Limits, Topic 3 of 8

Limits at Infinity

What happens to a function when x races toward +∞ or −∞? The answer reveals horizontal asymptotes, end behaviour, and the long-run story of every function you'll ever meet in calculus.

x → +∞ and x → −∞ Horizontal asymptotes Dominance hierarchy 10 worked examples
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§ 01What Does x → ∞ Actually Mean?

Infinity is not a number you can substitute into a function. It is a direction, a process of growing without any bound. Limits at infinity describe the destination of a function's output as its input takes that journey.

When we write limx→∞ f(x) = L, we are making a precise claim: for every positive tolerance ε, no matter how small, there exists a sufficiently large threshold M such that once x surpasses M, the value f(x) stays within ε of L forever. The function does not have to reach L: it has to approach it arbitrarily closely and stay close.

Similarly, limx→−∞ f(x) = L describes what happens as x decreases without bound: as x sprints off to the left along the number line without any floor to stop it.

The Foundational Fact For any positive integer n, limx→∞ 1/xⁿ = 0 and limx→−∞ 1/xⁿ = 0. This single fact is the engine behind virtually every limit-at-infinity calculation you will ever do. Everything else reduces to it.

This makes geometric sense: as x grows enormous, dividing 1 by a large number raised to a positive power produces something increasingly tiny. The reciprocal of a billion is essentially zero to any practical calculation, and the limit formalises this intuition exactly.

Horizontal Asymptotes — The Geometric Picture

When limx→∞ f(x) = L (a finite number), the line y = L is called a horizontal asymptote of f as x → +∞. Equivalently, if limx→−∞ f(x) = M, then y = M is a horizontal asymptote as x → −∞. A function can have at most two distinct horizontal asymptotes, one for each infinite direction, and it is entirely possible for both limits to equal the same value L (one asymptote) or for one or both limits to be infinite (no asymptote in that direction).

As x → +∞
limx→∞ f(x) = L

The graph hugs the line y = L from the right side of the picture.

As x → −∞
limx→−∞ f(x) = M

The graph hugs the line y = M from the left side of the picture. L and M may differ.

y = M y = L x y x→−∞ x→+∞

A function with two different horizontal asymptotes, y = M as x → −∞, and y = L as x → +∞

§ 02Rational Functions — The Divide by Highest Power Technique

Rational functions (polynomial over polynomial) are the most common setting for limits at infinity, and they all follow the same three-case rule that flows directly from comparing degrees.

The key technique is to divide every single term in both numerator and denominator by the highest power of x that appears anywhere in the denominator. After dividing, every term of the form c/xⁿ (with n ≥ 1) vanishes to zero as x → ±∞, leaving only the surviving constants.

The Three Cases for Rational Functions
degree(top) < degree(bottom)  →  limit = 0
degree(top) = degree(bottom)  →  limit = ratio of leading coefficients
degree(top) > degree(bottom)  →  limit = ±∞ (no horizontal asymptote)

Memorising these three cases is useful shorthand, but understanding why they work, by doing the division, builds the instincts you need for more complex functions later.

Example 01 Find limx→∞ (3x² + 7x − 2) / (5x² − x + 1)

Both numerator and denominator are degree 2. Divide every term by x² (highest power in denominator).

Divide numerator and denominator by x²
(3 + 7/x − 2/x²) / (5 − 1/x + 1/x²)
As x → ∞, all terms with x in denominator → 0
(3 + 0 − 0) / (5 − 0 + 0) = 3/5

The limit equals the ratio of the leading coefficients, exactly as the rule predicts.

Limit = 3/5  (horizontal asymptote y = 3/5)
Example 02 Find limx→∞ (4x − 9) / (x² + 1)

Degree of numerator (1) is less than degree of denominator (2). Divide by x².

Divide by x²
(4/x − 9/x²) / (1 + 1/x²)
As x → ∞
(0 − 0) / (1 + 0) = 0
Limit = 0  (horizontal asymptote y = 0)
Example 03 Find limx→∞ (2x³ − x) / (x² + 5)

Degree of numerator (3) exceeds degree of denominator (2). Divide by x².

Divide by x²
(2x − 1/x) / (1 + 5/x²)
As x → ∞: numerator → ∞, denominator → 1
∞ / 1 = ∞

No horizontal asymptote in this direction. The function grows without bound.

Limit = ∞  (no horizontal asymptote)
Example 04 Find limx→−∞ (6x² − 1) / (2x² + 3x)

Same degree. The sign of x doesn't matter for terms that end up as constants after dividing.

Divide by x²
(6 − 1/x²) / (2 + 3/x)
As x → −∞ (1/x² → 0, 3/x → 0)
(6 − 0) / (2 + 0) = 3
Limit = 3  (horizontal asymptote y = 3 on the left)

§ 03Radical Functions — The Sign Trap

When square roots appear, the divide-by-highest-power technique still works, but you must be careful about the sign of √(x²), which depends on the direction x is travelling.

