Source: Calc AB Unit 1 Lecture Notes, Advanced Calculus I (UCF)
Difficulty: Introductory. Prerequisites: the Limits material from Part 1 of this unit. You need to be comfortable evaluating limits and understanding when they exist before this section will make sense.
Tags: continuity, continuous function, discontinuity, removable discontinuity, jump discontinuity, infinite discontinuity, three conditions for continuity, IVT, Intermediate Value Theorem, Calc AB Unit 1
Once you know how limits work, the next question is: when does a function behave "nicely"? Continuity is the formal answer. A continuous function has no holes, jumps, or blow-ups at a given point, which means its limit and its actual value agree. This matters because most of the big theorems in calculus (the Intermediate Value Theorem, the Extreme Value Theorem, the Fundamental Theorem of Calculus) require continuity as a precondition. If you cannot check whether a function is continuous on an interval, those theorems are off-limits to you.
A function is continuous at a point when it is defined there, the limit exists there, and the limit equals the function's value. The Intermediate Value Theorem says that a continuous function on a closed interval hits every y-value between f(a) and f(b) at least once.
Continuity at a point
A function f(x) is continuous at x = c if and only if three conditions hold simultaneously:
f(c) is defined.
lim (x→c) f(x) exists.
lim (x→c) f(x) = f(c).
Think of it as: you can draw the graph through x = c without lifting your pen, and the point on the curve is exactly where you would expect it to be.
Removable discontinuity (hole)
A point where the limit exists but either f(c) is not defined or f(c) does not equal the limit. The discontinuity can be "removed" by redefining the function's value at that single point.
In simple terms, there is a hole in the graph that you could fill in with one dot.
Jump discontinuity
A point where the left-hand and right-hand limits both exist but are not equal. The graph makes a sudden jump.
Think of it as: the graph has a clean break, and the two pieces do not meet.
Infinite discontinuity
A point where the function heads to +∞ or −∞ (a vertical asymptote). The limit does not exist in the finite sense.
Intermediate Value Theorem (IVT)
If f is continuous on the closed interval [a, b], then for every value N between f(a) and f(b), there exists at least one c in [a, b] such that f(c) = N.
In simple terms, a continuous function cannot skip over any y-value. If it starts at 2 and ends at 10, it must pass through every number between 2 and 10 somewhere in that interval.
All three must hold at x = c. If any one fails, the function is discontinuous at that point:
Condition 1 – f(c) is defined. The function must produce an actual output at c. If there is a division by zero or the function simply is not defined there, continuity fails immediately.
Condition 2 – lim (x→c) f(x) exists. The left-hand and right-hand limits must agree. If they do not, you have a jump discontinuity and condition 2 fails.
Condition 3 – lim (x→c) f(x) = f(c). Even if the function is defined and the limit exists, the two values must match. If they do not, you have a removable discontinuity (the point exists but is in the "wrong" place).
Removable: Condition 2 passes, but condition 1 or 3 fails. The limit exists, so the gap could be patched.
Jump: Condition 2 fails because the one-sided limits differ. The function leaps from one value to another.
Infinite: The function heads towards ±∞ at that point. A vertical asymptote lives here.
The IVT has two requirements and one conclusion:
Requirement 1: f is continuous on [a, b]. No holes, jumps, or blow-ups on the entire closed interval.
Requirement 2: You pick a target value N that lies between f(a) and f(b).
Conclusion: There is guaranteed to be at least one point c in [a, b] where f(c) = N.
The theorem does not tell you what c is or how many such points exist. It only guarantees at least one.
The most common exam use: show that a root (zero) exists on an interval.
Confirm f is continuous on [a, b].
Evaluate f(a) and f(b).
If one is positive and the other is negative, then 0 lies between them.
By IVT, there must be some c in [a, b] where f(c) = 0.
Result | Statement |
|---|---|
Continuity at x = c | f(c) defined, lim (x→c) f(x) exists, and lim (x→c) f(x) = f(c) |
IVT | f continuous on [a, b] ⇒ f takes every value between f(a) and f(b) on that interval |
The IVT guarantees that if the temperature at 8 a.m. is 15 °C and at noon is 25 °C (and temperature varies continuously), then at some moment in between, the temperature was exactly 20 °C. Engineers use the IVT to confirm that solutions to equations exist within a given range before committing to a numerical method to find them.
