Exploring Types of Discontinuities
Why this matters
Imagine you're on a classic American road trip, driving from Seattle down to Portland. The GPS shows a straight shot on I-5 South. As you approach the bridge over the Columbia River, you see flashing lights. Traffic is stopped. There's a break in the road ahead!
Is it a small, single pothole you could easily patch over and continue? Is the drawbridge stuck open, with the road on the other side at a completely different height? Or has a section of the bridge collapsed entirely, leaving a massive, impassable chasm?
In calculus, a function's graph is like that road. When it's not perfectly smooth and connected, we say it has a "discontinuity." But just like with the bridge, it’s not enough to say "it's broken." A good engineer—and a good calculus student—needs to know what kind of break it is. Today, we'll become those engineers and learn to diagnose the three main types of discontinuities.
Concept overview
flowchart TD
A[Analyze f(x) at x=c] --> B{Does lim (x->c) f(x) exist and is finite?}
B -- No --> C{Are one-sided limits finite but unequal?}
C -- Yes --> D[Jump Discontinuity]
C -- No --> E[Infinite Discontinuity]
B -- Yes --> F{Is f(c) defined and f(c) = limit?}
F -- No --> G[Removable Discontinuity]
F -- Yes --> H[Continuous]
Core explanation
Hey there. It's Saavi. Let's talk about what happens when things fall apart—at least for our functions.
In the last topic, we defined continuity with a strict, three-part test. A function f(x) is continuous at a point x = c if and only if:
f(c)is defined (there's a solid point on the map).lim_{x->c} f(x)exists (the roads from both sides lead to the same spot).lim_{x->c} f(x) = f(c)(the spot the roads lead to is the same as the actual point on the map).
A discontinuity happens when at least one of these rules is broken. The specific way it breaks tells us the type of discontinuity.
Type 1: Removable Discontinuity (The Pothole)
A removable discontinuity is the most civil kind of break. It's like a tiny pothole in an otherwise perfect road.
How it happens: The limit as x approaches c exists, but either the function isn't defined at c, or it's defined at a different value.
- Rule that breaksRule #1 or Rule #3.
- The keyRule #2 is MET. The limit from the left and the limit from the right agree! They are heading to the same place.
Think of the function f(x) = (x^2 - 4) / (x - 2).
If you graph this, it looks exactly like the line y = x + 2. But, if you try to plug in x = 2, you get 0/0, which is undefined. There's a "hole" in the graph at (2, 4).
The limit as x approaches 2 exists and is 4. But the point f(2) doesn't exist. Because we could easily "fix" this by just defining f(2) = 4, we call this a removable discontinuity. You can remove the problem by filling in a single point.
// Example of a removable discontinuity
f(x) = (x^2 - 9) / (x + 3)
// At x = -3, the function is undefined.
// But the limit as x approaches -3 is -6.
// This is a hole at (-3, -6).Type 2: Jump Discontinuity (The Drawbridge)
A jump discontinuity is exactly what it sounds like. The function stops at one value and suddenly reappears at a different value.
How it happens: The limit from the left and the limit from the right both exist as finite numbers, but they are not equal.
- Rule that breaks: Rule #2. Because the one-sided limits don't agree, the overall limit doesn't exist.
Imagine a drawbridge over a river in Chicago. The road on the south side might end 20 feet above the water, while the road on the north side also ends 20 feet above the water. When the bridge is down, it's continuous. But if the bridge is up, the "function" of the road has a jump. You'd have to leap from one side to the other.
You'll see these most often in piecewise functions.
Consider this function, which could model cell phone plan costs:
f(x) = { $30, if x <= 2; $45, if x > 2 }
Here, x is the gigabytes of data used. For 2 GB or less, you pay $30. The moment you go over 2 GB, the price jumps to $45.
- The limit as
xapproaches 2 from the left is $30. - The limit as
xapproaches 2 from the right is $45. Since 30 ≠ 45, the limit atx = 2does not exist. This is a jump discontinuity.
Type 3: Infinite Discontinuity (The Canyon)
This is the most dramatic type of discontinuity. It occurs where a function has a vertical asymptote.
How it happens: As x approaches c from the left or the right (or both), the function's values shoot up to positive infinity or down to negative infinity.
- Rule that breaks: Rule #2 (the limit doesn't exist because it's not a real number) and usually Rule #1 (the function is undefined at the asymptote).
This is the collapsed bridge, the impassable canyon. There's no "jumping" across it, and you certainly can't patch it with a single point. The road on either side veers off into the sky or plummets into the earth.
