Textbook Question
1. Which of the following functions grow faster than e^x as x→∞? Which grow at the same rate as e^x? Which grow slower?
c. √x
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.8.10.c
10. True, or false? As x→∞,
c. 1/x - 1/x² = o(1/x)
Verified step by step guidance1
Recall the definition of the little-o notation: for functions f(x) and g(x), we say f(x) = o(g(x)) as x → ∞ if \( \lim_{x \to \infty} \frac{f(x)}{g(x)} = 0 \).
Identify the functions in the problem: \( f(x) = \frac{1}{x} - \frac{1}{x^2} \) and \( g(x) = \frac{1}{x} \).
Form the ratio \( \frac{f(x)}{g(x)} = \frac{\frac{1}{x} - \frac{1}{x^2}}{\frac{1}{x}} \).
Simplify the ratio: \( \frac{\frac{1}{x} - \frac{1}{x^2}}{\frac{1}{x}} = \frac{1}{x} \cdot \frac{x}{1} - \frac{1}{x^2} \cdot \frac{x}{1} = 1 - \frac{1}{x} \).
Evaluate the limit as \( x \to \infty \): \( \lim_{x \to \infty} \left(1 - \frac{1}{x}\right) = 1 \). Since this limit is not zero, conclude that \( \frac{1}{x} - \frac{1}{x^2} \neq o\left(\frac{1}{x}\right) \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Little-o Notation
Little-o notation, written as f(x) = o(g(x)) as x→∞, means that f(x) becomes insignificant compared to g(x) in the limit. Formally, f(x)/g(x) → 0 as x→∞, indicating f(x) grows much slower than g(x). It is used to describe the relative growth rates of functions.
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Sigma Notation
Asymptotic Behavior of Rational Functions
As x approaches infinity, terms like 1/x and 1/x² approach zero, but at different rates. Specifically, 1/x² tends to zero faster than 1/x, which affects how their differences behave asymptotically. Understanding these rates is key to comparing functions using little-o notation.
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Limit of a Quotient for Comparing Growth Rates
To determine if f(x) = o(g(x)), evaluate the limit of f(x)/g(x) as x→∞. If the limit is zero, f(x) is little-o of g(x). This method helps verify statements about asymptotic dominance, such as whether 1/x - 1/x² = o(1/x).
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Related Practice
Textbook Question
In Exercises 67–72, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:
c. Find the equation for the tangent line to f at the specified point (x_0, f(x_0)).
67. y= √(3x-2), 2/3 ≤ x ≤ 4, x_0=3
Textbook Question
In Exercises 1–4, show that each function y=f(x) is a solution of the accompanying differential equation.
1. 2y' + 3y = e^(-x)
c. y = e^(-x) + Ce^(-(3/2)x)
Textbook Question
80. Find all values of c that satisfy the conclusion of Cauchy's Mean Value Theorem for the given functions and interval.
c. f(x) = x³/ (3 - 4x), g(x) = x², (a, b) = (0, 3)
Textbook Question
2. Which of the following functions grow faster than e^x as x→∞? Which grow at the same rate as e^x? Which grow slower?
c. √(1+x^4)
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Textbook Question
88. Given that x>0, find the maximum value, if any, of
c. x^(1/x^n) (n a positive integer)