Types of Discontinuity

🕳️ Removable Discontinuities (Holes)

Part 1 of 4 — The "fixable" kind

Topics in This Part

🔑 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 $(x - a)$ is a factor of both $p$ and $q$ (and after cancelling, the denominator is nonzero at $a$).

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.

🩹 Filling the Hole

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

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.

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 $x = 5$ → height ?

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

↕️ Jump Discontinuities

Part 2 of 4 — When the function "jumps"

Topics in This Part

🔑 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 $\lim_{x \to a^+}$ exist (finite) but are different.

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 $1 \ne -1$ → jump at $x = 0$. Jump size: $|1 - (-1)| = 2$.

👁️ 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 $x = 1$?

⚡ Infinite (Essential) Discontinuities

Part 3 of 4 — Asymptote-style discontinuities

Topics in This Part

🔑 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 $\pm\infty$.

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 $x = 1$.
• One side $\to +\infty$, other $\to -\infty$.

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.

Infinite vs. Other 🎯

Classify 🔽

🧪 Mixed Practice — Classify & Diagnose

Part 4 of 4 — Putting it all together

Topics in This Part

🔑 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 $x = a$:

1. Does the two-sided limit exist (finite)?
   • Yes → Removable (hole). Can patch by setting $f(a) =$ limit.
   • No → continue.
2. Are both one-sided limits finite (just unequal)?
   • Yes → Jump.
   • No → continue.
3. 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

✏️ Worked Example

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

• $\lim_{x \to 1} f(x) = \lim_{x \to 1} (x + 1) = 2$.
• $f(1) = 5 \ne 2$.
• Limit exists → removable discontinuity at $x = 1$.

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

Classify 🎯

Exit Check ✅