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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.8.9.g

9. True, or false? As x→∞,
g. ln(x) = o(ln(2x))

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1
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: \( f(x) = \ln(x) \) and \( g(x) = \ln(2x) \). We need to analyze \( \lim_{x \to \infty} \frac{\ln(x)}{\ln(2x)} \).
Use logarithm properties to simplify the denominator: \( \ln(2x) = \ln(2) + \ln(x) \). So the limit becomes \( \lim_{x \to \infty} \frac{\ln(x)}{\ln(2) + \ln(x)} \).
Divide numerator and denominator by \( \ln(x) \) to simplify the expression: \( \lim_{x \to \infty} \frac{1}{\frac{\ln(2)}{\ln(x)} + 1} \).
Analyze the behavior as \( x \to \infty \): since \( \ln(x) \to \infty \), \( \frac{\ln(2)}{\ln(x)} \to 0 \), so the limit approaches \( \frac{1}{0 + 1} = 1 \). Since the limit is 1, not 0, the statement \( \ln(x) = o(\ln(2x)) \) is false.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Little-o Notation (Asymptotic Comparison)

Little-o notation, written as f(x) = o(g(x)), means that f(x) grows much slower than g(x) as x approaches infinity, specifically that the limit of f(x)/g(x) is zero. It is used to compare the relative growth rates of functions.
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Properties of Logarithmic Functions

Logarithmic functions like ln(x) grow without bound but very slowly as x approaches infinity. Key properties include ln(ab) = ln(a) + ln(b), which helps simplify expressions like ln(2x) into ln(2) + ln(x).
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Limits Involving Logarithmic Functions at Infinity

When evaluating limits of ratios of logarithmic functions as x→∞, constants become negligible. For example, ln(2x) = ln(2) + ln(x), so the ratio ln(x)/ln(2x) approaches 1, indicating similar growth rates.
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