Connecting Infinite Limits and Vertical Asymptotes - AP Calculus AB Study Guide

The Core Idea: Connecting Infinite Limits and Vertical Asymptotes

A vertical asymptote is a vertical line on a graph that a function approaches but never crosses. While we can often identify these visually, calculus provides a precise, analytical way to define and locate them using the concept of limits. This topic establishes the formal connection between the graphical feature of a vertical asymptote and the analytical behavior of an infinite limit.

An infinite limit occurs when the output values of a function, $f(x)$, grow without bound (approaching positive or negative infinity) as the input value, $x$, gets arbitrarily close to some number, $a$. The existence of such an infinite limit at $x=a$ is the definitive mathematical condition for the line $x=a$ to be a vertical asymptote of the function's graph. This topic provides the tools to move from an intuitive understanding of asymptotes to a rigorous, limit-based justification required in calculus.

Key Definitions

The formal definition of a vertical asymptote is based on one-sided infinite limits.

The vertical line $x=a$ is a vertical asymptote of the graph of a function $f$ if at least one of the following limit statements is true:

1. The limit as $x$ approaches $a$ from the right is positive infinity:

   $$\underset{x\to a^{+}}{\lim }f(x)=\infty$$

2. The limit as $x$ approaches $a$ from the right is negative infinity:

   $$\underset{x\to a^{+}}{\lim }f(x)=-\infty$$

3. The limit as $x$ approaches $a$ from the left is positive infinity:

   $$\underset{x\to a^{-}}{\lim }f(x)=\infty$$

4. The limit as $x$ approaches $a$ from the left is negative infinity:

   $$\underset{x\to a^{-}}{\lim }f(x)=-\infty$$

Only one of these conditions needs to be met to confirm the existence of a vertical asymptote at $x=a$.

Understanding Rational Functions and Asymptotes

For rational functions, which are functions of the form $f(x)=\frac{N(x)}{D(x)}$ where $N(x)$ and $D(x)$ are polynomials, we have a systematic process for identifying potential vertical asymptotes.

A vertical asymptote can only occur where the function is undefined, which for a rational function is at any $x$-value that makes the denominator zero.

1. Identify Candidates: First, find all values of $x=a$ such that the denominator $D(a)=0$. These are the only possible locations for vertical asymptotes.

2. Check the Numerator: For each candidate $x=a$:

   • If the numerator $N(a)\ne 0$ while the denominator $D(a)=0$, then the line $x=a$ is a vertical asymptote. The function's limit will approach $\pm \infty$ because you have a non-zero number divided by a quantity approaching zero.

   • If the numerator $N(a)=0$ and the denominator $D(a)=0$, the limit has the indeterminate form $\frac{0}{0}$. This indicates that the function may have either a hole (a removable discontinuity) or a vertical asymptote at $x=a$. To determine which it is, you must perform further algebraic analysis, typically by factoring both the numerator and denominator and canceling common factors.

Core Concepts & Rules

• The line $x=a$ is a vertical asymptote if the limit of $f(x)$ as $x$ approaches $a$ from the left or the right is $\infty$ or $-\infty$.

• For a rational function, potential vertical asymptotes exist at the $x$-values that make the denominator equal to zero.

• If the denominator of a rational function is zero at $x=a$ but the numerator is non-zero, then $x=a$ is a vertical asymptote.

• If both the numerator and denominator of a rational function are zero at $x=a$, the situation is inconclusive. You must simplify the function (e.g., by factoring) to determine if there is a hole or a vertical asymptote at $x=a$. After simplifying, if the denominator is still zero at $x=a$, it is a vertical asymptote; otherwise, it is a hole.

Step-by-Step Example 1: Basic Application

Find the vertical asymptotes of the function $f(x)=\frac{2x}{x-4}$ and justify your answer using limits.

Step 1: Identify potential vertical asymptotes.

A rational function can only have a vertical asymptote where its denominator is zero. Set the denominator equal to zero and solve for $x$.

$$x-4=0⟹x=4$$

The only candidate for a vertical asymptote is the line $x=4$.

Step 2: Check the numerator at the candidate value.

Evaluate the numerator at $x=4$.

$$2(4)=8$$

Since the numerator is $8$ (non-zero) and the denominator is $0$ at $x=4$, we can conclude that $x=4$ is a vertical asymptote.

Step 3: Justify the conclusion with a limit.

To provide a complete calculus justification, we must show that a one-sided limit is infinite. Let's evaluate the limit as $x$ approaches $4$ from the right.

$$\underset{x\to 4^{+}}{\lim }\frac{2x}{x-4}$$

• As $x\to 4^{+}$, the numerator $2x$ approaches $2(4)=8$.

• As $x\to 4^{+}$, the denominator $x-4$ approaches $0$ through small positive values (e.g., if $x=4.01$, then $x-4=0.01$). We denote this as $0^{+}$.

• Therefore, the limit is of the form $\frac{8}{0^{+}}$, which tends to $\infty$.

$$\underset{x\to 4^{+}}{\lim }\frac{2x}{x-4}=\infty$$

Conclusion:

Because ${\lim }_{x\to 4^{+}}f(x)=\infty$, the line $x=4$ is a vertical asymptote for the graph of $f(x)$.

