Find the limit or show that the limit does not exist. lim→-0 4x2 + 3x + 2x x→-∞

Calculus: Early Transcendentals
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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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## Finding Limits at Infinity

### Problem Statement
Find the limit or show that the limit does not exist.

\[ \lim_{x \to -\infty} \left( \sqrt{4x^2 + 3x + 2x} \right) \]

### Solution Steps

1. **Identify the Dominant Term:**
   - As \(x\) approaches \(-\infty\), the term with the highest power of \(x\) will dominate.

2. **Simplify the Expression:**
   \[ 4x^2 + 3x + 2x = 4x^2 + 5x \]
   - Here, \(4x^2\) is the dominant term.

3. **Approximate for Large Values of \(x\):**
   \[ \sqrt{4x^2 + 5x} \approx \sqrt{4x^2} \]
   
4. **Factor and Simplify:**
   \[ \sqrt{4x^2 + 5x} = \sqrt{4x^2 (1 + \frac{5x}{4x^2})} = \sqrt{4x^2} \sqrt{1 + \frac{5}{4x}} \]
   - For large negative \(x\), \(\frac{5}{4x}\) approaches 0.
   \[ \sqrt{4x^2} = 2|x| = 2(-x) \text{ for } x \to -\infty \]

5. **Calculate the Limit:**
   \[ \lim_{x \to -\infty} 2(-x) = \lim_{x \to -\infty} -2x \]
   - Since \(x \to -\infty\), \(-2x\) becomes \(+\infty\).

Thus:
\[ \lim_{x \to -\infty} \left( \sqrt{4x^2 + 5x} \right) = \infty \]

### Conclusion
The limit of the function as \(x\) approaches \(-\infty\) is \(\infty\).
Transcribed Image Text:## Finding Limits at Infinity ### Problem Statement Find the limit or show that the limit does not exist. \[ \lim_{x \to -\infty} \left( \sqrt{4x^2 + 3x + 2x} \right) \] ### Solution Steps 1. **Identify the Dominant Term:** - As \(x\) approaches \(-\infty\), the term with the highest power of \(x\) will dominate. 2. **Simplify the Expression:** \[ 4x^2 + 3x + 2x = 4x^2 + 5x \] - Here, \(4x^2\) is the dominant term. 3. **Approximate for Large Values of \(x\):** \[ \sqrt{4x^2 + 5x} \approx \sqrt{4x^2} \] 4. **Factor and Simplify:** \[ \sqrt{4x^2 + 5x} = \sqrt{4x^2 (1 + \frac{5x}{4x^2})} = \sqrt{4x^2} \sqrt{1 + \frac{5}{4x}} \] - For large negative \(x\), \(\frac{5}{4x}\) approaches 0. \[ \sqrt{4x^2} = 2|x| = 2(-x) \text{ for } x \to -\infty \] 5. **Calculate the Limit:** \[ \lim_{x \to -\infty} 2(-x) = \lim_{x \to -\infty} -2x \] - Since \(x \to -\infty\), \(-2x\) becomes \(+\infty\). Thus: \[ \lim_{x \to -\infty} \left( \sqrt{4x^2 + 5x} \right) = \infty \] ### Conclusion The limit of the function as \(x\) approaches \(-\infty\) is \(\infty\).
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