Difficulty: Intermediate | Prerequisites: limits, limit laws, continuity (see companion study notes).
Asymptotes describe the long-range and boundary behaviour of functions: what happens as x grows very large (horizontal asymptotes) or as x approaches a point where the function blows up (vertical asymptotes). The derivative, meanwhile, captures the instantaneous rate of change at a single point. Both topics rely heavily on limits and form the second half of the midterm material in MTH 132. If you can find asymptotes of rational functions and compute a derivative using the limit definition, you are well placed for the exam.
Horizontal asymptotes come from limits at infinity; vertical asymptotes come from one-sided limits that blow up to ±∞. For rational functions, compare the degrees of numerator and denominator to find horizontals, and set the denominator to zero to find verticals (but check for holes first). The derivative at a point is the limit of the difference quotient as h → 0, and it gives the slope of the tangent line.
Horizontal asymptote
A horizontal line y = L that the graph of f approaches as x → ∞ or x → −∞. Formally, y = L is a horizontal asymptote if lim(x→∞) f(x) = L or lim(x→−∞) f(x) = L.
In simple terms, it is the value the function settles towards in the long run.
Vertical asymptote
A vertical line x = a where f(x) → ∞ or f(x) → −∞ as x approaches a from one or both sides.
Think of it as: the function blows up near that x-value. The graph shoots upward or downward without bound.
Hole vs. vertical asymptote
When a factor cancels from both numerator and denominator, the zero in the denominator produces a hole (removable discontinuity), not a vertical asymptote. A vertical asymptote occurs only when the denominator is zero but the numerator is not (after cancellation).
Derivative (limit definition)
f'(x) = lim(h→0) [f(x+h) − f(x)] / h. This gives the instantaneous rate of change of f at x, or equivalently, the slope of the tangent line to the graph at that point.
In simple terms, zoom in on the curve until it looks like a straight line; the slope of that line is the derivative.
Tangent line
The line that just touches the curve at a single point and has the same slope as the curve there. Its equation: y − f(a) = f'(a)(x − a).
For f(x) = P(x)/Q(x) where P and Q are polynomials:
If degree(P) < degree(Q): horizontal asymptote at y = 0.
If degree(P) = degree(Q): horizontal asymptote at y = (leading coefficient of P) / (leading coefficient of Q).
If degree(P) > degree(Q): no horizontal asymptote (the function grows without bound).
Example: f(x) = (3x² + 5x)/(x² − 1).
Degrees are equal (both 2). Horizontal asymptote: y = 3/1 = 3.
Example: g(x) = (1 − x⁴)/(x⁴ + 2x).
Degrees are equal (both 4). Leading coefficients: −1 and 1. Horizontal asymptote: y = −1/1 = −1.
To justify using limits: divide every term by x⁴. As x → ∞, all terms with x in the denominator vanish, leaving −1/1 = −1.
Set the denominator equal to zero. For each solution x = a, check whether the numerator is also zero there.
Numerator ≠ 0 at x = a → vertical asymptote at x = a.
Numerator = 0 at x = a → factor cancels, giving a hole (removable discontinuity), not a vertical asymptote.
Example: f(x) = x/(x² − 1) = x/((x−1)(x+1)).
Denominator is zero at x = 1 and x = −1. The numerator is not zero at either point. Vertical asymptotes at x = 1 and x = −1.
Example: f(x) = (2x² − 2)/(x² − 3x + 2).
Factor: numerator = 2(x² − 1) = 2(x−1)(x+1). Denominator = (x−1)(x−2).
At x = 1: both numerator and denominator are zero. The (x−1) cancels, leaving 2(x+1)/(x−2). So x = 1 is a hole, not an asymptote. The y-value of the hole is 2(2)/(1−2) = −4.
At x = 2: denominator is zero but the numerator 2(2+1) = 6 ≠ 0. Vertical asymptote at x = 2.
Horizontal asymptote: degrees equal, leading coefficients 2 and 1, so y = 2.
The line x = a is a vertical asymptote of f(x) if at least one of the following holds:
lim(x→a⁺) f(x) = +∞ or −∞
lim(x→a⁻) f(x) = +∞ or −∞
f'(a) = lim(h→0) [f(a+h) − f(a)] / h
Setting up (not solving): f(x) = x².
f'(x) = lim(h→0) [(x+h)² − x²] / h
Full computation: f(x) = x² − 4x at x = 4.
f(4) = 16 − 16 = 0.
f(4+h) = (4+h)² − 4(4+h) = 16 + 8h + h² − 16 − 4h = 4h + h².
Difference quotient: [f(4+h) − f(4)] / h = (4h + h²) / h = 4 + h.
Take the limit: lim(h→0) (4 + h) = 4. So f'(4) = 4.
Once you have f'(a), the tangent line at x = a is:
y − f(a) = f'(a)(x − a)
Example continued: f(x) = x² − 4x at x = 4.
f(4) = 0, f'(4) = 4. Tangent line: y − 0 = 4(x − 4), i.e. y = 4x − 16.
