Limits by Algebra: Factor, Rationalize…

Part 1 · Hook

Why this matters

Substitute

x = 3

into

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

and you get

\frac{0}{0}

. It looks like a dead end, but it is the most common limit in calculus: every rate at an instant in Unit 2 starts out this way. Zero over zero is not an answer. It is a signal to do some algebra, and this topic shows exactly which algebra.

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 $x^2 - 9$ ?

$(x - 3)(x + 3)$
$(x - 3)^2$
$(x - 9)(x + 1)$
$x(x - 9)$

Show the answer

A difference of squares.

Correct: $(x - 3)(x + 3)$ :
$(x - 3)^2$ :
$(x - 9)(x + 1)$ :
$x(x - 9)$ :

2. When can $\lim_{x \to a} \frac{p(x)}{q(x)}$ (polynomials) be found by direct substitution?

When $q(a) \ne 0$
When $p(a) \ne 0$
Always
When $p(a) = q(a)$

Show the answer

The quotient law needs a nonzero denominator.

Correct: When $q(a) \ne 0$ :
When $p(a) \ne 0$ :
Always:
When $p(a) = q(a)$ :

3. What is $(\sqrt{x} - 2)(\sqrt{x} + 2)$ ?

$x - 4$
$x - 2$
$x + 4$
$x - 4\sqrt{x} + 4$

Show the answer

A conjugate pair: $(\sqrt{x})^2 - 4$ .

Correct: $x - 4$ :
$x - 2$ :
$x + 4$ :
$x - 4\sqrt{x} + 4$ :

Part 4 · See it

See it first

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. — $\frac{x^2 - 9}{x - 3}$ is the line $y = x + 3$ with one point missing. Factoring and canceling finds the height of the hole: the limit at $x = 3$ is 6. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Direct substitution gives $\frac{0}{0}$ an indeterminate form: the limit is not yet known
Top and bottom are both 0 at $x = a$ they share a factor $x - a$ (for polynomials), possibly hidden behind a square root or a fraction
Factor, rationalize or combine fractions, then cancel the shared factorthe new expression equals the old one for every $x \ne a$ , so it has the same limit
Substitute into the simplified expressionthe limit appears (or a nonzero over zero shows there is no finite limit)

Part 6 · Key ideas

Key ideas

$\frac{0}{0}$ is an indeterminate form: it means "rewrite first", never 0, 1 or "does not exist".
Canceling is legal in a limit because the limit never uses $x = a$ , and for $x \ne a$ the canceled factor is not 0.
Toolkit: factor and cancel; multiply by a conjugate to clear a square root; combine a complex fraction; expand powers completely.
After simplifying, a nonzero number over 0 means no finite limit (unbounded); check one-sided signs if asked.
Keep " $\lim_{x \to a}$ " on every line until you substitute; never write $\frac{0}{0} = \ldots$ .

Part 7 · Misconception

A common mistake

The wrong idea: Substitution gives $\frac{0}{0}$ , so the limit is 0 (or does not exist).

What actually happens: $\frac{0}{0}$ is indeterminate: it gives no information about the limit. Rewrite the expression; $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$ $= \lim_{x \to 3} (x + 3)$ $= 6$ .

Part 8 · Check yourself

Check yourself

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

Graph

The graph of $f(x)$ $= \frac{x^2 + x - 6}{x^2 - 4}$

The graph has a vertical asymptote at

x = -2

(dashed) and an open circle at

\left(2, \frac{5}{4}\right)

, where

f

is not defined.

1. Factor to find the exact value of $\lim_{x \to 2} f(x)$ .

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

Show the answer

$\lim_{x \to 2} \frac{(x + 3)(x - 2)}{(x + 2)(x - 2)}$ $= \lim_{x \to 2} \frac{x + 3}{x + 2}$ $= \frac{5}{4}$ . This matches the height of the hole on the graph.

Answer: $\frac{5}{4}$ ≈ 1.250

2. Substituting $x = 2$ into $f(x)$ gives $\frac{0}{0}$ . What does that tell you?

Not yet determined; rewrite $f$ and try again.
The limit is 0, because the numerator is 0.
The limit does not exist, because the denominator is 0.
The limit is 1, because the top and bottom are equal.

Show the answer

$\frac{0}{0}$ is an indeterminate form: the limit could be any number or might not exist. It is a signal to factor, rationalize or simplify. Here factoring gives $\frac{5}{4}$ .

Correct: Not yet determined; rewrite $f$ and try again.: Not yet determined; rewrite.
The limit is 0, because the numerator is 0.: A zero numerator alone settles nothing when the denominator also heads to 0.
The limit does not exist, because the denominator is 0.: A zero denominator alone settles nothing when the numerator also heads to 0.
The limit is 1, because the top and bottom are equal.: $\frac{0}{0}$ is not 1; zero over zero has no value.

3. What can you conclude about $\lim_{x \to -2} f(x)$ ?

No finite limit: the top heads to $-4$ , the bottom to 0.
It equals $\frac{1}{4}$ , found by factoring and then substituting $x = -2$ .
It equals 0, because $x = -2$ makes the denominator of $f$ equal to 0.
It equals $-\frac{1}{2}$ , the value of $\frac{x + 3}{x}$ at $x = -2$ .

