lim x→∞ f(x) = −∞ and lim x→−∞ f(x) = 3Which is bigger for large x: 0.001x³ or 1,000,000x²?
At x = 100: 0.001(10⁶) = 1,000 vs. 10⁶ · 10⁴ = 10¹⁰. The quadratic is winning by a landslide.
At x = 10⁹: 0.001(10²⁷) = 10²⁴ vs. 10⁶ · 10¹⁸ = 10²⁴. Dead heat.
At x = 10¹⁰: cubic 10²⁷, quadratic 10²⁶. The cubic has taken the lead — permanently.
Degree beats coefficient, always, eventually. That single idea decides every end-behavior question on the exam.
AP Precalculus uses limit notation to describe end behavior — no limit theory required:
lim x→∞ f(x) = ∞ — "as x increases without bound, f(x) increases without bound" (right end goes up)lim x→−∞ f(x) = −∞ — left end goes downlim x→∞ f(x) = 3 — the graph levels off at height 3 on the right: horizontal asymptote y = 3Read x→∞ as "far right," x→−∞ as "far left." You'll also see one-sided limits at vertical asymptotes (x→2⁺) in Lesson 6.
For large |x|, p(x) = aₙxⁿ + … behaves like its leading term aₙxⁿ. Everything else becomes a rounding error (that's what the warm-up demonstrates). Four cases:
| Degree | Leading coeff. | Left end (x→−∞) | Right end (x→∞) | Picture |
|---|---|---|---|---|
| even | aₙ > 0 | +∞ | +∞ | ⌣ both up |
| even | aₙ < 0 | −∞ | −∞ | ⌢ both down |
| odd | aₙ > 0 | −∞ | +∞ | ↗ like x³ |
| odd | aₙ < 0 | +∞ | −∞ | ↘ like −x³ |
Don't memorize the table — memorize x², −x², x³, −x³ and match.
A rational function r(x) = p(x)/q(x) (polynomial over polynomial) also obeys its leading terms: for large |x|,
r(x) ≈ (leading term of p) / (leading term of q)
Three regimes, with n = degree of numerator, d = degree of denominator:
lim x→±∞ r(x) = 0 — horizontal asymptote y = 0.lim x→±∞ r(x) = aₙ/b_d — horizontal asymptote y = aₙ/b_d.Example of the leading-term shortcut in regime 2:
r(x) = (6x³ − x + 2)/(2x³ + 9x²) ≈ 6x³/2x³ = 3 → lim x→±∞ r(x) = 3
Two features of horizontal asymptotes students get wrong:
The magic phrase pattern:
"Because the degree of the numerator equals the degree of the denominator, the ratio of leading terms is 6x³/2x³ = 3, so
lim x→∞ r(x) = 3andlim x→−∞ r(x) = 3; the graph has horizontal asymptote y = 3."
Calculator screenshots, tables of big inputs, or "it looks like it levels off" do not earn justification points. Leading terms do.
Problem: Describe the end behavior of p(x) = −4x⁶ + 50x⁵ − 3 using limit notation.
Solution: Leading term −4x⁶: even degree, negative coefficient → both ends down.
lim x→−∞ p(x) = −∞ lim x→∞ p(x) = −∞
Interpretation: The 50x⁵ term is huge for a while — but degree 6 beats degree 5, always, eventually. Coefficients never change the direction of end behavior, only how soon it kicks in.
Problem: Find lim x→∞ r(x) for:
(a) r(x) = (5x² + 1)/(x³ − 2) (b) r(x) = (5x³ + 1)/(x³ − 2) (c) r(x) = (5x⁴ + 1)/(x³ − 2)
Solution: (a) bottom-heavy (2 < 3): limit 0; HA y = 0. (b) balanced (3 = 3): ratio of leading coefficients 5/1: limit 5; HA y = 5. (c) top-heavy (4 > 3): behaves like 5x⁴/x³ = 5x → limit ∞; no HA.
Interpretation: One comparison — top degree vs. bottom degree — sorts every rational function into its regime before any algebra.
Problem: After a factory begins filtering its wastewater, the pollutant concentration in a holding pond is modeled by C(t) = (8t + 240)/(2t + 3) mg/L, t in hours. What happens to the concentration in the long run?
Solution: Degrees are equal (1 = 1), so lim t→∞ C(t) = 8/2 = 4.
In the long run, the concentration approaches (but never quite reaches) 4 mg/L. Note C(0) = 240/3 = 80 mg/L, and C decreases toward the asymptote — the filtration drives concentration down toward a floor of 4, perhaps the level entering from upstream.
Interpretation: Horizontal asymptotes are long-run values in context. FRQ 1 loves asking "interpret the meaning of this limit" — answer with the quantity, the value, the units, and "approaches in the long run."
Problem: A rational function g satisfies lim x→−∞ g(x) = −2, lim x→∞ g(x) = −2, and g(1) = −2. Sketch-level: what do these three facts say, and do they contradict each other?
