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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Question
![### Limit Evaluation
**Evaluate the following limit:**
\[
\lim_{x \to -\infty} \frac{3x^4 - 3x^2}{5x^6 + 8}
\]
**Limit =** [Input Box]
**Instructions:**
- Use "infinity" for \( \infty \) and "-infinity" for \( -\infty \).
**Explanation:**
To solve the limit \(\lim_{x \to -\infty} \frac{3x^4 - 3x^2}{5x^6 + 8}\), follow these steps:
1. **Identify the dominant terms** in the numerator and denominator as \( x \) approaches \(-\infty\).
2. **Factor out the highest power of \( x \)** in both the numerator and the denominator.
In this expression, \( x^6 \) is the highest power of \( x \) in the denominator, and \( x^4 \) is the highest power in the numerator. Factoring out \( x^6 \) from the denominator and \( x^4 \) from the numerator, you get:
\[
\lim_{x \to -\infty} \frac{3x^4(1 - \frac{1}{x^2})}{5x^6(1 + \frac{8}{5x^6})}
\]
Simplify this expression:
\[
\lim_{x \to -\infty} \frac{3 \left( 1 - \frac{1}{x^2} \right)}{5x^2 \left( 1 + \frac{8}{5x^6} \right)}
\]
As \( x \) approaches \(-\infty\), the terms \(\frac{1}{x^2}\) and \(\frac{8}{5x^6}\) approach 0:
\[
= \frac{3 \times 1}{5x^2 \times 1} = \frac{3}{5x^2}
\]
Since \( x^2 \) grows very large as \( x \) approaches \(-\infty\),
\[
\lim_{x \to -\infty} \frac{3}{5x^2} = 0
\]
Therefore, the limit is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3972ee88-18a6-4abd-967b-f522c80949cc%2Fa97faec2-13d2-4ee9-b6a3-9fff35ada2c6%2Fng8x6ul_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Limit Evaluation
**Evaluate the following limit:**
\[
\lim_{x \to -\infty} \frac{3x^4 - 3x^2}{5x^6 + 8}
\]
**Limit =** [Input Box]
**Instructions:**
- Use "infinity" for \( \infty \) and "-infinity" for \( -\infty \).
**Explanation:**
To solve the limit \(\lim_{x \to -\infty} \frac{3x^4 - 3x^2}{5x^6 + 8}\), follow these steps:
1. **Identify the dominant terms** in the numerator and denominator as \( x \) approaches \(-\infty\).
2. **Factor out the highest power of \( x \)** in both the numerator and the denominator.
In this expression, \( x^6 \) is the highest power of \( x \) in the denominator, and \( x^4 \) is the highest power in the numerator. Factoring out \( x^6 \) from the denominator and \( x^4 \) from the numerator, you get:
\[
\lim_{x \to -\infty} \frac{3x^4(1 - \frac{1}{x^2})}{5x^6(1 + \frac{8}{5x^6})}
\]
Simplify this expression:
\[
\lim_{x \to -\infty} \frac{3 \left( 1 - \frac{1}{x^2} \right)}{5x^2 \left( 1 + \frac{8}{5x^6} \right)}
\]
As \( x \) approaches \(-\infty\), the terms \(\frac{1}{x^2}\) and \(\frac{8}{5x^6}\) approach 0:
\[
= \frac{3 \times 1}{5x^2 \times 1} = \frac{3}{5x^2}
\]
Since \( x^2 \) grows very large as \( x \) approaches \(-\infty\),
\[
\lim_{x \to -\infty} \frac{3}{5x^2} = 0
\]
Therefore, the limit is
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