Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Problem Statement:**
e) \(\lim_{{x \to \infty}} \left( \log_4(x^2 + 3x) - \log_4(1 + 16x^2) \right)\) \_\_\_.
**Solution Steps:**
We are tasked with finding the limit of the given logarithmic expression as \(x\) approaches infinity. Here's a step-by-step approach on how to solve this problem:
1. **Apply Logarithmic Properties:**
Use the property of logarithms that states \(\log_b(A) - \log_b(B) = \log_b \left( \frac{A}{B} \right)\).
2. **Combine the Logarithms:**
Using the above property:
\[
\log_4(x^2 + 3x) - \log_4(1 + 16x^2) = \log_4 \left( \frac{x^2 + 3x}{1 + 16x^2} \right)
\]
3. **Simplify the Fraction:**
As \(x\) approaches infinity, the terms with the highest power of \(x\) in the numerator and the denominator will dominate. Let's simplify the fraction:
\[
\frac{x^2 + 3x}{1 + 16x^2} \approx \frac{x^2}{16x^2} = \frac{1}{16}
\]
As \(x\) gets very large, the impact of smaller powers of \(x\) diminishes, making the ratio approach \(\frac{1}{16}\).
4. **Find the Logarithm of the Simplified Fraction:**
Therefore, we need to find:
\[
\log_4 \left( \frac{1}{16} \right)
\]
Knowing that \(\frac{1}{16} = 4^{-2}\), we can rewrite the logarithm as:
\[
\log_4(4^{-2}) = -2
\]
Thus, the limit is:
\[
\boxed{-2}
\]
We have successfully evaluated the limit, which is \(-2\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9f8ea0c3-20cc-4167-a0d0-5d27a7bf8fa1%2F6edf37c4-5742-4730-9104-02f51fc61aa0%2Fqd0dclf_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
e) \(\lim_{{x \to \infty}} \left( \log_4(x^2 + 3x) - \log_4(1 + 16x^2) \right)\) \_\_\_.
**Solution Steps:**
We are tasked with finding the limit of the given logarithmic expression as \(x\) approaches infinity. Here's a step-by-step approach on how to solve this problem:
1. **Apply Logarithmic Properties:**
Use the property of logarithms that states \(\log_b(A) - \log_b(B) = \log_b \left( \frac{A}{B} \right)\).
2. **Combine the Logarithms:**
Using the above property:
\[
\log_4(x^2 + 3x) - \log_4(1 + 16x^2) = \log_4 \left( \frac{x^2 + 3x}{1 + 16x^2} \right)
\]
3. **Simplify the Fraction:**
As \(x\) approaches infinity, the terms with the highest power of \(x\) in the numerator and the denominator will dominate. Let's simplify the fraction:
\[
\frac{x^2 + 3x}{1 + 16x^2} \approx \frac{x^2}{16x^2} = \frac{1}{16}
\]
As \(x\) gets very large, the impact of smaller powers of \(x\) diminishes, making the ratio approach \(\frac{1}{16}\).
4. **Find the Logarithm of the Simplified Fraction:**
Therefore, we need to find:
\[
\log_4 \left( \frac{1}{16} \right)
\]
Knowing that \(\frac{1}{16} = 4^{-2}\), we can rewrite the logarithm as:
\[
\log_4(4^{-2}) = -2
\]
Thus, the limit is:
\[
\boxed{-2}
\]
We have successfully evaluated the limit, which is \(-2\).
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