# Connecting infinite limits and vertical asymptotes

> AP Calculus BC · Unit 1: Limits and Continuity
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u1-connecting-infinite-limits-and-vertical/

This sub-topic connects one-sided infinite limit behavior to vertical asymptotes. You will learn to locate vertical asymptotes for rational, logarithmic, and trigonometric functions, and correctly distinguish asymptotes from removable discontinuity holes.

**Prerequisites:** Evaluating one-sided limits; Factoring and simplifying rational functions; Definition of discontinuity

## Learning objectives

- Define one-sided and two-sided infinite limits
- Connect infinite limit behavior to vertical asymptotes
- Locate vertical asymptotes for rational, logarithmic, and trigonometric functions
- Distinguish vertical asymptotes from removable discontinuities (holes)

## Infinite Limits: Definition and One-Sided Behavior

**Infinite limit** — Notation describing unbounded growth or decay of $f(x)$ as $x$ approaches a constant $a$. The limit does not exist as a finite number; the notation only describes the function's behavior near $a$.

*Notation:* $\lim_{x \to a^\pm} f(x) = \pm\infty$

Infinite limits are almost always evaluated one-sided at a point, because the sign of the result depends which side of $a$ you approach from. For rational functions, if the numerator approaches a non-zero constant and the denominator approaches 0, the result is an infinite limit, with sign determined by the sign of the numerator and denominator near $a$.

**Worked example:** Evaluate $\lim_{x \to 3^-} \frac{2x + 1}{x - 3}$ and $\lim_{x \to 3^+} \frac{2x + 1}{x - 3}$

1. Evaluate the limit of the numerator as $x \to 3$:

   $$\lim_{x \to 3} (2x +1) = 2(3) + 1 = 7$$
2. This is a non-zero positive constant. For the left-hand limit ($x \to 3^-$), $x < 3$ so $x - 3 < 0$, approaching 0 from the negative side. A positive constant divided by a small negative number produces a large negative number, so:
3. $$\lim_{x \to 3^-} \frac{2x + 1}{x - 3} = -\infty$$
4. For the right-hand limit ($x \to 3^+$), $x > 3$ so $x - 3 > 0$, approaching 0 from the positive side. A positive constant divided by a small positive number is a large positive number, so:
5. $$\lim_{x \to 3^+} \frac{2x + 1}{x - 3} = +\infty$$

> **Exam tip:** Always evaluate one-sided limits separately when checking for infinite behavior. The AP exam frequently tests whether you recognize that a two-sided limit may not exist (because the two sides go to opposite infinities) but the one-sided behavior still creates a vertical asymptote.

## Core Connection: Vertical Asymptotes from Infinite Limits

**Vertical Asymptote** — The line $x=a$ is a vertical asymptote of $f(x)$ if at least one one-sided limit of $f(x)$ as $x$ approaches $a$ is infinite ($+\infty$ or $-\infty$). This is the core formal connection between infinite limits and asymptotes.

*Notation:* $x=a$ (a vertical line)

For rational functions, this gives a step-by-step rule to locate vertical asymptotes: (1) Factor numerator and denominator completely. (2) Cancel common factors to simplify; common factors create removable discontinuities (holes), not asymptotes. (3) Any $x=a$ that makes the simplified denominator zero is a vertical asymptote.

**Worked example:** Find all vertical asymptotes of $f(x) = \frac{x^2 - 9}{x^2 - 2x - 3}$

1. Factor numerator and denominator completely:

   $$x^2 - 9 = (x-3)(x+3), \quad x^2 - 2x -3 = (x-3)(x+1)$$
2. Simplify, noting the domain restriction $x \neq 3$:

   $$f(x) = \frac{x+3}{x+1} \quad \text{for } x \neq 3$$
3. Find zeros of the simplified denominator: $x+1 = 0 \implies x = -1$. Check the one-sided limit at $x=-1$:

   $$\lim_{x \to -1^-} \frac{x+3}{x+1} = \frac{2}{0^-} = -\infty$$
4. This confirms an infinite one-sided limit, so $x=-1$ is a vertical asymptote. At $x=3$, evaluate the limit of the simplified function:

   $$\lim_{x \to 3} \frac{x+3}{x+1} = \frac{6}{4} = \frac{3}{2}$$
5. The limit is finite, so $x=3$ is a hole (removable discontinuity), not a vertical asymptote. Final result: only $x=-1$ is a vertical asymptote.

> **Exam tip:** AP exam questions almost always include a common factor in rational function vertical asymptote problems to test if you confuse holes with vertical asymptotes. Always simplify first before identifying asymptotes.