"Continuous means the function is defined." Being defined is necessary but not sufficient. The limit must also exist and must equal the function's value. A piecewise function can be defined everywhere and still have a jump discontinuity.
"The IVT tells you where the value occurs." It does not. It guarantees existence, not location. You still need other methods (algebra, Newton's method, bisection) to find the actual point.
"The IVT works on open intervals." The standard theorem requires a closed interval [a, b]. On an open interval, the function might not attain its endpoint values, and the guarantee breaks down.
"If f(a) and f(b) are both positive, the IVT is useless." Not entirely. The IVT still guarantees every value between f(a) and f(b) is hit. It simply cannot guarantee a zero in that case.
⚠️ "State and verify the three conditions for continuity" is a standard exam question format. Practise writing out all three conditions explicitly for a given function.
⚠️ IVT problems almost always ask you to prove a root exists. The pattern is: check continuity, evaluate endpoints, note the sign change, cite IVT.
⚠️ Know how to identify the type of discontinuity from a graph (removable vs. jump vs. infinite). Multiple-choice questions test this regularly.
⚠️ Do not confuse the IVT with the Mean Value Theorem or the Extreme Value Theorem. They have different hypotheses and different conclusions.
True or False: If f(3) = 7, then f is continuous at x = 3.
Fill in the blank: A removable discontinuity occurs when the limit ____ but the function's value does not match (or is not defined).
True or False: The IVT can be applied on the open interval (a, b).
Fill in the blank: To use the IVT to show a root exists on [a, b], you need f(a) and f(b) to have ____ signs.
True or False: If all three continuity conditions hold at every point in [a, b], the IVT applies on that interval.
Answers: 1. False (the limit must also exist and equal 7). 2. Exists. 3. False (closed interval required). 4. Opposite. 5. True.
Q: State the three conditions for f(x) to be continuous at x = c.
A: (1) f(c) is defined. (2) lim (x→c) f(x) exists. (3) lim (x→c) f(x) = f(c).
Q: A function f is continuous on [1, 5], with f(1) = −3 and f(5) = 4. Explain why there must be a root on [1, 5].
A: Since f is continuous on the closed interval [1, 5] and f(1) is negative while f(5) is positive, the value 0 lies between f(1) and f(5). By the Intermediate Value Theorem, there exists at least one c in [1, 5] such that f(c) = 0.
Q: Identify the type of discontinuity: the left-hand limit at x = 2 is 5 and the right-hand limit at x = 2 is 8.
A: Jump discontinuity, because the one-sided limits exist but are not equal.
Q: A piecewise function is defined so that f(4) = 10, but lim (x→4) f(x) = 7. Is f continuous at x = 4? Which condition fails?
A: f is not continuous at x = 4. Conditions 1 and 2 pass (f(4) is defined and the limit exists), but condition 3 fails because lim (x→4) f(x) ≠ f(4).
Q: Can the IVT guarantee that f(x) = 3 has a solution on [a, b] if f(a) = 5 and f(b) = 9?
A: No. The value 3 does not lie between f(a) = 5 and f(b) = 9, so the IVT makes no guarantee for that target value on this interval.
Continuity is the gatekeeper for nearly every major theorem in calculus. The Extreme Value Theorem (a continuous function on a closed interval attains a maximum and minimum) and the Mean Value Theorem both require it. Differentiability, which you will study next, implies continuity, so understanding continuity now saves you from confusion later. The IVT also connects to numerical root-finding methods like the bisection method, which repeatedly applies the IVT logic to narrow down where a root lies.
continuity, continuous function, discontinuity, removable discontinuity, hole, jump discontinuity, infinite discontinuity, vertical asymptote, three conditions for continuity, defined, limit exists, limit equals function value, Intermediate Value Theorem, IVT, existence of roots, sign change, closed interval, Calc AB, AP Calculus AB, Unit 1, UCF Advanced Calculus I