The crucial fact: for any real x, √(x²) = |x|. When x → +∞, x is positive so |x| = x. When x → −∞, x is negative so |x| = −x. Forgetting this sign flip when x → −∞ is the single most common error students make with radical limits.

The Sign Trap — Never Forget This √(x²) = |x| = x when x > 0, but √(x²) = |x| = −x when x < 0. When you pull x outside a square root for a limit as x → −∞, you must write −x, not x.
Example 05 Find limx→+∞ (3x − 1) / √(x² + 4)

The highest power in the denominator is x (since √(x²) = x for large positive x). Divide numerator and denominator by x.

Divide numerator by x
3 − 1/x
Divide denominator: √(x²+4)/x = √((x²+4)/x²) = √(1 + 4/x²)
√(1 + 4/x²)
As x → +∞
(3 − 0) / √(1 + 0) = 3/1 = 3
Limit = 3
Example 06 Find limx→−∞ (3x − 1) / √(x² + 4)

Same function, opposite direction. Now x is large and negative, so √(x²) = −x.

Divide numerator by x (still x, since we're dividing both sides)
3 − 1/x
Divide denominator: √(x²+4)/x — but x < 0 here, so √(x²+4)/x = −√(1+4/x²)
−√(1 + 4/x²)
As x → −∞
(3 − 0) / (−√(1 + 0)) = 3 / (−1) = −3

This function has two different horizontal asymptotes: y = 3 as x → +∞ and y = −3 as x → −∞. The sign trap is why they differ.

Limit = −3  (note the sign flip!)

§ 04Exponential and Logarithmic Limits at Infinity

Exponential functions either explode to infinity or collapse to zero: never anything in between. Knowing which direction depends entirely on the sign of the exponent.

Key Exponential Limits
limx→+∞ eˣ = +∞  ·  limx→−∞ eˣ = 0
limx→+∞ e−x = 0  ·  limx→−∞ e−x = +∞
limx→+∞ ln x = +∞  ·  limx→0⁺ ln x = −∞

When exponentials and polynomials compete in a limit, the exponential always dominates. No polynomial, however large its degree, can outrun eˣ as x → ∞. This is the heart of the dominance hierarchy:

Rank Function type Example Beats…
Slowest Logarithms ln x, log₁₀ x Nothing grows slower (among common types)
Powers / polynomials x, x², x¹⁰⁰ Dominates logarithms
Exponentials eˣ, 2ˣ, 10ˣ Dominates all polynomials
Fastest Factorials / super-exp n!, xˣ Dominates exponentials
Example 07 Find limx→∞ x⁵ / eˣ

Polynomial x⁵ vs exponential eˣ. By dominance, eˣ wins: the denominator grows far faster than the numerator.

Apply L'Hôpital repeatedly (5 times, reducing x⁵ → 5x⁴ → 20x³ → 60x² → 120x → 120)
x⁵/eˣ → 5x⁴/eˣ → 20x³/eˣ → 60x²/eˣ → 120x/eˣ → 120/eˣ
Final limit
limx→∞ 120/eˣ = 120 · 0 = 0
Limit = 0  (exponential wins every time)
Example 08 Find limx→∞ ln x / x

Logarithm vs polynomial (degree 1). The polynomial dominates. Both → ∞, so apply L'Hôpital once.

L'Hôpital: differentiate top and bottom
(1/x) / 1 = 1/x
Limit
limx→∞ 1/x = 0
Limit = 0  (linear growth beats logarithm)

§ 05Trigonometric Functions at Infinity

Sine and cosine oscillate perpetually between −1 and 1. They never settle, so limx→∞ sin x does not exist, but that doesn't stop them appearing inside more complex limits.

When a bounded oscillating function (like sin x or cos x) is divided by a function that grows without bound, the Squeeze Theorem resolves the limit cleanly. The bounded numerator gets squeezed to zero by the growing denominator.

Example 09 Find limx→∞ sin x / x

sin x oscillates between −1 and 1, while x → ∞. Apply the Squeeze Theorem.

Bound sin x above and below
−1 ≤ sin x ≤ 1  for all x
Divide through by x (positive as x → ∞)
−1/x ≤ sin x / x ≤ 1/x
Both bounds → 0 as x → ∞
lim(−1/x) = 0 and lim(1/x) = 0
Squeeze Theorem
limx→∞ sin x / x = 0
Limit = 0
Example 10 Find limx→∞ (2x² + cos x) / (x² + 1)

The cos x term is bounded while x² grows. It will contribute nothing to the limit.

