Connecting infinite limits and vertical asymptotes — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Definition of one-sided and two-sided infinite limits, the core rule connecting infinite limits to vertical asymptotes, locating vertical asymptotes for rational, logarithmic, and trigonometric functions, and distinguishing vertical asymptotes from removable discontinuities (holes).

You should already know: How to evaluate one-sided limits, how to factor polynomials and simplify rational functions, the definition of discontinuity.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Connecting infinite limits and vertical asymptotes?

This topic, part of AP Calculus BC Unit 1: Limits and Continuity, makes up approximately 2-3% of the total AP exam score (as part of the 10-12% total exam weight for Unit 1), and appears in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam. At its core, this topic connects the unbounded behavior of functions near discontinuities to a key feature of their graphs: vertical asymptotes.

An infinite limit is notation that describes how a function behaves as it grows (or decreases) without bound as x approaches a constant value $a$, rather than approaching a finite limit. It does not mean the limit "exists" as a finite number; it is just a concise way to describe unbounded behavior near $x=a$. A vertical asymptote is the vertical line $x=a$ where this unbounded behavior occurs. The key AP CED learning outcome for this topic is the ability to use infinite limit behavior to locate vertical asymptotes, and vice versa, while correctly distinguishing vertical asymptotes from other discontinuities like holes. This is a foundational topic for all later graph analysis and integration work.

2. Infinite Limits: Definition and One-Sided Behavior

An infinite limit is written using the notation:

• ${\lim }_{x\to a^{+}}f(x)=\infty$: As $x$ approaches $a$ from the right (values slightly larger than $a$), $f(x)$ grows without bound (for any positive number $M$, you can find $f(x)$ larger than $M$ by taking $x$ close enough to $a$).
• ${\lim }_{x\to a^{-}}f(x)=-\infty$: As $x$ approaches $a$ from the left, $f(x)$ decreases without bound.
• The same notation extends to left-sided positive infinite limits and right-sided negative infinite limits.

To evaluate an infinite limit of a rational function as $x\to a$, you first evaluate the limit of the numerator and denominator separately at $a$. If the numerator approaches a non-zero constant, and the denominator approaches 0, the result is an infinite limit. The sign depends on the sign of the numerator and the sign of the denominator as $x$ approaches $a$ from the relevant side.

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

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.

3. Core Connection: Vertical Asymptotes from Infinite Limits

By definition, the line $x=a$ is a vertical asymptote of the function $f(x)$ if at least one of the one-sided limits of $f(x)$ as $x$ approaches $a$ is infinite (either $+\infty$ or $-\infty$). This is the full formal connection: infinite limit behavior at $x=a$ is exactly the condition for $x=a$ to be a vertical asymptote.

For rational functions, this gives a simple rule for locating vertical asymptotes:

1. Factor the numerator and denominator completely.
2. Cancel any common factors to get the simplified function (note that common factors create removable discontinuities, or holes, at the $x$-value that makes the factor zero).
3. Any $x=a$ that makes the simplified denominator equal to zero is a vertical asymptote, because the limit as $x\to a$ will be infinite.

Worked Example

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

1. Factor numerator and denominator: $x^2-9=(x-3)(x+3)$, $x^2-2x-3=(x-3)(x+1)$.
2. Simplify, noting the domain restriction $x\ne 3$: $f(x)=\frac{x+3}{x+1}$ for $x\ne 3$.
3. Find zeros of the simplified denominator: $x+1=0⟹x=-1$.
4. Check one-sided limits at $x=-1$: ${\lim }_{x\to -1^{-}}\frac{x+3}{x+1}=\frac{2}{0^{-}}=-\infty$, so we have an infinite one-sided limit.
5. At $x=3$, the limit is ${\lim }_{x\to 3}\frac{x+3}{x+1}=\frac{6}{4}=\frac{3}{2}$, which is finite, so $x=3$ is a hole, not a vertical asymptote.
6. 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.

4. Vertical Asymptotes of Non-Rational Functions

Rational functions are not the only functions with vertical asymptotes. For non-rational functions (trigonometric, logarithmic, etc.), you still follow the core rule: find all points where the function is undefined, then check if at least one one-sided limit at that point is infinite. Common cases tested on the AP exam:

• Logarithmic functions: $f(x)=\ln (g(x))$ is undefined when $g(x)\le 0$, so vertical asymptotes occur where $g(x)=0$ and $g(x)$ approaches 0 from the positive side (from within the domain of $f$).
• Trigonometric functions: $\tan x=\frac{\sin x}{\cos x}$, $\cot x=\frac{\cos x}{\sin x}$, $\sec x=\frac{1}{\cos x}$, $\csc x=\frac{1}{\sin x}$ all have vertical asymptotes where their denominators are zero, since the numerator is non-zero at those points.

