Difficulty: Intermediate | Prerequisites: Limits at Infinity study notes (companion document), factoring polynomials.
This picks up where the limits-at-infinity notes leave off. Asymptotes and holes describe the structural features of a rational function's graph, while continuity formalises the idea of a function having no breaks. Together, they form the toolkit you need for curve-sketching and for understanding where calculus techniques (derivatives, integrals) can and cannot be applied. You should be comfortable factoring quadratics and evaluating basic limits before reading this.
Factor both numerator and denominator of a rational function. Factors that cancel produce holes (with a specific (x, y) coordinate). Factors that remain in the denominator produce vertical asymptotes. Horizontal asymptotes come from the degree-comparison rule. Continuity at a point requires three things: the function is defined there, the limit exists, and the two are equal.
Horizontal asymptote
A horizontal line y = L that the graph approaches as x → ∞ or x → −∞. Found using the degree-comparison rule from the companion notes. Think of it as: the altitude the function cruises toward when x is very large.
Vertical asymptote
A vertical line x = a where the function's value blows up to ±∞. Occurs where the denominator equals zero after all common factors have been cancelled. In simple terms, a wall in the graph where the curve shoots up or down without bound.
Hole (removable discontinuity)
A point (a, L) missing from the graph because a factor cancels from both numerator and denominator. The function is undefined at x = a, but the limit as x → a still equals L. You "find" the hole by cancelling the common factor and then substituting x = a into the simplified function to get the y-coordinate.
Piecewise function
A function defined by different formulas on different intervals. For example, one rule for x < −1, another for x = −1, another for x > −1. Think of it as: a function built from separate pieces stitched together.
Three conditions for continuity at x = c
f(c) is defined (the function has a value at c).
lim (x → c) f(x) exists (the left-hand and right-hand limits agree).
lim (x → c) f(x) = f(c) (the limit equals the function's value).
All three must hold. If any one fails, f is discontinuous at c.
Consider f(x) = (x² + x − 12) / (x² − 9).
Step 1: Factor everything.
Numerator: x² + x − 12 = (x + 4)(x − 3)
Denominator: x² − 9 = (x + 3)(x − 3)
Step 2: Cancel common factors.
The factor (x − 3) appears in both. Cancel it to get the simplified form: (x + 4) / (x + 3), with the note that x ≠ 3.
Step 3: Identify the hole.
The cancelled factor (x − 3) means there is a hole at x = 3. To find the y-coordinate, substitute x = 3 into the simplified function: (3 + 4) / (3 + 3) = 7/6. So the hole is at (3, 7/6). Expressed using limits: lim (x → 3) f(x) = 7/6, but f(3) is undefined.
Step 4: Identify vertical asymptotes.
The remaining denominator factor is (x + 3), which equals zero at x = −3. Since this factor did not cancel, x = −3 is a vertical asymptote. Using limits: lim (x → −3) f(x) = ±∞ (the function blows up).
Step 5: Identify horizontal asymptotes.
The original numerator and denominator both have degree 2. Leading coefficients are both 1. So lim (x → ±∞) f(x) = 1/1 = 1. The horizontal asymptote is y = 1.
To check whether f(x) is continuous at x = c:
Verify f(c) is defined. If f(c) does not exist, stop: f is discontinuous at c.
Evaluate lim (x → c) f(x). Check that the left-hand limit and right-hand limit agree. If they do not, the limit does not exist and f is discontinuous at c.
Check that f(c) equals the limit. If they match, f is continuous at c.
Given:
f(x) = x² − 6 for x < −1
f(x) = Ax − 12 for x = −1
f(x) = −5 for x > −1
Find A so that f is continuous everywhere. The only point to worry about is x = −1, where the pieces meet.
Step 1: f(−1) must be defined.
f(−1) = A(−1) − 12 = −A − 12. This is defined for any value of A, so no problem here.
Step 2: The limit as x → −1 must exist.
Left-hand limit: lim (x → −1⁻) (x² − 6) = (−1)² − 6 = 1 − 6 = −5.
Right-hand limit: lim (x → −1⁺) (−5) = −5.
Both sides give −5, so the limit exists and equals −5.
Step 3: f(−1) must equal the limit.
Set −A − 12 = −5. Solving: −A = 7, so A = −7.
Students often assume that if the denominator is zero at a point, there must be a vertical asymptote. This is only true if the numerator is not also zero there (or if the common factor does not fully cancel). Always factor first.
