Difficulty: Introductory to Intermediate | Prerequisites: Part 1 of these notes (two-sided and one-sided limits from graphs).
This picks up where Part 1 left off. Once you can read limits from a plain curve, the next challenge is graphs that are not one smooth line: step functions that jump at fixed intervals, absolute value functions with sharp corners, and piecewise functions stitched from different formulas. These are the graph types most likely to produce one-sided limits, DNE results, and exam questions asking you to solve for a constant.
Step functions (like taxi fares) jump at set intervals, so their one-sided limits almost always disagree at the jump points, giving DNE. Absolute value functions have a sharp V-shape at the vertex, but limits there usually exist because both sides head to the same point. Piecewise functions are where it all comes together: you evaluate each piece's formula separately at the boundary, and if your exam asks you to "find C so the limit exists," you set the two pieces equal.
Step function (greatest integer function / floor function)
A function whose graph looks like a staircase: flat horizontal segments that jump to the next level at certain x-values. Each step is a half-open interval, closed on one end and open on the other. In simple terms, think of it as a pricing table: the cost stays the same across a range, then jumps.
Piecewise function (piecewise-defined function)
A function defined by two or more formulas, each applying on a different interval. For example, f(x) = 2x + 5 for x < −3 and f(x) = x for x ≥ −3. The boundary between pieces is where limit questions focus. Think of it as a function that changes its rules at a specific x-value.
Absolute value function
A function of the form f(x) = |expression|, whose graph forms a V-shape (or a transformed version of one). The vertex of the V is where the expression inside the absolute value equals zero. In simple terms, absolute value makes everything non-negative, so the graph bounces upward at the vertex instead of crossing below the x-axis.
Vertex (of an absolute value function)
The point at the bottom of the V-shape, found by setting the expression inside the absolute value equal to zero and solving for x. For f(x) = |4x + 4|, the vertex is at x = −1, giving the point (−1, 0).
Continuity condition for a piecewise function
A piecewise function is continuous at the boundary x = a if the limit from the left equals the limit from the right equals the actual function value at a. When the exam asks "find C so the limit exists," you are being asked to choose C so that the left-hand and right-hand limits agree.
The source material includes two taxi fare problems that are classic step function limit exercises. The setup is the same each time: a cab charges a base fare for entering, then adds a fixed amount for each 1/5 of a mile. The cost function C(x) is constant on each interval and jumps at x = 0.2, 0.4, 0.6, 0.8, etc.
How step function limits work:
At every jump point, the left-hand and right-hand limits will differ. The graph has a closed dot on one end of each step and an open dot on the other. You read the limit by checking which y-value the graph heads toward from each direction.
Worked example (Problem 17):
Base fare: $3.20, plus $0.30 per 1/5 mile.
C(x) = $3.20 if x = 0, C(x) = $3.50 if 0 < x < 0.2, and so on.
At x = 0.73 (which falls in the middle of a step, not at a jump point):
lim x→0.73⁻ C(x) = $4.40
lim x→0.73⁺ C(x) = $4.40
lim x→0.73 C(x) = $4.40 (both sides agree, so the limit exists)
Worked example (Problem 18):
Base fare: $2.40, plus $0.40 per 1/5 mile.
At x = 0.8 (this is a jump point):
lim x→0.8⁻ C(x) = $4.00 (approaching from the left, the graph is on the lower step)
lim x→0.8⁺ C(x) = $4.40 (approaching from the right, the graph is on the upper step)
lim x→0.8 C(x) = DNE (4.00 ≠ 4.40)
The pattern: If x falls in the middle of a step, both sides agree and the limit exists. If x is exactly at a jump point, the two sides disagree and the limit is DNE.
Absolute value functions produce a V-shaped graph. The vertex (the bottom of the V) is the interesting point for limits, because it is the sharpest change in direction on the graph. Despite the sharp corner, limits at the vertex typically exist because both sides of the V head toward the same y-value.
How to find the vertex: Set the expression inside the absolute value equal to zero and solve.
Worked example (Problem 19): f(x) = |4x + 4|
Vertex: 4x + 4 = 0, so x = −1. The vertex is at (−1, 0).
The graph opens upward (V-shape).
To find lim x→−1 f(x), use a table of values:
From the left: f(−1) = 0.4, f(−1.01) = 0.04, f(−1.001) = 0.004 → approaching 0.
From the right: the values also approach 0.
lim x→−1 f(x) = 0. Both sides agree.
To find lim x→−2 f(x), use a table of values:
From the left: f(−2) = 4, f(−2.01) = 4.04, f(−2.001) = 4.004 → approaching 4.
From the right: f(−1.9) = 3.6, f(−1.99) = 3.96, f(−1.999) = 3.996 → approaching 4.
lim x→−2 f(x) = 4. Both sides agree.
