Topic 1.10

Rational Functions and Holes

6 MIN READ7 IDEAS23 PROBLEMS14 flashcards

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01
Compare the power of a shared factor in the numerator vs. the denominator.

Numerator's power $\ge$ denominator's power → the denominator's copies all cancel → hole.

Denominator's power higher → a copy remains underneath → vertical asymptote.

Careful with the first case: even when the numerator has more copies, the point is still a hole, never a zero. The original denominator is $0$ there, so the function is undefined at that input — and an undefined input cannot be a zero.

Practice 23 problems Next: Chapter 1.11A Equivalent Representations & Binomial Theorem

Recap: What Is a Hole?

A hole happens when a factor cancels completely between the numerator and denominator. The function is undefined at that single x-value, but everywhere else nearby, it behaves exactly like the simplified version.

The Limit Notation for a Hole

In 1.9, you saw that a vertical asymptote has one-sided limits that go to $\pm \infty$ . A hole is completely different — the two-sided limit actually EXISTS and is a finite number — it's just that the function itself is undefined there.

CONCEPT

Comparing a Hole to a Vertical Asymptote

At a vertical asymptote: $lim(x \to a)$ $f(x) = \pm \infty$ (the limit is infinite, or the two sides don't even match).

At a hole: $lim(x \to a)$ $f(x) = L$ , some ordinary finite number — even though $f(a)$ itself is undefined.

Quick check

$f(x) = \frac{x - 5}{x^{2} - 25}$ . What is $\lim_{x \to 5} f(x)$ , and is $f(5)$ defined?

Step-by-Step: Finding a Hole with Limit Notation

Worked example

For $v(x) = (x^{2} + x - 6) / (x - 2)$ , find the domain, zero, and hole — and express the hole using limit notation.

v(x) = \frac{x^2 + x - 6}{x - 2}

01
Factor the numerator:
$x^2 + x - 6 = (x - 2)(x + 3)$
02
The factor $(x-2)$ cancels completely:
$v(x) = \frac{(x - 2)(x + 3)}{x - 2} = x + 3 \quad (x \neq 2)$
03
Domain: all reals except $x=2$ . Zero (from the simplified form): $x=-3$ .
04
Express the hole using limit notation — plug $x=2$ into the SIMPLIFIED expression:
$\lim_{x \to 2} v(x) = 2 + 3 = 5$
05
The hole is at $(2, 5)$ . Even though $v(2)$ is undefined, the LIMIT as $x$ approaches 2 is a perfectly ordinary number: 5.

The two-sided limit exists and equals 5 — that's exactly where the open circle sits.

A Trickier Case: A Squared Factor That Fully Cancels

In 1.9, you saw a case where a repeated factor only PARTIALLY canceled, leaving a genuine vertical asymptote. Here's the contrast: what if the SAME power appears in both the numerator and denominator?

Worked example

For $z(x) = [(x+7)(x-3)^{2}] / [(x-3)^{2}(x-5)]$ , find the domain, hole, and vertical asymptote.

z(x) = \frac{(x + 7)(x - 3)^2}{(x - 3)^2 (x - 5)}

01
The factor $(x-3)^{2}$ appears with the EXACT SAME power in both the numerator and denominator — it cancels completely, leaving nothing behind:
$z(x) = \frac{x + 7}{x - 5} \quad (x \neq 3)$
02
Since nothing is left over from $(x-3)$ , $x=3$ is a HOLE, not a vertical asymptote. Domain: all reals except $x=3$ and $x=5$ .
03
Find the hole's location using limit notation:
$\lim_{x \to 3} z(x) = \frac{3 + 7}{3 - 5} = \frac{10}{-2} = -5$
04
Hole at $(3, -5)$ . The remaining factor $(x-5)$ in the denominator gives a genuine vertical asymptote at $x=5$ .

A squared factor CAN still produce a hole — the deciding factor is whether it fully cancels, not whether it's squared.

KEY RULE

Compare the POWER of a shared factor in the numerator vs. the denominator.

Numerator's power $\ge$ denominator's power → the denominator's copies all cancel → HOLE.

Denominator's power higher → a copy remains underneath → VERTICAL ASYMPTOTE.

Careful with the first case: even when the numerator has MORE copies, the point is still a hole, never a zero. The ORIGINAL denominator is $0$ there, so the function is undefined at that input — and an undefined input cannot be a zero.

