#AP Calculus AB/BC: Connecting Infinite Limits and Vertical Asymptotes 🚀

Hey there! Let's dive into a crucial topic that bridges limits and discontinuities: vertical asymptotes. This is a key area, so buckle up and let's make sure you've got it down pat!

#🔗 Review: Essential Pre-requisites

Before we get started, make sure you're comfortable with these topics:

• 1.2: Defining Limits and Using Limit Notation
• 1.5: Determining Limits Using Algebraic Properties of Limits
• 1.6: Determining Limits Using Algebraic Manipulation

#⚠️ Understanding Discontinuities

Discontinuities are points where a function is either undefined or behaves erratically. Recognizing these is super important because many calculus theorems only apply to continuous functions. Let's focus on one specific type:

#📏 Vertical Asymptotes

Remember from Algebra II? Vertical asymptotes are those vertical lines that a function gets super close to but never actually touches. They're like invisible barriers! 🚧

Graph of a function with a vertical asymptote at $x = 2$.

#♾️ Infinite Limits: The Key Connection

Infinite limits are limits that evaluate to either positive or negative infinity. These occur when a function's value grows or shrinks without any bound. This is exactly what happens near a vertical asymptote!

#🔗 Connecting Infinite Limits to Vertical Asymptotes

If any of the following conditions are true, then there's a vertical asymptote at x = a:

$$lim_{x\to a} f(x) = \pm\infty$$

$$lim_{x\to a^+} f(x) = \pm\infty$$

$$lim_{x\to a^-} f(x) = \pm\infty$$

#📝 Practice Time: Infinite Limits and Vertical Asymptotes

Let's solidify these concepts with some examples.

#Example 1: Showing a Vertical Asymptote with Limits

❓ Show that $x = -3$ is a vertical asymptote for $f(x) = \frac{1}{x+3}$ using limits.

#Solution to Example 1

To show that $x = -3$ is a vertical asymptote, we need to demonstrate that the limit as $x$ approaches $-3$ is either positive or negative infinity.

Let's first examine the limit from the right:

$$\lim_{x\to -3^+} \frac{1}{x+3} = \infty$$

As $x$ approaches $-3$ from the right (e.g., $-2.999$), the denominator approaches $0$ from the positive side, causing the fraction to approach positive infinity. 📈

Now, let's look at the limit from the left:

$$\lim_{x\to -3^-} \frac{1}{x+3} = -\infty$$

As $x$ approaches $-3$ from the left (e.g., $-3.001$), the denominator approaches $0$ from the negative side, causing the fraction to approach negative infinity. 📉

#Example 2: Finding Vertical Asymptotes of $ln(x)$

❓ Find the vertical asymptote for the function $f(x) = ln(x)$.

Natural logarithm (ln) graph.

#Solution to Example 2

By looking at the graph of $ln(x)$, we can see that there's a vertical asymptote at $x = 0$. Let's confirm this using an infinite limit:

$$\lim_{x\to 0^+} ln(x) = -\infty$$

The natural logarithm of 0 is undefined, and as $x$ gets closer to 0 from the right, $ln(x)$ approaches negative infinity. This confirms the vertical asymptote at $x = 0$. 💡

#🎯 Final Exam Focus

• High-Value Topics: Vertical asymptotes and infinite limits are often tested together. Expect to see questions that ask you to identify asymptotes using limits.
• Common Question Types: You'll likely see questions that require you to:
  • Evaluate limits that approach infinity.
  • Identify vertical asymptotes from a function's equation or graph.
  • Connect the behavior of a function near a vertical asymptote to its infinite limits.
• Time Management: Be quick with your limit evaluations. Practice recognizing common patterns that lead to infinite limits. ⏰
• Common Pitfalls: Don't forget to check both left-hand and right-hand limits when evaluating limits at potential asymptotes. Also, be careful with the signs (+/-) when dealing with infinity. Always double-check your work!

#🚀 Practice Questions

Practice Question

Multiple Choice Questions

1. The function $f(x) = \frac{x-2}{x^2-4}$ has a vertical asymptote at: (A) $x = 2$ only (B) $x = -2$ only (C) $x = 2$ and $x = -2$ (D) No vertical asymptotes

2. The limit $\lim_{x\to 1^-} \frac{1}{x-1}$ is: (A) $\infty$ (B) $-\infty$ (C) 0 (D) 1

Free Response Question

Consider the function $g(x) = \frac{x+1}{x^2-1}$.

(a) Find all vertical asymptotes of $g(x)$. Justify your answer using limits.

(b) Evaluate $\lim_{x\to 1} g(x)$.

(c) Sketch a graph of $g(x)$ near its vertical asymptote(s), showing the behavior of the function using the limits you calculated.

Answer Key

Multiple Choice

1. (B) $x = -2$ only (Note that the factor $(x-2)$ cancels out, leaving a removable discontinuity at $x=2$.)
2. (B) $-\infty$

Free Response

(a) Vertical Asymptotes (3 points)

• Factor the denominator: $g(x) = \frac{x+1}{(x+1)(x-1)}$
• Simplify: $g(x) = \frac{1}{x-1}$ for $x\neq -1$
• Vertical asymptote at $x=1$ because the denominator is 0 and the numerator is not 0. * Show limit: $\lim_{x\to 1^+} g(x) = \infty$ and $\lim_{x\to 1^-} g(x) = -\infty$ (1 point for correct factorization, 1 point for identifying x=1, 1 point for limits)

(b) Limit at x=1 (1 point)

• The limit does not exist because the left and right limits do not agree.

(c) Sketch (2 points)

• The graph should show a vertical asymptote at x=1. * The graph should approach positive infinity as x approaches 1 from the right and negative infinity as x approaches 1 from the left.

You've got this! Keep practicing, and you'll be a master of infinite limits and vertical asymptotes in no time! 🎉