Algebra for Limits: Factoring and… — AP® Calculus AB Notes

Here is the kind of limit you will meet again and again: $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$ . Substitute $x = 3$ and both the top and the bottom are 0. The limit laws cannot help, because the denominator heads to 0. The way forward is to rewrite the fraction into a form where the trouble disappears. That takes three precalculus moves, and this page drills each one: factoring, rationalizing and simplifying complex fractions.

Factoring

To factor is to rewrite a sum as a product. A factored form shows what each piece is multiplied by, and that is what lets pieces cancel. The patterns you need:

Pattern	Example
Common factor	$x^2 - 4x = x(x - 4)$
Difference of squares: $a^2 - b^2 = (a - b)(a + b)$	$x^2 - 16 = (x - 4)(x + 4)$
Trinomial $x^2 + bx + c$ : two numbers that multiply to $c$ and add to $b$	$x^2 + 2x - 8$ $= (x + 4)(x - 2)$
Difference of cubes: $a^3 - b^3$ $= (a - b)(a^2 + ab + b^2)$	$x^3 - 8$ $= (x - 2)(x^2 + 2x + 4)$
Sum of cubes: $a^3 + b^3$ $= (a + b)(a^2 - ab + b^2)$	$x^3 + 1$ $= (x + 1)(x^2 - x + 1)$

A useful shortcut: if a polynomial is 0 at $x = a$ , then $x - a$ is one of its factors. So when substitution gives 0 on top at $x = 3$ , you already know to look for the factor $x - 3$ .

Worked example 1. Factor $2x^2 + 5x - 3$ . The leading coefficient is not 1, which is where most errors happen.

Multiply $a \cdot c = 2 \cdot (-3) = -6$ . Find two numbers that multiply to $-6$ and add to $b = 5$ : they are 6 and $-1$ .

Split the middle term and group: $2x^2$ $+ 6x - x - 3$ $= 2x(x + 3) - 1(x + 3)$ $= (2x - 1)(x + 3)$ .

Check by expanding: $(2x - 1)(x + 3)$ $= 2x^2$ $+ 6x - x - 3$ $= 2x^2$ $+ 5x - 3$ .

Cancel factors, never terms

Canceling divides the top and bottom by the same nonzero factor. A term that is added cannot be canceled. This is the single most common algebra error on limit questions.

Wrong	Right
$\frac{x^2 - 4x}{x^2 - 16}$ $= \frac{-4x}{-16}$	$\frac{x(x - 4)}{(x - 4)(x + 4)}$ $= \frac{x}{x + 4}$
$\frac{x^2 + 9}{x + 3} = x + 3$	does not simplify: $x^2 + 9$ has no real factors
$\frac{x + 6}{x + 3} = 2$	does not simplify

When in doubt, test a number. At $x = 1$ , $\frac{x^2 - 4x}{x^2 - 16}$ $= \frac{-3}{-15}$ $= \frac{1}{5}$ . The wrong answer $\frac{x}{4}$ gives $\frac{1}{4}$ ; the right answer $\frac{x}{x + 4}$ gives $\frac{1}{5}$ .

Rationalizing with a conjugate

When a square root sits next to a number, as in $\sqrt{x} - 2$ , factoring does not apply directly. Instead multiply by its conjugate, the same expression with the opposite sign in the middle: $\sqrt{x} + 2$ . A conjugate pair multiplies to a difference of squares, and the square root disappears: $(\sqrt{x} - 2)(\sqrt{x} + 2) = (\sqrt{x})^2 - 2^2 = x - 4$ To keep the fraction's value, multiply the top and the bottom by the conjugate. This is called rationalizing.

Worked example 2. Simplify $\frac{\sqrt{x} - 2}{x - 4}$ for $x \ne 4$ .

$\frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \frac{x - 4}{(x - 4)(\sqrt{x} + 2)} = \frac{1}{\sqrt{x} + 2}$

Notice the bottom was not multiplied out. Leaving it as $(x - 4)(\sqrt{x} + 2)$ is what makes the factor $x - 4$ visible for canceling.

Near $x = 4$ , the simplified form heads to $\frac{1}{\sqrt{4} + 2} = \frac{1}{4}$ .

The graph of y = (square root of x minus 2) divided by (x minus 4) for x from 0 to 12: it starts at (0, 0.5), falls slowly, and has an open circle at (4, 1/4). A panel shows the algebra: multiplying by the conjugate gives 1 over (square root of x plus 2), which is 1/4 at x = 4. — **Figure 1.** The original fraction and its simplified form share a graph, except for the hole at $x = 4$ . LevlPrep original diagram.

Complex fractions

A complex fraction has fractions inside its numerator or denominator. Combine the small fractions over a common denominator first, then divide.

Worked example 3. Simplify $\frac{\frac{1}{x + 2} - \frac{1}{2}}{x}$ for $x \ne 0, -2$ .

Top, over the common denominator $2(x + 2)$ : $\frac{2 - (x + 2)}{2(x + 2)}$ $= \frac{-x}{2(x + 2)}$ . The parentheses around $x + 2$ matter: the minus sign applies to both terms.

Divide by $x$ , which is the same as multiplying by $\frac{1}{x}$ : $\frac{-x}{2(x + 2)} \cdot \frac{1}{x}$ $= \frac{-1}{2(x + 2)}$ .

Near $x = 0$ , this heads to $\frac{-1}{4}$ .

Expanding powers

Limits often contain expressions like $(3 + h)^2 - 9$ . Expand every power completely before simplifying: $(3 + h)^2 - 9 = 9 + 6h + h^2 - 9 = 6h + h^2 = h(6 + h)$ Forgetting the middle term, $6h$ , is the classic slip; $(a + h)^2$ is never $a^2 + h^2$ . For a cube, $(a + h)^3$ $= a^3 + 3a^2h + 3ah^2 + h^3$ .

In every one of these moves, the goal is the same: produce a factor that matches the troublesome one in the denominator, so it can cancel. The next topic puts these moves to work on limits.