Connecting Limits at Infinity and Horizontal Asymptotes: Study Guide (Calculus AB 1.15)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer, so limits are written in a compact form:

lim (x → ∞) f(x) means “the limit as x increases without bound of f(x)”. lim (x → −∞) f(x) means “the limit as x decreases without bound of f(x)”.

On paper, write “x → ∞” under “lim” as usual. Keep “lim” on every line until the last step.

What “x → ∞” means

In Topic 1.14 you let x approach a number and allowed the outputs to grow without bound. That gave vertical asymptotes. In this topic it is the other way round: the inputs grow without bound, and you ask what the outputs do. This is called a limit at infinity.

Look at f(x) = (3x + 1)/x, which equals 3 + 1/x for x ≠ 0.

x	10	100	1000	−10	−100	−1000
f(x)	3.1	3.01	3.001	2.9	2.99	2.999

As x gets larger, 1/x shrinks towards 0, so f(x) gets closer and closer to 3. The same happens as x becomes very negative. We write

lim (x → ∞) (3x + 1)/x = 3 and lim (x → −∞) (3x + 1)/x = 3

Meaning. lim (x → ∞) f(x) = L says that you can make f(x) as close to L as you like by taking x large enough. It does not say f(x) ever equals L.

∞ is not a number. You cannot substitute it. You use a few basic facts instead:

• lim (x → ±∞) 1/xⁿ = 0 for any positive integer n. A fixed number divided by a huge number is tiny.
• lim (x → −∞) eˣ = 0, but lim (x → ∞) eˣ = ∞.
• lim (x → ∞) x² = ∞. The outputs grow without bound, so there is no finite limit.

End behaviour and horizontal asymptotes

Limits at infinity describe the end behaviour of a function: what the graph does at the far right (x → ∞) and the far left (x → −∞). There are three possibilities at each end.

What the limit does	Example	What the graph does at that end
Approaches a number L	lim (x → ∞) (3x + 1)/x = 3	Levels off towards the line y = L
Grows without bound	lim (x → ∞) x² = ∞	Rises (or falls) forever; no horizontal asymptote
Neither	lim (x → ∞) sin x does not exist	Keeps oscillating; no single level

Horizontal asymptote. The line y = L is a horizontal asymptote of the graph of f if lim (x → ∞) f(x) = L or lim (x → −∞) f(x) = L.

Each end can give at most one horizontal asymptote, so a graph has at most two. They can be the same line or different lines. For example, arctan x approaches π/2 as x → ∞ and −π/2 as x → −∞, so its graph has two horizontal asymptotes, y = π/2 and y = −π/2.

A horizontal asymptote says nothing about the middle of the graph. The graph may cross it. The function (sin x)/x is a good example. For x > 0, −1/x ≤ (sin x)/x ≤ 1/x, and both outer functions approach 0, so by the squeeze theorem (Topic 1.8) lim (x → ∞) (sin x)/x = 0. The line y = 0 is a horizontal asymptote, yet the graph crosses it at every multiple of π.

Rational functions: divide by the highest power

For a rational function, divide every term on the top and the bottom by the highest power of x in the denominator. Most terms then become a number over a power of x, which goes to 0.

For example, as x → ∞:

(6x² − 5)/(2x² + x) = (6 − 5/x²)/(2 + 1/x), which approaches (6 − 0)/(2 + 0) = 3.

Doing this many times shows a pattern that depends only on the degrees (n on top, m on the bottom):

Degrees	Limit as x → ±∞	Horizontal asymptote	Example
n < m	0	y = 0	(4x + 1)/(x² + 3) → 0
n = m	ratio of leading coefficients	y = (leading coefficient of top)/(leading coefficient of bottom)	(6x² − 5)/(2x² + x) → 3
n > m	∞ or −∞ (no finite limit)	none	(x³ + 1)/(x² − 2) → ∞ as x → ∞

For a rational function the limit is the same at both ends when it is finite, so there is at most one horizontal asymptote. Use the table to predict, but show the division when you are asked to justify a limit.

Worked example 1: a rational function and where it meets its asymptote

Question. Let f(x) = (6x² − x + 4)/(3x² + 5x − 2). Find lim (x → ∞) f(x) and lim (x → −∞) f(x). State the horizontal asymptote, and find where the graph meets it.

1. Identify the highest power in the denominator. It is x².
2. Divide every term by x². For x ≠ 0, f(x) = (6 − 1/x + 4/x²)/(3 + 5/x − 2/x²)
3. Take the limit. As x → ∞, each of 1/x, 4/x², 5/x and 2/x² approaches 0: lim (x → ∞) f(x) = (6 − 0 + 0)/(3 + 0 − 0) = 6/3 = 2
4. Repeat for x → −∞. The same terms still approach 0, so lim (x → −∞) f(x) = 2 as well.
5. State the asymptote. The line y = 2 is the only horizontal asymptote.
6. Find where f(x) = 2. Set 6x² − x + 4 = 2(3x² + 5x − 2) = 6x² + 10x − 4. The x² terms cancel: −x + 4 = 10x − 4, so 11x = 8 and x = 8/11. The denominator there is 390/121, not 0, so f(8/11) = 2 really is a point on the graph.

Answer. Both limits equal 2. The horizontal asymptote is y = 2, and the graph crosses it once, at x = 8/11.

Check. f(1000) ≈ 1.9963 and f(−1000) ≈ 2.0037. The graph approaches 2 from below on the right and from above on the left.

