Types of Discontinuities: Infinite and Other

TL;DR

Discontinuities are points where a function "breaks." An infinite discontinuity occurs when the function value shoots off to positive or negative infinity at a specific point. Other common types include removable discontinuities (holes) and jump discontinuities (sudden shifts).

1. The Mental Model

Imagine tracing a function's graph with your pencil. If you have to lift your pencil at any point, that's a discontinuity. Different reasons for lifting your pencil lead to different types of discontinuities.

2. The Core Material

When we talk about continuity, we're essentially asking if a function is "smooth" and connected everywhere. If it's not, it's discontinuous. There are three main types you'll encounter:

a. Infinite Discontinuities (Vertical Asymptotes)

Photo by Steve A Johnson on Pexels

An infinite discontinuity happens when your function's value approaches positive or negative infinity as you get closer and closer to a certain x-value. Think of it like a wall that the graph can never quite touch, but gets infinitely close to. This "wall" is called a vertical asymptote.

Mathematically, this means:
$\lim_{x \to a^-} f(x) = \pm\infty$ or $\lim_{x \to a^+} f(x) = \pm\infty$

Where $a$ is the x-value where the discontinuity occurs. A classic example is $f(x) = \frac{1}{x}$ at $x=0$. As $x$ gets very close to 0 from the positive side, $f(x)$ gets very large and positive. As $x$ gets very close to 0 from the negative side, $f(x)$ gets very large and negative.

b. Removable Discontinuities (Holes)

Photo by Вадим Биць on Pexels

A removable discontinuity is like a tiny hole in the graph. The function is undefined at a single point, but if you could just "fill in" that one point, the function would become continuous there. These often occur when you have a common factor in the numerator and denominator of a rational function that cancels out.

Mathematically, this means:
$\lim_{x \to a} f(x)$ exists, but $f(a)$ is undefined or $f(a) \neq \lim_{x \to a} f(x)$.

You can "remove" the discontinuity by defining or redefining $f(a)$ to be equal to the limit.

c. Jump Discontinuities

Photo by Serg Alesenko on Pexels

A jump discontinuity occurs when the function "jumps" from one value to another at a specific point. The limit from the left side of the point is different from the limit from the right side.

Mathematically, this means:
$\lim_{x \to a^-} f(x)$ exists and $\lim_{x \to a^+} f(x)$ exists, but $\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)$.

Piecewise functions are a common place to see jump discontinuities.

graph TD
    A["Function Discontinuity?"] --> B{Does a limit exist at 'a'?};
    B -- Yes --> C{Is f(a) defined and equal to the limit?};
    B -- No --> D{Are both one-sided limits +/- infinity?};
    D -- Yes --> E["Infinite Discontinuity (Vertical Asymptote)"];
    D -- No --> F{Are both one-sided limits finite but different?};
    F -- Yes --> G["Jump Discontinuity"];
    F -- No --> H["Other/More Complex Discontinuity (e.g., oscillating)"];
    C -- No --> I["Removable Discontinuity (Hole)"];
    C -- Yes --> J["Continuous at 'a'"];

3. Worked Example

Let's analyze the discontinuities of the function $f(x) = \frac{x^2 - 4}{x - 2}$.

First, try to simplify the function. Notice that $x^2 - 4$ is a difference of squares: $(x-2)(x+2)$.

So, $f(x) = \frac{(x-2)(x+2)}{x-2}$.

For any $x \neq 2$, we can cancel out the $(x-2)$ terms, leaving $f(x) = x+2$.

However, at $x=2$, the original function is undefined because the denominator becomes zero.
Let's look at the limit as $x$ approaches 2:
$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} (x+2) = 2+2 = 4$.

Since the limit exists (it's 4) but the function is undefined at $x=2$, this is a removable discontinuity. There's a "hole" in the graph at the point $(2, 4)$.

Now consider $g(x) = \frac{1}{x-3}$.
Here, the denominator is zero when $x=3$. Let's check the limits around $x=3$:
$\lim_{x \to 3^-} \frac{1}{x-3} = -\infty$ (e.g., if $x=2.99$, $x-3$ is a small negative number, so $1/(x-3)$ is a large negative number).
$\lim_{x \to 3^+} \frac{1}{x-3} = +\infty$ (e.g., if $x=3.01$, $x-3$ is a small positive number, so $1/(x-3)$ is a large positive number).
Since the function approaches positive or negative infinity, $x=3$ is an infinite discontinuity.

4. Key Takeaways

• Infinite discontinuities occur when a function's value goes to $\pm\infty$ at a point, creating a vertical asymptote.
• Removable discontinuities are "holes" in the graph where the limit exists, but the function isn't defined or doesn't match the limit at that point.
• Jump discontinuities happen when the function's value suddenly "jumps" from one value to another, meaning the left and right limits are different.
• Factoring rational functions often helps identify removable discontinuities.
• Denominators equalling zero are key places to check for either infinite or removable discontinuities.
• Piecewise functions are common sources of jump discontinuities.

Common mistakes to avoid:
- Confusing a vertical asymptote with a hole: check if the limit is finite (hole) or infinite (asymptote).
- Forgetting to check the original function's domain when simplifying, especially for removable discontinuities.
- Assuming all points where the denominator is zero are infinite discontinuities; some might be removable.
- Not checking both one-sided limits for jump discontinuities; they both need to exist but be different.

5. Now Try It

For the function $h(x) = \frac{x^2 + x - 6}{x-2}$:
1. Identify any potential points of discontinuity.
2. Determine the type of discontinuity (infinite, removable, or jump) at each point.
3. If it's a removable discontinuity, what value would you need to define $h(x)$ as at that point to make it continuous there?

Success looks like clearly identifying the point(s) of discontinuity, correctly classifying them, and, if removable, stating the value that would "fill the hole."