Show the answer

At $x = -2$ the numerator is $4 - 2 - 6 = -4$ , not 0, while the denominator heads to 0. A nonzero number over a number near 0 is unbounded, as the vertical asymptote shows.

Correct: No finite limit: the top heads to $-4$ , the bottom to 0.: $-4$ over a quantity near 0 is unbounded.
It equals $\frac{1}{4}$ , found by factoring and then substituting $x = -2$ .: After canceling, $\frac{x + 3}{x + 2}$ still has a zero denominator at $-2$ .
It equals 0, because $x = -2$ makes the denominator of $f$ equal to 0.: A denominator near 0 makes the quotient large in size, not 0.
It equals $-\frac{1}{2}$ , the value of $\frac{x + 3}{x}$ at $x = -2$ .: The simplified form is $\frac{x + 3}{x + 2}$ , not $\frac{x + 3}{x}$ .

4. Which line of work is written correctly?

$\lim_{x \to 2} \frac{(x + 3)(x - 2)}{(x + 2)(x - 2)}$ $= \lim_{x \to 2} \frac{x + 3}{x + 2}$ $= \frac{5}{4}$
$\frac{x^2 + x - 6}{x^2 - 4}$ $= \frac{0}{0}$ $= \frac{5}{4}$
$\lim_{x \to 2} \frac{(x + 3)(x - 2)}{(x + 2)(x - 2)}$ $= \frac{(x + 3)}{(x + 2)}$ $= \frac{2 + 3}{2 + 2}$ $= \frac{5}{4}$
$\lim_{x \to 2} \frac{x^2 + x - 6}{x^2 - 4}$ $= \frac{x + 6}{4}$ $= 2$

Show the answer

Each "=" joins equal things, and "lim" stays on every line until the number appears.

Correct: $\lim_{x \to 2} \frac{(x + 3)(x - 2)}{(x + 2)(x - 2)}$ $= \lim_{x \to 2} \frac{x + 3}{x + 2}$ $= \frac{5}{4}$ : Every equals sign is true.
$\frac{x^2 + x - 6}{x^2 - 4}$ $= \frac{0}{0}$ $= \frac{5}{4}$ : $\frac{0}{0} = \frac{5}{4}$ is false: "=" links unequal things.
$\lim_{x \to 2} \frac{(x + 3)(x - 2)}{(x + 2)(x - 2)}$ $= \frac{(x + 3)}{(x + 2)}$ $= \frac{2 + 3}{2 + 2}$ $= \frac{5}{4}$ : The limit (a number) is set equal to an expression in $x$ in the middle step; keep "lim" until you substitute.
$\lim_{x \to 2} \frac{x^2 + x - 6}{x^2 - 4}$ $= \frac{x + 6}{4}$ $= 2$ : The $x^2$ terms were canceled; only factors can cancel.

5. Find $\lim_{x \to -3} \frac{x^2 + 5x + 6}{x + 3}$ .

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

Show the answer

$\lim_{x \to -3} \frac{(x + 2)(x + 3)}{x + 3}$ $= \lim_{x \to -3} (x + 2)$ $= -1$ .

Answer: -1

6. Find $\lim_{h \to 0} \frac{(4 + h)^2 - 16}{h}$ .

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

Show the answer

$\frac{16 + 8h + h^2 - 16}{h}$ $= \frac{h(8 + h)}{h}$ $= 8$ $+ h$ for $h \ne 0$ , so the limit is 8.

Answer: 8

7. What is $\lim_{x \to 3} \frac{|x - 3|}{x - 3}$ ?

It does not exist: $-1$ from the left, $1$ from the right.
$1$ , because near $x = 3$ the top and bottom have equal sizes, so they cancel.
$0$ , because the top becomes 0.
$-1$ , because $x - 3$ is negative.

Show the answer

For $x > 3$ , $|x - 3| = x - 3$ , so the quotient is 1. For $x < 3$ , $|x - 3| = -(x - 3)$ , so the quotient is $-1$ . The one-sided limits differ.

Correct: It does not exist: $-1$ from the left, $1$ from the right.: Left $-1$ , right $1$ .
$1$ , because near $x = 3$ the top and bottom have equal sizes, so they cancel.: Only for $x > 3$ ; to the left the quotient is $-1$ .
$0$ , because the top becomes 0.: The bottom also heads to 0.
$-1$ , because $x - 3$ is negative.: Only for $x < 3$ .

Part 9 · Summary

Summary

When direct substitution gives the indeterminate form

\frac{0}{0}

, rewrite: factor and cancel, multiply by a conjugate, or simplify a complex fraction. The simplified expression matches the original except at

x = a

, so it has the same limit, which substitution now finds. Write every step with "lim" and true equals signs.

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

The graph of f(x)f(x)=x2+x−6x2−4= \frac{x^2 + x - 6}{x^2 - 4}

Summary

What comes next

Connections

The graph of $f(x)$ $= \frac{x^2 + x - 6}{x^2 - 4}$