the limit, so choose $c=5$ .

Your next revision session

On your next rational-function problem, write the original restrictions first. Then keep three questions separate: What is the simplified formula? What is the value at the input? What is the limit near that input?
Keep a short record of domain mistakes. When correcting a problem, identify where a restriction was lost and rewrite the result with every restriction.

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These sources establish course context. The exercises in this guide are original, and the courses are not presented as equivalent.

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Cancelling Factors: Why the Missing Input Stays Missing

1. Keep the original domain

Worked example

Try it yourself

2. A limit asks about nearby inputs

Worked example

Try it yourself

3. Filling the hole defines an extension

Worked example

Try it yourself

Your next revision session

Official course outlines and descriptions

Connect these skills to your course

Work on the step you’re missing.

Free resources to work through

Before Calculus: 10 Algebra Skills You Should Know

Concordia MATH 203 Readiness Checklist