Difficulty: Intermediate | Prerequisites: Limits (Sections 1.1–1.3)
Continuity formalises the intuitive idea that you can draw a function's graph without lifting your pen. It ties together everything from the limits sections: a function is continuous at a point when the limit exists and equals the function's actual value there. This section also introduces the Intermediate Value Theorem, a powerful result that lets you prove a function must hit a particular y-value somewhere on an interval, even if you cannot find the exact x. You need to be comfortable with limits (Sections 1.1 to 1.3) before tackling this material.
A function is continuous at x = c when three conditions hold: f(c) is defined, the limit exists, and the limit equals f(c). When any condition fails, you get a discontinuity (hole, jump, or asymptote). The Intermediate Value Theorem says that a continuous function on [a, b] must take every y-value between f(a) and f(b).
Continuity at a point
A function f is continuous at x = c when all three of these hold: (1) f(c) is defined, (2) lim (x→c) f(x) exists (both one-sided limits exist and are equal), and (3) lim (x→c) f(x) = f(c). In simple terms, the function arrives where it is supposed to be, with no gap, jump, or blowup.
Discontinuity
A point where at least one of the three continuity conditions fails. There are three types: vertical asymptote, hole (removable), and jump.
Removable discontinuity (hole)
The limit exists, but either f(c) is not defined or f(c) does not equal the limit. Think of it as: the graph has a missing dot that you could fill in to make the function continuous. On a graph, it appears as an open circle.
Jump discontinuity
The left-hand and right-hand limits both exist but are not equal. The graph literally jumps from one y-value to another at x = c.
Vertical asymptote (infinite discontinuity)
The function grows without bound near x = c. The limit from at least one side is ±∞. The graph shoots up or down near the asymptote.
Intermediate Value Theorem (IVT)
If f is continuous on the closed interval [a, b] and m is any number between f(a) and f(b), then there exists at least one c in (a, b) such that f(c) = m. In simple terms, a continuous function cannot skip over a y-value. If f(a) is negative and f(b) is positive, the function must cross zero somewhere between a and b.
Continuity on an interval
A function f is continuous on a closed interval [a, b] if it is continuous at every point in (a, b), continuous from the right at a, and continuous from the left at b.
A function f is continuous at x = c if and only if:
f(c) is defined (there is a y-value at x = c).
lim (x→c) f(x) exists (the left-hand and right-hand limits are equal).
f(c) = lim (x→c) f(x) (the actual value matches the limiting value).
When lim (x→c) f(x) = f(c), the function is continuous. The discrete value from direct substitution equals the limit value.
Vertical asymptote: The function blows up. At least one side goes to ±∞. Condition 2 fails (the limit does not exist as a finite number).
Hole (removable discontinuity): The limit exists, but either f(c) is undefined or f(c) ≠ the limit. Condition 1 or Condition 3 fails. You could "fix" the function by redefining f(c) to equal the limit.
Jump: Both one-sided limits exist but are different. Condition 2 fails. Common in piecewise functions where the pieces do not meet.
If f is continuous on [a, b] and m is a number between f(a) and f(b), then there exists at least one c in (a, b) where f(c) = m.
The most common application: proving a function has a root (an x-intercept) on an interval. If f(a) and f(b) have opposite signs (one positive, one negative), then f must equal zero somewhere between a and b. The IVT guarantees it.
The IVT does not tell you what c is. It only tells you it exists.
A function is continuous on a closed interval [a, b] if:
It is continuous at every point in the open interval (a, b).
It is continuous from the right at a (the right-hand limit at a equals f(a)).
It is continuous from the left at b (the left-hand limit at b equals f(b)).
As long as c is a number in the natural domain of the trig function, the function is continuous at c. For example:
sin(x) and cos(x) are continuous everywhere (domain is all real numbers).
tan(x) is continuous at every point in its domain (everywhere except x = π/2 + nπ).
The same logic applies to cot, sec, and csc: continuous wherever they are defined.
Example: lim (x→π) cos(x/(x−1)) = cos(lim (x→π) x/(x−1)), because cosine is continuous.
If a one-to-one function f is continuous for all values in its domain, then its inverse f⁻¹ is continuous for each point in the range of f. One-to-one functions pass both the vertical and horizontal line tests.
