Cancellation and Excluded Values, foundations for calculus lesson

Simplifying a fraction feels like tidying up, and most of the time it is. But the fraction $\frac{x^2 - 4}{x - 2}$ has no value at $x = 2$ , while its tidier form $x + 2$ gives $4$ there. Calculus lives exactly in that gap. A limit asks what happens near $x = 2$ , and the honest answer depends on knowing that the original expression still has a hole at $x = 2$ after you simplify.

Key terms

Factor

Definition A polynomial $F$ is a factor of a polynomial $E$ when $E$ can be written as a product $E = F \cdot G$ for some polynomial $G$ . In a rational expression, the factors that matter are those of the whole numerator and of the whole denominator after each has been written as a single product.

In plain words A factor is something the whole expression is multiplied by. A term is something that is added. Only factors can later be cancelled.

Example Factoring gives

x^2 - 5x + 6 = (x - 2)(x - 3),

so $x - 2$ and $x - 3$ are factors. In $x + 3$ the $x$ is a term, not a factor, because $x + 3$ is a sum and not a product with $x$ as one piece.

Denominator and division by zero

Definition In a quotient $\frac{A}{B}$ the expression $B$ is the denominator. Division by zero is undefined, so $\frac{A}{B}$ has a value at a given input only when $A$ and $B$ both have values there and $B \ne 0$ . There is no number $c$ with $0 \cdot c = 5$ , and every number $c$ satisfies $0 \cdot c = 0$ , which is why neither $\frac{5}{0}$ nor $\frac{0}{0}$ is assigned a value.

In plain words A zero in the denominator is never allowed, whatever the numerator is.

Example The expression

\frac{x + 1}{x^2 - 4}

has denominator $(x - 2)(x + 2)$ , which is zero at $x = 2$ and at $x = -2$ . At $x = 3$ it has the value $\frac{4}{5}$ , but at $x = 2$ it has no value at all.

Excluded value

Definition An excluded value of an expression is a real number at which the expression is not defined. For a rational expression $\frac{P(x)}{Q(x)}$ with polynomials $P$ and $Q$ , the excluded values are exactly the real zeros of the denominator $Q$ , found before any simplification.

In plain words Before you change anything, find every input that makes the denominator zero and write those inputs down.

Example For

\frac{x^2 - 9}{x^2 - 3x}

the denominator factors as $x(x - 3)$ , so the excluded values are $x = 0$ and $x = 3$ . The numerator plays no part in this decision.

Cancellation

Definition If $C \ne 0$ and $B \ne 0$ , then

\frac{A \cdot C}{B \cdot C} = \frac{A}{B}.

Cancellation removes a factor $C$ that multiplies the entire numerator and the entire denominator, and it is valid only at inputs where $C \ne 0$ . A term that is merely added in the numerator or denominator can never be cancelled.

In plain words You may cross out the same factor above and below, because dividing by a nonzero number and multiplying by it again changes nothing. You may not cross out pieces of sums.

Example Factor fully first.

\begin{aligned} \frac{x^2 - 9}{x^2 - 3x} &= \frac{(x - 3)(x + 3)}{x(x - 3)} \\[0.4em] &= \frac{x + 3}{x}, \\[0.4em] &x \ne 0, \ x \ne 3. \end{aligned}

The common factor $x - 3$ cancels where it is not zero. Compare a tempting mistake. Crossing out the $x$ in $\frac{x + 6}{x + 2}$ would give $3$ , yet at $x = 1$ the expression equals $\frac{7}{3}$ . The $x$ there is a term, not a factor.

Equivalent expressions on the original domain

Definition The original domain of an expression is the set of real inputs at which the expression, exactly as first written, has a real value. Two expressions are equivalent on a set $D$ when they have the same value at every input in $D$ . After cancellation the simplified formula is equivalent to the original on the original domain, and the excluded values of the original remain excluded. Writing a simpler formula does not add the inputs that were excluded before.

In plain words The simple formula is a faster way to compute the same values, but only at the inputs the original allowed. The missing inputs stay missing.

Example Let

f(x) = \frac{x^2 - 4}{x - 2}, \qquad g(x) = x + 2.

For every $x \ne 2$ the two agree, since $x^2 - 4 = (x - 2)(x + 2)$ . Yet $g(2) = 4$ while $f(2)$ is undefined. So $f(x) = x + 2$ for $x \ne 2$ , and the graph of $f$ is the line $y = x + 2$ with an open circle at $(2, 4)$ .

$x$	$1.9$	$1.99$	$2$	$2.01$	$2.1$
$f(x)$	$3.9$	$3.99$	undefined	$4.01$	$4.1$

The values close in on $4$ from both sides, which is the kind of fact a limit records, but $f$ itself never takes the value $4$ .

Zero numerator

Definition A quotient $\frac{A}{B}$ equals $0$ exactly when $A = 0$ and $B \ne 0$ . If both $A = 0$ and $B = 0$ at some input, the quotient is undefined there.

In plain words A zero on top gives the answer zero, as long as the bottom is not also zero. A zero on the bottom always means no value.

Example Take

\frac{x^2 - x}{x^2 - 1} = \frac{x(x - 1)}{(x - 1)(x + 1)}.

At $x = 0$ the numerator is $0$ and the denominator is $-1$ , so the value is $0$ . At $x = 1$ both are $0$ , so the expression is undefined, even though the simplified form $\frac{x}{x + 1}$ would give $\frac{1}{2}$ . At $x = -1$ only the denominator is $0$ , so it is undefined there as well. The expression equals zero only at $x = 0$ .

Check yourself

Simplify the expression below and list its excluded values.

\frac{2x^2 + 6x}{x^2 - 9}

Show the answer

The denominator $(x - 3)(x + 3)$ gives the excluded values $3$ and $-3$ . Then

\begin{aligned} \frac{2x(x + 3)}{(x - 3)(x + 3)} &= \frac{2x}{x - 3}, \\[0.4em] &x \ne 3, \ x \ne -3. \end{aligned}

The value $-3$ stays excluded even though its factor cancelled.

Where is the expression below equal to zero, and where is it undefined?

\frac{x^2 - 4x}{x^2 - 16}

Show the answer

It factors as $\frac{x(x - 4)}{(x - 4)(x + 4)}$ . It is undefined at $x = 4$ and $x = -4$ . The numerator is zero at $x = 0$ and $x = 4$ , but at $x = 4$ the denominator is zero too. So the expression equals zero only at $x = 0$ .

A classmate simplifies $\frac{x + 4}{x}$ to $4$ by crossing out $x$ . Show with one input that this is wrong.

Show the answer

At $x = 2$ the expression equals $\frac{6}{2} = 3$ , not $4$ . The $x$ in the numerator is a term of a sum, so it is not a common factor and cannot be cancelled.

Cancellation and Excluded Values

Key terms

Factor

Denominator and division by zero

Excluded value

Cancellation

Equivalent expressions on the original domain

Zero numerator

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus