Piecewise functions require you to check continuity at the boundary where the formula changes: compute the left-hand limit, the right-hand limit, and the function value, then see if they all agree. When they do not, the mismatch tells you which type of discontinuity you have (removable, jump, or infinite), and sometimes you can solve for a constant that forces the pieces to line up.
Difficulty: Introductory to Intermediate
Prerequisites: Part 1 study notes (continuity definitions, interval notation). You should already know how to evaluate one-sided limits and write intervals of continuity for basic function types.
Once you know the rules for where standard functions are continuous, the next question is: what happens when a function is defined by different formulas on different intervals? Piecewise functions are everywhere in applied maths (tax brackets, shipping rates, grading scales), and they are the primary setting where continuity can fail in interesting ways. This section teaches you to diagnose exactly what goes wrong at a boundary point and, when possible, to fix it by choosing the right constant. These skills feed directly into the Intermediate Value Theorem and later into integration of piecewise functions.
Piecewise function
A function defined by different formulas on different intervals of its domain. Think of it as a function that changes its rule depending on where x is.
Left-hand limit
The value f(x) approaches as x comes in from the left (from smaller values). Written lim(x→a⁻) f(x). In simple terms, this is what the graph is heading toward just before the boundary.
Right-hand limit
The value f(x) approaches as x comes in from the right (from larger values). Written lim(x→a⁺) f(x). This is what the graph is heading toward just after the boundary.
Removable discontinuity (hole)
The two-sided limit exists, but either the function is not defined there or its value does not match the limit. Think of it as a single point that could be "filled in" to make the function continuous.
Jump discontinuity
The left-hand and right-hand limits both exist but are not equal. The graph makes an abrupt step. In simple terms, the two pieces do not meet.
Infinite discontinuity
At least one one-sided limit is ±∞. The graph shoots off toward infinity, typically at a vertical asymptote.
Continuous real-world function
A quantity that changes smoothly without instantaneous jumps. Gasoline in a tank drains continuously; the gear your car is in jumps between discrete values.
Discontinuous real-world function
A quantity that changes in discrete steps or sudden jumps. Student enrolment, parking fees charged per hour, and gear position are all discontinuous.
The method is always the same at a boundary point x = a:
Find the left-hand limit: use the formula that applies when x < a.
Find the right-hand limit: use the formula that applies when x > a.
If both one-sided limits are equal, the two-sided limit exists and equals that common value.
Check f(a): use whichever formula applies at x = a (watch for a separately defined value).
Compare: if lim(x→a) f(x) = f(a), the function is continuous at a.
Worked example (Problem 7 style):
f(x) = { −5x if x < 4; 1 if x = 4; 5x if x > 4 }
lim(x→4⁻) f(x) = −5(4) = −20
lim(x→4⁺) f(x) = 5(4) = 20
The one-sided limits are not equal (−20 ≠ 20), so lim(x→4) f(x) does not exist.
f(4) = 1 (given directly).
Classification: jump discontinuity at x = 4, because the left and right limits exist but differ.
Worked example (Problem 8, Part 1 style):
f(x) = { (x² − 16)/(x − 4) if x ≠ 4; 17 if x = 4 }
For x ≠ 4, factor: (x² − 16)/(x − 4) = (x − 4)(x + 4)/(x − 4) = x + 4.
lim(x→4) f(x) = 4 + 4 = 8.
f(4) = 17.
The limit exists (8) but does not equal f(4) (17), so this is a removable discontinuity. The hole could be filled by redefining f(4) = 8.
When reading a graph at a point x = a:
Continuous: the curve passes through the point with no break, hole, or asymptote.
Removable discontinuity: there is an open circle (hole) at a point, but the curve approaches the same y-value from both sides. The function is either undefined there or defined at a different y-value (a filled dot elsewhere).
Jump discontinuity: the curve approaches different y-values from the left and right. You will see the graph split into two pieces at that x-value, each ending at a different height.
Infinite discontinuity: the graph shoots off toward +∞ or −∞ on at least one side. Typically a vertical asymptote.
Ask yourself: does the quantity change smoothly, or does it jump?
Continuous: gasoline in a tank (drains smoothly as you drive), income tax as a function of income (the tax formula is piecewise-linear but the pieces connect, so no jumps).
Discontinuous: engine gear (jumps between integers 0, 1, 2, 3, 4, 5), number of enrolled students (whole numbers that change at registration events), parking fee at $2 per hour or fraction thereof (jumps at every whole hour, like a step function).
Worked example (Problem 11 style):
f(x) = { x − 1 if x < 1; 2x + c if x ≥ 1 }
For continuity at x = 1, the left-hand limit must equal the right-hand value:
Left-hand limit: lim(x→1⁻) (x − 1) = 1 − 1 = 0.
Right-hand value: f(1) = 2(1) + c = 2 + c.
Set them equal: 0 = 2 + c, so c = −2.
The technique is always: evaluate the limit from the side that does not include the boundary, evaluate the formula that does include the boundary at that point, and set them equal.
Piecewise continuity condition at x = a:
\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)All three must be equal. If any one differs, the function is discontinuous at a.
Solving for a constant c: set the limit from the open side equal to the formula value at the boundary, then solve for c.
