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

Part 1 · Hook

Why this matters

Type

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

into a calculator at

x = 3

and it shows an error. Yet the values near 3 settle down on 6. The error is a message, not an answer: it tells you which tool to pick next. This lesson turns the result of one substitution into a plan.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Which is the factored form of $x^2 - 5x + 6$ ?

$(x - 2)(x - 3)$
$(x + 2)(x - 3)$
$(x - 1)(x - 6)$
$(x + 2)(x + 3)$

Show the answer

$(x - 2)(x - 3)$ $= x^2 - 5x + 6$ : the numbers multiply to 6 and add to $-5$ .

Correct: $(x - 2)(x - 3)$ :
$(x + 2)(x - 3)$ :
$(x - 1)(x - 6)$ :
$(x + 2)(x + 3)$ :

2. If $\lim_{x \to 2^-} f(x) = 5$ and $\lim_{x \to 2^+} f(x) = 1$ , then $\lim_{x \to 2} f(x)$

does not exist
equals 3
equals 5
equals 1

Show the answer

A two-sided limit exists only when the one-sided limits are equal.

Correct: does not exist:
equals 3:
equals 5:
equals 1:

3. For a polynomial $P$ , $\lim_{x \to c} P(x)$ is

$P(c)$
$0$
$P(0)$
not known without a graph

Show the answer

The limit laws give direct substitution for polynomials.

Correct: $P(c)$ :
$0$ :
$P(0)$ :
not known without a graph:

Part 4 · See it

See it first

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. — Substitute first; the result picks the next step. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

You substitute $x = c$ and get a numberthat number is the limit, as long as the expression is defined near $c$
You get $\frac{0}{0}$ the numerator and denominator share a hidden zero factor, so rewrite (factor, conjugate, common denominator) and substitute again
You get a nonzero number over 0the denominator shrinks while the numerator does not, so the function is unbounded near $c$ and the limit does not exist
The function changes formula at $x = c$ (piecewise or absolute value)find the left-hand and right-hand limits separately; equal means the limit exists

Part 6 · Key ideas

Key ideas

Substitute first. It is the fastest test, and its result tells you what to do next.
$\frac{0}{0}$ is an indeterminate form: not an answer, a signal to rewrite.
A nonzero number over 0 means the function is unbounded near $c$ ; no algebra will make the limit a number.
At a break point, the two sides can follow different rules. Find both one-sided limits.
Write the limit symbol on every line until you substitute: the simplified expression equals the original only for $x \ne c$ .

Part 7 · Misconception

A common mistake

The wrong idea: Substituting gives $\frac{0}{0}$ , so the limit does not exist.

What actually happens: $\frac{0}{0}$ is indeterminate: the limit may be any number or may not exist. Rewrite the expression and substitute again. For $\frac{x^2 - 9}{x - 3}$ the limit is 6.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Four functions near $x = 3$

Four functions, each to be studied as x approaches 3
Function	Formula
$f$	$f(x)$ $= \frac{x^2 - 5x + 6}{x - 3}$
$g$	$g(x)$ $= \frac{\sqrt{x + 1} - 2}{x - 3}$
$h$	$h(x) = \frac{x + 3}{x - 3}$
$k$	$k(x) = \frac{x^2 + 1}{x + 3}$

1. Substituting $x = 3$ into which function gives a nonzero number divided by $0$ ?

$h$
$f$
$g$
$k$

Show the answer

At $x = 3$ , $h$ gives $\frac{6}{0}$ : a nonzero numerator over a zero denominator.

Correct: $h$ : $h(3)$ would be $\frac{3 + 3}{3 - 3} = \frac{6}{0}$ , so $h$ is unbounded near 3.
$f$ : $f$ gives $\frac{9 - 15 + 6}{0} = \frac{0}{0}$ , an indeterminate form, not a nonzero number over 0.
$g$ : $g$ gives $\frac{\sqrt{4} - 2}{0} = \frac{0}{0}$ , an indeterminate form.
$k$ : $k$ gives $\frac{10}{6}$ ; its denominator is not 0 at $x = 3$ .

2. Find $\lim_{x \to 3} f(x)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Substitution gives $\frac{0}{0}$ , so factor: $\frac{(x - 2)(x - 3)}{x - 3}$ $= x - 2$ for $x \ne 3$ . Then $\lim_{x \to 3} (x - 2)$ $= 1$ .

