Difficulty: Intermediate | Prerequisites: Chapter 2.1 (Introduction to Limits)
This chapter extends the limit concept into two directions: what happens when a function's output blows up toward infinity near a particular x-value, and what happens to a function's output when x itself heads toward positive or negative infinity. Both ideas are essential for sketching rational functions accurately and for understanding long-run behaviour in business models (cost per unit as production scales, revenue as market size grows). You should already be comfortable evaluating basic limits and know how to factor polynomials.
Infinite limits describe vertical asymptotes, where a function's value shoots to positive or negative infinity near a specific x-value. Limits at infinity describe horizontal asymptotes, where the function settles toward a finite value (or diverges) as x grows without bound. For polynomials and rational functions, the leading term is what matters in the long run.
Infinite limit
A limit where f(x) increases or decreases without bound as x approaches a particular value a. Written as lim(x→a) f(x) = ∞ or lim(x→a) f(x) = −∞. Think of it as the function "blowing up" near that x-value.
Vertical asymptote
A vertical line x = a where the function's value heads toward positive or negative infinity. Vertical asymptotes come from zeros of the denominator (after cancelling any common factors with the numerator). In simple terms, this is where the graph shoots straight up or straight down.
Limit at infinity
The value that f(x) approaches as x grows arbitrarily large (x → ∞) or arbitrarily large in the negative direction (x → −∞). Think of it as asking "where does the function eventually settle?"
Horizontal asymptote
A horizontal line y = L that the graph of f(x) approaches as x → ∞ or x → −∞. In simple terms, this is the long-run value the function flattens out toward.
Leading term
The term with the highest power of x in a polynomial. This is the term that dominates the function's behaviour for very large or very small values of x. In simple terms, when x is enormous, only the biggest power matters.
One-sided limit
A limit evaluated from only the left side (x → a⁻) or only the right side (x → a⁺) of a point. Think of it as approaching from one direction only, which matters when the function behaves differently on each side of a vertical asymptote.
Infinite limits describe the behaviour of functions that are unbounded near x = a
Vertical asymptotes come from zeros of the denominator
Factor the numerator and denominator first
Cancel any common factors
The remaining zeros of the denominator are where the vertical asymptotes live
To determine the direction (up or down) of a vertical asymptote, check one-sided limits
From the left: lim(x → a⁻) f(x)
From the right: lim(x → a⁺) f(x)
These can go in different directions (one side to +∞, the other to −∞)
Worked example:
f(x) = (x² + x − 2) / (x² − 1)
Factor: f(x) = (x + 2)(x − 1) / ((x + 1)(x − 1))
Cancel (x − 1): f(x) = (x + 2) / (x + 1), with x ≠ 1
Vertical asymptote at x = −1 (the remaining zero of the denominator):
lim(x → −1⁻) f(x) = −∞ (from the left, the function drops to negative infinity)
lim(x → −1⁺) f(x) = +∞ (from the right, the function rises to positive infinity)
Note: x = 1 is a hole, not a vertical asymptote, because the (x − 1) factor cancelled.
Limits at infinity describe what happens to f(x) as x becomes arbitrarily large (positive or negative)
These give the long-run, end behaviour of the function
The key principle: for large x, only the leading term matters
Right-hand behaviour (as x → +∞):
Look at what happens as x grows large and positive
Example: lim(x → ∞) (3x³ − 500x²) = lim(x → ∞) 3x³ = ∞
The −500x² looks large, but 3x³ grows faster and dominates
Left-hand behaviour (as x → −∞):
Look at what happens as x grows large and negative
Example: lim(x → −∞) (3x³ − 500x²) = lim(x → −∞) 3x³ = −∞
A negative number cubed stays negative, so the cube term pulls everything to −∞
Caution: The leading term is the term with the highest power of x, even if it does not appear first in the expression. In 3x³ − 500x², the leading term is 3x³, not −500x².
When you have polynomials dividing (a rational function), transform it by dividing numerator and denominator by the highest power of x in the denominator.
Example: 5x⁴ / 2x³ simplifies to (5/2)x as x → ∞, so the limit is ∞
This technique is the standard method for finding horizontal asymptotes of rational functions
\lim_{x \to a^-} f(x) = -\infty \quad \text{(left-sided infinite limit)}\lim_{x \to a^+} f(x) = +\infty \quad \text{(right-sided infinite limit)}\lim_{x \to \infty} f(x) = L \quad \text{(horizontal asymptote at } y = L\text{)}\lim_{x \to \pm\infty} p(x) = \lim_{x \to \pm\infty} a_n x^n \quad \text{(leading term dominance for polynomial } p\text{)}Factoring shortcut from the notes:
x² + x − 2 = (x + 2)(x − 1)
x² − 1 = (x + 1)(x − 1)
These are standard factorisations you should be able to do quickly on an exam.