Piecewise models appear constantly in everyday life. Tax brackets are piecewise-linear functions of income, and a well-designed tax schedule is continuous (no sudden jump in the amount you owe at a bracket boundary). Shipping cost schedules, on the other hand, often have jump discontinuities at weight thresholds. Recognising which real-world quantities are continuous and which are not is a practical skill that shows up well beyond calculus exams.
Students often think that if the two-sided limit does not exist, the discontinuity must be infinite. Not so. A jump discontinuity also has no two-sided limit, because the one-sided limits disagree. The difference: for a jump, both one-sided limits are finite; for an infinite discontinuity, at least one is ±∞.
Students sometimes classify a removable discontinuity as a jump because "the function value jumps to a different point." The test is whether the two-sided limit exists. If it does, it is removable, full stop, regardless of what f(a) equals.
Students forget that the income tax function (0% on the first $10,000, then 10% on the next $40,000, then 25% beyond) is continuous. Because the tax is computed on slices of income and the pieces connect at the bracket boundaries, there is no jump. The rate changes, but the total tax does not leap.
When solving for c, students occasionally set the two formulas equal to each other in general (for all x) instead of evaluating each side at the boundary point. You only need the values to match at the single boundary x-value.
⚠️ Expect at least one piecewise function on the exam where you must compute both one-sided limits, state f(a), and classify the discontinuity. This is a four-part question worth easy marks if you follow the steps.
⚠️ "Find c so that f is continuous" is a classic exam question. The technique is mechanical: left-hand limit = right-hand value at the boundary. Practise it until it is automatic.
⚠️ Real-world continuous vs. discontinuous judgements appear as conceptual multiple-choice. The key question: can the quantity take every value in between, or does it jump? Gasoline level: continuous. Gear number: discontinuous. Parking fee per hour: discontinuous.
⚠️ On graph-reading questions, watch for open vs. filled circles carefully. An open circle means the function is not defined at that y-value; a filled circle means it is.
True or false: If the left-hand and right-hand limits at a point are both 5 but f(a) = 7, the discontinuity is removable. _____
Fill in the blank: For f(x) = { x − 1 if x < 1; 2x + c if x ≥ 1 }, the value of c that makes f continuous is _____.
True or false: The parking fee function ($2 per hour or fraction) is continuous. _____
True or false: Income tax computed on slices (0% then 10% then 25%) is continuous as a function of income. _____
Fill in the blank: When the left-hand limit is −20 and the right-hand limit is 20, the discontinuity is a _____ discontinuity.
Answers: 1. True. 2. c = −2. 3. False. 4. True. 5. Jump.
Q: Let f(x) = { −5x if x < 4; 1 if x = 4; 5x if x > 4 }. Evaluate lim(x→4⁻) f(x), lim(x→4⁺) f(x), lim(x→4) f(x), and f(4). Classify the discontinuity.
A: lim(x→4⁻) f(x) = −20. lim(x→4⁺) f(x) = 20. The two-sided limit does not exist because −20 ≠ 20. f(4) = 1. This is a jump discontinuity.
Q: Let f(x) = { (x² − 16)/(x − 4) if x ≠ 4; 17 if x = 4 }. Find lim(x→4) f(x) and classify the discontinuity.
A: Factor: (x² − 16)/(x − 4) = x + 4. So lim(x→4) f(x) = 8. Since f(4) = 17 ≠ 8, the limit exists but does not equal the function value. This is a removable discontinuity.
Q: Is the function f(x) = gallons of gasoline in your tank after driving x miles continuous or discontinuous?
A: Continuous. The fuel level decreases smoothly as you drive; it does not jump.
Q: Is the function f(t) = engine gear (0, 1, 2, 3, 4, 5) at time t continuous or discontinuous?
A: Discontinuous. The gear is always an integer and jumps between values.
Q: For f(x) = { x − 1 if x < 1; 2x + c if x ≥ 1 }, find c so that f is continuous.
A: Left-hand limit at x = 1: lim(x→1⁻) (x − 1) = 0. Right-hand value: f(1) = 2(1) + c = 2 + c. Set equal: 0 = 2 + c, so c = −2.
Q: Is the income tax function (0% on first $10,000, 10% on next $40,000, 25% on the rest) continuous or discontinuous?
A: Continuous. The tax is computed on slices of income, and the total tax amount connects smoothly at each bracket boundary. At $10,000 the tax is $0 from both sides. At $50,000 the tax is $4,000 from both sides.
Piecewise analysis leads directly into the Intermediate Value Theorem (IVT), which requires a function to be continuous on a closed interval. If you can show continuity by finding the right constant c, you unlock IVT as a tool for proving roots exist. Later in the course, when you integrate piecewise functions, you will split the integral at the boundary points, and knowing which pieces connect continuously affects whether the antiderivative is itself continuous. Classifying discontinuities also feeds into improper integrals in Calculus II, where infinite discontinuities require special treatment.
piecewise function, piecewise continuity, left-hand limit, right-hand limit, one-sided limit, removable discontinuity, hole, jump discontinuity, step function, infinite discontinuity, vertical asymptote, classifying discontinuities, find c for continuity, continuous real-world function, discontinuous real-world function, gasoline tank, engine gear, parking fee, income tax brackets, MTH 132, Calculus I, Michigan State University, Section 1.8