Worked Example

Find all vertical asymptotes of $f(x)=\ln (x^2-4x+3)$.

1. First find the domain: the argument of the log must be positive: $x^2-4x+3=(x-1)(x-3)>0$, so domain is $(-\infty ,1)\cup (3,\infty )$.
2. Candidates for vertical asymptotes are the boundaries of the domain: $x=1$ and $x=3$, where the argument equals zero.
3. Check $x=1$: as $x\to 1^{-}$ (from the domain side), $(x-1)(x-3)=(\text{negative})(\text{negative})=\text{positive}$ approaching 0. So ${\lim }_{x\to 1^{-}}\ln ((x-1)(x-3))=-\infty$, so $x=1$ is a vertical asymptote.
4. Check $x=3$: as $x\to 3^{+}$ (from the domain side), $(x-1)(x-3)=(\text{positive})(\text{positive})=\text{positive}$ approaching 0. So ${\lim }_{x\to 3^{+}}f(x)=-\infty$, so $x=3$ 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.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming $x=2$ is a vertical asymptote of $f(x)=\frac{x^2-4}{x-2}$ because it makes the original denominator zero. Why: Students confuse undefined points with asymptotes, forgetting to check for common factors that create removable discontinuities. Correct move: 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 move: Concluding $x=a$ is not a vertical asymptote because the two one-sided limits go to opposite infinities. Why: Students incorrectly believe both one-sided limits must go to the same infinity for an asymptote to exist. Correct move: 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 move: 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: Students confuse "the limit does not exist as a finite number" with "no infinite behavior that creates an asymptote". Correct move: 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 move: Stating that $x=0$ is a vertical asymptote of $f(x)=\frac{\sin x}{x}$ because the denominator is zero at $x=0$. Why: Students memorize "denominator zero means vertical asymptote" without checking the limit. Correct move: 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 move: For $f(x)=\ln (x-4)$, claiming $x=0$ is a vertical asymptote because $\ln (0)$ is undefined. Why: Students forget to check what input makes the argument of the logarithm zero. Correct move: 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.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

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? 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 $x=4$

Worked Solution: First, factor the numerator and denominator completely: the numerator $3x^2-10x-8=(3x+2)(x-4)$, and the denominator $x(x^2-16)=x(x-4)(x+4)$. Cancel the common $(x-4)$ factor, giving the simplified function $\frac{3x+2}{x(x+4)}$ for $x\ne 4$. The simplified denominator is zero at $x=0$ and $x=-4$, and one-sided limits at both points are infinite, so both are vertical asymptotes. At $x=4$, the limit is $\frac{3(4)+2}{4(4+4)}=\frac{14}{32}=\frac{7}{16}$, a finite value, so $x=4$ is a hole, not an asymptote. The correct answer is B.

Question 2 (Free Response)

Let $f(x)=\frac{e^{2x}}{(x-2)(x+4)}$. (a) Find all candidate points for vertical asymptotes of $f(x)$, and justify why each is a candidate. (b) For each candidate point, confirm whether it is a vertical asymptote by evaluating one-sided limits. (c) State all vertical asymptotes of $f(x)$, and explain why there are no other vertical asymptotes.

Worked Solution: (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. (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. (c) All vertical asymptotes are $x=2$ and $x=-4$. $f(x)$ is defined and continuous everywhere else on $(-\infty ,-4)\cup (-4,2)\cup (2,\infty )$, so there are no other points where $f$ is undefined, hence no other points with infinite limit behavior, so no additional vertical asymptotes.

Question 3 (Application / Real-World Style)

In a mixing tank, the concentration of salt in a brine solution $t$ minutes after a drain is opened is given by $C(t)=\frac{15t}{10-t}$ measured in grams per liter, for $0\le t<10$. Identify the vertical asymptote of $C(t)$, evaluate the one-sided limit as $t$ approaches the asymptote from the domain side, and interpret the result in context.

Worked Solution: $C(t)$ is undefined at $t=10$, which is the only candidate for a vertical asymptote. Evaluate the one-sided limit as $t\to 10^{-}$ (from the domain side): the numerator $15t$ approaches $150$, a positive constant, and the denominator $10-t$ approaches 0 from the positive side. So ${\lim }_{t\to 10^{-}}\frac{15t}{10-t}=+\infty$, so the vertical asymptote is $t=10$. In context, this means that as 10 minutes approaches, the concentration of salt in the tank grows without bound as all the water drains out, leaving nearly pure salt in the tank.

7. Quick Reference Cheatsheet

8. 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.

Follow on topics: Connecting limits at infinity and horizontal asymptotes Removable vs non-removable discontinuities Improper integrals Curve sketching