Vertical asymptotes model situations where a quantity spikes toward infinity near a critical threshold, such as the cost per item when production approaches zero, or the concentration of a drug as the volume of solution approaches zero.
Horizontal asymptotes describe long-run behaviour in business: average cost per unit often approaches a floor as production scales up, and market-saturation models flatten out as a product reaches its natural ceiling.
Students often think the largest coefficient determines the leading term. It does not. The leading term is the one with the highest power of x, regardless of coefficient size. In 3x³ − 500x², the leading term is 3x³, even though 500 is much larger than 3.
Students sometimes assume a zero of the denominator is always a vertical asymptote. It is not. If the same factor cancels with the numerator, the result is a hole (a removable discontinuity), not an asymptote. Always factor and simplify first.
A common error is evaluating lim(x → −∞) of an odd-power polynomial and getting a positive result. Remember: a negative number raised to an odd power stays negative. So lim(x → −∞) 3x³ = −∞, not +∞.
Students sometimes forget to check both one-sided limits at a vertical asymptote. The function can approach +∞ from one side and −∞ from the other, which changes how you sketch the graph.
⚠️ Expect a question that gives you a rational function and asks you to identify vertical and horizontal asymptotes. Factor first, cancel common factors, then find the remaining denominator zeros.
⚠️ Leading-term dominance is a favourite exam shortcut question. You will likely see a polynomial with a large lower-order coefficient designed to distract you from the actual leading term.
⚠️ One-sided limits at vertical asymptotes are commonly tested. Be ready to determine whether the function goes to +∞ or −∞ from each side.
⚠️ The "divide by the highest power" technique for rational functions at infinity is a standard exam method. Practise it until it is automatic.
True or false: The leading term of 2x² − 1000x⁵ + 7 is 2x².
Fill in the blank: Vertical asymptotes come from zeros of the __________ (after cancelling common factors).
True or false: If (x − 3) cancels from both numerator and denominator, x = 3 is a vertical asymptote.
Fill in the blank: lim(x → −∞) 4x³ = __________.
True or false: A function can approach +∞ from the left of a vertical asymptote and −∞ from the right.
Answers: 1. False (it is −1000x⁵). 2. Denominator. 3. False (it is a hole). 4. −∞. 5. True.
Q: Given f(x) = (x² − 4) / (x² − x − 2), find the vertical asymptote(s) and any holes.
A: Factor: f(x) = (x − 2)(x + 2) / ((x − 2)(x + 1)). Cancel (x − 2). There is a hole at x = 2 and a vertical asymptote at x = −1.
Q: Evaluate lim(x → ∞) (3x³ − 500x²).
A: The leading term is 3x³. As x → ∞, 3x³ → ∞, so the limit is ∞.
Q: Evaluate lim(x → −∞) (3x³ − 500x²).
A: The leading term is still 3x³. As x → −∞, a negative number cubed is negative, so 3x³ → −∞. The limit is −∞.
Q: For f(x) = (x + 2) / (x + 1), determine lim(x → −1⁻) f(x) and lim(x → −1⁺) f(x).
A: As x → −1⁻, the numerator → 1 and the denominator → 0⁻ (small negative), so f(x) → −∞. As x → −1⁺, the numerator → 1 and the denominator → 0⁺ (small positive), so f(x) → +∞.
Q: What is the horizontal asymptote of f(x) = 5x⁴ / (2x⁴ + 3x)?
A: Divide numerator and denominator by x⁴. As x → ∞, f(x) → 5/2. The horizontal asymptote is y = 5/2.
This material connects directly to graphing rational functions (Chapter 2.3 and beyond), where you need asymptotes to sketch accurate graphs. Vertical and horizontal asymptotes are the skeleton of any rational function's graph.
The idea of leading-term dominance reappears in later calculus when you study growth rates, L'Hôpital's Rule, and comparing functions for convergence. Understanding which term "wins" at infinity is a skill that carries through the rest of the course.
In business applications, horizontal asymptotes model diminishing returns: average cost, marginal revenue, and learning curves all eventually flatten, and limits at infinity are how you describe that mathematically.
Infinite limits, limits at infinity, vertical asymptote, horizontal asymptote, end behaviour, leading term, leading term dominance, one-sided limits, left-hand limit, right-hand limit, rational functions, asymptotic behaviour, removable discontinuity, hole in a graph, factoring rational expressions, denominator zeros, unbounded functions, polynomial end behaviour, long-run behaviour, calculus for business, FIU calculus, Chapter 2.2