Infinite Limits & Asymptotes

You’ll be able to

Identify vertical asymptotes from limits that grow without bound
Evaluate limits at infinity by comparing degrees of numerator and denominator
Connect end behavior to horizontal asymptotes

Vertical asymptotes: outputs blow up

When the outputs of f grow without bound as x approaches a, we write lim(x→a) f(x) = ∞ (or −∞) and say the line x = a is a vertical asymptote. This typically happens where a denominator goes to zero but the numerator does not. Note the "= ∞" is a description of how the limit fails to exist as a finite number — the function is running off the top or bottom of the graph, and the one-sided limits may even have opposite signs.

Limits at infinity: end behavior

A limit at infinity, lim(x→∞) f(x), asks what value the outputs settle toward as x runs far to the right (or left, for −∞). If the outputs approach a finite number L, the line y = L is a horizontal asymptote. This describes the long-run end behavior of the graph, which is a completely different question from what happens at any finite point.

Rational functions: compare the degrees

For a rational function p(x)/q(x), the limit at infinity is decided by the leading terms. If the bottom degree is bigger, the fraction → 0 (horizontal asymptote y = 0). If the degrees are equal, the limit is the ratio of the leading coefficients. If the top degree is bigger, the outputs grow without bound (no horizontal asymptote — there may be a slant asymptote instead). A quick way to see this is to divide every term by the highest power of x in the denominator.

End behavior of a rational function

lim(x→±∞) (aₙxⁿ + … )/(bₘxᵐ + … ) = 0 if n < m; aₙ/bₘ if n = m; ±∞ if n > m

Only the leading terms survive at infinity. Degrees equal ⇒ divide the leading coefficients.

Worked example

Find lim(x→∞) (3x² − 5x + 1)/(6x² + 2x) and state the horizontal asymptote.

1.Compare degrees: numerator and denominator are both degree 2 — equal degrees.
2.For equal degrees, the limit is the ratio of the leading coefficients, 3/6.
3.Alternatively, divide every term by x²: (3 − 5/x + 1/x²)/(6 + 2/x). As x → ∞ the 1/x and 1/x² terms vanish.
4.That leaves 3/6 = 1/2.

Answer: lim(x→∞) = 1/2, so the graph has the horizontal asymptote y = 1/2.

Tip

Do not confuse the two asymptote questions. Vertical asymptotes come from x-values that make the function blow up (denominator → 0). Horizontal asymptotes come from letting x → ±∞ and seeing where the outputs level off. Different question, different method.

Checkpoint

What is lim(x→∞) (4x + 7)/(x² − 3)?

Checkpoint

The function f(x) = (x + 1)/(x − 4) has a vertical asymptote where?

On the exam

A horizontal asymptote describes end behavior only — a graph is allowed to cross its horizontal asymptote at finite x-values. Do not claim a curve "can never touch" y = L; that restriction applies to vertical asymptotes, not horizontal ones.

Answer the 2 checkpoints as you read.