Solution: The two limits say y = −2 is a horizontal asymptote — the graph levels off at −2 on both far ends. g(1) = −2 says the graph crosses (or touches) its own asymptote at x = 1. No contradiction: the asymptote constrains eventual behavior, not mid-graph behavior. A graph can wander through y = −2 any finite number of times before settling.
Interpretation: "Graphs never touch asymptotes" is an Algebra-2 myth the AP explicitly punishes. Horizontal: crossable. (Vertical asymptotes are the uncrossable kind — the function isn't even defined there. Lesson 6.)
1000x² loses to 0.001x³ eventually. Fix: compare degrees first; coefficients only matter when degrees tie.−2(x+5)(x−1)²(x−4)³ has leading term −2·x·x²·x³ = −2x⁶.lim x→−∞ (−3x⁵ + 7x² − 1) =1. (B). Leading term −3x⁵; as x → −∞, x⁵ → −∞, so −3x⁵ → +∞. (A) forgets the negative coefficient; (C)/(D) treat lower-order terms or coefficients as limits.
r(x) = (2x³ − x)/(5x³ + 4) is2. (A). Equal degrees (3 = 3) → HA at ratio of leading coefficients: y = 2/5. (B) is the bottom-heavy rule; (C) inverts the ratio; (D) is the top-heavy rule.
lim x→∞ (x² + 1)/(x⁴ − 3) =3. (C). Bottom-heavy (2 < 4) → limit 0. (A) ratio-of-coefficients error (that needs equal degrees); (B) inverts the comparison; (D) uses the constant terms.
f(x) = x³?4. (D). Same degree (3) and same sign of leading coefficient (positive) → same end behavior; the −100x² can't change the tails. (A) flips both ends; (B)/(C) are even-degree patterns (both ends same direction).
r(x) = (3x⁴ + 2)/(x² + 1), which is true?5. (B). Top-heavy: r behaves like 3x⁴/x² = 3x², which → +∞ in both directions (even power). (C) would fit odd top-heavy growth like 3x; (A)/(D) wrongly grant an HA.
lim x→−∞ p(x) =6. (A). Even degree → both ends same direction; negative leading coefficient → both down. Left end: −∞. (D) is the classic "lower terms matter" myth — they don't, for end behavior.
lim x→∞ f(x) = 7. Which statement MUST be true?7. (B). That's what the limit statement means: outputs eventually stay arbitrarily close to 7. (A)/(C) are the uncrossable-asymptote myth; (D) f could approach 7 from above, decreasing, or oscillate while settling.
r(x) = ((x − 2)(x + 5)) / ((x − 2)(x − 7)). As x → ∞, r(x) approaches8. (C). Leading terms x²/x² → 1. (The common factor (x − 2) creates a hole at x = 2 — Lesson 6 — but holes don't affect end behavior.) (A)/(B) misapply rules; (D) wrongly reads top-heavy.
c(t) = (5t + 3)/(2t + 1) mg/L. In the long run the concentration approaches9. (D). Equal degrees → 5/2 = 2.5 mg/L. (A)/(B) read off constants or the numerator only; (C) inverts to 3/2.
p(x) = 4x⁴ − x³?10. (A). Leading term 4x⁴: even degree, positive coefficient → both ends +∞. The −x³ term is irrelevant to the tails. (B) is the odd-positive pattern; (D) needs a negative leading coefficient.
11. (B). Degree wins eventually: 0.1x³ outgrows every quadratic and linear option for large enough x. (A)/(C) lead for moderate x but lose the long game; (D) is degree 1.
12. (FRQ-style, 6 points)
(i) [2 pts] Degrees equal (2 = 2); leading-term ratio 6x²/2x² = 3, so lim x→∞ r(x) = 3 and lim x→−∞ r(x) = 3.
(ii) [2 pts] Horizontal asymptote y = 3. Crossing: solve r(x) = 3 → 6x² − 5x − 6 = 3(2x² − 8) = 6x² − 24 → −5x − 6 = −24 → −5x = −18 → x = 18/5 = 3.6. Since x = 3.6 doesn't make the denominator zero (2(3.6)² − 8 = 25.92 − 8 = 17.92 ≠ 0), the graph crosses its horizontal asymptote at (3.6, 3).
(iii) [2 pts] In the long run, the company's cost-to-revenue ratio approaches 3 — costs approach three times revenue. (Any interpretation naming the quantity, the value 3, and long-run/approach language earns the points; noting that a ratio of 3 means persistent unprofitability is a bonus, not required.)
12. (FRQ-style) 🚫 Let r(x) = (6x² − 5x − 6)/(2x² − 8).
(i) Find lim x→∞ r(x) and lim x→−∞ r(x), with reasoning based on leading terms.
(ii) State the horizontal asymptote and determine whether the graph of r crosses it. (Solve r(x) = your asymptote value.)
(iii) Interpret: if r(x) models a company's cost-to-revenue ratio in year x, what does your answer to (i) mean in context?
Consider the functions f(x) = −2x⁵ + 18x³ and g(x) = (4x³ + x)/(x³ + 2x + 1).
