DWG NO. 10.3 — Lesson 3 of 5

Limits at Infinity and Asymptotic Behavior

Unit 10: Limits and an Introduction to Calculus · ~20 min

Objective Evaluate limits as x approaches infinity to determine the horizontal asymptotes of a rational function.

Lesson 2.5 introduced horizontal asymptotes with a rule of thumb based on comparing degrees. That rule wasn't arbitrary — it's a shortcut for a limit. li\(m_{x→∞}\) f(x) asks what f(x) approaches as x grows without bound in the positive direction; li\(m_{x→-∞}\) f(x) asks the same as x grows without bound in the negative direction. When either limit equals a finite number L, y = L is a horizontal asymptote.

The technique: divide by the highest power

To evaluate a limit at infinity for a rational function, divide every term in the numerator and denominator by the highest power of x that appears in the denominator. Then use the fact that li\(m_{x→\pm∞}\) c/\(x^{n}\) = 0 for any constant c and any n > 0 — a constant divided by an ever-growing power of x shrinks to nothing.

Three cases, by comparing degrees

Comparison	li\(m_{x→\pm∞}\) f(x)	Horizontal asymptote
deg(numerator) < deg(denominator)	0	y = 0
deg(numerator) = deg(denominator)	ratio of leading coefficients	y = that ratio
deg(numerator) > deg(denominator)	±∞ (no finite limit)	none (end behavior dominates instead)

f(x) = (2x+1)/(x+3) — as x→±∞, f(x) gets arbitrarily close to 2, so y = 2 is a horizontal asymptote

Worked Example 1 · Equal degrees

ProblemEvaluate li\(m_{x→∞}\) (3x²+2)/(5x²−x+1).

1Numerator and denominator both have degree 2. Divide every term by x²: (3 + 2/x²)/(5 − 1/x + 1/x²).

2As x→∞, every term with x in the denominator vanishes: (3+0)/(5−0+0).

lim = 3/5 (horizontal asymptote y = 3/5)

Worked Example 2 · Numerator degree less than denominator

ProblemEvaluate li\(m_{x→∞}\) (2x+1)/(x³−4).

1Denominator degree (3) exceeds numerator degree (1). Divide every term by x³: (2/x²+1/x³)/(1−4/x³).

2Every term in the numerator vanishes as x→∞, while the denominator approaches 1.

lim = 0 (horizontal asymptote y = 0)

Worked Example 3 · Numerator degree greater than denominator

ProblemEvaluate li\(m_{x→∞}\) (x³+1)/(x²−3).

1Numerator degree (3) exceeds denominator degree (2), so the fraction behaves like x³/x² = x for large x.

2As x→∞, x grows without bound — the ratio has no finite limit.

lim = ∞ (does not exist as a finite number; no horizontal asymptote)

Guided practice

1.Evaluate li\(m_{x→∞}\) (4x−7)/(2x+3).
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Equal degrees — ratio of leading coefficients 4/2. lim = 2.
2.Evaluate li\(m_{x→-∞}\) 5/(x²+1).
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Denominator degree (2) exceeds numerator degree (0). lim = 0.
3.Evaluate li\(m_{x→∞}\) (x²−1)/(3x+5).
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Numerator degree (2) exceeds denominator degree (1) — grows without bound. lim = ∞ (DNE, no horizontal asymptote).
4.Evaluate li\(m_{x→∞}\) (7x³−2x)/(2x³+x²−1).
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Equal degrees (3) — ratio of leading coefficients 7/2. lim = 7/2.
5.Find the horizontal asymptote of f(x) = (6x²+x)/(2x²−9) using a limit at infinity.
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Equal degrees (2) — ratio of leading coefficients 6/2 = 3. y = 3.

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