Warmup 2.3, Calculus for Business (FIU), Section 2.3
Source: Warmup 2.3 worksheet | Tags: continuity, limits, piecewise functions, rational functions, calculus for business
Difficulty: Introductory to Intermediate | Prerequisites: Limits (Section 2.1 and 2.2), one-sided limits, evaluating basic polynomial and rational expressions.
Continuity is the bridge between limits and everything that follows in calculus, especially derivatives. A function is continuous at a point when its graph passes through that point without a hole, jump, or vertical asymptote. You need to be comfortable evaluating limits (including one-sided limits) before this section will make sense. If you struggled with Section 2.2, revisit it first. This topic sets up the formal language you will use when discussing differentiability later in the course.
A function is continuous at x = a when three things are true at once: the limit exists, the function value exists, and they are equal. Most exam questions test whether you can check all three conditions, spot which one fails, and identify where rational or piecewise functions break continuity.
Continuity at a point
A function f is continuous at x = a if and only if all three conditions hold: the limit of f(x) as x approaches a exists, f(a) is defined, and the limit equals f(a). In simple terms, you can draw through the point without lifting your pen.
One-sided limit
The value a function approaches from one direction only. The left-hand limit (x → a⁻) comes from values less than a; the right-hand limit (x → a⁺) comes from values greater than a. Think of it as checking what the graph does from each side separately.
Limit does not exist (DNE)
A limit fails to exist at a point when the left-hand and right-hand limits are not equal, or when the function grows without bound. In simple terms, the two sides of the graph disagree about where they are heading.
Removable discontinuity (hole)
A point where the limit exists but the function either is not defined there or takes a different value. Think of it as a single missing dot on an otherwise smooth curve.
Jump discontinuity
A point where both one-sided limits exist but are not equal. The graph literally jumps from one height to another.
Infinite discontinuity (vertical asymptote)
A point where the function grows without bound, typically caused by division by zero. The graph shoots off toward positive or negative infinity.
Piecewise function
A function defined by different formulas on different intervals. Continuity questions almost always target the boundary points where the formula changes.
Rational function
A function written as one polynomial divided by another. These are continuous everywhere except where the denominator equals zero.
A function f(x) is continuous at x = a if and only if all three of the following hold:
The limit exists. lim(x→a) f(x) must exist. This means the left-hand limit and right-hand limit must both exist and be equal to each other.
The function value exists. f(a) must be defined. The function must actually produce a value at x = a (no hole, no division by zero).
The limit equals the function value. lim(x→a) f(x) = f(a). The value the function is approaching must match the value the function actually takes.
All three must hold simultaneously. If even one fails, the function is not continuous at that point. The order matters for efficiency on exams: check whichever is easiest to evaluate first, and stop as soon as one fails.
When given a graph, you check continuity at each point of interest by reading off the one-sided limits visually and comparing them to the function value (the solid dot).
At x = -3
Left-hand limit: as x → -3⁻, f(x) → 3
Right-hand limit: as x → -3⁺, f(x) → 3
The one-sided limits agree, so lim(x→-3) f(x) = 3 at first glance. However, the worksheet marks the limit as DNE here, which indicates a subtlety in the graph (such as the function value not matching, or an open circle). Always check the original graph carefully.
At x = -1
Left-hand limit: as x → -1⁻, f(x) → 1
Right-hand limit: as x → -1⁺, f(x) → 2
The one-sided limits disagree (1 ≠ 2), so lim(x→-1) f(x) DNE. This is a jump discontinuity.
At x = 2
Left-hand limit: as x → 2⁻, f(x) → 5
Right-hand limit: as x → 2⁺, f(x) → 5
Both sides approach 5, but the worksheet notes lim(x→2) f(x) = 3 = DNE, suggesting the overall limit is evaluated differently from the one-sided values shown (possibly f(2) = 3 with a removable discontinuity, or a notation concern). The key takeaway: even when one-sided limits seem to agree, check f(a) separately.
At x = 4
Left-hand limit: as x → 4⁻, f(x) → 4
Right-hand limit: as x → 4⁺, f(x) → 2
The one-sided limits disagree (4 ≠ 2), so lim(x→4) f(x) DNE. Another jump discontinuity.
Reading strategy for graph problems
Trace the curve from the left to find the left-hand limit.
Trace the curve from the right to find the right-hand limit.
Look for solid dots (defined values) vs. open circles (undefined or different values).
Compare: do the two one-sided limits match each other, and do they match the solid dot?
Piecewise functions are defined by different rules on different intervals. The boundary point (where the rule changes) is where continuity can break.
Worked example from Warmup 2.3:
f(x) = (-x - 4)/3 when x < 1, and f(x) = x² - 3x when x ≥ 1.
Check continuity at x = 1:
Left-hand limit (x → 1⁻): Use the first piece. Plug in x = 1: (-1 - 4)/3 = -5/3.
Right-hand limit (x → 1⁺): Use the second piece. Plug in x = 1: 1² - 3(1) = 1 - 3 = -2.
Comparison: -5/3 ≠ -2, so the limit does not exist. The function is not continuous at x = 1.
Note: the worksheet also checks (-(-2) - 4)/3 = -2/3 ≠ -2. The exact computation depends on how you read the handwritten formula, but the conclusion is the same: the two sides produce different values, so continuity fails.
General method for piecewise continuity checks
Identify the boundary value (here, x = 1).
Compute the left-hand limit using the formula that applies for x < boundary.
Compute the right-hand limit using the formula that applies for x ≥ boundary (or x > boundary).
If left ≠ right, the limit DNE and the function is discontinuous.