The classic example is f(x) = 1/x. At x = 0, the graph runs up along the y-axis on the right and down along the y-axis on the left. This is a vertical asymptote.
Any time you have a rational function where the denominator is zero but the numerator is not zero at x=c, you have a vertical asymptote, which means you have an infinite discontinuity.
For g(x) = 5 / (x - 4), at x = 4 the denominator is zero but the numerator is 5. This creates a vertical asymptote at x=4. It's an infinite discontinuity.
Justifying Your Conclusion
On the AP Exam, you can't just say "it's a jump." You have to prove it using the language of calculus. Always go back to the three-part definition of continuity. State which condition fails and why. For example: "The function has a jump discontinuity at x=2 because the limit from the left is 30 and the limit from the right is 45. Since the one-sided limits are not equal, the overall limit does not exist." That's a perfect, point-scoring justification.
Worked examples
Let's walk through a few problems together. I'll show you how to methodically analyze a function and justify your conclusion—no guessing!
Removable Discontinuity
Problem: Analyze the continuity of the function f(x) = (x^2 + x - 6) / (x - 2) at x = 2. If it is discontinuous, classify the type.
Step-by-Step Solution:
-
Check the 3 continuity conditions at
c = 2.- Condition 1: Is
f(2)defined? If we plug inx = 2, we getf(2) = (2^2 + 2 - 6) / (2 - 2) = (4 + 2 - 6) / 0 = 0/0. This is an indeterminate form, which means the function is undefined atx = 2.- Right away, we know the function is discontinuous at
x = 2because it fails the first condition. But we need to know the type, so we must press on.
- Right away, we know the function is discontinuous at
- Condition 1: Is
-
Check the limit.
- Condition 2: Does
lim_{x->2} f(x)exist? Since we got0/0, that's a huge hint that we can simplify the expression. Let's factor the numerator:x^2 + x - 6 = (x + 3)(x - 2). Now, let's re-evaluate the limit:lim_{x->2} ( (x + 3)(x - 2) ) / (x - 2)Since the limit only cares about values nearx=2, not atx=2, we knowx - 2is not zero, so we can cancel it out.lim_{x->2} (x + 3) = 2 + 3 = 5The limit exists and is equal to 5.
- Condition 2: Does
-
Classify the discontinuity.
- The limit exists (it's 5), but the function is undefined at the point. This is the classic signature of a removable discontinuity. It's a hole in the graph at the point
(2, 5).
- The limit exists (it's 5), but the function is undefined at the point. This is the classic signature of a removable discontinuity. It's a hole in the graph at the point
Justification: The function f(x) has a removable discontinuity at x = 2 because lim_{x->2} f(x) = 5, but f(2) is undefined.
Jump Discontinuity
Problem: Priya is tracking her phone's battery. The charging rate is modeled by the piecewise function B(t), where t is hours.
B(t) = { 10t + 20, if 0 <= t < 3; 80, if t >= 3 }
Is the function B(t) continuous at t = 3? If not, classify the discontinuity.
Step-by-Step Solution:
-
Check the 3 continuity conditions at
c = 3.- Condition 1: Is
B(3)defined? We look at the second piece of the function,80, if t >= 3. The "or equal to" tells us to use this piece. So,B(3) = 80. The function is defined. So far, so good.
- Condition 1: Is
-
Check the limit.
- Condition 2: Does
lim_{t->3} B(t)exist? Because the rule changes att=3, we MUST check the one-sided limits.- Left-hand limit
lim_{t->3^-} B(t). We use the first piece,10t + 20.lim_{t->3^-} (10t + 20) = 10(3) + 20 = 30 + 20 = 50. - Right-hand limit
lim_{t->3^+} B(t). We use the second piece,80.lim_{t->3^+} 80 = 80.
- Left-hand limit
- Condition 2: Does
-
Compare and classify.
- The left-hand limit (50) does not equal the right-hand limit (80). Therefore, the overall limit
lim_{t->3} B(t)does not exist. - Since the function fails the second condition of continuity this way, we have a jump discontinuity.
- The left-hand limit (50) does not equal the right-hand limit (80). Therefore, the overall limit
Justification: The function B(t) is discontinuous at t = 3. It is a jump discontinuity because lim_{t->3^-} B(t) = 50 while lim_{t->3^+} B(t) = 80. Since the one-sided limits exist but are not equal, the overall limit does not exist.
Try it yourself
Ready to try diagnosing a couple on your own? Remember to use the three-part test for continuity to justify your answer.
Problem 1:
Consider the function g(x) = (x - 4) / (x^2 - 16).
Find all points of discontinuity for g(x) and classify each one (removable, jump, or infinite).
