A function is continuous at a point if three things hold: the limit exists, the function is defined there, and the limit equals the function's value. Polynomials and exponentials are continuous everywhere; rational, radical, and logarithmic functions are continuous only on their domains. When continuity breaks, you get a jump, a removable hole, or an infinite (vertical asymptote) discontinuity.
Difficulty: Introductory | Prerequisites: Section 1.2 (Limits), basic understanding of function types (polynomial, rational, exponential, logarithmic)
Continuity sits between limits (Section 1.2) and derivatives (Section 1.4). You have already learnt to evaluate limits; now you need to know when a function behaves "nicely" enough that the limit just equals the function value. That property is continuity, and it is a prerequisite for defining the derivative (the instantaneous rate of change). If you are comfortable finding limits by substitution and you know the basic function families, you are ready for this section.
Continuous function
A function that has no breaks, jumps, or gaps in its graph. Formally, f(x) is continuous at x = a when the limit exists, f(a) exists, and the limit equals f(a). In simple terms, it is a curve you can draw without lifting your pencil.
Continuity at a point
The property that all three continuity conditions hold at a single value x = a. Think of it as the function arriving smoothly at the value it is supposed to have there.
Discontinuity
Any point where at least one of the three continuity conditions fails. The function breaks its smooth behaviour at that point.
Jump discontinuity
A break where the left-hand limit and right-hand limit both exist but are not equal. The graph "jumps" from one value to another. Think of it as a staircase step: the function lands at one level from the left and a different level from the right.
Removable discontinuity
A break where the limit exists but the function is either undefined at the point or defined at a different value, leaving a "hole" in the graph. In simple terms, the curve has a pinprick gap that could be patched by redefining one point.
Infinite discontinuity
A break where the function grows without bound (towards positive or negative infinity) near the point, producing a vertical asymptote. Think of it as the graph shooting off the page.
Domain
The set of all input values for which a function is defined. Continuity questions only apply at points inside (or on the boundary of) the domain.
Direct substitution
The method of evaluating a limit by plugging the target value straight into the function. This works precisely when the function is continuous at that point.
Polynomial function
A function of the form f(x) = a_n x^n + a_(n-1) x^(n-1) + ... + a_1 x + a_0, where every exponent is a non-negative integer. Continuous everywhere on the real line.
Rational function
A ratio of two polynomials, f(x) = P(x)/Q(x). Continuous everywhere except where the denominator equals zero.
Radical function
A function involving an nth root of an expression, f(x) = nth root of g(x). With an even root, it is continuous only where the expression inside the root is non-negative.
Logarithmic function
A function of the form f(x) = ln[g(x)]. Continuous only where the expression inside the logarithm is strictly positive.
Vertical asymptote
A vertical line x = a that the graph approaches but never crosses, where the function values grow without bound. The hallmark of an infinite discontinuity.
A function f(x) is continuous at x = a when all three of the following hold:
Condition 1 – The limit exists. lim (x->a) f(x) must exist. Both one-sided limits must agree.
Condition 2 – The function is defined. f(a) must exist. There has to be an actual y-value at that point.
Condition 3 – They match. lim (x->a) f(x) = f(a). The value the function approaches is the value it takes.
If even one condition fails, the function is discontinuous at x = a. Each type of failure gives a different kind of discontinuity (covered below).
The informal version: a continuous function is one you can draw without lifting your pencil from the paper.
Continuity at a point is a necessary (but not sufficient) condition for the derivative to exist at that point. A function must be continuous at x = a for the instantaneous rate of change to be defined there, but continuity alone does not guarantee the derivative exists.
Polynomial functions – any function of the form f(x) = a_n x^n + ... + a_1 x + a_0 with non-negative integer exponents. Example: f(x) = 0.01x^3 - 2x - 3.
Exponential functions of the form f(x) = e^(g(x)), where g(x) is a polynomial. Example: g(x) = 5e^(0.02x).
