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Exploring Types of Discontinuities

Lesson ~11 min read

In simple terms: In simple terms, this lesson explains the different ways a function's graph can be "broken" and how to use the formal definition of continuity to name each type of break: removable, jump, or infinite.

Why this matters

Imagine you're on a cross-country road trip from Boston to Seattle. The road, your function f(x), is supposed to be a smooth, connected path. But what happens when it isn't?

You might hit a single, jarring pothole. The road exists on either side, but there's a tiny spot missing. Or, you could reach a river where a bridge has collapsed. The road continues on the other side, but at a totally different level—you can't just drive across. Worst of all, you might find the road leads to the edge of a massive canyon, plunging infinitely downward.

These three scenarios—the pothole, the collapsed bridge, and the canyon—are like the three main types of discontinuities we see in calculus. They are breaks in our function's path. In this lesson, we'll learn the official names for these breaks and, more importantly, how to use calculus to prove exactly which type of break you're looking at.

Concept overview

flowchart TD
    A[Start: Analyze f(x) at x=c] --> B{Is f(c) defined?};
    B -->|No| C{Does lim x->c f(x) exist?};
    B -->|Yes| D{Does lim x->c f(x) exist?};
    C -->|Yes| E[Removable Discontinuity];
    C -->|No| F{Do left/right limits exist but differ?};
    F -->|Yes| G[Jump Discontinuity];
    F -->|No| H[Infinite Discontinuity];
    D -->|No| F;
    D -->|Yes| I{Does lim x->c f(x) == f(c)?};
    I -->|Yes| J[Function is Continuous];
    I -->|No| E;

Core explanation

Hey there. It's Saavi. Let's talk about what happens when functions aren't continuous. We call these breaks "discontinuities," and our job is to be like detectives, using evidence to classify them.

The Continuity Checklist

First, let's quickly recap the formal definition of continuity. For a function f to be continuous at a point x = c, it must meet all three of these conditions:

  1. 1
    The Point Exists
    f(c) is defined. (There's a dot on the graph at x=c.)
  2. 2
    The Limit Exists
    lim x→c f(x) exists. (The graph approaches the same height from both the left and the right.)
  3. 3
    The Point and Limit Agree
    lim x→c f(x) = f(c). (The place the graph is heading is the same place where the dot is.)

A discontinuity happens when one or more of these rules are broken. The specific rule that fails tells us what kind of discontinuity we have.

Type 1: Removable Discontinuity (The Pothole)

A removable discontinuity is the most polite type of break. It's like a single missing point on an otherwise perfect road.

How it happens: The limit as x approaches c exists (Rule #2 is met!), but either the point f(c) doesn't exist (Rule #1 fails) or the point exists in a different spot than the limit (Rule #3 fails).

Think of the function f(x) = (x² - 9) / (x - 3).

If you try to plug in x = 3, you get 0/0, which is undefined. So, f(3) is not defined, and Rule #1 fails. The function is discontinuous at x = 3.

But what's the limit? lim x→3 (x² - 9) / (x - 3) = lim x→3 (x - 3)(x + 3) / (x - 3) = lim x→3 (x + 3) = 3 + 3 = 6

The limit exists and is equal to 6! The graph is heading toward the point (3, 6), but there's a hole right there. Because we could "remove" the discontinuity by simply defining f(3) = 6 (like filling a pothole), we call it removable.

Graphically, this always looks like a hole in the curve.

Type 2: Jump Discontinuity (The Collapsed Bridge)

A jump discontinuity is exactly what it sounds like. The function stops at one value and suddenly reappears at another.

How it happens: The limit as x approaches c does not exist (Rule #2 fails). Specifically, it fails because the left-hand limit and the right-hand limit are different finite numbers.

lim x→c⁻ f(x) ≠ lim x→c⁺ f(x)

Consider this piecewise function, which might model different pricing tiers for a data plan:

        / 30,  if x ≤ 100  (cost for up to 100 GB)
g(x) = {
        \ 50,  if x > 100  (cost for over 100 GB)

Let's check the continuity at x = 100.

  • The limit from the left: lim x→100⁻ g(x) = 30
  • The limit from the right: lim x→100⁺ g(x) = 50

Since 30 ≠ 50, the overall limit lim x→100 g(x) does not exist. The function "jumps" from a height of 30 to a height of 50. This is a jump discontinuity.

This is where many students get confused: They see that g(100) is defined (it's 30), but that doesn't make it continuous. The failure of Rule #2 (the limit not existing) is what matters here and what defines the jump.

Type 3: Infinite Discontinuity (The Canyon)

This is the most dramatic break. The function flies off toward positive or negative infinity.

