Difficulty: Intermediate | Prerequisites: Basic limit evaluation (Section 2.1), polynomial and rational function algebra.
Limits at infinity describe what happens to a function's output as the input grows without bound (positive or negative). This sits at the heart of curve-sketching and asymptote analysis, which you will use constantly through differential calculus and optimisation problems later in the course. You should already be comfortable evaluating simple limits by substitution and factoring. If Section 2.1 felt shaky, revisit it before diving in here.
Limits at infinity tell you the long-run behaviour of a function. For polynomials, the leading term decides everything. For rational functions, compare the degree of the numerator and denominator: same degree gives a finite ratio, higher numerator degree gives infinity, higher denominator degree gives zero.
Limit at infinity
The value a function f(x) approaches as x increases or decreases without bound. Written as lim (x → ∞) f(x) or lim (x → −∞) f(x). Think of it as: where does the graph settle (or head) when you zoom out far to the right or left?
Leading term
The term in a polynomial with the highest power of x. In −5x⁴ − 3x + 1, the leading term is −5x⁴. In simple terms, this is the term that dominates all the others when x is very large.
Degree of a polynomial
The highest exponent on x. For −5x⁴ − 3x + 1, the degree is 4. Think of it as: how fast this polynomial grows compared to others.
Horizontal asymptote
A horizontal line y = L that the graph of f(x) approaches as x → ∞ or x → −∞. In simple terms, it is the long-run ceiling or floor the function settles toward but never quite reaches.
Vertical asymptote
A vertical line x = a where f(x) blows up to ±∞. This happens when the denominator of a rational function equals zero and the numerator does not. Think of it as: a wall the graph cannot cross, shooting upward or downward instead.
Hole (removable discontinuity)
A single missing point on the graph, caused by a factor that cancels from both the numerator and denominator. The function is undefined there, but the limit still exists. In simple terms, the graph has a tiny gap you could "fill in" to make it continuous.
Rational function
A function written as one polynomial divided by another, f(x) = p(x) / q(x). Think of it as: a fraction whose top and bottom are both polynomials.
Continuity at a point
A function f(x) is continuous at x = c when three conditions hold: f(c) is defined, the limit as x → c exists, and f(c) equals that limit. In simple terms, you can draw through the point without lifting your pen.
The leading term controls the limit. Every other term becomes negligible as x → ±∞.
To evaluate lim (x → ∞) of a polynomial, look only at the sign and degree of the leading term.
If the leading coefficient is positive and the degree is even: limit is +∞.
If the leading coefficient is negative and the degree is even: limit is −∞.
If the leading coefficient is positive and the degree is odd: limit is +∞ as x → ∞, and −∞ as x → −∞.
If the leading coefficient is negative and the degree is odd: limit is −∞ as x → ∞, and +∞ as x → −∞.
Worked example: lim (x → ∞) (−5x⁴ − 3x + 1). The leading term is −5x⁴. Since the coefficient is negative and the degree is even, as x → ∞ the function heads to −∞. Answer: −∞.
The rule depends on comparing the degree of the numerator (top) with the degree of the denominator (bottom).
Degree of numerator < degree of denominator: the limit is 0. The denominator grows faster and squashes the fraction toward zero.
Example: lim (x → ∞) (x + 1) / (x² + 3) = 0.
Degree of numerator = degree of denominator: the limit is the ratio of the leading coefficients.
Worked example: lim (x → −∞) (2 + x − 3x²) / (6x² + 1). Both top and bottom have degree 2. Leading coefficients are −3 (top) and 6 (bottom). So the limit is −3/6 = −1/2.
Degree of numerator > degree of denominator: the limit is ±∞ (the fraction blows up).
Worked example: lim (x → −∞) 2x⁷ / (x + 2)². The numerator has degree 7, the denominator has degree 2. Since 7 > 2, the magnitude goes to infinity. Determining the sign: as x → −∞, x⁷ is negative (odd power), so 2x⁷ is negative, and (x + 2)² is positive. Negative divided by positive gives negative. Answer: −∞.
When direct substitution gives a non-zero number divided by zero, the limit is ±∞ (vertical asymptote).
Worked example: lim (x → 4) (x + 6) / (x − 4)². Substituting x = 4 gives 10/0. Since (x − 4)² is always positive near x = 4, and the numerator is positive (10), the function heads to +∞ from both sides. Answer: +∞.
When both numerator and denominator are zero at the same point, factor and cancel first. The cancelled factor reveals a hole, not an asymptote.
To build a polynomial f(x) with lim (x → −∞) f(x) = −∞, choose a leading term that is negative when x is large and negative. For instance, f(x) = x³ + 2x works: odd degree, positive leading coefficient, so as x → −∞ the function goes to −∞. Another valid choice: f(x) = −x² + 5, since −x² dominates and is always negative for large |x|.