Step-by-Step Example 2: Exam-Style Application

Find all vertical asymptotes for the function $g(x)=\frac{x^2-9}{x^2+x-6}$.

Step 1: Identify potential vertical asymptotes.

Set the denominator equal to zero and solve for $x$.

$$x^2+x-6=0$$

$$(x+3)(x-2)=0$$

The candidates for vertical asymptotes are $x=-3$ and $x=2$.

Step 2: Analyze the candidate $x=-3$.

• Evaluate the denominator at $x=-3$: $(-3+3)(-3-2)=(0)(-5)=0$.

• Evaluate the numerator at $x=-3$: $(-3{)}^2-9=9-9=0$.

• Since we have the indeterminate form $\frac{0}{0}$, we must simplify the function by factoring.

Step 3: Simplify the function.

$$g(x)=\frac{x^2-9}{x^2+x-6}=\frac{(x-3)(x+3)}{(x+3)(x-2)}$$

For $x\ne -3$, we can cancel the $(x+3)$ terms:

$$g(x)=\frac{x-3}{x-2},x\ne -3$$

To determine the behavior at $x=-3$, we evaluate the limit of the simplified function:

$$\underset{x\to -3}{\lim }\frac{x-3}{x-2}=\frac{-3-3}{-3-2}=\frac{-6}{-5}=\frac{6}{5}$$

Since the limit exists and is a finite number, there is a hole (removable discontinuity) at $x=-3$, not a vertical asymptote.

Step 4: Analyze the candidate $x=2$.

• Evaluate the denominator at $x=2$: $(2+3)(2-2)=(5)(0)=0$.

• Evaluate the numerator at $x=2$: $(2{)}^2-9=4-9=-5$.

• Since the numerator is non-zero ($-5$) and the denominator is zero, $x=2$ is a vertical asymptote.

Conclusion:

The only vertical asymptote for the graph of $g(x)$ is the line $x=2$.

Using Your Calculator

The process of finding vertical asymptotes is primarily analytical (algebraic). A graphing calculator is best used as a tool to verify your conclusions, not to find them.

**To verify an asymptote for $f(x)=\frac{x^2-9}{x^2+x-6}‘:\ast \ast 1.\ast \ast GraphtheFunction:\ast \ast \ast Press‘Y=‘.\ast Enterthefunctioninto‘Y1‘:$(X^2 - 9) / (X^2 + X - 6). Use parentheses for the numerator and denominator. * Press `GRAPH`. You should see the graph approaching a vertical line at $x=2. You will likely not "see" the hole at $x=-3$, as it is a single point.

2. Use the Table to Investigate Limits:

   • Press 2nd then TBLSET(Table Setup). Set TblStart to a value near your suspected asymptote (e.g., $1.9$) and ΔTbl to a small number (e.g., $0.01$).

   • Press 2nd then TABLE.

   • Scroll through the values as $x$ approaches $2$. You will see the Y1 values become very large negative numbers as $x$ approaches $2$ from the left (e.g., at $x=1.99$) and very large positive numbers as $x$ approaches $2$ from the right (e.g., at $x=2.01$). At $x=2$, the table will show an ERROR. This behavior supports the conclusion that $x=2$ is a vertical asymptote.

   • To investigate the hole at $x=-3$, you can see that values around $x=-3$ (e.g., $-3.01$, $-2.99$) are close to $1.2$ ($6/5$), while at $x=-3$ itself, there is an ERROR. This supports the existence of a hole.

AP Exam Quick Hit

Common Question Types

• Finding Asymptotes from an Equation: A multiple-choice question will provide a rational function and ask for the equations of its vertical asymptotes. Example: "Find all vertical asymptotes of $f(x)=\frac{x-5}{x^2-25}$." (The answer is $x=-5$ only, as $x=5$ is a hole).

• Justifying Asymptotes in a Free-Response Question (FRQ): A part of an FRQ may state: "Find the equation of each vertical asymptote of the graph of $f$. Justify your answer." A full-credit justification requires stating the infinite limit. Example: "The line $x=a$ is a vertical asymptote because ${\lim }_{x\to a^{+}}f(x)=-\infty$."

• Interpreting Limits from a Graph: A question may show a graph with a vertical asymptote and ask to identify the value of a limit. Example: "The graph of $f$ is shown above. What is ${\lim }_{x\to 2^{-}}f(x)$?" (The answer would be $\infty$ or $-\infty$ depending on the graph).

Common Mistakes

• Confusing Holes and Asymptotes: Automatically assuming that any value making the denominator zero corresponds to a vertical asymptote. Always check if the numerator is also zero, which signals the need for further investigation (factoring).

• Incomplete Justification: Stating that $x=a$ is a vertical asymptote "because the denominator is zero" is not a valid justification on the AP Exam. The justification must be based on the definition: a one-sided limit must be $\infty$ or $-\infty$.

• Algebraic Errors: Simple mistakes in factoring the numerator or denominator can lead to incorrect conclusions about which discontinuities are holes and which are asymptotes.

• Forgetting Negative Signs: When analyzing a limit like ${\lim }_{x\to 3^{-}}\frac{1}{x-3}$, students might forget that $x-3$ is a small negative number, leading them to conclude the limit is $\infty$ when it is actually $-\infty$.