(a) (i) Write f in factored form. (ii) Find all real zeros of f with multiplicities and describe the graph's behavior at each intercept.
(b) (i) Using limit notation, describe the end behavior of f. (ii) Explain how the degree and leading coefficient determine your answer.
(c) (i) Using limit notation, describe the end behavior of g. (ii) A student claims g has no horizontal asymptote "because the numerator's coefficients are bigger." Identify and correct the error in this reasoning.
(a) [2 pts]
(i) [1 pt] f(x) = −2x³(x² − 9) = −2x³(x − 3)(x + 3)
(ii) [1 pt] Zeros: x = 0 (multiplicity 3 — crosses, flattening through the origin), x = 3 (multiplicity 1 — crosses), x = −3 (multiplicity 1 — crosses).
(b) [2 pts]
(i) [1 pt] lim x→−∞ f(x) = ∞ and lim x→∞ f(x) = −∞.
(ii) [1 pt] For large |x|, f behaves like its leading term −2x⁵: odd degree makes the two ends point opposite directions, and the negative coefficient flips x⁵'s pattern, giving up-on-the-left, down-on-the-right.
(c) [2 pts]
(i) [1 pt] Numerator and denominator both have degree 3, so g behaves like 4x³/x³ = 4: lim x→−∞ g(x) = 4 and lim x→∞ g(x) = 4 (horizontal asymptote y = 4).
(ii) [1 pt] The error: comparing coefficients instead of degrees. Whether a horizontal asymptote exists depends on comparing the polynomials' degrees; here the degrees are equal, so a horizontal asymptote exists at the ratio of leading coefficients, y = 4 — regardless of how large any coefficient is.
1. (B). Leading term −3x⁵; as x → −∞, x⁵ → −∞, so −3x⁵ → +∞. (A) forgets the negative coefficient; (C)/(D) treat lower-order terms or coefficients as limits.
2. (A). Equal degrees (3 = 3) → HA at ratio of leading coefficients: y = 2/5. (B) is the bottom-heavy rule; (C) inverts the ratio; (D) is the top-heavy rule.
3. (C). Bottom-heavy (2 < 4) → limit 0. (A) ratio-of-coefficients error (that needs equal degrees); (B) inverts the comparison; (D) uses the constant terms.
4. (D). Same degree (3) and same sign of leading coefficient (positive) → same end behavior; the −100x² can't change the tails. (A) flips both ends; (B)/(C) are even-degree patterns (both ends same direction).
5. (B). Top-heavy: r behaves like 3x⁴/x² = 3x², which → +∞ in both directions (even power). (C) would fit odd top-heavy growth like 3x; (A)/(D) wrongly grant an HA.
6. (A). Even degree → both ends same direction; negative leading coefficient → both down. Left end: −∞. (D) is the classic "lower terms matter" myth — they don't, for end behavior.
7. (B). That's what the limit statement means: outputs eventually stay arbitrarily close to 7. (A)/(C) are the uncrossable-asymptote myth; (D) f could approach 7 from above, decreasing, or oscillate while settling.
8. (C). Leading terms x²/x² → 1. (The common factor (x − 2) creates a hole at x = 2 — Lesson 6 — but holes don't affect end behavior.) (A)/(B) misapply rules; (D) wrongly reads top-heavy.
9. (D). Equal degrees → 5/2 = 2.5 mg/L. (A)/(B) read off constants or the numerator only; (C) inverts to 3/2.
10. (A). Leading term 4x⁴: even degree, positive coefficient → both ends +∞. The −x³ term is irrelevant to the tails. (B) is the odd-positive pattern; (D) needs a negative leading coefficient.
11. (B). Degree wins eventually: 0.1x³ outgrows every quadratic and linear option for large enough x. (A)/(C) lead for moderate x but lose the long game; (D) is degree 1.
12. (FRQ-style, 6 points)
(i) [2 pts] Degrees equal (2 = 2); leading-term ratio 6x²/2x² = 3, so lim x→∞ r(x) = 3 and lim x→−∞ r(x) = 3.
(ii) [2 pts] Horizontal asymptote y = 3. Crossing: solve r(x) = 3 → 6x² − 5x − 6 = 3(2x² − 8) = 6x² − 24 → −5x − 6 = −24 → −5x = −18 → x = 18/5 = 3.6. Since x = 3.6 doesn't make the denominator zero (2(3.6)² − 8 = 25.92 − 8 = 17.92 ≠ 0), the graph crosses its horizontal asymptote at (3.6, 3).
(iii) [2 pts] In the long run, the company's cost-to-revenue ratio approaches 3 — costs approach three times revenue. (Any interpretation naming the quantity, the value 3, and long-run/approach language earns the points; noting that a ratio of 3 means persistent unprofitability is a bonus, not required.)
🎯 Exam tip: Every end-behavior justification on the FRQ is the same two-sentence template: "For large |x|, [function] behaves like [leading term or leading-term ratio]. Therefore lim x→±∞ … = [value/±∞]." Learn the template; swap the nouns.