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Connecting Limits at Infinity and Horizontal Asymptotes

Samuel Baker

Samuel Baker

7 min read

Next Topic - Working with the Intermediate Value Theorem (IVT Calc)

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Study Guide Overview

This study guide covers limits at infinity and horizontal asymptotes. It reviews the concept of limits, explains how to find horizontal asymptotes using the degree rules for rational functions, discusses function growth rates (logarithmic, root, polynomial, exponential), and provides practice examples and questions focused on evaluating limits at infinity and finding horizontal asymptotes. Key terms include: horizontal asymptotes, limits at infinity, degree rules, and growth rates. The guide also includes common pitfalls and strategies for the AP exam.

#AP Calculus AB/BC: Limits at Infinity & Horizontal Asymptotes πŸš€

Hey there, future AP Calculus master! Let's nail down limits at infinity and horizontal asymptotes. This is a crucial topic, and by the end of this guide, you'll be feeling super confident. Let's dive in!

#1.15 Connecting Limits at Infinity and Horizontal Asymptotes

This section is all about how functions behave as x gets super big (or super small). We're talking about what happens way, way out on the graph.

# πŸ” Review: Limits to Infinity

Remember, a limit is what a function's y-value approaches as x approaches a certain value. When we say "x approaches infinity (∞) or negative infinity (-∞)," we're looking at the function's end behavior.

lim⁑xβ†’βˆžf(x)\lim_{x \to \infty} f(x)xβ†’βˆžlim​f(x) lim⁑xβ†’βˆ’βˆžf(x)\lim_{x \to -\infty} f(x)xβ†’βˆ’βˆžlim​f(x)

Key Concept

These limits tell us where the function is heading as x goes way out to the right or left.

# 🧐 Finding Limits to Infinity

The key to finding limits at infinity? Horizontal Asymptotes (HAs)! Think of HAs as the "speed limit" for a function's y-values. 🚦

# πŸš₯ Horizontal Asymptotes

A horizontal asymptote is a y-value that the graph approaches but never actually touches or crosses (at least not at the extreme ends). It's like a finish line the function tries to reach but never quite gets to. πŸ›‘

https://zupay.blob.core.windows.net/resources/files/0baca4f69800419293b4c75aa2870acd_2ff9d2_2743.png

Image Courtesy of MATH.net

Caption: The graph shows a function with a horizontal asymptote at y = 5. Notice how the function gets closer and closer to y = 5 as x goes to infinity, but never actually touches it.

In this graph, the horizontal asymptote is at y = 5. So, we can say: lim⁑xβ†’βˆžf(x)=5\lim_{x \to \infty} f(x) = 5limxβ†’βˆžβ€‹f(x)=5.

# πŸ” Finding Horizontal Asymptotes

Here's the lowdown on finding HAs for rational functions (functions that are fractions of polynomials): Let's think of a function as p(x)q(x)\frac{p(x)}{q(x)}q(x)p(x)​.

Memory Aid

HA Rules - Think "Degree Detectives"

πŸ₯‡ Bottom Heavy: If the degree of p(x) < degree of q(x), the HA = 0. (The bottom grows faster)

πŸ₯ˆ Top Heavy: If the degree of p(x) > degree of q(x), there's no HA. (The top grows faster)

πŸ₯‰ Equal Degree: If the degree of p(x) = degree of q(x), the HA is the ratio of the leading coefficients. (They grow at the same rate)

Examples:

πŸ₯‡ lim⁑xβ†’βˆž4x+23x2+1=0\lim_{x \to \infty} \frac{4x+2}{3x^2+1} = 0limxβ†’βˆžβ€‹3x2+14x+2​=0 (Bottom wins!)

πŸ₯ˆ lim⁑xβ†’βˆž4x2+23x+1=∞\lim_{x \to \infty} \frac{4x^2+2}{3x+1} = \inftylimxβ†’βˆžβ€‹3x+14x2+2​=∞ (Top wins!)

πŸ₯‰ lim⁑xβ†’βˆž3x2+24x2+1=34\lim_{x \to \infty} \frac{3x^2+2}{4x^2+1} = \frac{3}{4}limxβ†’βˆžβ€‹4x2+13x2+2​=43​ (Coefficients battle it out!)

Exam Tip

When you see ∞∞\frac{\infty}{\infty}βˆžβˆžβ€‹, think HAs!

#Exceptions to These Rules

  • Oscillating Functions: The limit of an oscillating function at infinity does not exist. 🎒
    • Ex: lim⁑xβ†’βˆžsin⁑(x)=DNE\lim_{x \to \infty} \sin(x) = DNElimxβ†’βˆžβ€‹sin(x)=DNE
  • Squeeze Theorem: Any function that is part of the squeeze theorem will equal 0 as it approaches infinity. πŸ₯ͺ
    • Ex: lim⁑xβ†’βˆžsin⁑(x)x=0\lim_{x \to \infty} \frac{\sin(x)}{x} = 0limxβ†’βˆžβ€‹xsin(x)​=0

# 🌱 The Growth Rates of Functions

Different functions grow at different speeds! Knowing this can be super helpful. Here's the growth rate hierarchy from slowest to fastest: πŸͺ΄

log < root < polynomial < exponential

Quick Fact

Exponential functions are the Usain Bolts of the function world – they grow incredibly fast!


# ✏️ Infinite Limits & Horizontal Asymptotes Practice

Let's put this knowledge to work with some examples!

#Infinite Limits: Example 1

Evaluate: lim⁑xβ†’βˆž3xβˆ’1ex\lim_{x \to \infty} \frac{3x-1}{e^x}limxβ†’βˆžβ€‹ex3xβˆ’1​

Common Mistake

Don't just plug in infinity! Think about growth rates.