Connection to Topic 1.14. The denominator factors as (3x − 1)(x + 2), and the top is not 0 at x = 1/3 or x = −2. So the graph also has vertical asymptotes x = 1/3 and x = −2. Vertical asymptotes come from limits equal to infinity; horizontal ones come from limits at infinity.

Worked example 2: two different horizontal asymptotes

Question. Let g(x) = (5x + 3)/√(4x² + 1). Find the limits of g as x → ∞ and as x → −∞, and state every horizontal asymptote.

The key fact is √(x²) = |x|, not x. So √(4x² + 1) = |x|·√(4 + 1/x²) for x ≠ 0.

1. As x → ∞, x is positive, so |x| = x. Divide top and bottom by x: g(x) = (5 + 3/x)/√(4 + 1/x²), which approaches 5/√4 = 5/2.
2. As x → −∞, x is negative, so |x| = −x. Divide top and bottom by −x. The top becomes −5 − 3/x and the bottom becomes √(4 + 1/x²): g(x) = (−5 − 3/x)/√(4 + 1/x²), which approaches −5/2.
3. State the asymptotes. The graph has two: y = 5/2 on the right and y = −5/2 on the left.

Check. g(1000) ≈ 2.5015 and g(−1000) ≈ −2.4985. This makes sense: the bottom is always positive, so g(x) has the same sign as 5x + 3, which is negative for very negative x.

Different asymptotes at the two ends usually come from |x| hiding inside a square root or from exponential terms such as eˣ, which behave very differently at −∞ and at ∞.

Comparing how fast functions grow

Limits at infinity also let you compare two functions that both grow without bound. Look at the limit of their ratio as x → ∞, where f(x) and g(x) are both positive for large x:

lim (x → ∞) f(x)/g(x)	Conclusion
∞	f grows faster than g; f dominates
0	g grows faster than f; g dominates
a positive number	f and g grow at the same rate

For example, lim (x → ∞) (x² + 1)/(3x²) = 1/3, so x² + 1 and 3x² grow at the same rate. Constant multiples and lower-order terms do not change the rate.

Tables show the pattern for the three main families:

x	5	10	20
x³	125	1000	8000
2ˣ	32	1024	1 048 576
x³/2ˣ	3.906	0.977	0.00763

At first x³ is ahead, but the exponential overtakes it and the ratio collapses towards 0. In the same way, ln 100 ≈ 4.6 while √100 = 10, and ln 1 000 000 ≈ 13.8 while √1 000 000 = 1000.

For large x, the order from slowest to fastest is:

logarithms (ln x) < positive powers (xᵖ, p > 0) < exponentials (bˣ, b > 1)

So, for example, lim (x → ∞) (ln x)/√x = 0 and lim (x → ∞) x¹⁰⁰/1.01ˣ = 0, even though x¹⁰⁰ is enormous for moderate x.

Background. In this topic you use this order as a known fact, supported by tables and graphs. In Unit 4 (Topic 4.7), L’Hospital’s rule gives a way to prove these limits.

Worked example 3: using growth rates to find a limit

Question. Find lim (x → ∞) (2ˣ + x⁵)/(3·2ˣ − x²). Then describe the behaviour as x → −∞.

1. Identify the fastest-growing term in the denominator. As x → ∞, 2ˣ dominates x², so divide every term by 2ˣ: (1 + x⁵/2ˣ)/(3 − x²/2ˣ)
2. Use relative magnitudes. Exponentials beat powers, so x⁵/2ˣ → 0 and x²/2ˣ → 0.
3. Take the limit. (1 + 0)/(3 − 0) = 1/3. So y = 1/3 is a horizontal asymptote on the right.
4. As x → −∞, 2ˣ → 0, so the expression behaves like x⁵/(−x²) = −x³. As x → −∞, −x³ → ∞. The limit is ∞, so there is no horizontal asymptote on the left.

Answer. lim (x → ∞) = 1/3; lim (x → −∞) = ∞. The only horizontal asymptote is y = 1/3.

Check. At x = 30 the expression is about 0.341, close to 0.333. (The denominator is also 0 at one negative value, x ≈ −1.16, which gives a vertical asymptote. That does not affect end behaviour.)

Common misconceptions

• “Substitute ∞.” Expressions such as ∞/∞ or ∞ − ∞ are not numbers. Rewrite first, then use facts like 1/x → 0.
• “A graph can never cross its horizontal asymptote.” It can, as Figure 1 and (sin x)/x show. The asymptote only describes the far ends of the graph.
• “The function eventually equals L.” A limit at infinity describes approach. In Worked example 1, f(x) is never 2 for large x.
• “√(x²) = x.” It is |x|. Forgetting this gives the wrong sign as x → −∞.
• “Both ends always give the same asymptote.” True for rational functions, but not for functions with roots, eˣ or arctan x.
• “y = ∞ is a horizontal asymptote.” An asymptote is a line y = L with L a real number. If the limit is ∞, there is no horizontal asymptote at that end.
• “x³ is bigger than 2ˣ, so it grows faster.” Compare the ratio as x → ∞, not values at small x.
• Mixing up the two asymptote types. Vertical: x → a and f(x) → ±∞ (Topic 1.14). Horizontal: x → ±∞ and f(x) → L (this topic).

Where this leads

Limits at infinity return in Unit 5 when you sketch curves, and in Unit 4 when L’Hospital’s rule handles ∞/∞. BC students use the same growth-rate ideas when they test series for convergence. If vertical asymptotes still feel shaky, look back at Topic 1.14, Connecting Infinite Limits and Vertical Asymptotes. Next comes Topic 1.16, Working with the Intermediate Value Theorem. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.