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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![**Problem: Continuity of a Two-Variable Function**
**Objective:** Determine whether the function \( z = g(x, y) \) is continuous at \((0,0)\).
The function is defined as:
\[
g(x, y) =
\begin{cases}
\frac{x^3 - y^3}{x^2 + y^2} & \text{for } (x, y) \neq (0,0) \\
0 & \text{for } (x, y) = (0,0)
\end{cases}
\]
**Steps to Determine Continuity:**
1. **Evaluate the function at the point \((0,0)\):**
\[ g(0, 0) = 0 \]
2. **Check the limit of the function as \((x,y)\) approaches \((0,0)\):**
\[ \lim_{(x,y) \to (0,0)} \frac{x^3 - y^3}{x^2 + y^2} \]
3. **Convert to polar coordinates** for simplicity, where \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \):
\[
\lim_{r \to 0} \frac{(r \cos(\theta))^3 - (r \sin(\theta))^3}{(r \cos(\theta))^2 + (r \sin(\theta))^2} = \lim_{r \to 0} \frac{r^3 (\cos^3(\theta) - \sin^3(\theta))}{r^2 (\cos^2(\theta) + \sin^2(\theta))}
\]
Simplifying further:
\[
= \lim_{r \to 0} r \cdot \frac{\cos^3(\theta) - \sin^3(\theta)}{\cos^2(\theta) + \sin^2(\theta)}
\]
Since \( \cos^2(\theta) + \sin^2(\theta) = 1 \), the expression becomes:
\[
\lim_{r \to 0} r (\cos^3(\theta) - \sin^3(\theta)) = 0
\]
4. **Conclusion:**
Since the limit](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7e32939d-01c6-47f4-9969-27c7a171091a%2F499e19ef-081f-4d46-be00-58fa3cc5b611%2F5yn71kx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem: Continuity of a Two-Variable Function**
**Objective:** Determine whether the function \( z = g(x, y) \) is continuous at \((0,0)\).
The function is defined as:
\[
g(x, y) =
\begin{cases}
\frac{x^3 - y^3}{x^2 + y^2} & \text{for } (x, y) \neq (0,0) \\
0 & \text{for } (x, y) = (0,0)
\end{cases}
\]
**Steps to Determine Continuity:**
1. **Evaluate the function at the point \((0,0)\):**
\[ g(0, 0) = 0 \]
2. **Check the limit of the function as \((x,y)\) approaches \((0,0)\):**
\[ \lim_{(x,y) \to (0,0)} \frac{x^3 - y^3}{x^2 + y^2} \]
3. **Convert to polar coordinates** for simplicity, where \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \):
\[
\lim_{r \to 0} \frac{(r \cos(\theta))^3 - (r \sin(\theta))^3}{(r \cos(\theta))^2 + (r \sin(\theta))^2} = \lim_{r \to 0} \frac{r^3 (\cos^3(\theta) - \sin^3(\theta))}{r^2 (\cos^2(\theta) + \sin^2(\theta))}
\]
Simplifying further:
\[
= \lim_{r \to 0} r \cdot \frac{\cos^3(\theta) - \sin^3(\theta)}{\cos^2(\theta) + \sin^2(\theta)}
\]
Since \( \cos^2(\theta) + \sin^2(\theta) = 1 \), the expression becomes:
\[
\lim_{r \to 0} r (\cos^3(\theta) - \sin^3(\theta)) = 0
\]
4. **Conclusion:**
Since the limit
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