For b > 0 and b ≠ 1: bˣ is continuous on (−∞, +∞).
log_b(x) is continuous on (0, +∞).
Example: f(x) = ln|x| − 2ln(x + 3). The domain is (−3, 0) ∪ (0, +∞), and f is continuous on that domain.
Students often think "continuous" means "the function is defined." Being defined is only the first of three conditions. A function can be defined at x = c but still be discontinuous there if the limit does not equal f(c).
Students confuse holes with vertical asymptotes. Both are discontinuities, but they are different: a hole has a finite limit (you could patch it), while a vertical asymptote has no finite limit (the function blows up).
Students sometimes apply the IVT without checking that the function is continuous on the interval. The IVT requires continuity. If there is a discontinuity between a and b, the theorem does not apply.
Students often think the IVT tells you exactly where f(c) = m. It only guarantees existence. Finding the actual value of c requires separate work (such as algebra or numerical methods).
⚠️ The three-condition checklist for continuity is heavily tested. Be able to evaluate each condition separately and identify which one fails.
⚠️ Classifying discontinuities (hole vs jump vs asymptote) is a common exam question format. Know how to read each from a graph and from an algebraic expression.
⚠️ IVT problems typically give you a continuous function and two x-values, then ask you to prove that a root or a particular y-value must exist on the interval. Show your work: state that f is continuous, compute f(a) and f(b), note that the target value lies between them, and invoke the IVT.
⚠️ Continuity of composed functions: if g is continuous at c and f is continuous at g(c), then f(g(x)) is continuous at c. This lets you pull limits inside continuous functions like cos, ln, and eˣ.
Fill in the blank: The three conditions for continuity at x = c are: f(c) is __________, lim (x→c) f(x) __________, and f(c) = __________.
True or False: A hole (removable discontinuity) means the limit does not exist.
True or False: The IVT can be applied to any function on a closed interval, continuous or not.
Fill in the blank: If f(2) = −3 and f(5) = 4 and f is continuous on [2, 5], then by the IVT there exists some c in (2, 5) where f(c) = __________. (Give one specific value that must be hit.)
True or False: sin(x) is continuous everywhere.
Answers: 1. Defined, exists, lim (x→c) f(x). 2. False (the limit does exist at a hole; that is what makes it removable). 3. False (continuity on the interval is required). 4. 0 (or any number between −3 and 4). 5. True.
Q: Is f(x) = (x² − 1)/(x − 1) continuous at x = 1? Why or why not?
A: f(1) is undefined (0/0), so Condition 1 fails. The function has a removable discontinuity (hole) at x = 1. The limit exists (factor to get x + 1, so the limit is 2), but since f(1) is not defined, the function is not continuous there.
Q: A function has a left-hand limit of 4 and a right-hand limit of 7 at x = 3. What type of discontinuity is this?
A: A jump discontinuity. The one-sided limits both exist but are not equal (4 ≠ 7), so the two-sided limit does not exist.
Q: f is continuous on [1, 5], f(1) = −2, and f(5) = 6. Prove that f has a root on (1, 5).
A: Since f is continuous on [1, 5] and f(1) = −2 < 0 < 6 = f(5), the value 0 lies between f(1) and f(5). By the Intermediate Value Theorem, there exists some c in (1, 5) such that f(c) = 0.
Q: Is tan(x) continuous at x = π/2?
A: No. x = π/2 is not in the domain of tan(x) (tan is undefined there), so the function cannot be continuous at that point. There is a vertical asymptote.
Q: f(x) = eˣ. Is f continuous at x = 0? Verify the three conditions.
A: (1) f(0) = e⁰ = 1 (defined). (2) lim (x→0) eˣ = 1 (exists). (3) f(0) = 1 = lim (x→0) eˣ. All three conditions hold, so f is continuous at x = 0.
Continuity is a prerequisite for the derivative. The definition of the derivative as a limit of a difference quotient assumes that you are working with a function that behaves well enough for that limit to exist. Discontinuities break derivatives.
The IVT connects to root-finding algorithms like the bisection method, which you may encounter in applied maths or numerical analysis courses. The bisection method is essentially the IVT applied repeatedly to narrow down where a root is.
Continuity of compositions (f(g(x))) connects to the chain rule in differentiation (Ch. 3), where you differentiate composite functions.
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