## Vertical Asymptotes of Non-Rational Functions

Rational functions are not the only functions with vertical asymptotes. For any function, find points where the function is undefined, then check if at least one one-sided limit at that point is infinite. Two common cases tested on the AP exam are:

- **Logarithmic functions**: For $f(x) = \ln(g(x))$, $f$ is undefined when $g(x) \leq 0$. Vertical asymptotes occur at $x=a$ where $g(x)$ approaches 0 from the positive side (within the domain of $f$).
- **Trigonometric functions**: Reciprocal trigonometric functions like $\tan x$, $\cot x$, $\sec x$, and $\csc x$ have vertical asymptotes where their denominators are zero, since the numerator is non-zero at these points.

**Worked example:** Find all vertical asymptotes of $f(x) = \ln(x^2 - 4x + 3)$

1. First find the domain of $f(x)$: the argument of the logarithm must be positive:

   $$x^2 - 4x + 3 = (x-1)(x-3) > 0 \implies \text{Domain: } (-\infty, 1) \cup (3, \infty)$$
2. Candidates for vertical asymptotes are the domain boundaries: $x=1$ and $x=3$, where the argument equals zero.
3. Check $x=1$: as $x \to 1^-$ (from the domain side), $(x-1)(x-3)$ is positive and approaches 0. So:

   $$\lim_{x \to 1^-} \ln((x-1)(x-3)) = -\infty$$
4. Thus $x=1$ is a vertical asymptote. Check $x=3$: as $x \to 3^+$ (from the domain side), $(x-1)(x-3)$ is positive and approaches 0. So:

   $$\lim_{x \to 3^+} \ln((x-1)(x-3)) = -\infty$$
5. Thus $x=3$ is also a vertical asymptote.

> **Exam tip:** For logarithmic functions, only check boundaries of the domain where the argument approaches 0 from the positive side. Points where the argument approaches 0 from the negative side are outside the domain, so no asymptote exists there.

## AP-Style Worked Practice

**Worked example:** Which of the following gives all values of $x$ at which the graph of $f(x) = \frac{3x^2 - 10x - 8}{x(x^2 - 16)}$ has a vertical asymptote?<br>Options: (A) $x=4$ only, (B) $x=-4$ and $x=0$ only, (C) $x=-4$, $x=0$, and $x=4$, (D) $x=-\frac{2}{3}$ and $x=4$

1. Factor numerator and denominator completely:

   $$3x^2 - 10x -8 = (3x + 2)(x - 4), \quad x(x^2 -16) = x(x-4)(x+4)$$
2. Cancel the common $(x-4)$ factor, giving the simplified function:

   $$f(x) = \frac{3x+2}{x(x+4)} \quad \text{for } x \neq 4$$
3. The simplified denominator equals zero at $x=0$ and $x=-4$. One-sided limits at both points are infinite, so both are vertical asymptotes. At $x=4$, the limit is finite:

   $$\lim_{x \to 4} \frac{3x+2}{x(x+4)} = \frac{14}{32} = \frac{7}{16}$$
4. So $x=4$ is a hole, not an asymptote. The correct answer is (B).

**Worked example:** Let $f(x) = \frac{e^{2x}}{(x-2)(x+4)}$. (a) Find all candidate points for vertical asymptotes, and justify why each is a candidate. (b) Confirm whether each candidate is a vertical asymptote by evaluating one-sided limits. (c) State all vertical asymptotes and explain why there are no others.

1. Part (a): $f(x)$ is a quotient of continuous functions, so it is only undefined where the denominator equals zero. Set $(x-2)(x+4) = 0$, so candidates are $x=2$ and $x=-4$. These are candidates because $f$ is undefined at both points, so infinite limit behavior is possible.
2. Part (b): At $x=2$: $\lim_{x \to 2^-} \frac{e^{2x}}{(x-2)(x+4)} = \frac{e^4}{(\text{negative})(\text{positive})} = -\infty$, so $x=2$ is a vertical asymptote. At $x=-4$: $\lim_{x \to -4^-} \frac{e^{2x}}{(x-2)(x+4)} = \frac{e^{-8}}{(\text{negative})(\text{negative})} = +\infty$, so $x=-4$ is also a vertical asymptote.
3. Part (c): All vertical asymptotes are $x=2$ and $x=-4$. $f(x)$ is defined and continuous everywhere else on its domain, so there are no other points with possible infinite limit behavior, hence no additional vertical asymptotes.