Worked example (Problem 20): f(x) = |3x|
Vertex: 3x = 0, so x = 0. The vertex is at (0, 0).
lim x→0 f(x) = 0. Both sides of the V head to the origin.
lim x→−2 f(x) = 6. At x = −2 the graph is smooth (no corner), so you simply evaluate: |3(−2)| = |−6| = 6.
Key insight: The table-of-values method works well here. Pick x-values very close to a from both sides (e.g. a − 0.1, a − 0.01, a − 0.001 from the left, and similar from the right), evaluate f(x), and see what the outputs converge toward.
Piecewise functions are the bread and butter of limit exam questions. You have two (or more) formulas that apply on different intervals, and the exam asks what happens at the boundary where the rules change.
Step-by-step method:
Identify the boundary x-value (where the conditions switch).
Plug the boundary value into the left-side formula to get the left-hand limit.
Plug the boundary value into the right-side formula to get the right-hand limit.
If the two results are equal, the two-sided limit is that value. If they differ, the limit is DNE.
Graphing tip: When graphing piecewise functions, calculate the y-value each formula gives at the boundary. One formula gives an open dot (the side that does not include the boundary) and the other gives a closed dot (the side that does include the boundary).
Worked example (Problem 22):
f(x) = 2x + 5 for x < −3, and f(x) = x for x ≥ −3. Boundary at x = −3.
Left piece at boundary: 2(−3) + 5 = −1. This gives the point (−3, −1) with an open dot (because the left piece uses strict inequality x < −3).
Right piece at boundary: x = −3. This gives the point (−3, −3) with a closed dot (because x ≥ −3 includes the boundary).
lim x→−3⁻ f(x) = −1 (following the line 2x + 5 from the left).
lim x→−3⁺ f(x) = −3 (following the line x from the right).
lim x→−3 f(x) = DNE, because −1 ≠ −3.
Worked example (Problem 23):
f(x) = 3x − 3 for x < 3, and f(x) = x for x ≥ 3. Boundary at x = 3.
Left piece: 3(3) − 3 = 6. Open dot at (3, 6).
Right piece: x = 3. Closed dot at (3, 3).
lim x→3⁻ f(x) = 6.
lim x→3⁺ f(x) = 3.
lim x→3 f(x) = DNE, because 6 ≠ 3.
This is a common exam question: you are given a piecewise function with an unknown constant C in one of the pieces, and you must find the value of C that makes the overall limit exist at the boundary.
The method:
Evaluate the piece without C at the boundary (this gives you a definite number).
Evaluate the piece with C at the boundary (this gives you an expression involving C).
Set them equal and solve for C.
The logic is straightforward: for the two-sided limit to exist, the left-hand and right-hand limits must be equal. One piece gives a fixed value, the other gives a value that depends on C. Setting them equal forces the two sides to agree.
Worked example (Problem 24):
f(x) = (1/5)x + C for x < 5, and f(x) = −x + 10 for x > 5.
Right piece at boundary: g(x) = −x + 10. Build a table for x-values approaching 5 from the right (x = 5.1, 5.01, 5.001): g(x) = 4.9, 4.99, 4.999 → approaching 5.
Left piece at boundary: h(x) = (1/5)x + C. Build a table for x-values approaching 5 from the left (x = 4.9, 4.99, 4.999): h(x) = 0.98 + C, 0.998 + C, 1 + C → approaching 1 + C.
Set them equal: 1 + C = 5, so C = 4.
Worked example (Problem 25):
f(x) = (1/3)x + C for x < 3, and f(x) = −x + 7 for x > 3.
Right piece: approaching 3 from the right, −x + 7 → 4.
Left piece: approaching 3 from the left, (1/3)x + C → 1 + C.
Set equal: 1 + C = 4, so C = 3.
The source material also works through additional boundary values (x = 4, x = 5, x = 2) with different piecewise formulas, all following the same pattern. In each case the method is identical: find the numerical limit from one side, find the expression with C from the other side, and solve.
Evaluating a piecewise limit at the boundary x = a:
Left-hand limit: substitute x = a into the formula that applies for x < a.
Right-hand limit: substitute x = a into the formula that applies for x > a (or x ≥ a).
Compare: if the results match, the two-sided limit equals that value. If not, DNE.
Finding the vertex of f(x) = |mx + b|:
Set mx + b = 0 and solve for x. The vertex is at (x, 0). The graph opens upward with slope m on the right side and slope −m on the left side.
Solving for C in a piecewise continuity problem:
Set lim x→a⁻ f(x) = lim x→a⁺ f(x). One side gives a number, the other gives an expression in C. Solve the resulting equation.
Table-of-values approach:
When in doubt, pick x-values that get progressively closer to a from each side (e.g. a ± 0.1, a ± 0.01, a ± 0.001). Evaluate f(x) at each. The pattern in the outputs tells you what the limit is.