Horizontal asymptote (equal degrees): y = (leading coeff. of numerator) / (leading coeff. of denominator)
Vertical asymptote: x = a where the denominator is zero and the numerator is not (after cancellation)
Derivative: f'(a) = lim(h→0) [f(a+h) − f(a)] / h
Tangent line: y − f(a) = f'(a)(x − a)
Horizontal asymptotes describe long-run equilibria: the steady-state temperature of a cooling object, the carrying capacity in a population model, or the terminal velocity of a falling object. The derivative measures how fast any quantity is changing at a given instant, which is the basis of velocity, marginal cost, and reaction rates in chemistry.
Students sometimes write x = 1 as a horizontal asymptote when they mean y = 1. Horizontal asymptotes are horizontal lines (y = constant); vertical asymptotes are vertical lines (x = constant).
A zero in the denominator does not always produce a vertical asymptote. If the same factor cancels from the numerator, you get a hole instead.
Students sometimes think a function cannot cross a horizontal asymptote. It can, particularly for non-monotonic functions; the asymptote describes end behaviour only.
When computing the derivative from the limit definition, students often forget to expand (x+h)² fully or drop the h² term too early.
⚠️ Finding horizontal and vertical asymptotes of a rational function is a reliable long-answer question (Question 15 style). Know the degree-comparison rule and always factor to check for holes.
⚠️ The limit definition of the derivative appears both as a "set up the expression" short answer and as a full computation with tangent line equation in the long-answer section.
⚠️ The formal definition of a vertical asymptote using one-sided limits is tested as a short-answer definition question.
⚠️ Distinguishing a hole from a vertical asymptote is a common trap. Factor first, cancel, then check what remains in the denominator.
Fill in the blank: For f(x) = (5x³ + 1)/(2x³ − x), the horizontal asymptote is y = ______.
True or False: If the denominator of a rational function is zero at x = 3 and the numerator is also zero at x = 3, there is always a vertical asymptote at x = 3.
Fill in the blank: The derivative f'(a) gives the slope of the ______ line at x = a.
True or False: A horizontal asymptote describes behaviour as x → ±∞.
Fill in the blank: The tangent line equation at x = a is y − f(a) = ______ · (x − a).
Answers: 1. 5/2. 2. False (it may be a hole if the factor cancels). 3. tangent. 4. True. 5. f'(a).
Q: Find the horizontal asymptote of f(x) = (3x² + 5x)/(x² − 1).
A: Degrees of numerator and denominator are both 2. Horizontal asymptote: y = 3/1 = 3.
Q: Identify all vertical asymptotes of f(x) = x/(x² − 1).
A: Denominator factors as (x−1)(x+1). Both give zero, and the numerator x is nonzero at x = 1 and x = −1. Vertical asymptotes at x = 1 and x = −1.
Q: For f(x) = (2x² − 2)/(x² − 3x + 2), find all asymptotes and holes.
A: Factor: 2(x−1)(x+1) / (x−1)(x−2). Cancel (x−1) → hole at x = 1. Remaining denominator zero at x = 2, numerator 2(3) = 6 ≠ 0 → vertical asymptote at x = 2. Degrees equal → horizontal asymptote y = 2.
Q: Using the limit definition, set up f'(x) for f(x) = x².
A: f'(x) = lim(h→0) [(x+h)² − x²] / h.
Q: Compute the slope of the tangent line to f(x) = x² − 4x at x = 4 using the limit definition, then find the tangent line equation.
A: f'(4) = lim(h→0) [(4+h)² − 4(4+h) − 0] / h = lim(h→0) (4h + h²)/h = lim(h→0) (4+h) = 4. Since f(4) = 0, the tangent line is y = 4(x − 4), i.e. y = 4x − 16.
Q: Define "vertical asymptote" formally using one-sided limits.
A: The line x = a is a vertical asymptote of f if lim(x→a⁺) f(x) = ±∞ or lim(x→a⁻) f(x) = ±∞ (or both).
Q: Determine lim(x→1⁺) 1/(x−1).
A: As x → 1⁺, (x−1) is a small positive number, so 1/(x−1) → +∞.
The derivative introduced here through the limit definition will be computed using shortcut rules (power rule, product rule, chain rule) in the next unit. Asymptotes reappear when you sketch curves using derivatives: horizontal asymptotes define the end behaviour of the sketch, and vertical asymptotes mark where the function is undefined. The tangent line is the simplest linear approximation of a function near a point, which leads to linearisation and differentials.
horizontal asymptote, vertical asymptote, limits at infinity, end behaviour, rational function asymptotes, hole vs asymptote, removable discontinuity, degree comparison, derivative, limit definition of the derivative, difference quotient, tangent line, tangent line equation, slope of tangent, instantaneous rate of change, f'(x), MTH 132, Calculus I, Michigan State University