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

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: definitions of one-sided and two-sided infinite limits, rules for locating vertical asymptotes of common functions, matching infinite limit behavior to vertical asymptote placement, and interpreting vertical asymptotes in applied contexts.

You should already know: How to evaluate one-sided and two-sided limits, how to factor polynomials and identify domain discontinuities, basic properties of rational and logarithmic functions.

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 AB 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 establishes the formal graphical and analytical connection between unbounded function behavior (infinite limits) and vertical asymptotes, a core concept in Unit 1: Limits and Continuity, which accounts for ~10-12% of the total AP Calculus AB exam weight per the course CED. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a foundational step for graphing, continuity classification, and differential equation solution curve sketching. An infinite limit occurs when as $x$ approaches a constant value $c$ from the left, right, or both sides, the value of $f(x)$ grows without bound (positive or negative infinity), which we write as ${\lim }_{x\to c^{-}}f(x)=\pm \infty$, ${\lim }_{x\to c^{+}}f(x)=\pm \infty$, or ${\lim }_{x\to c}f(x)=\pm \infty$. By definition, the line $x=c$ is a vertical asymptote of $f(x)$ if at least one one-sided infinite limit exists at $x=c$. Unlike removable discontinuities (holes), which occur when the limit is finite but the function is undefined at $c$, vertical asymptotes correspond to permanent unbounded behavior. This topic teaches you to identify them analytically, rather than relying only on graph reading.

2. Infinite Limits: Definition and One-Sided Sign Testing

Unlike finite limits, where $f(x)$ approaches a constant number as $x$ approaches $c$, infinite limits describe unbounded growth (or decay toward negative infinity) of the function value near $x=c$. It is critical to note that ${\lim }_{x\to c}f(x)=\infty$ does not mean the limit exists; infinity is not a number, it is just notation to describe how $f(x)$ grows without any upper bound. For a vertical asymptote to exist at $x=c$, we only need at least one one-sided infinite limit at $c$, not that both sides go to the same infinity. For example, the left-hand limit could go to $-\infty$ and the right-hand limit could go to $+\infty$, but $x=c$ is still a vertical asymptote. To evaluate the sign of an infinite limit, we use the test point method: after confirming the function is undefined at $x=c$ and no common factor cancels the $(x-c)$ term in the denominator, we pick a value of $x$ very close to $c$ on the side we are testing, plug it into the simplified function, and check the sign of the output. If the output is a large positive number, the limit is $+\infty$; if large negative, it is $-\infty$.

Worked Example

Evaluate ${\lim }_{x\to 3^{-}}\frac{x+2}{x^2-2x-3}$ and state what the result tells us about vertical asymptotes.

1. Factor the denominator: $x^2-2x-3=(x-3)(x+1)$. There are no common factors between the numerator $(x+2)$ and the denominator, so no removable discontinuity exists at $x=3$.
2. We are approaching $x=3$ from the left, so we pick a test point $x=2.9$, which is just less than 3.
3. Evaluate the sign of each factor: numerator $2.9+2=4.9>0$, $(x-3)=2.9-3=-0.1<0$, $(x+1)=3.9>0$.
4. Multiply signs: $\frac{(+)}{(-)(+)}=-$, so the output is a large negative number. This gives ${\lim }_{x\to 3^{-}}\frac{x+2}{x^2-2x-3}=-\infty$.
5. Because at least one one-sided infinite limit exists at $x=3$, the line $x=3$ is a vertical asymptote of the function.

Exam tip: When testing the sign, you only need to track the sign of each factor, not the actual numerical value. This saves significant time on MCQs, where you do not need to show intermediate calculations.

3. Locating Vertical Asymptotes Analytically

For most functions you will encounter on the AP exam, vertical asymptotes occur at values excluded from the domain of the function where the function approaches $\pm \infty$. The process for finding vertical asymptotes depends on the type of function, but follows a consistent core workflow: first simplify the function, then check for values that cause unbounded behavior. For rational functions (ratio of two polynomials): Simplify the function by canceling any common factors between numerator and denominator. Any value $x=c$ that makes the simplified denominator equal to zero is a vertical asymptote at $x=c$. Removable discontinuities (holes) occur at roots of the original denominator that cancel out, so they are not vertical asymptotes. For logarithmic functions of the form $f(x)=\ln (g(x))$: The argument $g(x)$ must be positive, so vertical asymptotes occur where $g(c)=0$ and $g(x)$ is positive on at least one side of $x=c$, because as $x$ approaches $c$ from that side, $\ln (g(x))$ approaches $-\infty$.

