Limits by Algebra: Factor… — AP® Calculus AB Notes

Try direct substitution on $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$ and you get $\frac{0}{0}$ . This is not a broken problem. It is the most important kind of limit in the course, because every rate at an instant in Unit 2 begins exactly this way: a change in output over a change in input, both shrinking to 0. This page shows what $\frac{0}{0}$ means and the algebra that gets past it.

What $\frac{0}{0}$ means

When substitution makes both the numerator and the denominator 0, the result $\frac{0}{0}$ is called an indeterminate form. "Indeterminate" means the limit is not determined yet. Depending on the function, the true limit could be 6, or $-\frac{1}{16}$ , or it might not exist. These three limits all start as $\frac{0}{0}$ :

Limit	Substitution gives	Actual limit
$\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$	$\frac{0}{0}$	6
$\lim_{x \to 0} \frac{x^2}{x}$	$\frac{0}{0}$	0
$\lim_{x \to 0} \frac{x}{x^2}$	$\frac{0}{0}$	does not exist

So $\frac{0}{0}$ is never an answer: not 0, not 1, not "does not exist." It is an instruction: rewrite the expression and try again.

Compare a different result: substitution gives a nonzero number over 0, such as $\frac{-4}{0}$ . That one is not indeterminate. The outputs are unbounded, and there is no finite limit.

Why canceling is allowed

The graph of y = (x squared minus 9) over (x minus 3) is the line y = x + 3 with an open circle at (3, 6). A panel shows the algebra: factor x squared minus 9 as (x - 3)(x + 3), cancel x - 3 for x not equal to 3, and get x + 3, so the limit at x = 3 is 6. — **Figure 1.** $\frac{x^2 - 9}{x - 3}$ and $x + 3$ have the same graph, except that the first has a hole at $x = 3$ . LevlPrep original diagram.

For every $x \ne 3$ , $\frac{x^2 - 9}{x - 3}$ $= \frac{(x - 3)(x + 3)}{x - 3}$ $= x$ $+ 3$ , because the factor $x - 3$ is not zero and can be divided out. The two functions differ only at $x = 3$ , where the first is undefined. A limit never looks at $x = 3$ itself, so the two functions have the same limit there: $\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} (x + 3) = 6$ That is the whole strategy: rewrite into a function that agrees everywhere except at $a$ , then substitute.

Strategy 1: factor and cancel

If a polynomial is 0 at $x = a$ , it has a factor $x - a$ . So when both top and bottom are 0 at $a$ , they share that factor.

Worked example 1. Find $\lim_{x \to 2} \frac{x^2 + x - 6}{x^2 - 4}$ .

Substitution: $\frac{4 + 2 - 6}{4 - 4}$ $= \frac{0}{0}$ . Indeterminate, so factor.

$\lim_{x \to 2} \frac{x^2 + x - 6}{x^2 - 4} = \lim_{x \to 2} \frac{(x + 3)(x - 2)}{(x + 2)(x - 2)} = \lim_{x \to 2} \frac{x + 3}{x + 2} = \frac{5}{4}$

Worked example 2. Find $\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$ .

Difference of cubes: $x^3 - 8$ $= (x - 2)(x^2 + 2x + 4)$ .

$\lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{x - 2} = \lim_{x \to 2} (x^2 + 2x + 4) = 4 + 4 + 4 = 12$

Strategy 2: rationalize

When a square root is involved, multiply the top and bottom by the conjugate. The square root disappears and a common factor appears.

Worked example 3. Find $\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}$ .

Substitution gives $\frac{0}{0}$ . Multiply by $\frac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3}$ :

$\lim_{x \to 0} \frac{(x + 9) - 9}{x(\sqrt{x + 9} + 3)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x + 9} + 3)} = \lim_{x \to 0} \frac{1}{\sqrt{x + 9} + 3} = \frac{1}{6}$

Leave the denominator in factored form; multiplying it out hides the $x$ you need to cancel.

Strategy 3: combine fractions, or expand

Worked example 4. Find $\lim_{h \to 0} \frac{\frac{1}{2 + h} - \frac{1}{2}}{h}$ .

Combine the top over $2(2 + h)$ : $\frac{2 - (2 + h)}{2(2 + h)}$ $= \frac{-h}{2(2 + h)}$ .

$\lim_{h \to 0} \frac{-h}{2(2 + h)} \cdot \frac{1}{h} = \lim_{h \to 0} \frac{-1}{2(2 + h)} = -\frac{1}{4}$

Worked example 5. Find $\lim_{h \to 0} \frac{(3 + h)^2 - 9}{h}$ .

Expand: $(3 + h)^2 - 9$ $= 6h + h^2$ $= h(6 + h)$ .

$\lim_{h \to 0} \frac{h(6 + h)}{h} = \lim_{h \to 0} (6 + h) = 6$

This is the same calculation as the falling stone in Topic 1.1, now in limit notation.

When the simplified form still has a zero denominator

After canceling, substitute again. If the denominator is still 0 but the numerator is not, there is no finite limit. For example, $\frac{x^2 - 25}{x^2 - 10x + 25}$ $= \frac{(x - 5)(x + 5)}{(x - 5)^2}$ $= \frac{x + 5}{x - 5}$ , and at $x = 5$ that is $\frac{10}{0}$ : unbounded, so the limit does not exist.

Absolute values call for one-sided limits. $\frac{|x - 3|}{x - 3}$ equals 1 for $x > 3$ and $-1$ for $x < 3$ , so $\lim_{x \to 3} \frac{|x - 3|}{x - 3}$ does not exist.

Writing the work

Loses credit	Earns credit
$\frac{x^2 - 9}{x - 3}$ $= \frac{0}{0}$ $= 6$	$\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$ $= \lim_{x \to 3} (x + 3)$ $= 6$
$\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$ $= x$ $+ 3$ $= 6$	keep " $\lim_{x \to 3}$ " until the line where you substitute

The equals sign may only join things that are equal. A limit is a number, $x + 3$ is an expression, and $\frac{0}{0}$ is not a number at all, so none of them can be chained with the others.

What 00\frac{0}{0} means