Students plug the boundary value into the wrong piece. For a left-hand limit, use the formula that governs x-values less than the boundary. For a right-hand limit, use the formula that governs x-values greater than the boundary. The strict vs. non-strict inequality in the definition tells you which is which.
Students assume absolute value functions always have limits that exist. They do at the vertex and at points away from it, but only because both sides of the V converge. It is the convergence that matters, not the function type. If an absolute value function were embedded in a piecewise definition, a DNE result would be possible.
Students think finding C makes the function continuous. Finding C so the limit exists is a necessary step, but continuity also requires f(a) to be defined and equal to the limit. Some exam questions distinguish between these. At this stage you are usually just asked for the limit to exist, which is the left = right condition.
Students evaluate the step function at the exact jump point instead of taking one-sided limits. At a jump, the function value is whatever the closed dot says, but the limit question asks about what the graph approaches from each side. These are not the same thing.
⚠️ Piecewise limit problems appear on nearly every Brief Calculus exam. You will be given a piecewise function, asked for one or both one-sided limits, the two-sided limit, and sometimes asked to graph the function with correct open/closed dots.
⚠️ "Find C so the limit exists" is a favourite exam question because it tests whether you understand the left = right condition. Expect it.
⚠️ Step function (taxi cab) problems test whether you can distinguish between a limit at a jump point (DNE) and a limit in the middle of a step (exists). Read the graph carefully.
⚠️ On absolute value problems, the exam often asks for the limit at the vertex and at some other point. The vertex is the one that requires more thought. Away from the vertex, just evaluate the function directly.
True or False: For a step function, the limit always exists at every x-value. Answer: False. At each jump point, the left-hand and right-hand limits disagree, so the two-sided limit is DNE.
Fill in the blank: The vertex of f(x) = |4x + 4| is at the point ____. Answer: (−1, 0). Set 4x + 4 = 0, so x = −1, and f(−1) = 0.
True or False: To find C so that lim x→5 f(x) exists for a piecewise function, you set the left-hand limit equal to the right-hand limit and solve. Answer: True.
Fill in the blank: For f(x) = 3x − 3 when x < 3 and f(x) = x when x ≥ 3, the left-hand limit at x = 3 is ____. Answer: 6. Plug x = 3 into the left piece: 3(3) − 3 = 6.
True or False: If a piecewise function has f(x) = x + 2 for x < 4 and f(x) = x + 2 for x ≥ 4, the limit at x = 4 does not exist because the function changes rules. Answer: False. Both pieces give f(4) = 6, so both one-sided limits are 6, and the two-sided limit is 6.
Q: A taxi charges $2.50 plus $0.50 for each 1/4 mile. The cost function C(x) is a step function. What is lim x→0.5 C(x)?
A: x = 0.5 is a jump point (it falls exactly at 2/4 of a mile). The left-hand limit gives the cost for the interval just below 0.5, and the right-hand limit gives the cost for the interval just above 0.5. Since these differ, lim x→0.5 C(x) = DNE.
Q: For f(x) = |2x − 6|, find lim x→3 f(x).
A: The vertex is at x = 3 (set 2x − 6 = 0). Both sides of the V head toward f(3) = 0, so the limit is 0.
Q: f(x) = x + 1 for x < 2, and f(x) = 5 − x for x ≥ 2. Find lim x→2 f(x).
A: Left-hand limit: plug x = 2 into x + 1 = 3. Right-hand limit: plug x = 2 into 5 − x = 3. Both sides give 3, so lim x→2 f(x) = 3.
Q: f(x) = (1/4)x + C for x < 4, and f(x) = −x + 7 for x ≥ 4. Find C so that lim x→4 f(x) exists.
A: Right-hand limit: −(4) + 7 = 3. Left-hand limit: (1/4)(4) + C = 1 + C. Set 1 + C = 3, so C = 2.
Q: A piecewise function has lim x→5⁻ f(x) = 7 and lim x→5⁺ f(x) = 7, but f(5) = 10. Does the limit exist, and if so, what is it?
A: Yes, the limit exists and equals 7. Both one-sided limits agree, so the two-sided limit is 7 regardless of f(5).
This connects to continuity directly: a function is continuous at x = a precisely when lim x→a f(x) = f(a). Piecewise continuity problems ("find C") are really continuity problems in disguise, and understanding them here prepares you for the formal definition.
This connects to differentiability because a function must be continuous to be differentiable. Absolute value functions are continuous at the vertex but not differentiable there (the sharp corner means no single tangent line exists). That distinction comes up in derivative chapters.
This connects to Part 1 of these notes, which covers the foundational concepts of two-sided and one-sided limits from graphs. If any of the material here felt unfamiliar, revisit Part 1 first.
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