Difficulty: Intermediate | Prerequisites: Limits study notes (evaluating one-sided and two-sided limits, piecewise functions)
Continuity formalises a simple idea: can you draw the graph without lifting your pen? It bridges the gap between limits (which you have just learned) and derivatives (which come next). You cannot differentiate a function at a point where it is not continuous, so this section is the gatekeeper for everything that follows. You should be comfortable evaluating one-sided limits and two-sided limits before tackling continuity.
A function is continuous at a point if three conditions hold: the function is defined there, the limit exists there, and the limit equals the function value. Most exam questions ask you to verify all three conditions explicitly, find where they fail, or solve for a constant that forces continuity in a piecewise function.
Continuity at a point
A function f is continuous at x = c if all three of the following conditions are met: (I) f(c) is defined, (II) lim (x→c) f(x) exists, (III) lim (x→c) f(x) = f(c). Think of it as: no gaps, no jumps, no blowups at that point.
Discontinuity
A point where at least one of the three continuity conditions fails. There are several types, each corresponding to a different kind of failure.
Removable discontinuity (hole)
The limit exists at x = c, but either f(c) is not defined or f(c) does not equal the limit. You could "fix" the function by redefining f(c) to match the limit. On a graph, this looks like an open circle with a gap or a filled dot at the wrong height.
Jump discontinuity
The left-hand limit and right-hand limit both exist but are not equal, so the two-sided limit does not exist (Condition II fails). On a graph, the function "jumps" from one value to another.
Infinite discontinuity
The function blows up to ±∞ at x = c (a vertical asymptote). The limit does not exist as a finite number. In simple terms, the graph shoots off the page.
Continuous everywhere
A function is continuous everywhere if it is continuous at every point in its domain. Polynomials are continuous everywhere. Rational functions are continuous everywhere except where the denominator is zero.
For f to be continuous at x = c, you must verify:
Condition I: f(c) is defined. The function must have a value at x = c. If you get division by zero or the point is not in the domain, this fails immediately.
Condition II: lim (x→c) f(x) exists. The two-sided limit must exist, which means both one-sided limits must be equal. For piecewise functions, check left and right separately.
Condition III: lim (x→c) f(x) = f(c). The limit and the function value must match.
On the exam, you must state and check all three, in order. If any condition fails, the function is not continuous at that point, and you should identify which condition fails.
Example: Show f(x) = (x+2)/(x−7) is continuous at x = 6.
I) f(6) = (6+2)/(6−7) = 8/(−1) = −8. Defined.
II) lim (x→6) (x+2)/(x−7) = −8. The limit exists.
III) lim (x→6) f(x) = f(6) = −8. They match.
All three conditions are satisfied, so f is continuous at x = 6.
Example: f(x) = (x+4)/(x−2). Is f continuous everywhere?
At x = 2, f(2) is not defined (denominator is zero). Condition I fails. f is not continuous at x = 2.
Example: f(x) = { x−5 if x < 1; 3 if x = 1; x²−2x−3 if x > 1 }. Test continuity at x = 1.
I) f(1) = 3. Defined.
II) Left: lim (x→1⁻) (x−5) = −4. Right: lim (x→1⁺) (x²−2x−3) = 1−2−3 = −4. Both sides give −4, so lim (x→1) f(x) = −4. Exists.
III) lim (x→1) f(x) = −4 but f(1) = 3. These are not equal. Condition III fails.
f is NOT continuous at x = 1.
This is a standard exam question. You have a piecewise function with an unknown constant A in one piece, and you need all three conditions to hold at the boundary.
Example: f(x) = { x³ − x if x < 2; Ax + 3 if x ≥ 2 }. Find A so f is continuous.
I) f(2) = 2A + 3 (using the ≥ piece). Defined for any A.
II) Left: lim (x→2⁻) (x³−x) = 8−2 = 6. Right: lim (x→2⁺) (Ax+3) = 2A+3. For the limit to exist, set 6 = 2A+3, giving A = 3/2.
III) With A = 3/2: f(2) = 2(3/2)+3 = 6 and lim (x→2) f(x) = 6. They match.
f is continuous only if A = 3/2.
Example: f(x) = { x² − 3 if x < 1; Ax + 5 if x ≥ 1 }. Find A.
Left: lim (x→1⁻) (x²−3) = 1−3 = −2. Right: lim (x→1⁺) (Ax+5) = A+5.
Set A+5 = −2, so A = −7.
With A = −7: f(1) = −7+5 = −2 and lim f(x) = −2. Checks out.
A filled dot at (c, y) means f(c) = y.
An open circle at (c, y) means the function approaches y but f(c) is either undefined or defined at a different value.