Worked Example

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

1. Factor the numerator and denominator: $x^2-4=(x-2)(x+2)$, and $x^2-3x+2=(x-1)(x-2)$.
2. Simplify the function by canceling the common $(x-2)$ factor, leaving $f(x)=\frac{x+2}{x-1}$ for all $x\ne 2$.
3. Find roots of the simplified denominator: $x-1=0⟹x=1$. The root $x=2$ canceled out, so it is a removable discontinuity, not a vertical asymptote.
4. Confirm infinite limit behavior: ${\lim }_{x\to 1^{-}}\frac{x+2}{x-1}=-\infty$ and ${\lim }_{x\to 1^{+}}\frac{x+2}{x-1}=+\infty$, so both one-sided limits are infinite. The only vertical asymptote is $x=1$.

Exam tip: Always simplify the function first before identifying vertical asymptotes. AP exam questions deliberately include common factors to test if you can distinguish holes from asymptotes.

4. Matching Infinite Limit Behavior to Graph Shape

Once you have found a vertical asymptote at $x=c$, the sign of the left-hand and right-hand infinite limits tells you the shape of the graph near the asymptote, which is frequently tested in graph-sketching FRQs and graph-identification MCQs. If ${\lim }_{x\to c^{-}}f(x)=+\infty$ and ${\lim }_{x\to c^{+}}f(x)=+\infty$, both sides of the graph shoot upward toward the asymptote. If one side is $+\infty$ and the other is $-\infty$, the graph goes up on one side and down on the other. For functions that are only defined on one side of the asymptote (such as $\ln (x)$ at $x=0$), you only need to describe the behavior on the side where the function exists. Never assume both sides of an asymptote go to the same sign of infinity—always confirm with the test point method.

Worked Example

Describe the shape of the graph of $f(x)=\ln (4-x^2)$ near its vertical asymptotes.

1. Find the domain of the function: $4-x^2>0⟹x^2<4⟹-2<x<2$, so the function is undefined at $x=2$ and $x=-2$.
2. Evaluate limits on the defined sides: As $x\to 2^{-}$ (the only side of $x=2$ with function values), $4-x^2\to 0^{+}$, so $\ln (4-x^2)\to -\infty$. As $x\to -2^{+}$ (the only side of $x=-2$ with function values), $4-x^2\to 0^{+}$, so $\ln (4-x^2)\to -\infty$.
3. Describe shape: Vertical asymptotes are at $x=2$ and $x=-2$. The graph approaches negative infinity as it approaches $x=2$ from the left, and approaches negative infinity as it approaches $x=-2$ from the right.

Exam tip: For functions defined on only one side of a vertical asymptote, you only need an infinite limit on that side for it to count as a vertical asymptote, per AP exam definition.

5. Common Pitfalls (and how to avoid them)

• Wrong move: After factoring a rational function, counting $x=c$ as a vertical asymptote even though the $(x-c)$ factor cancels with the numerator. Why: Students forget that canceled factors correspond to removable discontinuities (holes), not unbounded behavior. Correct move: Always cancel common factors first, then only count roots of the simplified denominator as vertical asymptotes.
• Wrong move: Assuming that if a function is undefined at $x=c$, then $x=c$ must be a vertical asymptote. Why: Students confuse domain restrictions with unbounded behavior, forgetting that discontinuities can be removable or jump. Correct move: Always confirm that at least one one-sided limit at $x=c$ is infinite before labeling it a vertical asymptote.
• Wrong move: Claiming that ${\lim }_{x\to c}f(x)=\infty$ means the limit exists and equals infinity. Why: The notation for infinite limits is just a description of behavior, not a finite limit value. Correct move: When asked if the limit exists, state that infinite limits do not exist (DNE), and the notation $\infty$ only describes the direction of unbounded growth.
• Wrong move: Assuming both sides of a vertical asymptote must go to the same sign of infinity. Why: Students generalize from examples like $\frac{1}{x^2}$ where both sides are positive, and forget that common functions like $\frac{1}{x}$ have opposite signs. Correct move: Always test the sign of the one-sided limit separately for the left and right side of the asymptote.
• Wrong move: For $f(x)=\ln ((x-3{)}^2)$, claiming that there is no vertical asymptote at $x=3$ because $(x-3{)}^2$ never changes sign. Why: Students confuse the requirement for a sign change across $x=c$ with the requirement for the argument to be positive near $c$. Correct move: For any $x=c$ where the argument of the logarithm approaches zero from the positive side (on at least one side), $x=c$ is a vertical asymptote, regardless of sign change on both sides.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following gives the number of vertical asymptotes of $f(x)=\frac{3x(x-4)}{x^3-4x^2-9x+36}$? (A) 0 (B) 1 (C) 2 (D) 3