Rewrite by separating
(2x² + cos x) / (x² + 1) = 2x²/(x²+1) + cos x/(x²+1)
First term: divide by x²
lim 2/(1 + 1/x²) = 2
Second term: |cos x| ≤ 1 and x²+1 → ∞, so by squeeze:
lim cos x/(x²+1) = 0
Combine
2 + 0 = 2
Limit = 2  (the oscillation is swamped)

§ 06Common Mistakes

Mistake 1 — Forgetting the sign flip when x → −∞ through a square root When x → −∞, √(x²) = −x (not x). Forgetting this produces the wrong sign and makes both limits appear equal when they are actually negatives of each other.
Mistake 2 — Dividing by the wrong power Always divide by the highest power of x in the denominator, not the numerator. Dividing by a different power will leave surviving terms with positive powers of x that prevent the limit from resolving.
Mistake 3 — Concluding that degree(top) > degree(bottom) always gives ∞ It gives ±∞ depending on the signs of the leading coefficients and the direction x is travelling. limx→−∞ 2x³/x² = lim −∞, not +∞. Always check the sign.
Mistake 4 — Claiming limx→∞ sin x = 0 sin x does NOT have a limit as x → ∞. It oscillates forever. The limit does not exist. Only sin x divided by a growing function (like x or eˣ) collapses to zero.

§ 07Practice Quiz — 10 Questions

Test your grasp of limits at infinity, rational functions, exponentials, radicals, and oscillating functions.

Score: 0 / 10
Question 1 of 10 — Fill in the Blank

limx→∞ 1/x³ = ?

Answer:

As x grows without bound, dividing 1 by an ever-larger number approaches…
As x → ∞, x³ → ∞, so 1/x³ → 0. This is the foundational fact of all infinity limits.
Question 2 of 10 — Multiple Choice

limx→∞ (5x² − 3) / (2x² + 7x) = ?

B — 5/2. Equal degrees: divide by x². (5 − 3/x²)/(2 + 7/x) → 5/2.
Question 3 of 10 — Multiple Choice

limx→∞ (x + 1) / (x² − 4) = ?

C — 0. Degree of top (1) is less than degree of bottom (2). Divide by x²: (1/x + 1/x²)/(1 − 4/x²) → 0/1 = 0.
Question 4 of 10 — Multiple Choice

What is limx→+∞ eˣ?

D — +∞. eˣ grows without bound as x → +∞. Compare with limx→−∞ eˣ = 0. The two directions give completely different results.
Question 5 of 10 — Fill in the Blank

limx→+∞ (2x − 5) / √(4x² + 1) = ?

Answer:

Divide numerator and denominator by x. Since x → +∞, √(x²) = x, so √(4x²+1)/x = √(4 + 1/x²).
Divide by x: (2 − 5/x)/√(4 + 1/x²) → 2/√4 = 2/2 = 1.
Question 6 of 10 — Multiple Choice

limx→−∞ (2x − 5) / √(4x² + 1) = ?

B — −1. Same function as Q5 but x → −∞. Now √(x²) = −x (sign flip!). Divide by x: (2 − 5/x) / (−√(4 + 1/x²)) → 2/(−2) = −1. This shows the two horizontal asymptotes y = 1 and y = −1.
Question 7 of 10 — Multiple Choice

Which function grows fastest as x → ∞?

C — eˣ. The dominance hierarchy: ln x ≪ polynomials ≪ exponentials. eˣ eventually overtakes every polynomial, no matter how high the degree, and dwarfs ln x completely.
Question 8 of 10 — Multiple Choice

limx→∞ cos x / x = ?

C — 0. Since −1 ≤ cos x ≤ 1, we have −1/x ≤ cos x/x ≤ 1/x. Both bounds → 0 by Squeeze Theorem. Even though cos x oscillates, dividing it by the growing x forces it to zero.
Question 9 of 10 — Fill in the Blank

limx→∞ (7x³ − 2x + 1) / (3x³ + 5) = ?

Answer (as a fraction, enter decimal):

Same degree: the limit equals the ratio of leading coefficients.
Leading coefficients are 7 and 3, so the limit is 7/3 ≈ 2.333. Confirm by dividing by x³: (7 − 2/x² + 1/x³)/(3 + 5/x³) → 7/3.
Question 10 of 10 — Multiple Choice

A function has limx→∞ f(x) = 4. What does this mean geometrically?

C. limx→∞ f(x) = 4 means the graph has a horizontal asymptote y = 4 on the right side. The graph gets arbitrarily close to — but doesn't necessarily reach — the line y = 4 as you move further right.

Next StepsContinue the Limits Series

Limits at infinity underpin horizontal asymptotes in curve sketching and appear throughout integration and series. Continue with:

  • Indeterminate Forms, when both numerator and denominator grow together in an ambiguous competition.
  • Continuity, how limits at ordinary points define the seamless behaviour of functions.
  • Curve Sketching, where limits at infinity tell you the right-hand and left-hand tails of every graph.

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