Topic 1.9

Rational Functions and Vertical Asymptotes

7 MIN READ6 IDEAS23 PROBLEMS14 flashcards

Practice 23 problems Next: Chapter 1.10 Rational Functions and Holes

Vertical Asymptotes and One-Sided Limits

You already know vertical asymptotes happen where the denominator is zero (and doesn't cancel). Now let's describe them more precisely — the function often behaves differently on each side of the asymptote, and we use one-sided limit notation to capture that.

CONCEPT

One-Sided Limit Notation

$lim(x \to a⁻) f(x)$ means "as $x$ approaches a from the LEFT (values slightly less than a)."

$lim(x \to a⁺) f(x)$ means "as $x$ approaches a from the RIGHT (values slightly greater than a)."

Worked Example

Worked example

For $f(x) = (x+1) / (x-3)$ , find the vertical asymptote and describe the behavior on each side using limit notation.

f(x) = \frac{x + 1}{x - 3}

01
The denominator is zero at $x=3$ , and the numerator isn't zero there — so $x=3$ is a vertical asymptote.
02
Test a value just LEFT of 3, like $x=2.9$ : $f(2.9) = 3.9/(-0.1) = -39$ — a large negative number.
03
Test a value just RIGHT of 3, like $x=3.1$ : $f(3.1) = 4.1/(0.1) = 41$ — a large positive number.
04
Write the one-sided limits:
$\lim_{x \to 3^-} f(x) = -\infty \qquad \lim_{x \to 3^+} f(x) = +\infty$

The graph confirms it: heading to $-\infty$ on the left side, $+\infty$ on the right side.

Quick check

$f(x) = \frac{2x - 1}{x + 5}$ . What is $\lim_{x \to -5^{-}} f(x)$ ?

The Formal Vertical Asymptote Theorem

Here's the general statement:

CONCEPT

Vertical Asymptotes (General Statement)

Write $f(x) = N(x)/D(x)$ with any common factors canceled.

The graph of $f$ has a vertical asymptote at every input where $D(x) = 0$ (and $N(x)

e 0$).

Numerically Approaching a Vertical Asymptote

You can also SEE a vertical asymptote happen in a table of values — watch what happens to the output as the input creeps closer and closer to the asymptote from each side.

Worked example

For $h(s) = (s^{2} + 3s) / (s - 4)$ , complete the tables below as $s$ approaches 4 from the left and right. What do you notice?

h(s) = \frac{s^2 + 3s}{s - 4}

s (from left)	h(s)
3.9	−269.1
3.99	−2,789
3.999	−27,989
3.9999	−279,989

s (from right)	h(s)
4.1	291.1
4.01	2,811
4.001	28,011
4.0001	280,011

CONCEPT

What the Table Shows

As $s$ gets closer to 4 from the LEFT, $h(s)$ plunges toward more and more negative numbers — heading to $-\infty$ .

As $s$ gets closer to 4 from the RIGHT, $h(s)$ shoots up toward larger and larger positive numbers — heading to $+\infty$ .

This numerical pattern is just another way of confirming the same one-sided limits you'd find algebraically or graphically.

Multiplicity's Effect: When a Factor Partially Cancels

Here's a subtle case worth practicing: what if the SAME factor appears in both the numerator and denominator, but the denominator has a higher power? It's tempting to assume this always creates a hole — but watch closely.

Worked example

For $k(x) = (x^{2} + 2x - 8) / (x + 4)^{2}$ , find the domain, hole(s), zero(s), vertical asymptote(s) with limit notation, and horizontal asymptote.

k(x) = \frac{x^2 + 2x - 8}{(x + 4)^2}

01
Factor the numerator:
$x^2 + 2x - 8 = (x + 4)(x - 2)$
02
One copy of $(x+4)$ cancels — but the denominator has TWO copies, so one remains:
$k(x) = \frac{(x + 4)(x - 2)}{(x + 4)^2} = \frac{x - 2}{x + 4} \quad (x \neq -4)$
03
Since a factor of $(x+4)$ is still left in the denominator after simplifying, $x=-4$ is a VERTICAL ASYMPTOTE, not a hole! Domain: all reals except $x=-4$ .
04
Zero: from the simplified form, $x=2$ (check: it doesn't make the original denominator zero, so it's valid).
05
Because only ONE copy of $(x+4)$ remains (an odd amount), the sign FLIPS across the asymptote, just like a normal linear factor:
$\lim_{x \to -4^-} k(x) = +\infty \qquad \lim_{x \to -4^+} k(x) = -\infty$
06
Horizontal asymptote: both original numerator and denominator have degree 2, so:
$\text{degree num} = \text{degree denom} = 2 \;\Rightarrow\; y = \frac{1}{1} = 1$