How it happens: The limit as x approaches c does not exist (Rule #2 fails). It fails because the function's values increase or decrease without bound as you get closer to c. This happens at a vertical asymptote.

lim x→c⁻ f(x) = ±∞ or lim x→c⁺ f(x) = ±∞

Consider the function h(x) = 1 / (x - 2). At x = 2, the function is undefined (Rule #1 fails). Let's check the limits:

  • As x approaches 2 from the right (e.g., 2.1, 2.01), x-2 is a small positive number, so h(x) goes to +∞.
  • As x approaches 2 from the left (e.g., 1.9, 1.99), x-2 is a small negative number, so h(x) goes to -∞.

Since the function values shoot off to infinity, the limit does not exist. This is an infinite discontinuity at the vertical asymptote x = 2.

Justifying Your Conclusion

On the AP exam, you can't just say "it's a jump." You have to prove it. To justify your classification, you must appeal to the three-part definition of continuity and use limits.

  • For a removable discontinuity at c, you must show that lim x→c f(x) exists but is not equal to f(c) (or f(c) is undefined).
  • For a jump discontinuity at c, you must show that the left and right-hand limits exist but are not equal.
  • For an infinite discontinuity at c, you must show that the left or right-hand limit (or both) goes to ±∞.

Mastering this justification is key to showing you truly understand the concepts.

Worked examples

Let's walk through a few problems together. The goal isn't just to get the answer, but to build the logical argument for why it's the answer.

Example 1

Classifying a Hole

Problem: Let f(x) = (x² + x - 6) / (x - 2). Determine if f(x) is continuous at x = 2. If not, classify the discontinuity and justify your answer.

Step-by-Step Solution:

  1. Check the 3-part definition of continuity at c = 2.

    • Rule 1: Is f(2) defined? Plugging in x = 2, we get f(2) = (2² + 2 - 6) / (2 - 2) = (4 + 2 - 6) / 0 = 0/0. This is an indeterminate form, which means the function is undefined at x = 2. Conclusion: Since Rule #1 fails, the function is discontinuous at x = 2. Now we need to classify it.

  2. Find the limit to classify the discontinuity. Since we got 0/0, this hints at a removable discontinuity (a hole). Let's find the limit by factoring. lim x→2 (x² + x - 6) / (x - 2) = lim x→2 (x - 2)(x + 3) / (x - 2)

    This is the key step. We can cancel the (x - 2) terms because when we take a limit, we are concerned with values near x=2, not at x=2. = lim x→2 (x + 3) = 2 + 3 = 5

  3. Formulate the justification. We have all the evidence we need.

    • The limit as x approaches 2 exists and is 5.
    • The function value f(2) is undefined.

    Justification: The function f(x) has a removable discontinuity at x = 2 because lim x→2 f(x) = 5, but f(2) is undefined.

    Common Mistake Alert: Students often stop after finding f(2) is undefined and just write "discontinuous." You must proceed to find the limit to correctly classify the type of discontinuity.

Example 2

Analyzing a Piecewise Function

Problem: Let h(x) be defined as:

        / x² - 1,   if x < 1
h(x) = {
        \ 2x + 1,   if x ≥ 1

Is h(x) continuous at x = 1? If not, classify the discontinuity.

Step-by-Step Solution:

  1. Check the 3-part definition at c = 1.

    • Rule 1: Is h(1) defined? Yes. For x ≥ 1, we use the bottom piece: h(1) = 2(1) + 1 = 3.
    • Rule 2: Does lim x→1 h(x) exist? For piecewise functions, we MUST check the limit from both the left and the right.
      • Left-hand limit (x &lt; 1)
        lim x→1⁻ h(x) = lim x→1⁻ (x² - 1) = 1² - 1 = 0
      • Right-hand limit (x ≥ 1)
        lim x→1⁺ h(x) = lim x→1⁺ (2x + 1) = 2(1) + 1 = 3 Since the left-hand limit (0) does not equal the right-hand limit (3), the overall limit lim x→1 h(x) does not exist.

  2. Classify and Justify. Since the limit does not exist because the left and right limits approach different finite values, we have a jump.

    • Rule 3 is irrelevant. We don't even need to check if the limit equals the point value, because the limit doesn't exist in the first place!

    Justification: The function h(x) has a jump discontinuity at x = 1 because lim x→1⁻ h(x) = 0 and lim x→1⁺ h(x) = 3. Since the left and right-hand limits exist but are not equal, the overall limit does not exist.

Try it yourself

Time to try a couple on your own. Remember to use the full definition of continuity to justify your answers.

Problem 1: Consider the function g(x):

        / sin(x)/x,   if x < 0
g(x) = {
        \ x - 1,    if x ≥ 0

Determine the type of discontinuity at x = 0. Justify your answer using limits.

Hint: Remember the special trigonometric limit lim x→0 sin(x)/x = 1. Check the left and right-hand limits separately.

Problem 2: Find and classify all points of discontinuity for the function f(x) = (x + 1) / (x² - 4x - 5).

Hint: Start by factoring the denominator. You'll find two points where the denominator is zero. One will lead to a 0/0 situation (a hole), and the other will lead to a (non-zero)/0 situation (a vertical asymptote).

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