Infinite Limits & Asymptotes

You’ll be able to

Connect infinite limits to vertical asymptotes
Evaluate limits at infinity by comparing degrees of numerator and denominator
Identify horizontal and vertical asymptotes of rational functions

Vertical asymptotes: outputs blow up

A vertical asymptote at x = a occurs when a one-sided limit is infinite: lim(x→a) f(x) = ±∞. For a rational function in lowest terms, this happens at values of x that make the denominator zero but not the numerator — substitution yields a nonzero-over-zero form. The sign of the result (±∞) depends on the side of approach and the signs of the factors, so vertical asymptotes are naturally analyzed one side at a time.

Limits at infinity: the tug-of-war of degrees

A limit at infinity, lim(x→∞) f(x), asks about end behavior. For a rational function, compare the degree of the numerator to the degree of the denominator. If the denominator wins (higher degree), the limit is 0. If degrees are equal, the limit is the ratio of leading coefficients. If the numerator wins, the outputs grow without bound (±∞) and there is no horizontal asymptote. A quick way is to divide every term by the highest power of x in the denominator.

End behavior of rational functions

deg(bottom) > deg(top): limit 0 · equal degrees: ratio of leading coefficients · deg(top) > deg(bottom): ±∞

These three cases decide the horizontal asymptote (or its absence) as x → ±∞.

Horizontal asymptotes describe the far ends

A horizontal asymptote y = L exists when lim(x→∞) f(x) = L or lim(x→−∞) f(x) = L. A function may approach different horizontal asymptotes at the two ends, and — unlike a vertical asymptote — a graph is allowed to cross a horizontal asymptote in the middle; the asymptote only governs the tails. When the numerator degree exceeds the denominator degree by exactly one, there is a slant (oblique) asymptote instead, found by polynomial long division.

Worked example

Find lim(x→∞) (3x² + 5)/(6x² − x).

1.Both numerator and denominator have degree 2, so this is the equal-degree case.
2.Divide every term by x²: (3 + 5/x²)/(6 − 1/x).
3.As x → ∞, the terms 5/x² and 1/x both approach 0.
4.The expression approaches (3 + 0)/(6 − 0) = 3/6 = 1/2.

Answer: The limit is 1/2, the ratio of the leading coefficients. So y = 1/2 is a horizontal asymptote.

Watch out

Do not confuse the two "infinity" questions. A vertical asymptote comes from x approaching a finite value where the function blows up. A horizontal asymptote comes from x running to ±∞. They ask different things and use different techniques.

Checkpoint

What is lim(x→∞) (4x + 7)/(2x² + 1)?

Checkpoint

The function h(x) = (x + 1)/(x − 3) has which vertical asymptote?

On the exam

For limits at infinity of rational functions, memorize the three-case degree rule — it turns most such problems into a one-line answer without any algebra. Reserve the divide-by-highest-power method for showing work when the exam demands justification.

Answer the 2 checkpoints as you read.