To build a rational function with lim (x → ∞) f(x) = 0, make the degree of the denominator larger than the degree of the numerator. For instance, f(x) = (x + 1) / (x² + 3).
\lim_{x \to \pm\infty} (a_n x^n + \cdots + a_0) = \lim_{x \to \pm\infty} a_n x^n(Only the leading term matters for polynomials at infinity.)
\lim_{x \to \pm\infty} \frac{a_n x^n + \cdots}{b_m x^m + \cdots} = \begin{cases} 0 & \text{if } n < m \\ \frac{a_n}{b_m} & \text{if } n = m \\ \pm\infty & \text{if } n > m \end{cases}(The "degree comparison" rule for rational functions. When n > m, the sign depends on the signs of the leading coefficients and the direction of x.)
Students often think "infinity" is a number you can substitute in. It is not. You cannot plug in x = ∞. Limits at infinity describe a trend, not a substitution.
Students sometimes divide every term by the highest power of x in the denominator when evaluating rational-function limits. This works, but the degree-comparison shortcut (comparing the degrees of numerator and denominator) is faster and less error-prone on exams. Know both approaches.
A common mistake is forgetting to track the sign when the degree of the numerator exceeds the degree of the denominator. The limit is ±∞, and which sign you get depends on the leading coefficients and the direction x is heading. Always check the sign by considering whether the dominant terms are positive or negative.
Students frequently confuse holes and vertical asymptotes. Both involve the denominator being zero, but a hole occurs only when the same factor cancels from both numerator and denominator. If it does not cancel, you have a vertical asymptote.
⚠️ The degree-comparison rule for rational functions is tested constantly. Be able to state the three cases (n < m, n = m, n > m) from memory.
⚠️ "Show your work" means writing the leading-term reasoning explicitly, not just the final answer. On this homework, skipping steps costs marks.
⚠️ Questions asking you to construct a function with a given limit behaviour (Questions 2 and 3 on the homework) test whether you understand the rules in reverse. Pick a simple example and verify it.
⚠️ When a limit at a finite point gives non-zero over zero, state clearly that the limit is +∞ or −∞ and explain the sign. Examiners expect you to check the sign from both sides if asked.
True or false: lim (x → ∞) (3x² − x + 7) = ∞. (True. Positive leading coefficient, even degree.)
Fill in the blank: When the degree of the numerator equals the degree of the denominator, the limit at infinity equals ________. (The ratio of the leading coefficients.)
True or false: If both the numerator and denominator are zero at x = a, there is always a vertical asymptote at x = a. (False. It could be a hole if the common factor cancels.)
Fill in the blank: lim (x → ∞) (5x³) / (2x³ + 1) = ________. (5/2.)
True or false: lim (x → −∞) x⁵ = +∞. (False. Odd power with positive coefficient: as x → −∞, x⁵ → −∞.)
Q: Evaluate lim (x → ∞) (−2x⁶ + 4x² − 1).
A: The leading term is −2x⁶. Degree 6 (even), negative coefficient. As x → ∞, this heads to −∞. Answer: −∞.
Q: Evaluate lim (x → −∞) (4 + 3x − x²) / (2x² − 5).
A: Degree of numerator = 2, degree of denominator = 2. Leading coefficients: −1 (top) and 2 (bottom). The limit is −1/2.
Q: Evaluate lim (x → ∞) (x + 3) / (x³ + 1).
A: Degree of numerator (1) < degree of denominator (3). The limit is 0.
Q: Evaluate lim (x → 2) (x + 5) / (x − 2)².
A: Substituting x = 2 gives 7/0. Since (x − 2)² > 0 near x = 2 and the numerator is positive (7), the function goes to +∞ from both sides. Answer: +∞.
Q: Give an example of a polynomial f(x) with more than one term such that lim (x → −∞) f(x) = +∞.
A: f(x) = x⁴ + 3x. The leading term x⁴ has even degree and positive coefficient, so it heads to +∞ regardless of the direction of x.
Limits at infinity model long-run trends in business: average cost per unit as production scales up, or revenue growth that levels off. When economists say a market "approaches saturation," they are describing a horizontal asymptote. Vertical asymptotes appear in models where a quantity becomes unbounded at a threshold, such as waiting time as a queue reaches capacity.
This material connects directly to derivatives (Chapter 3), because asymptotic behaviour tells you the shape of a curve before you differentiate. It also feeds into curve-sketching, where you combine limits, derivatives, and intercepts to draw a complete graph. The continuity concepts here (holes vs. asymptotes) reappear when you study the Intermediate Value Theorem and optimisation on closed intervals.
Limits at infinity, end behaviour, horizontal asymptote, vertical asymptote, hole in a graph, removable discontinuity, leading term test, degree comparison rule, rational function limits, polynomial end behaviour, limits to infinity calculus for business, FIU calculus, Section 2.2, Section 2.3, long-run behaviour, asymptotic behaviour