Choosing a Way to Find a Limit — AP® Calculus AB Notes

By now you have several tools for limits: reading a graph, building a table, the limit laws, factoring and rationalizing. A real question does not say which one to use. This page gives you a plan. One substitution, read carefully, tells you which tool comes next.

Substitute first

Start every limit $\lim_{x \to c} f(x)$ by putting $x = c$ into the expression. This is fast, and it is often the whole job. The limit laws say that for a polynomial, or a quotient of polynomials whose denominator is not 0 at $c$ , the limit equals the value: $\lim_{x \to c} f(x)$ $= f(c)$ . The same holds for roots, as long as the expression is defined on both sides of $c$ .

Worked example 1. Find $\lim_{x \to 2} \frac{x^2 + 1}{x + 3}$ .

Substitute: the denominator is $2 + 3 = 5$ , not 0. So the limit is the value.

$\lim_{x \to 2} \frac{x^2 + 1}{x + 3} = \frac{4 + 1}{5} = 1$

Read the result

Substitution can come out three ways. Each one sends you somewhere different.

A decision map for limits. Start by substituting x = c. A number means that is the limit. Zero over zero means rewrite by factoring, using a conjugate or combining fractions, then substitute again. A nonzero number over zero means the function is unbounded near c, so check the sign on each side. At a break point of a piecewise or absolute value function, find both one-sided limits and compare them. — **Figure 1.** A decision map: the result of substituting picks the next step. LevlPrep original diagram.

Substituting gives	What it means	What to do
a number, like $7$	the expression is defined near $c$	that number is the limit
$\frac{0}{0}$	top and bottom share a zero factor	rewrite, then substitute again
nonzero over 0, like $\frac{6}{0}$	the bottom shrinks, the top does not	the function is unbounded; check each side's sign

One more case sits beside the table. If the function changes formula at $c$ (a piecewise function, or an absolute value), the two sides follow different rules. Then find the left-hand and right-hand limits separately and compare them.

Zero over zero: rewrite

$\frac{0}{0}$ is called an indeterminate form. It is not a number and it is not an answer. It says the numerator and the denominator both reach 0 at $c$ , which usually means they share a factor that is 0 there. Remove that factor and the rest of the expression tells you the limit. Three moves cover almost every case.

Factor and cancel when you see polynomials.

Worked example 2. Find $\lim_{x \to 3} \frac{x^2 - x - 6}{x - 3}$ .

Substitute: $\frac{9 - 3 - 6}{0} = \frac{0}{0}$ . Rewrite.

Factor the top: $x^2 - x - 6$ $= (x - 3)(x + 2)$ . For every $x \ne 3$ ,

$\frac{(x - 3)(x + 2)}{x - 3} = x + 2$

A limit never looks at $x = 3$ itself, so the two expressions have the same limit: $\lim_{x \to 3} (x + 2)$ $= 5$ .

Multiply by the conjugate when a square root sits next to a subtraction.

Worked example 3. Find $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$ .

Substitute: $\frac{2 - 2}{0} = \frac{0}{0}$ . Multiply top and bottom by the conjugate $\sqrt{x + 4} + 2$ :

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

Now substitute: $\frac{1}{2 + 2} = \frac{1}{4}$ .

Combine fractions when small fractions sit inside a big one.

Worked example 4. Find $\lim_{x \to 3} \frac{\frac{1}{x} - \frac{1}{3}}{x - 3}$ .

Substitute: $\frac{0}{0}$ . Put the top over one denominator: $\frac{1}{x} - \frac{1}{3} = \frac{3 - x}{3x}$ . Then

$\frac{3 - x}{3x(x - 3)} = \frac{-(x - 3)}{3x(x - 3)} = -\frac{1}{3x} \quad (x \ne 3)$

Substitute: $-\frac{1}{9}$ .

Keep the limit symbol on every line until the final substitution. The rewritten expression equals the original only for $x \ne c$ , and the limit symbol is what makes that enough. Never write $\frac{0}{0} = 5$ : the 5 comes from the rewritten expression, not from $\frac{0}{0}$ .

Nonzero over zero: unbounded

Suppose substitution gives $\frac{6}{0}$ . Near $c$ the numerator stays close to 6 while the denominator gets as small as you like. Dividing 6 by a tiny number gives a huge number. So the function is unbounded near $c$ , and the limit does not exist. No rewriting changes this, because there is no common zero factor to cancel.

The sign on each side tells the shape. For $h(x) = \frac{x + 3}{x - 3}$ near 3: just left of 3, $x - 3$ is a tiny negative number, so $h$ is a huge negative number. Just right of 3, $h$ is a huge positive number. The graph has a vertical asymptote at $x = 3$ . A later topic gives this behavior its own notation.

Break points: one side at a time

For a piecewise function, use the formula that applies on each side. For an absolute value, remember $|u| = u$ when $u \ge 0$ and $|u| = -u$ when $u < 0$ , so it is a piecewise function in disguise.

Worked example 5. Let $p(x) = 2x + 1$ for $x < 2$ and $p(x) = x^2 + 1$ for $x \ge 2$ . Find $\lim_{x \to 2} p(x)$ .

From the left, use $2x + 1$ : $\lim_{x \to 2^-} p(x) = 5$ . From the right, use $x^2 + 1$ : $\lim_{x \to 2^+} p(x) = 5$ .

The one-sided limits are equal, so $\lim_{x \to 2} p(x) = 5$ . Had they differed, the limit would not exist.

A checklist

Substitute $x = c$ .
A number? That is the limit.
$\frac{0}{0}$ ? Factor, use a conjugate, or combine fractions. Then substitute again.
Nonzero over 0? The function is unbounded near $c$ ; the limit does not exist. Check the sign on each side.
A formula change at $c$ ? Find both one-sided limits and compare.
Still stuck, or no formula at all? Use a graph or a table of values.