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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![The image contains a mathematical limit expression, which is often covered in calculus courses. Here is the transcription for educational purposes:
---
**Problem 29: Evaluate the Limit**
\[ \lim_{x \to 1^{+}} \frac{\sqrt{3(x-1)}}{|x-1|} \]
This problem involves evaluating the limit of a function as \( x \) approaches 1 from the right (\( x \to 1^{+} \)).
---
In this limit problem, you are dealing with the behavior of the function as \( x \) gets arbitrarily close to 1 from the right side. This involves understanding the numerator and the absolute value function in the denominator and how they affect the evaluation near the point of interest. For \( x \to 1^{+} \), the expression inside the square root is always non-negative, and the absolute value expression in the denominator simplifies differently based on the sign of \( x-1 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F76c35efe-76b6-4dad-9a58-d1a8ec4fa095%2Ffa50b3e6-84fe-488e-8dbf-bfa1a3eb7fbf%2Fkwdj4ea.jpeg&w=3840&q=75)
Transcribed Image Text:The image contains a mathematical limit expression, which is often covered in calculus courses. Here is the transcription for educational purposes:
---
**Problem 29: Evaluate the Limit**
\[ \lim_{x \to 1^{+}} \frac{\sqrt{3(x-1)}}{|x-1|} \]
This problem involves evaluating the limit of a function as \( x \) approaches 1 from the right (\( x \to 1^{+} \)).
---
In this limit problem, you are dealing with the behavior of the function as \( x \) gets arbitrarily close to 1 from the right side. This involves understanding the numerator and the absolute value function in the denominator and how they affect the evaluation near the point of interest. For \( x \to 1^{+} \), the expression inside the square root is always non-negative, and the absolute value expression in the denominator simplifies differently based on the sign of \( x-1 \).
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