**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 $-2$.

Answered: Image /qna-images/answer/6edf37c4-5742-4730-9104-02f51fc61aa0.jpg

Answered: lim (log, (x +3x) –-log, (1+16x²))

Math

Calculus

lim (log, (x +3x) –-log, (1+16x²))

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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Concept explainers

Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topics, such as continuity and differentiability better. In a broader sense, every calculus notion is a limit. Other branches of calculus have also been derived from limits.

Question

**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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