Answer: 1

3. What can you conclude about $\lim_{x \to 3} h(x)$ ?

It equals $6$ , the value of the numerator at $x = 3$
It does not exist; $h$ is unbounded near $x = 3$
It equals $1$ , after the $x - 3$ terms cancel
It can be found by factoring and cancelling first

Show the answer

A nonzero number over 0 means the denominator shrinks while the numerator stays near 6, so $|h(x)|$ grows without bound. Left of 3, $h$ is very negative; right of 3, very positive. The limit does not exist.

It equals $6$ , the value of the numerator at $x = 3$ : The numerator's value alone ignores the denominator, which is heading to 0.
Correct: It does not exist; $h$ is unbounded near $x = 3$ : Near 3 the numerator is about 6 and the denominator is tiny, so $h$ is unbounded and the limit does not exist.
It equals $1$ , after the $x - 3$ terms cancel: Nothing cancels: $x + 3$ and $x - 3$ are different factors.
It can be found by factoring and cancelling first: Factoring helps with $\frac{0}{0}$ . Here the numerator is not 0 at $x = 3$ , so there is no common factor to cancel.

4. Let $q(x) = \frac{|x - 5|}{x - 5}$ . Which is true about $\lim_{x \to 5} q(x)$ ?

It equals $1$ , because $|x - 5|$ is not negative for any $x$
It equals $0$ , because the numerator approaches $0$
It does not exist, because $q$ grows without bound near $x = 5$
It does not exist; $q$ is $-1$ on the left, $1$ on the right

Show the answer

For $x > 5$ , $|x - 5| = x - 5$ and $q(x) = 1$ . For $x < 5$ , $|x - 5| = -(x - 5)$ and $q(x) = -1$ . The one-sided limits, $-1$ and $1$ , differ, so the limit does not exist.

It equals $1$ , because $|x - 5|$ is not negative for any $x$ : The numerator is not negative, but the denominator $x - 5$ is negative left of 5, which makes $q = -1$ there.
It equals $0$ , because the numerator approaches $0$ : Both numerator and denominator approach 0; that is $\frac{0}{0}$ , which needs more work, not the answer 0.
It does not exist, because $q$ grows without bound near $x = 5$ : The limit does fail to exist, but not by growing: $q$ only takes the values $-1$ and $1$ .
Correct: It does not exist; $q$ is $-1$ on the left, $1$ on the right: Split at the break: $q = -1$ for $x < 5$ and $q = 1$ for $x > 5$ . Unequal one-sided limits, so no limit.

5. A student writes: " $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$ : substituting gives $\frac{0}{0}$ , so the limit does not exist." Which correction is right?

Rewrite: $\frac{x^2 - 1}{x - 1} = x + 1$ , so the limit is 2
$\frac{0}{0}$ equals 1, so the limit of the quotient is 1
$\frac{0}{0}$ equals 0, so the limit of the quotient is 0
It does not exist, but because of a jump at $x = 1$

Show the answer

$\frac{0}{0}$ is an indeterminate form: it tells you nothing yet. Here $\frac{(x - 1)(x + 1)}{x - 1}$ $= x + 1$ for $x \ne 1$ , so the limit is 2.

Correct: Rewrite: $\frac{x^2 - 1}{x - 1} = x + 1$ , so the limit is 2: Factor and cancel: $x + 1 \to 2$ . Writing " $\frac{0}{0} = 2$ " would be wrong too; the 2 comes from the rewritten expression.
$\frac{0}{0}$ equals 1, so the limit of the quotient is 1: $\frac{0}{0}$ is not a number. Treating it as 1 skips the rewriting step.
$\frac{0}{0}$ equals 0, so the limit of the quotient is 0: $\frac{0}{0}$ is not 0. A zero numerator only gives 0 when the denominator is not also 0.
It does not exist, but because of a jump at $x = 1$ : The expression equals $x + 1$ on both sides of 1, so there is no jump and the limit exists.

6. Find $\lim_{x \to -1} \frac{x^2 + 3x + 2}{x^2 - 1}$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Both parts are 0 at $x = -1$ . Factor: $\frac{(x + 1)(x + 2)}{(x + 1)(x - 1)}$ $= \frac{x + 2}{x - 1}$ for $x \ne -1$ . Substitute: $\frac{1}{-2} = -\frac{1}{2}$ .

Answer: $\frac{-1}{2}$ ≈ -0.500

Part 9 · Summary

Summary

Substitute first and read the result. A number is the limit.

\frac{0}{0}

means rewrite: factor and cancel, multiply by a conjugate, or combine fractions, then substitute again. A nonzero number over 0 means the function is unbounded there. At a break point, compare the one-sided limits.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Four functions near x=3x = 3

Summary

What comes next

Connections

Four functions near $x = 3$