  • The denominator, e^x, is an exponential function, which grows much faster than the numerator, 3x - 1. So, we have a smallΒ numberlargeΒ number\frac{\text{small number}}{\text{large number}}largeΒ numbersmallΒ number​ situation.
  • Therefore, lim⁑xβ†’βˆž3xβˆ’1ex=0\lim_{x \to \infty} \frac{3x-1}{e^x} = 0limxβ†’βˆžβ€‹ex3xβˆ’1​=0.

#Infinite Limits: Example 2

Evaluate: lim⁑xβ†’βˆž2xxβˆ’5\lim_{x \to \infty} \frac{2^x}{x-5}limxβ†’βˆžβ€‹xβˆ’52x​

  • Here, the exponential function, 2^x, is in the numerator, which grows much faster than the denominator, x - 5. So, we have a largeΒ numbersmallΒ number\frac{\text{large number}}{\text{small number}}smallΒ numberlargeΒ number​ situation.
  • Therefore, lim⁑xβ†’βˆž2xxβˆ’5=∞\lim_{x \to \infty} \frac{2^x}{x-5} = \inftylimxβ†’βˆžβ€‹xβˆ’52x​=∞.

# 🎯 Final Exam Focus

  • High-Priority Topics: Limits at infinity, horizontal asymptotes, growth rates of functions. These appear in both MCQs and FRQs.
  • Common Question Types: Finding HAs of rational functions, evaluating limits involving exponential and logarithmic functions, applying the squeeze theorem.
  • Time Management: Quickly identify the degrees of polynomials and growth rates of functions to save time on MCQs.
  • Common Pitfalls: Forgetting to consider growth rates, misapplying HA rules, not recognizing oscillating functions.
  • Strategies: Always simplify before evaluating limits, think about the relative growth rates of functions, and remember your HA rules!

# πŸ“ Practice Questions

Practice Question

Multiple Choice Questions

  1. lim⁑xβ†’βˆž5x3βˆ’2x+12x3+7x2βˆ’4\lim_{x \to \infty} \frac{5x^3 - 2x + 1}{2x^3 + 7x^2 - 4}limxβ†’βˆžβ€‹2x3+7x2βˆ’45x3βˆ’2x+1​ is equal to: (A) 0 (B) 5/2 (C) ∞ (D) 2/5

  2. lim⁑xβ†’βˆž4x+sin⁑(x)x\lim_{x \to \infty} \frac{4x + \sin(x)}{x}limxβ†’βˆžβ€‹x4x+sin(x)​ is equal to: (A) 0 (B) 4 (C) ∞ (D) DNE

  3. lim⁑xβ†’βˆžln⁑(x)x2\lim_{x \to \infty} \frac{\ln(x)}{x^2}limxβ†’βˆžβ€‹x2ln(x)​ is equal to: (A) 0 (B) 1 (C) ∞ (D) DNE

Free Response Question

Consider the function f(x)=3x2βˆ’5x2+2xf(x) = \frac{3x^2 - 5}{x^2 + 2x}f(x)=x2+2x3x2βˆ’5​.

(a) Find all horizontal asymptotes of f(x).

(b) Evaluate lim⁑xβ†’βˆžf(x)\lim_{x \to \infty} f(x)limxβ†’βˆžβ€‹f(x).

(c) Evaluate lim⁑xβ†’βˆ’βˆžf(x)\lim_{x \to -\infty} f(x)limxβ†’βˆ’βˆžβ€‹f(x).

(d) Does the function have any vertical asymptotes? If so, find them.

Answer Key

Multiple Choice

  1. (B) The degrees are equal, so the limit is the ratio of the coefficients: 5/2. 2. (B) Divide each term by x, lim⁑xβ†’βˆž4+sin⁑(x)x=4+0=4\lim_{x \to \infty} 4 + \frac{\sin(x)}{x} = 4 + 0 = 4limxβ†’βˆžβ€‹4+xsin(x)​=4+0=4.
  2. (A) The denominator grows faster than the numerator, so the limit is 0. Free Response

(a) The horizontal asymptote is y = 3, since the degrees are equal, the limit is the ratio of the coefficients: 3/1 = 3. (b) lim⁑xβ†’βˆžf(x)=3\lim_{x \to \infty} f(x) = 3limxβ†’βˆžβ€‹f(x)=3 (same as the horizontal asymptote).

(c) lim⁑xβ†’βˆ’βˆžf(x)=3\lim_{x \to -\infty} f(x) = 3limxβ†’βˆ’βˆžβ€‹f(x)=3 (same as the horizontal asymptote).

(d) Yes, vertical asymptotes occur where the denominator is zero: x2+2x=0β€…β€ŠβŸΉβ€…β€Šx(x+2)=0x^2 + 2x = 0 \implies x(x+2) = 0x2+2x=0⟹x(x+2)=0. Thus, the vertical asymptotes are at x = 0 and x = -2. Scoring Breakdown for FRQ

  • (a) 1 point for stating the horizontal asymptote y = 3. * (b) 1 point for stating the correct limit as x approaches infinity.
  • (c) 1 point for stating the correct limit as x approaches negative infinity.
  • (d) 1 point for identifying the vertical asymptotes at x = 0 and x = -2.

You got this! Keep practicing, and you'll be a pro in no time. Let's ace that exam! πŸ’ͺ

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Previous Topic - Connecting Infinite Limits and Vertical AsymptotesNext Topic - Working with the Intermediate Value Theorem (IVT Calc)

Question 1 of 11

What is the value of lim⁑xβ†’βˆž5x2\lim_{x \to \infty} \frac{5}{x^2}limxβ†’βˆžβ€‹x25​?

5

∞\infty∞

0

1