Even though $(x+4)$ appeared in both, it only PARTLY cancels — one copy remains, so it's a genuine asymptote with a sign flip.

COMMON MISTAKE

A shared factor doesn't automatically mean a hole — check whether it FULLY cancels. If the denominator's power is higher, some of that factor survives, and you still get a vertical asymptote.

The remaining power after cancellation (not the original power) determines whether the sign flips (odd remaining power) or stays the same (even remaining power) across the asymptote.

Always simplify completely before reading off the horizontal asymptote.

Careful with the reason: canceling a common factor DOES lower the degree of the numerator and of the denominator. What it leaves alone is the DIFFERENCE between the two degrees, and the ratio of the leading coefficients — and those are the only two things the horizontal asymptote depends on.

Quick check

$r(x) = \frac{(x + 5)(x - 4)}{(x - 4)^{2}}$ . Hole or vertical asymptote at $x = 4$ ?

Practice Problems

Worked example

Problem 1. For $f(x) = 5 / (x + 2)$ , write the one-sided limit notation describing behavior at the vertical asymptote.

01
Vertical asymptote at $x=-2$ . Test $x=-2.1$ : $f=5 / (-0.1)=-50$ (negative). Test $x=-1.9$ : $f=5 / (0.1)=50$ (positive).
02
$lim(x \to -2⁻)$ $f(x) = -\infty$ and $lim(x \to -2⁺)$ $f(x) = +\infty$ .

Worked example

Problem 2. For $g(x) = (x-5) / (x-1)^{2}$ , what happens to the sign on each side of $x=1$ ?

01
The factor $(x-1)$ is squared (even power) and does NOT cancel with the numerator — so the denominator is always positive near $x=1$ (a square is never negative).
02
Since the numerator $(x-5)$ is negative near $x=1$ (as 1−5=−4), the function is negative on BOTH sides — the sign does NOT flip: $lim(x \to 1⁻)$ $g(x) = -\infty$ and $lim(x \to 1⁺)$ $g(x) = -\infty$ .

Common slips

A shared factor doesn't automatically mean a hole — check whether it fully cancels. If the denominator's power is higher, some of that factor survives, and you still get a vertical asymptote.

The remaining power after cancellation (not the original power) determines whether the sign flips (odd remaining power) or stays the same (even remaining power) across the asymptote.

Always simplify completely before reading off the horizontal asymptote.

Careful with the reason: canceling a common factor does lower the degree of the numerator and of the denominator. What it leaves alone is the difference between the two degrees, and the ratio of the leading coefficients — and those are the only two things the horizontal asymptote depends on.

Lock it in

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14 cards · Rational functions, Hole or asymptote?

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Recap card

5 lines to re-read the night before.

01
One-sided limit notation — $lim(x \to a⁻)$ and $lim(x \to a⁺)$ — describes behavior approaching a vertical asymptote from each side separately.
02
Vertical asymptotes occur at zeros of $D(x)$ that don't fully cancel with the numerator.
03
Numerical tables approaching from each side confirm the same behavior you'd find algebraically or graphically.
04
Partial cancellation: if the denominator's power is higher than the numerator's for a shared factor, some of it survives — still a vertical asymptote, not a hole.
05
The remaining power after cancellation (odd or even) determines whether the sign flips or stays the same across the asymptote.

Practice 23 problems Next: Chapter 1.10 Rational Functions and Holes

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Rational Functions and Vertical Asymptotes

Vertical Asymptotes and One-Sided Limits

Worked Example

The Formal Vertical Asymptote Theorem

Numerically Approaching a Vertical Asymptote

Multiplicity's Effect: When a Factor Partially Cancels

Practice Problems

Recap card

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