A hole and a vertical asymptote cannot occur at the same x-value for the same factor. If a factor cancels completely, it is a hole. If it remains, it is a vertical asymptote.
When asked to "use limits" to find asymptotes and holes, students sometimes just state the answer without writing the limit notation. The homework warns that failure to use limits results in zero credit. Write lim (x → a) f(x) = L explicitly.
In the continuity test, students sometimes check only one or two of the three conditions and declare the function continuous. All three must hold. A function can have a defined value at a point but still be discontinuous if the limit disagrees with that value.
⚠️ Question 4 on the homework requires limit notation for every asymptote and hole. Write the limit expression, evaluate it, and then state the conclusion ("so y = 1 is a horizontal asymptote"). Skipping the limit expression means zero credit.
⚠️ The three-step continuity test is a standard exam question. Be able to state the three conditions from memory and apply them to a piecewise function.
⚠️ When solving for a constant like A, show all three steps separately, not just the algebra. The grading rubric rewards the structure.
⚠️ Holes require both coordinates: (x, y). Stating only x = 3 without computing y = 7/6 is incomplete.
True or false: If (x − 5) cancels from both the numerator and denominator of a rational function, x = 5 is a vertical asymptote. (False. It is a hole.)
Fill in the blank: The three conditions for continuity at x = c are: f(c) is ________, the limit as x → c ________, and ________. (Defined; exists; the limit equals f(c).)
True or false: A function can have at most one horizontal asymptote. (False. A function can have two, one as x → ∞ and a different one as x → −∞, though for the rational functions in this course, both directions give the same value.)
Fill in the blank: To find the y-coordinate of a hole at x = a, substitute x = a into the ________ form of the function. (Simplified / cancelled.)
Q: Find all asymptotes and holes of g(x) = (x² − 4) / (x² + x − 6), using limits.
A: Factor: numerator = (x − 2)(x + 2), denominator = (x − 2)(x + 3). The factor (x − 2) cancels, so there is a hole at x = 2. Substituting into the simplified form (x + 2) / (x + 3): (2 + 2) / (2 + 3) = 4/5. Hole at (2, 4/5). The remaining denominator factor (x + 3) = 0 at x = −3, so x = −3 is a vertical asymptote: lim (x → −3) g(x) = ±∞. Both numerator and denominator have degree 2 with leading coefficient 1, so lim (x → ±∞) g(x) = 1. Horizontal asymptote: y = 1.
Q: State the three conditions for f(x) to be continuous at x = c.
A: (1) f(c) is defined. (2) lim (x → c) f(x) exists. (3) lim (x → c) f(x) = f(c).
Q: Given f(x) = {3x + 1 for x < 2, Bx² for x ≥ 2}, find B so that f is continuous at x = 2.
A: Left-hand limit: lim (x → 2⁻) (3x + 1) = 7. Right-hand limit: lim (x → 2⁺) (Bx²) = 4B. For the limit to exist, 4B = 7, so B = 7/4. Check: f(2) = B(4) = 7. The limit equals f(2), so f is continuous at x = 2 when B = 7/4.
Q: Does f(x) = (x² + x − 12) / (x² − 9) have a vertical asymptote at x = 3?
A: No. Factor: (x + 4)(x − 3) / (x + 3)(x − 3). The (x − 3) cancels, so x = 3 is a hole, not a vertical asymptote. The hole is at (3, 7/6).
Continuity matters in business modelling whenever a function must be smooth across thresholds. Tax brackets, tiered pricing, and insurance deductibles are all piecewise functions, and checking continuity at each breakpoint ensures the model does not produce jumps or undefined values. Asymptotes show up in cost-per-unit analysis: as production volume grows, the average cost per unit approaches a floor (horizontal asymptote) but never quite reaches it.
Continuity is a prerequisite for the Intermediate Value Theorem, which you will meet in the next section. It is also required for differentiability: a function that is not continuous at a point cannot have a derivative there. The factoring techniques used to find holes are the same ones you will use when simplifying derivatives of rational functions later in Chapter 3.
Asymptotes, vertical asymptote, horizontal asymptote, hole in graph, removable discontinuity, continuity test, three conditions for continuity, piecewise function continuity, finding A for continuity, rational function asymptotes, factoring to find holes, calculus for business FIU, Section 2.2, Section 2.3, continuous everywhere, limit notation for asymptotes