Products of polynomials and exponentials. Example: h(x) = 5e^(0.02x)(0.01x^3 - 2x - 3).
All of these have no jumps, gaps, or breaks. The limit exists at every point, the function is defined everywhere, and the function value equals the limit at each point.
Rational functions – f(x) = P(x)/Q(x). Continuous everywhere except where the denominator Q(x) = 0. Example: q(t) = (2t + 3)/(4 - t) is continuous for all t except t = 4.
Radical functions (even root) – f(x) = sqrt(g(x)). Continuous where the radicand g(x) >= 0. Example: r(z) = sqrt(2z - 6) is continuous for z >= 3.
Logarithmic functions – f(x) = ln(g(x)). Continuous where the argument g(x) > 0. Example: s(x) = ln(x - 5) is continuous for x > 5.
The key rule: rational, radical, and logarithmic functions are continuous wherever they are defined. Outside their domain, the question of continuity does not apply.
When f is continuous at x = a and a is in the domain, the limit is just the function value:
lim (x->a) f(x) = f(a)
This is the direct substitution method. It follows straight from the definition of continuity: if the limit equals the function value, then plugging in gives you the answer.
Polynomial: lim (x->2) (x^2 + 5x - 1) = (2)^2 + 5(2) - 1 = 13. Domain: all real numbers.
Rational: lim (t->-1) (2t + 3)/(4 - t) = (2(-1) + 3)/(4 - (-1)) = 1/5. Domain: all reals except t = 4. Since t = -1 is in the domain, substitution works.
Exponential: lim (z->5) e^(0.02z) = e^(0.02 x 5) = e^0.1 ≈ 1.105. Domain: all real numbers.
Radical: lim (z->4) sqrt(2z - 6) = sqrt(2(4) - 6) = sqrt(2) ≈ 1.41. Domain: z >= 3. Since 4 >= 3, substitution is valid.
Direct substitution only works for continuous functions at the target point. If the function is not continuous there, substitution may give an incorrect or undefined result.
A function is discontinuous at a point if any of the three continuity conditions fails. Each failure mode has a name.
The left-hand and right-hand limits both exist, but they are not equal. The overall limit does not exist (DNE).
Example: rental car pricing. A car costs $50 per day (no proration). At t = 3 days:
lim (t->3-) C(t) = 150 (approaching from the left, the cost is still $150)
lim (t->3+) C(t) = 200 (as soon as the fourth day begins, the cost is $200)
lim (t->3) C(t) = DNE, because 150 ≠ 200
C(t) is not continuous at t = 3
The graph shows a step function with a visible break at every integer.
The limit exists, but the function is defined at a different value (or not defined at all), so the limit does not equal the function value.
Example: piecewise function. f(x) = (x^2 + x - 2)/(x - 1) for x ≠ 1, and f(1) = 4.
lim (x->1-) f(x) = 3
lim (x->1+) f(x) = 3
lim (x->1) f(x) = 3, but f(1) = 4
3 ≠ 4, so f is not continuous at x = 1
The graph shows an open circle (hole) at (1, 3) and a filled dot at (1, 4). This is called "removable" because you could fix it by redefining f(1) = 3.
The function grows without bound near the point, producing a vertical asymptote. The limit does not exist.
Example: rational function. f(x) = 5x/(x - 3).
As x approaches 3 from one side, f(x) -> +∞
As x approaches 3 from the other side, f(x) -> -∞
lim (x->3) f(x) = DNE
f is not continuous at x = 3; the graph has a vertical asymptote there
Look for three visual cues:
A step or break between two pieces of the curve (jump)
An open circle with the filled dot at a different height, or just an open circle (removable)
The curve shooting off to ±∞ near a vertical line (infinite)
\text{Continuity at } x = a: \quad \lim_{x \to a} f(x) = f(a)\text{Polynomial: } f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0\text{Rational: } f(x) = \frac{P(x)}{Q(x)}, \quad Q(x) \neq 0\text{Radical: } f(x) = \sqrt[n]{g(x)}\text{Logarithmic: } f(x) = \ln[g(x)], \quad g(x) > 0Continuity matters wherever you model a real situation with a function. Profit, revenue, and cost functions in business are typically modelled as continuous (polynomial or exponential) so that rates of change (marginal cost, marginal revenue) can be computed. Pricing schedules like flat per-day rental fees or tiered shipping rates are naturally discontinuous, which is why calculus tools do not apply directly at the breakpoints. Understanding where a model is continuous tells you where you can safely use derivatives to optimise.
