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What does an infinite limit tell you about a function, and how does it locate a vertical asymptote?

Topic 1.14 Connecting Infinite Limits and Vertical Asymptotes: use infinite limits to identify vertical asymptotes and describe behavior near them with correct sign analysis.

A focused answer to AP Calculus AB Topic 1.14, connecting infinite one-sided limits to vertical asymptotes, with sign analysis to determine whether the function goes to plus or minus infinity, and worked examples.

Generated by Claude Opus 4.88 min answer

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  1. What this topic is asking
  2. Infinite limits
  3. Sign analysis for the direction
  4. Where asymptotes come from
  5. Holes versus asymptotes: factor first
  6. Reporting the result

What this topic is asking

The College Board (Topic 1.14) wants you to connect an infinite limit to a vertical asymptote. When a function's outputs grow without bound near a point, the line x=ax = a is a vertical asymptote, and you must use sign analysis to say whether the function heads to +∞+\infty or βˆ’βˆž-\infty on each side.

Infinite limits

This happens for a rational function where the denominator approaches 00 but the numerator approaches a nonzero number. (If both approach 00, you have a 00\frac{0}{0} form to resolve first - that may be a hole instead.)

Sign analysis for the direction

Where asymptotes come from

A vertical asymptote of a rational function sits at each zero of the denominator that does not cancel with the numerator. If a factor cancels, that point is a removable hole, not an asymptote (Topic 1.10). So always factor first, cancel removable factors, and the remaining denominator zeros are the asymptotes.

Holes versus asymptotes: factor first

The single most important habit before declaring a vertical asymptote is to factor and cancel first. A zero of the denominator only produces an asymptote if it survives cancellation; if the same factor appears in the numerator and cancels, that point is a removable hole instead, where the limit is finite. For example, xβˆ’2x2βˆ’4=xβˆ’2(xβˆ’2)(x+2)=1x+2\frac{x - 2}{x^2 - 4} = \frac{x - 2}{(x-2)(x+2)} = \frac{1}{x+2} has a hole at x=2x = 2 (limit 14\frac14) but a genuine asymptote at x=βˆ’2x = -2. Skipping the factoring step leads to the common error of calling every denominator zero an asymptote. After cancelling, the surviving denominator zeros are the asymptotes, and at each you do sign analysis. This connects the topic tightly to the classification of discontinuities and to evaluating 00\frac{0}{0} forms by algebraic manipulation.

Reporting the result

When a limit is infinite, state the direction with the ±∞\pm\infty symbol and, if asked, note that the (finite) limit does not exist. On a graph, an infinite limit shows the curve hugging the vertical line x=ax = a, shooting up or down. Pairing the one-sided behaviors gives the full picture near the asymptote. A useful shortcut for the direction: an odd-power factor in the denominator (like a single (xβˆ’a)(x - a)) changes sign across aa, so the two sides give opposite infinities, whereas an even-power factor (like (xβˆ’a)2(x - a)^2) keeps the same sign on both sides, so both one-sided limits share the same infinity. Recognizing the parity of the factor lets you predict the behavior before doing the full sign table.

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 2021 (style)1 marksSection I (multiple choice, no calculator). Evaluate lim⁑xβ†’2+1xβˆ’2\lim_{x \to 2^+} \frac{1}{x - 2}. (A) 00 (B) +∞+\infty (C) βˆ’βˆž-\infty (D) 11
Show worked answer β†’

The correct answer is (B), +∞+\infty.

As xβ†’2+x \to 2^+, xx is slightly greater than 22, so xβˆ’2x - 2 is a small positive number, and 1smallΒ positiveβ†’+∞\frac{1}{\text{small positive}} \to +\infty. The numerator 11 is positive and the denominator approaches 00 through positive values, so the quotient grows without bound positively. (From the left, xβˆ’2x - 2 would be small negative, giving βˆ’βˆž-\infty.)

AP 2023 (style)3 marksSection II (free response, no calculator). For f(x)=x+1xβˆ’3f(x) = \frac{x + 1}{x - 3}: (a) Identify the vertical asymptote. (b) Find lim⁑xβ†’3βˆ’f(x)\lim_{x \to 3^-} f(x) and lim⁑xβ†’3+f(x)\lim_{x \to 3^+} f(x), with sign analysis. (c) Describe the behavior of the graph near the asymptote.
Show worked answer β†’

A 3-point infinite-limit question.

(a) (1 point) The denominator is zero at x=3x = 3 and the numerator 4β‰ 04 \neq 0 there, so x=3x = 3 is a vertical asymptote.
(b) (1 point) Near x=3x = 3 the numerator β†’4\to 4 (positive). As xβ†’3βˆ’x \to 3^-, xβˆ’3x - 3 is small negative, so fβ†’40βˆ’=βˆ’βˆžf \to \frac{4}{0^-} = -\infty. As xβ†’3+x \to 3^+, xβˆ’3x - 3 is small positive, so fβ†’40+=+∞f \to \frac{4}{0^+} = +\infty.
(c) (1 point) The graph drops to βˆ’βˆž-\infty just left of x=3x = 3 and rises to +∞+\infty just right of it, hugging the vertical line x=3x = 3.

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