Quick check

$q(x) = \frac{(x - 4)^{3}(x + 1)}{(x - 4)^{2}(x + 3)}$ . Hole or asymptote at $x = 4$ , and at what height?

Domain Notation with Holes and Asymptotes Together

When writing domain in interval notation, both holes AND vertical asymptotes get excluded — they just look identical in interval notation, even though they mean different things graphically.

CONCEPT

Example: Writing the Domain for $z(x)$ Above

$z(x)$ has a hole at $x=3$ and a vertical asymptote at $x=5$ .

Domain in interval notation: $(-\infty, 3)$ ∪ $(3, 5)$ ∪ $(5, \infty)$

Notice the interval notation alone can't tell you which excluded point is a hole and which is an asymptote — you have to say that separately.

COMMON MISTAKE

A hole is NOT visible in interval notation any differently than a vertical asymptote — always state which is which separately, in words.

Don't forget to exclude the hole's x-value from the domain, even though the function 'looks fine' there on a simplified expression — the ORIGINAL function was undefined there.

Practice Problems

Worked example

Problem 1. For $w(x) = (x^{2} - 2x - 8) / (x + 2)$ , find the hole using limit notation.

01
Factor: $x^{2}-2x-8 = (x+2)(x-4)$ . The factor $(x+2)$ cancels, leaving $w(x) = x-4$ (x≠−2).
02
$lim(x \to -2)$ $w(x) = -2-4 = -6$ . Hole at $(-2, -6)$ .

Worked example

Problem 2. Does $m(x) = (x-1)^{2}(x+3) / [(x-1)(x+3)^{2}]$ have a hole or an asymptote at $x=1$ ? At $x=-3$ ?

01
At $x=1$ : numerator has power $2$ , denominator has power $1$ . The denominator's single copy cancels and one copy is left over in the NUMERATOR.

That leftover makes the SIMPLIFIED expression equal $0$ at $x=1$ — but the ORIGINAL denominator is $0$ there too, so $m(1)$ is undefined. An undefined input is not a zero.

So $x = 1$ is a HOLE, and it sits at $(1, 0)$ — on the $x$ -axis, which is what makes it easy to mistake for a zero.
02
At $x=-3$ : numerator has power 1, denominator has power 2 — the denominator's power is HIGHER, so one copy remains in the denominator after canceling. This makes $x=-3$ a VERTICAL ASYMPTOTE.

Key Takeaways

A hole occurs where a factor cancels completely — the two-sided limit exists ( $lim(x \to a) f(x) = L$ ), even though $f(a)$ is undefined.
Contrast with a vertical asymptote: there, the limit is infinite ( $\pm \infty$ ), not a finite number.
Compare powers of a shared factor: numerator's power ≥ denominator's → HOLE; denominator higher → vertical asymptote.
A hole can sit ON the $x$ -axis. If the numerator had more copies, the hole lands at $(a, 0)$ — it still is not a zero, because the original function is undefined there.
In interval notation, holes and vertical asymptotes both get excluded — but they aren't visually distinguishable in that notation, so always clarify separately.

Common slips

A hole is not visible in interval notation any differently than a vertical asymptote — always state which is which separately, in words.

Don't forget to exclude the hole's x-value from the domain, even though the function 'looks fine' there on a simplified expression — the original function was undefined there.

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14 cards · Rational functions, Hole or asymptote?

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Recap card

One line to re-read the night before.

01
Compare the power of a shared factor in the numerator vs. the denominator.

Numerator's power $\ge$ denominator's power → the denominator's copies all cancel → hole.

Denominator's power higher → a copy remains underneath → vertical asymptote.

Careful with the first case: even when the numerator has more copies, the point is still a hole, never a zero. The original denominator is $0$ there, so the function is undefined at that input — and an undefined input cannot be a zero.

Practice 23 problems Next: Chapter 1.11A Equivalent Representations & Binomial Theorem

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Rational Functions and Holes

Recap: What Is a Hole?

The Limit Notation for a Hole

Step-by-Step: Finding a Hole with Limit Notation

A Trickier Case: A Squared Factor That Fully Cancels

Domain Notation with Holes and Asymptotes Together

Practice Problems

Key Takeaways

Recap card

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