Problem 2:
For what value of the constant k is the function h(x) continuous everywhere?
h(x) = { x^2 - 1, if x < 2; kx - 5, if x >= 2 }
If k = 5, what type of discontinuity would exist at x = 2?
In simple terms, this topic is about the different ways a function's graph can be broken (discontinuous) and what we, as mathematicians, call each type of break: a hole, a jump, or a canyon.
// Example of a removable discontinuity
f(x) = (x^2 - 9) / (x + 3)
// At x = -3, the function is undefined.
// But the limit as x approaches -3 is -6.
// This is a hole at (-3, -6).- LIM-2.A: Justify conclusions about continuity at a point using the definition.
- LIM-2.A.1
- Types of discontinuities include removable discontinuities, jump discontinuities, and discontinuities due to vertical asymptotes.
flowchart TD
A[Analyze f(x) at x=c] --> B{Does lim (x->c) f(x) exist and is finite?}
B -- No --> C{Are one-sided limits finite but unequal?}
C -- Yes --> D[Jump Discontinuity]
C -- No --> E[Infinite Discontinuity]
B -- Yes --> F{Is f(c) defined and f(c) = limit?}
F -- No --> G[Removable Discontinuity]
F -- Yes --> H[Continuous]
Read what Saavi narrates
(gentle, warm intro music fades)
Hey everyone. It's Saavi from Shrutam. Let's talk about something we can all relate to: hitting a roadblock.
Imagine you're on a road trip, maybe driving from Seattle to Portland. The GPS shows a clear path, but then you see flashing lights. There's a break in the road ahead. Is it a small pothole you can easily patch? Is the drawbridge stuck open, with the road on the other side at a totally different height? Or has the bridge collapsed into a giant canyon?
In calculus, a function's graph is like that road. When it's not perfectly connected, we have a discontinuity. But our job is to be like a good engineer and diagnose the *type* of break.
Today, we're looking at the three main types: removable, jump, and infinite. Let's look at an example you'll see all the time.
Imagine a piecewise function. Let's say Priya is tracking her phone's battery charge. The function is called B of t, where t is time in hours. From time zero up to three hours, the charge is modeled by the line ten t plus twenty. At and after three hours, the charge is a flat eighty percent.
Is this charging process continuous at the three-hour mark? Let's check.
First, what is the charge *at* three hours? The rule says for t greater than or equal to three, the charge is eighty. So, B of three is eighty. The point exists.
Second, what's the limit as we approach three hours? Since the rule changes, we have to check from both sides. The limit from the left, using the ten t plus twenty rule, is ten times three, plus twenty... which is fifty. The limit from the right, using the eighty rule, is just... eighty.
Uh oh. The left side is heading to fifty, but the right side is heading to eighty. They don't meet! This is a classic jump discontinuity. The limit does not exist because the two sides don't agree.
This is a really common mistake students make. They see that the function has a value at the point, and get confused. But for a jump, the key is that the limits from the left and right don't match. It's the drawbridge stuck open.
So, when you see a piecewise function, always check both one-sided limits. It’s the key to telling a jump apart from a simple hole.
You can do this. Just take it one step at a time, apply the definitions, and you'll be able to spot any kind of break in the road. Keep up the great work.
(warm, encouraging music fades in)
The `0/0` form is indeterminate, which often signals a *removable* discontinuity (a hole) that can be found by factoring and canceling. An infinite discontinuity (vertical asymptote) usually comes from `(non-zero number)/0`.
When you see `0/0`, immediately think "factor and cancel" to find the true limit.
The entire definition of a jump discontinuity rests on the left- and right-hand limits being different. If you only plug in the value, you're not checking the limits at all.
If `c` is the point where a piecewise function changes its rule, you *must* calculate the limit from the left and the limit from the right separately.
A function can be undefined at a point, but the limit can still exist (this is a removable discontinuity). Or, the function can be undefined and the limit also doesn't exist (this is an infinite discontinuity). They are separate conditions.
Always follow the three-part test in order. 1. Check the point. 2. Check the limit. 3. Compare them.
The very definition of a removable discontinuity (a hole) is that the limit *does* exist. It's the function's value at that point that is either missing or misplaced.
Remember the pothole analogy. You can clearly see where the road *should* be. That location is the limit.
Rational functions can have removable discontinuities (holes) if the factor that causes the zero in the denominator can be canceled out by a matching factor in the numerator.
Always try to factor the numerator and denominator of a rational function first to see what cancels. Canceled factors lead to holes; leftover factors in the denominator lead to vertical asymptotes.