Worked Solution: First factor the denominator by grouping: $x^3-4x^2-9x+36=x^2(x-4)-9(x-4)=(x^2-9)(x-4)=(x-3)(x+3)(x-4)$. The numerator is $3x(x-4)$, so we cancel the common $(x-4)$ factor, leaving $f(x)=\frac{3x}{(x-3)(x+3)}$ for $x\ne 4$. The simplified denominator has roots at $x=3$ and $x=-3$, neither of which cancel with the numerator. The root $x=4$ cancels, so it is a hole, not an asymptote. This leaves two vertical asymptotes. The correct answer is (C).

Question 2 (Free Response)

Let $f(x)=\frac{e^x}{x^2-4}$. (a) Find all values of $x$ where vertical asymptotes occur, and justify your answer. (b) Evaluate the one-sided limits ${\lim }_{x\to 2^{+}}f(x)$ and ${\lim }_{x\to -2^{-}}f(x)$. (c) Explain why $f(x)$ does not have a vertical asymptote at $x=0$, even though $f(0)$ is defined.

Worked Solution: (a) Factor the denominator: $x^2-4=(x-2)(x+2)$. The numerator $e^x$ is never zero for real $x$, so there are no common factors to cancel. The denominator is zero at $x=2$ and $x=-2$, and at least one one-sided limit at each value is infinite. By definition, the vertical asymptotes are $\boxed{x=2}$ and $\boxed{x=-2}$. (b) For ${\lim }_{x\to 2^{+}}f(x)$: $e^x>0$ for all $x$, $(x-2)>0$ for $x>2$, $(x+2)>0$ near $x=2$, so $\frac{(+)}{(+)(+)}=+$, giving ${\lim }_{x\to 2^{+}}f(x)=\boxed{+\infty }$. For ${\lim }_{x\to -2^{-}}f(x)$: $e^x>0$, $(x-2)<0$ for $x<-2$, $(x+2)<0$ near $x=-2$, so $\frac{(+)}{(-)(-)}=+$, giving ${\lim }_{x\to -2^{-}}f(x)=\boxed{+\infty }$. (c) At $x=0$, $f(x)$ is defined, and the limit as $x\to 0$ is finite: ${\lim }_{x\to 0}\frac{e^x}{x^2-4}=\frac{1}{-4}=-\frac{1}{4}$, which matches the function value. There is no unbounded behavior at $x=0$, so it cannot be a vertical asymptote.

Question 3 (Application / Real-World Style)

A chemical mixing tank initially contains 10 liters of pure water. A salt solution with concentration 2 kg/L is pumped in at a rate of 1 L/min, and the well-mixed solution is pumped out at a rate of 2 L/min. The concentration of salt in the tank after $t$ minutes is given by $C(t)=\frac{2(10-t)-2(10-t)\ln (10-t)}{10-t}$, for $0<t<10$ (concentration in kg/L). Find the vertical asymptote of $C(t)$ on $(0,10)$ and interpret what it means in context.

Worked Solution: First simplify $C(t)$: the $(10-t)$ factor cancels for all $t\ne 10$, leaving $C(t)=2-2\ln (10-t)$ for $0<t<10$. Evaluate the limit as $t$ approaches 10 from the left: ${\lim }_{t\to 10^{-}}(10-t)=0^{+}$, so ${\lim }_{t\to 10^{-}}\ln (10-t)=-\infty$, which gives ${\lim }_{t\to 10^{-}}C(t)=2-2(-\infty )=+\infty$. The vertical asymptote is at $\boxed{t=10}$ minutes. Context interpretation: After 10 minutes, the tank is completely empty, so the concentration of salt grows without bound as the tank approaches emptiness, which matches the physical behavior of the system.

7. Quick Reference Cheatsheet

8. What's Next

This topic is a foundational prerequisite for all subsequent work on graphing functions, analyzing continuity, and sketching solution curves for differential equations, all of which are heavily tested on the AP Calculus AB exam. Next you will connect vertical asymptotes to horizontal asymptotes by studying limits at infinity, which extends the idea of unbounded behavior to $x$ growing without bound, rather than $x$ approaching a constant. Without correctly identifying vertical asymptotes, you will not be able to correctly classify discontinuities, answer graph interpretation questions, or draw accurate solution curves for differential equations on FRQs.

• Connecting limits at infinity and horizontal asymptotes
• Continuity of functions at a point
• Classifying discontinuities
• Sketching graphs of functions