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What happens to a function as the input grows without bound, and how does that reveal a horizontal asymptote?

Topic 1.15 Connecting Limits at Infinity and Horizontal Asymptotes: evaluate limits as x approaches plus or minus infinity and use them to identify horizontal asymptotes, especially for rational functions.

A focused answer to AP Calculus AB Topic 1.15, evaluating limits as x approaches infinity, the degree rule for rational functions, and identifying horizontal asymptotes, with worked examples.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. Limits at infinity
  3. The degree rule for rational functions
  4. The reliable technique: divide by the highest power
  5. Why the degree rule works
  6. A function can have two horizontal asymptotes

What this topic is asking

The College Board (Topic 1.15) wants you to evaluate limits at infinity - the end behavior of a function as xβ†’+∞x \to +\infty or xβ†’βˆ’βˆžx \to -\infty - and connect them to horizontal asymptotes. For rational functions, a quick degree comparison settles the limit, and you must handle roots and the sign of x2=∣x∣\sqrt{x^2} = |x| carefully.

Limits at infinity

A key building block is lim⁑xβ†’Β±βˆž1xn=0\lim_{x \to \pm\infty} \frac{1}{x^n} = 0 for any positive power nn: as xx explodes, 1xn\frac{1}{x^n} vanishes.

The degree rule for rational functions

The reliable technique: divide by the highest power

Rather than memorizing only the rule, the safe method is to divide every term, top and bottom, by the highest power of xx in the denominator, then send the 1xn\frac{1}{x^n} terms to 00. This also handles roots if you remember x2=∣x∣\sqrt{x^2} = |x|, which equals xx for xβ†’+∞x \to +\infty but βˆ’x-x for xβ†’βˆ’βˆžx \to -\infty.

Why the degree rule works

The degree rule is not a separate fact to memorize alongside the dividing-through technique; it is the result of that technique applied in general. When you divide every term by the highest power of xx in the denominator, each term becomes a constant plus pieces of the form cxk\frac{c}{x^k} that vanish at infinity. If the numerator's degree is smaller, its leading term ends up divided by a higher power and goes to zero, leaving 0constant=0\frac{0}{\text{constant}} = 0. If the degrees match, the leading terms of top and bottom both survive as their coefficients, giving the ratio of leading coefficients. If the numerator's degree is larger, its leading term is divided by a smaller power and grows without bound, giving ±∞\pm\infty. Understanding the rule as the outcome of dividing through means you can always fall back on the reliable method when a function is not a simple polynomial ratio, such as one involving roots or different powers.

A function can have two horizontal asymptotes

The limits as xβ†’+∞x \to +\infty and xβ†’βˆ’βˆžx \to -\infty can differ, especially with roots like x2+1\sqrt{x^2 + 1} where x2=∣x∣\sqrt{x^2} = |x| flips sign. When the two end limits are different finite numbers, the graph has two distinct horizontal asymptotes. Always evaluate both ends if the question asks for all asymptotes. Polynomial ratios without roots usually give the same horizontal asymptote at both ends, so the two-asymptote behavior is the signature of an absolute value or an even root in the expression. A horizontal asymptote also describes only end behavior: unlike a vertical asymptote, a curve is allowed to cross its horizontal asymptote in the middle of the graph and only has to settle toward it as xx grows large in magnitude, a subtlety worth remembering when sketching.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)1 marksSection I (multiple choice, no calculator). Evaluate lim⁑xβ†’βˆž3x2+52x2βˆ’x\lim_{x \to \infty} \frac{3x^2 + 5}{2x^2 - x}. (A) 00 (B) 32\frac{3}{2} (C) ∞\infty (D) 5βˆ’x\frac{5}{-x}
Show worked answer β†’

The correct answer is (B), 32\frac{3}{2}.

The numerator and denominator have the same degree (22), so the limit at infinity is the ratio of the leading coefficients: 32\frac{3}{2}. Equivalently, divide top and bottom by x2x^2: 3+5/x22βˆ’1/xβ†’3+02βˆ’0=32\frac{3 + 5/x^2}{2 - 1/x} \to \frac{3 + 0}{2 - 0} = \frac{3}{2}. The line y=32y = \frac{3}{2} is the horizontal asymptote.

AP 2023 (style)3 marksSection II (free response, no calculator). For f(x)=4x+1x2+3f(x) = \frac{4x + 1}{\sqrt{x^2 + 3}}: (a) Evaluate lim⁑xβ†’βˆžf(x)\lim_{x \to \infty} f(x). (b) Evaluate lim⁑xβ†’βˆ’βˆžf(x)\lim_{x \to -\infty} f(x). (c) State all horizontal asymptotes and justify.
Show worked answer β†’

A 3-point limits-at-infinity question with a root.

(a) (1 point) For large positive xx, x2+3β‰ˆβˆ£x∣=x\sqrt{x^2 + 3} \approx |x| = x, so f(x)β‰ˆ4xx=4f(x) \approx \frac{4x}{x} = 4; more carefully, divide by xx: 4+1/x1+3/x2β†’41=4\frac{4 + 1/x}{\sqrt{1 + 3/x^2}} \to \frac{4}{1} = 4.
(b) (1 point) For large negative xx, x2+3β‰ˆβˆ£x∣=βˆ’x\sqrt{x^2 + 3} \approx |x| = -x, so dividing by xx flips the sign: 4+1/xβˆ’1+3/x2β†’4βˆ’1=βˆ’4\frac{4 + 1/x}{-\sqrt{1 + 3/x^2}} \to \frac{4}{-1} = -4.
(c) (1 point) Horizontal asymptotes y=4y = 4 (as xβ†’βˆžx \to \infty) and y=βˆ’4y = -4 (as xβ†’βˆ’βˆžx \to -\infty). The two differ because x2=∣x∣\sqrt{x^2}=|x| changes sign with xx.

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