If the limit exists (both sides approach the same y-value) but does not equal the filled dot, Condition III fails.
If the left and right sides approach different y-values, Condition II fails.
A rational function f(x) = p(x)/q(x) is continuous at every x where q(x) ≠ 0. To find discontinuities, solve q(x) = 0.
Example: f(x) = (x−3)/((x+1)(x−9)) is continuous everywhere except at x = −1 and x = 9, where the denominator is zero.
Students often think that if the limit exists, the function must be continuous. It is not enough. You also need f(c) to be defined and to equal the limit (Condition III). A hole is the classic counterexample: the limit exists, but the function value is missing or wrong.
Students often skip checking Condition I and jump straight to the limit. On the exam, you will lose marks for this. Always start by verifying that f(c) is defined.
Students sometimes think that if Condition II fails (limit DNE), they still need to check Condition III. They do not. If the limit does not exist, Condition III automatically fails as well, because you cannot compare a nonexistent limit to f(c). But you should still mention this on the exam for full marks.
Students often confuse the piecewise boundary rule. At x = c, the function value comes from whichever piece includes x = c (look for the ≥ or ≤ sign). The limit, however, requires checking both pieces.
⚠️ The exam requires you to show all three conditions explicitly, labelled I, II, III. Stating just the answer without the conditions earns zero credit.
⚠️ "Find A so the piecewise function is continuous" is a guaranteed question. Set left-hand limit = right-hand limit, solve for A, then verify Condition III. Write a continuity statement at the end.
⚠️ Graph-based continuity questions test whether you can identify which condition fails. Know the difference between a hole (Condition III fails) and a jump (Condition II fails).
⚠️ For rational functions, the exam asks "where is f(x) continuous?" The answer is everywhere except where the denominator is zero. State this clearly.
True or False: A function can have a limit at x = c but still be discontinuous there. True. If f(c) is undefined or does not match the limit, Condition I or III fails.
Fill in the blank: The three conditions for continuity at x = c are (I) f(c) is ______, (II) lim (x→c) f(x) ______, and (III) lim (x→c) f(x) = ______. Defined; exists; f(c).
True or False: If the left-hand and right-hand limits are different, then Condition III also fails. True. If the two-sided limit does not exist, it cannot equal f(c).
Fill in the blank: To make a piecewise function continuous, set the left-hand limit equal to the ______ limit and solve for the unknown constant. Right-hand.
True or False: f(x) = (x−3)/((x+1)(x−9)) is continuous at x = 3. True. The denominator is not zero at x = 3, so f(3) is defined and all three conditions hold.
Q: Let f(x) = (x+4)/(x−2). Is f continuous everywhere? If not, explain which condition fails.
A: f(2) is not defined because the denominator is zero. Condition I fails. f is not continuous at x = 2.
Q: For f(x) = { −3+x if x < 1; 3−x if x ≥ 1 }, locate all values of x where f is NOT continuous.
A: At x = 1: Left limit = −3+1 = −2. Right limit = 3−1 = 2. Since −2 ≠ 2, the two-sided limit DNE. Condition II fails. f is not continuous at x = 1. (f is continuous everywhere else.)
Q: Let f(x) = { x² − 3 if x < 1; Ax + 5 if x ≥ 1 }. Find A so that f is continuous.
A: I) f(1) = A + 5. II) Left limit = 1 − 3 = −2. Right limit = A + 5. Set A + 5 = −2, so A = −7. III) With A = −7: f(1) = −2 and lim f(x) = −2. They match. f is continuous only if A = −7.
Q: Where is f(x) = (x−3)/((x+1)(x−9)) continuous?
A: The denominator is zero at x = −1 and x = 9. f is continuous everywhere except at x = −1 and x = 9.
Q: A graph shows f(3) = 5 (filled dot) and the graph approaches y = 6 from both sides as x → 3. Which condition of continuity fails?
A: Condition III fails. lim (x→3) f(x) = 6 but f(3) = 5, so the limit does not equal the function value.
Continuity depends entirely on the limit concepts from the previous section (Ch. 2.1–2.3). It also sets the stage for derivatives: the definition of the derivative is itself a limit, and that limit can only exist at points where the function behaves smoothly. Later in the course, the Intermediate Value Theorem (which guarantees that continuous functions hit every value between their endpoints) will also rely on this foundation.
Continuity, continuous function, three conditions of continuity, discontinuity, removable discontinuity, hole, jump discontinuity, infinite discontinuity, vertical asymptote, piecewise continuity, finding A for continuity, domain, rational function continuity, MAC2233, calculus for business, FIU, exam 1 review, section 2.4