If left = right, the limit exists. Then check that f(boundary) equals that limit.
Rational functions (a polynomial divided by a polynomial) are continuous everywhere their denominator is nonzero. The only points of discontinuity are where the denominator equals zero.
Example (a): h(x) = 11 - 7/x
The denominator is x. Set x = 0: division by zero.
Discontinuous at x = 0. Continuous everywhere else.
Example (b): g(x) = (x + 17) / (4 - 9x²)
Set the denominator equal to zero: 4 - 9x² = 0 → 9x² = 4 → x² = 4/9 → x = ±2/3.
Discontinuous at x = -2/3 and x = 2/3. Continuous at all other real numbers.
Example (c): f(x) = (x² - x - 12) / (x² + 4)
Set the denominator equal to zero: x² + 4 = 0 → x² = -4. No real solutions.
Continuous for all real numbers. The denominator is always positive (at least 4), so it never equals zero.
Quick method
Write the denominator.
Solve denominator = 0.
If real solutions exist, the function is discontinuous at those x-values.
If no real solutions exist, the function is continuous everywhere.
Students often think that if f(a) exists, the function must be continuous at a. It does not follow. You still need the limit to exist and to equal f(a). A function can be defined at a point and still have a hole or a jump there.
Students often confuse "the limit exists" with "the function is defined." These are independent conditions. A limit can exist at a point where the function is undefined (removable discontinuity), and a function can be defined at a point where the limit does not exist (jump discontinuity).
Students sometimes check only one side of a piecewise boundary. You must evaluate both the left-hand and right-hand limits, then compare. Checking only the piece that includes the boundary point is not enough.
Students often assume that if both one-sided limits exist, the overall limit must also exist. The overall limit exists only when the two one-sided limits are equal to each other.
⚠️ The three conditions for continuity are almost guaranteed to appear on the exam in some form. Know them cold, in order, with the correct notation.
⚠️ Graph-reading questions test whether you can distinguish open circles from closed circles, and whether you check both sides before claiming a limit exists.
⚠️ Piecewise function problems at boundary points are a staple. The exam will likely give you a piecewise function and ask you to determine continuity at the boundary. Always evaluate both pieces at that point.
⚠️ For rational functions, be ready to factor a denominator and solve for zeros quickly. The question "where is this function continuous?" is really asking "where does the denominator equal zero?"
Real-world application: Continuity matters in business modelling because discontinuous functions produce unreliable predictions. A cost function with a jump discontinuity, for instance, means a small change in production volume causes a sudden, disproportionate change in cost, which complicates budgeting and forecasting.
True or False: If lim(x→a) f(x) exists, then f is continuous at x = a.
Answer: False. The limit existing is only one of the three conditions. You also need f(a) to be defined and equal to the limit.
Fill in the blank: A rational function is continuous everywhere except where the ______ equals zero.
Answer: denominator.
True or False: If f(3) = 7, then lim(x→3) f(x) = 7.
Answer: False. The function value and the limit are independent. The limit depends on what the function approaches, not the value assigned at that point.
Fill in the blank: When the left-hand limit and right-hand limit at a point are not equal, the point has a ______ discontinuity.
Answer: jump.
True or False: The function f(x) = (x² - 9)/(x² + 1) is continuous for all real numbers.
Answer: True. The denominator x² + 1 is always at least 1, so it never equals zero.
Q: State the three conditions required for a function f to be continuous at x = a.
A: (1) lim(x→a) f(x) exists. (2) f(a) is defined. (3) lim(x→a) f(x) = f(a).
Q: A function is defined by f(x) = 2x + 1 for x < 3 and f(x) = x² - 2 for x ≥ 3. Is f continuous at x = 3?
A: Left-hand limit: 2(3) + 1 = 7. Right-hand limit: 3² - 2 = 7. f(3) = 7. All three conditions are met, so yes, f is continuous at x = 3.
Q: Where is g(x) = (x - 5) / (x² - 16) discontinuous?
A: Set the denominator to zero: x² - 16 = 0 → (x - 4)(x + 4) = 0 → x = 4 or x = -4. The function is discontinuous at x = 4 and x = -4.
Q: A graph shows f approaching 3 from the left of x = 2 and 5 from the right of x = 2, with a solid dot at (2, 4). Is f continuous at x = 2? Why or why not?
A: No. The left-hand limit (3) does not equal the right-hand limit (5), so the overall limit does not exist. Condition 1 fails.
Q: Is h(x) = (x² + 3x + 2) / (x² + 5) continuous for all real numbers?
A: The denominator x² + 5 is always at least 5 (never zero), so h is continuous for all real numbers.
This connects to derivatives because a function must be continuous at a point to be differentiable there (though continuity alone is not sufficient for differentiability). When you reach Section 2.4 or Chapter 3, the continuity check becomes a prerequisite step.
This also connects to the Intermediate Value Theorem (IVT), which requires continuity on a closed interval. If you are asked to show a function has a root in some interval, you will need to confirm continuity first.
In business applications, continuity assumptions underpin most optimisation models. If a cost, revenue, or profit function has a discontinuity, standard calculus techniques (finding critical points, second derivative test) may not apply at that point.
Continuity, continuous function, discontinuity, removable discontinuity, jump discontinuity, infinite discontinuity, vertical asymptote, three conditions for continuity, one-sided limits, left-hand limit, right-hand limit, limit does not exist, DNE, piecewise function, piecewise continuity, rational function continuity, denominator equals zero, hole in a graph, calculus for business, FIU, Section 2.3, Warmup 2.3