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 6**: Find the limit or explain why it does not exist (and if it is ±∞).
(i) \[ \lim_{x \to \infty} \frac{\sqrt{16x^4 + 7x}}{8x^2 + 5} \]
---
On an Educational website, the transcription and explanation for this question would appear as follows:
---
**Question 6**: Find the limit or explain why it does not exist (and if it is ±∞).
(i) \[ \lim_{x \to \infty} \frac{\sqrt{16x^4 + 7x}}{8x^2 + 5} \]
**Explanation**:
To solve for the limit as \( x \to \infty \), we can simplify the expression inside the limit.
1. **Numerator Analysis**:
The term inside the square root in the numerator is \( 16x^4 + 7x \). When \( x \) is very large, the \( 16x^4 \) term will dominate over the \( 7x \) term. So, \( \sqrt{16x^4 + 7x} \approx \sqrt{16x^4} \). Simplifying further, we get \( \sqrt{16x^4} = 4x^2 \). Therefore, as \( x \to \infty \):
\[
\sqrt{16x^4 + 7x} \approx 4x^2.
\]
2. **Denominator Analysis**:
In the denominator, the expression is \( 8x^2 + 5 \). Similarly, for very large \( x \), the \( 8x^2 \) term will dominate over the constant \( 5 \) term. Hence:
\[
8x^2 + 5 \approx 8x^2.
\]
3. **Combining and Simplifying**:
\[
\lim_{x \to \infty} \frac{\sqrt{16x^4 + 7x}}{8x^2 + 5} \approx \lim_{x \to \infty} \frac{4x^2}{8x^2} = \lim_{x \to \infty} \frac{4}{8} = \frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbca33481-84f0-4d0e-9458-2fdcb949c0ec%2F0d3f3d0e-9a22-458c-97a5-a315f521ccc3%2Firq3tdc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 6**: Find the limit or explain why it does not exist (and if it is ±∞).
(i) \[ \lim_{x \to \infty} \frac{\sqrt{16x^4 + 7x}}{8x^2 + 5} \]
---
On an Educational website, the transcription and explanation for this question would appear as follows:
---
**Question 6**: Find the limit or explain why it does not exist (and if it is ±∞).
(i) \[ \lim_{x \to \infty} \frac{\sqrt{16x^4 + 7x}}{8x^2 + 5} \]
**Explanation**:
To solve for the limit as \( x \to \infty \), we can simplify the expression inside the limit.
1. **Numerator Analysis**:
The term inside the square root in the numerator is \( 16x^4 + 7x \). When \( x \) is very large, the \( 16x^4 \) term will dominate over the \( 7x \) term. So, \( \sqrt{16x^4 + 7x} \approx \sqrt{16x^4} \). Simplifying further, we get \( \sqrt{16x^4} = 4x^2 \). Therefore, as \( x \to \infty \):
\[
\sqrt{16x^4 + 7x} \approx 4x^2.
\]
2. **Denominator Analysis**:
In the denominator, the expression is \( 8x^2 + 5 \). Similarly, for very large \( x \), the \( 8x^2 \) term will dominate over the constant \( 5 \) term. Hence:
\[
8x^2 + 5 \approx 8x^2.
\]
3. **Combining and Simplifying**:
\[
\lim_{x \to \infty} \frac{\sqrt{16x^4 + 7x}}{8x^2 + 5} \approx \lim_{x \to \infty} \frac{4x^2}{8x^2} = \lim_{x \to \infty} \frac{4}{8} = \frac
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