Students often think "continuous" means "smooth" or "has no sharp corners." A function with a sharp corner (like |x|) is still continuous; the break in smoothness affects differentiability, not continuity.
Students sometimes assume that if a function is defined at a point, it must be continuous there. Being defined is only one of three conditions. The limit must also exist and match.
A common error is applying direct substitution to evaluate a limit at a point where the function is not continuous. Substitution only works when all three continuity conditions hold.
Students may confuse "the limit does not exist" with "the function is not defined." These are different failures. A function can be defined at a point where the limit does not exist (jump discontinuity), or the limit can exist at a point where the function is not defined (removable discontinuity).
⚠️ You will almost certainly be asked to state the three conditions for continuity at a point. Know them cold.
⚠️ Expect a piecewise or graphical question where you must identify the type of discontinuity (jump, removable, or infinite) and explain which condition fails.
⚠️ Direct substitution problems are common. The exam will test whether you check that the function is continuous at the target point before substituting.
⚠️ Continuity is a prerequisite for the derivative. Questions may ask: "Is the function differentiable at x = a?" The first check is always whether it is continuous there.
True or false: A polynomial function is continuous everywhere. (True)
True or false: If f(a) exists, then f is continuous at a. (False, all three conditions must hold)
Fill in the blank: A rational function is continuous everywhere except where the ______ equals zero. (denominator)
True or false: A removable discontinuity means the limit does not exist at that point. (False, the limit exists but does not equal the function value)
Fill in the blank: Direct substitution works for evaluating limits only when the function is ______ at the target point. (continuous)
Q: State the three conditions required for a function f(x) to be continuous at x = a.
A: (1) lim (x->a) f(x) exists, (2) f(a) exists, (3) lim (x->a) f(x) = f(a).
Q: Is the function f(x) = (x^2 - 9)/(x - 3) continuous at x = 3? Why or why not?
A: No. The function is not defined at x = 3 (division by zero), so Condition 2 fails. The limit exists (it equals 6, since (x^2 - 9)/(x - 3) simplifies to x + 3 for x ≠ 3), so this is a removable discontinuity.
Q: A step function charges $20 for 0 < t <= 1, $40 for 1 < t <= 2, and $60 for 2 < t <= 3. Is it continuous at t = 1?
A: No. The left-hand limit is 20 and the right-hand limit is 40. Since these are not equal, the overall limit does not exist. This is a jump discontinuity.
Q: Evaluate lim (x->4) sqrt(3x - 8). Is direct substitution valid here?
A: sqrt(3(4) - 8) = sqrt(4) = 2. Direct substitution is valid because the square root function is continuous at x = 4 (the radicand 3(4) - 8 = 4 > 0, so x = 4 is in the domain).
Q: Classify the discontinuity of f(x) = 7/(x + 2) at x = -2.
A: Infinite discontinuity. As x approaches -2, f(x) grows without bound (vertical asymptote). The limit does not exist.
Continuity connects directly to derivatives (Section 1.4). A function must be continuous at a point for the derivative to exist there, so every differentiability question starts with a continuity check. This also connects back to limits (Section 1.2): the three-condition definition of continuity is built entirely on the limit concept. Later in the course, the Intermediate Value Theorem (which guarantees a continuous function hits every value between f(a) and f(b)) depends on continuity as its central hypothesis.
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