## Common pitfalls

- **Wrong:** Claiming $x=2$ is a vertical asymptote of $f(x) = \frac{x^2 - 4}{x-2}$ because it makes the original denominator zero.
  - Why it fails: Students confuse undefined points with asymptotes, forgetting to check for common factors that create removable discontinuities.
  - Correct: Always simplify the function first, then check if the limit as $x$ approaches the undefined point is infinite; finite limits mean holes, not asymptotes.
- **Wrong:** Concluding $x=a$ is not a vertical asymptote because the two one-sided limits go to opposite infinities.
  - Why it fails: Students incorrectly believe both one-sided limits must go to the same infinity for an asymptote to exist.
  - Correct: Recall that any infinite one-sided limit (one or two sides) is enough to confirm a vertical asymptote at $x=a$, regardless of whether the two sides match.
- **Wrong:** Claiming $x=0$ is not a vertical asymptote of $f(x) = \frac{1}{x^2}$ because $\lim_{x \to 0} \frac{1}{x^2} = +\infty$ (so the limit does not exist as a finite number).
  - Why it fails: Students confuse 'the limit does not exist as a finite number' with 'no infinite behavior that creates an asymptote'.
  - Correct: Remember infinite limit notation describes unbounded behavior, not an existing finite limit; if $\lim_{x \to a} f(x) = \pm\infty$, $x=a$ is a vertical asymptote.
- **Wrong:** Stating that $x=0$ is a vertical asymptote of $f(x) = \frac{\sin x}{x}$ because the denominator is zero at $x=0$.
  - Why it fails: Students memorize 'denominator zero means vertical asymptote' without checking the limit.
  - Correct: Always evaluate the limit as $x$ approaches the undefined point; $\lim_{x \to 0} \frac{\sin x}{x} = 1$, so this is a removable discontinuity, not an asymptote.
- **Wrong:** For $f(x) = \ln(x-4)$, claiming $x=0$ is a vertical asymptote because $\ln(0)$ is undefined.
  - Why it fails: Students forget to check what input makes the argument of the logarithm zero.
  - Correct: For logarithmic functions, set the argument equal to zero to find the candidate vertical asymptote, then confirm the limit is infinite from the domain side.

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Right-sided infinite limit | $\lim_{x \to a^+} f(x) = \infty$ | $f(x)$ grows without bound as $x$ approaches $a$ from the right; limit does not exist as a finite number |
| Left-sided infinite limit | $\lim_{x \to a^-} f(x) = -\infty$ | $f(x)$ decreases without bound as $x$ approaches $a$ from the left |
| Vertical asymptote definition | $x=a$ is a VA if at least one $\lim_{x \to a^\pm} f(x) = \pm\infty$ | Only requires one infinite one-sided limit; both sides do not need to match |
| Rational function VAs | After canceling common factors, $x=a$ is a VA if simplified denominator $= 0$ at $x=a$ | Canceled common factors create holes (removable discontinuities), not VAs |
| Logarithm VAs | $y = \ln(g(x))$ has VA at $x=a$ if $\lim_{x \to a^\pm} g(x) = 0^+$ | Only check boundaries of the domain of $f$ |
| Tangent VAs | $y = \tan x$ has VAs at $x = \frac{\pi}{2} + k\pi, k \in \mathbb{Z}$ | Follows from $\tan x = \frac{\sin x}{\cos x}$; VAs where $\cos x = 0$ |
| Hole vs Vertical Asymptote | If $\lim_{x \to a} f(x)$ is finite, $x=a$ is a hole; if infinite, it is a VA | All holes in rational functions come from common factors |

## What's next

This topic is a core prerequisite for the rest of Unit 1: Limits and Continuity, and for key topics later in the AP Calculus BC course. Immediately next, you will connect infinite limits at infinity to horizontal asymptotes, then use your understanding of asymptotes for full curve sketching of functions, including derivative and second derivative graphs. This topic is also critical for the BC-exclusive topic of improper integrals, where you must identify vertical asymptotes to correctly classify and evaluate improper integrals of functions with discontinuities. Without correctly identifying vertical asymptotes from infinite limits, you will misclassify discontinuities for integration and miss points on FRQ questions.

- [Improper integrals](https://www.owlsprep.com/study/ap-calculus-bc-u6-improper-integrals/)
- [Connecting limits at infinity and horizontal asymptotes](https://www.owlsprep.com/study/ap-calculus-bc-u1-connecting-limits-at-infinity-and/)
- [Working with the Intermediate Value Theorem (IVT)](https://www.owlsprep.com/study/ap-calculus-bc-u1-working-with-the-intermediate-value/)

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