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Types of Discontinuity

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Types of Discontinuity - Complete Interactive Lesson

Part 1: Removable Discontinuities

🕳️ Removable Discontinuities (Holes)

Part 1 of 4 — The "fixable" kind

Topics in This Part

Section
What Is a Removable Discontinuity?
🔑 Spotting Holes Algebraically
Filling the Hole

🔑 Why this matters: Holes happen whenever a factor cancels in a rational expression. They're the " $0/0$ that simplifies" case.

📐 Definition

A discontinuity at $x = a$ is removable if $\lim_{x \to a} f(x)$ exists (finite), but either $f(a)$ is undefined or $f(a) \ne$ the limit.

The discontinuity is "removable" because we could redefine $f(a)$ to equal the limit and the function would become continuous there.

Visually: a single missing or out-of-place point on an otherwise smooth curve — a hole.

🔑 Spotting Holes Algebraically

For a rational function $\dfrac{p(x)}{q(x)}$ , a hole at $x = a$ occurs when $($ is a factor of and (and after cancelling, the denominator is nonzero at ).

🩹 Filling the Hole

Define a "fixed" function: $\tilde f(x) = \begin{cases} f(x), & x \ne 2 \\ 4, & x = 2 \end{cases}$

Hole Spotting 🎯

Find the Hole Height 🧮

For each function, compute the limit at the indicated point (the height of the hole).

1) $f(x) = \dfrac{x^2 - 25}{x - 5}$ at → height ?

Part 2: Jump Discontinuities

↕️ Jump Discontinuities

Part 2 of 4 — When the function "jumps"

Topics in This Part

Section
What Is a Jump?
🔑 Diagnosing from Piecewise Definitions
Reading Jumps from Graphs

🔑 Why this matters: Jumps occur in piecewise functions and step functions — common in real-world models (taxes, shipping costs, etc.).

📐 Definition

A jump discontinuity at $x = a$ occurs when both one-sided limits $\lim_{x \to a^-}$ and exist (finite) but are .

Part 3: Infinite Discontinuities

⚡ Infinite (Essential) Discontinuities

Part 3 of 4 — Asymptote-style discontinuities

Topics in This Part

Section
What Is an Infinite Discontinuity?
🔑 Spotting Them
Why They're Not Removable

🔑 Why this matters: This is the "vertical asymptote" type from your infinite-limits lesson, viewed through the discontinuity lens.

📐 Definition

A function has an infinite discontinuity (also called essential discontinuity) at $x = a$ if at least one one-sided limit at $a$ is .

Part 4: Mixed Practice

🧪 Mixed Practice — Classify & Diagnose

Part 4 of 4 — Putting it all together

Topics in This Part

Section
🔑 Classification Flowchart
Common AP Setups
Worked Examples

🔑 Why this matters: AP free-response problems often ask you to identify the type of discontinuity and justify with limits.

🔑 Classification Flowchart

For a discontinuity at $x = a$ :

Does the two-sided limit exist (finite)?
- Yes → Removable (hole). Can patch by setting $f(a) =$ limit.

Worked example. $f(x) = \dfrac{x^2 - 4}{x - 2}$ .

Direct sub: $0/0$ .
Factor: $\dfrac{(x-2)(x+2)}{x-2} = x + 2$ for $x \ne 2$ .
$\lim_{x \to 2} f(x) = 4$ .
$f(2)$ undefined → hole at $x = 2$ , height 4.

This $\tilde f$ is continuous everywhere — we removed the discontinuity by plugging in the limit value.

💡 The phrase "removable" means the limit exists; the discontinuity can be patched.

2) $g(x) = \dfrac{x^2 - x - 6}{x - 3}$ at $x = 3$ → height ?

So the function "jumps" from one value to another at $a$ . The two-sided limit DNE, and the discontinuity is not removable (no single value can patch it).

🔑 Diagnosing from Piecewise Definitions

For a piecewise function, check the boundary $x = a$ :

Compute $\lim_{x \to a^-} f(x)$ from the left piece.
Compute $\lim_{x \to a^+} f(x)$ from the right piece.
If they're finite but unequal → jump discontinuity.

Worked example. $f(x) = \begin{cases} x + 1, & x < 0 \\ x^2 - 1, & x \ge 0 \end{cases}$

Left limit at 0: $0 + 1 = 1$ .
Right limit at 0: $0 - 1 = -1$ .
Both finite, but $1 \ne -1$ → at . Jump size: .

👁️ Reading Jumps from Graphs

A jump shows as a vertical gap between two pieces, often with:

An open circle (○) on one side (the value the function doesn't take), and
A closed circle (●) on the other side (the value the function does take).

Example sketch. A graph that has a closed point at $(2, 1)$ and an open point at $(2, 4)$ with the curve continuing differently on each side has a jump of size $|4 - 1| = 3$ at $x = 2$ .

💡 The closed circle is $f(2)$ ; the open circle is the value the function approaches from the other side.

Jump Practice 🎯

Compute Jump Sizes 🧮

1) $f(x) = \begin{cases} 3x, & x < 2 \\ x + 7, & x \ge 2 \end{cases}$ — jump size at $x = 2$ ?

2) $g(x) = \begin{cases} x^2, & x < 1 \\ 5, & x \ge 1 \end{cases}$ — jump size at ?

The function blows up near $a$ — there's a vertical asymptote there.

💡 The connection: infinite discontinuity ⇔ vertical asymptote at that $x$ -value.

🔑 Spotting Infinite Discontinuities

For a rational function $\dfrac{p(x)}{q(x)}$ :

$q(a) = 0$ AND $p(a) \ne 0$ → infinite discontinuity at $a$ .
$q(a) = 0$ AND $p(a) = 0$ → factor and check (could be hole or asymptote).

Worked example. $f(x) = \dfrac{x + 5}{x - 1}$ .

$x = 1$ : $q(1) = 0$ , $p(1) = 6 \ne 0$ → infinite discontinuity at .

Other common cases: $\ln x$ at $x = 0$ , $\tan x$ at $x = \pi/2 + k\pi$ .

🚫 Why They're Not Removable

For a discontinuity to be removable, the limit must exist as a finite number. With an infinite discontinuity, the function values aren't bounded near $a$ — no finite value can "fix" the gap. The discontinuity is essential.

Type	Two-sided limit	Removable?
Removable (hole)	Finite, exists	✅ Yes
Jump	DNE (sides finite, disagree)	❌ No
Infinite	DNE / $\pm\infty$	❌ No

Infinite vs. Other 🎯

Are both one-sided limits finite (just unequal)?

Yes → Jump.
No → continue.

Is at least one one-sided limit $\pm\infty$ ?

Yes → Infinite (essential).

💡 Three categories cover essentially every case you'll see in AP Calc AB.

📋 Common AP Setups

Setup	Type
$\dfrac{p(x)}{q(x)}$ where both vanish & factors cancel	Removable
$\dfrac{p(x)}{q(x)}$ where only denominator vanishes	Infinite
Piecewise with mismatched pieces at boundary	Jump
Piecewise where pieces match at boundary	Continuous (no discontinuity)
$\ln x$ at 0; $\tan x$ at $\pi/2 + k\pi$	Infinite

✏️ Worked Example

Classify the discontinuities of $f(x) = \begin{cases} \dfrac{x^2 - 1}{x - 1}, & x \ne 1 \\ 5, & x = 1 \end{cases}$

$\lim_{x \to 1} f(x) = \lim_{x \to 1} (x + 1) = 2$ .

(You could "fix" it by redefining $f(1) = 2$ .)

One side

\to +\infty

, other

\to -\infty

Limit exists → removable discontinuity at

x = 1