Math Problem Statement

solve q1

Is the function f(x,y) = (xy(x² - y²))/(x² + y²) for (x,y) ≠ (0,0), 4 for (x,y) = (0,0) continuous at the origin, and can it be redefined to make it continuous?

Solution

Let's address Question 1 from the image step by step:

The function is defined as:

\begin{cases} \frac{xy(x^2 - y^2)}{x^2 + y^2}, & (x, y) \neq (0, 0) \\ 4, & (x, y) = (0, 0) \end{cases}$$ ### Part (a): Is $$f(x, y)$$ continuous at the origin? To determine the continuity of $$f(x, y)$$ at the origin, we must check if the following limit exists and if it equals the value of the function at the origin: $$\lim_{(x, y) \to (0, 0)} f(x, y).$$ We need to check if the limit of the function as $$(x, y) \to (0, 0)$$ is equal to 4 (the value of the function at $$(0, 0)$$). #### Approach along different paths: - **Path 1**: Along the line $$y = 0$$: $$f(x, 0) = \frac{x \cdot 0 \cdot (x^2 - 0^2)}{x^2 + 0^2} = 0.$$ The limit along this path is 0. - **Path 2**: Along the line $$x = 0$$: $$f(0, y) = \frac{0 \cdot y \cdot (0^2 - y^2)}{0^2 + y^2} = 0.$$ The limit along this path is also 0. - **Path 3**: Along the line $$y = x$$: $$f(x, x) = \frac{x \cdot x \cdot (x^2 - x^2)}{x^2 + x^2} = 0.$$ The limit along this path is 0 as well. - **Path 4**: Along the parabola $$y = x^2$$: $$f(x, x^2) = \frac{x \cdot x^2 \cdot (x^2 - x^4)}{x^2 + x^4} = \frac{x^3 \cdot (x^2 - x^4)}{x^2(1 + x^2)} = \frac{x^5(1 - x^2)}{x^2(1 + x^2)}.$$ As $$x \to 0$$, this expression tends to 0. In all these cases, the limit of $$f(x, y)$$ as $$(x, y) \to (0, 0)$$ is 0, but the function value at $$(0, 0)$$ is 4. Therefore, the function is **not continuous** at the origin. ### Part (b): Redefine if necessary to make it continuous at $$(0, 0)$$. To make the function continuous at the origin, we need the limit of the function as $$(x, y) \to (0, 0)$$ to equal the function value at the origin. Since the limit is 0, we can redefine the function at $$(0, 0)$$ to be 0, i.e., set: $$f(0, 0) = 0.$$ With this redefinition, the function would be continuous at the origin. ### Final Answers: - **(a)** The function is **not continuous** at the origin. - **(b)** The function can be made continuous by redefining $$f(0, 0) = 0$$. --- Would you like a detailed explanation of the differentiability of this function or any other clarifications? Here are 5 related questions: 1. How do you determine the limit of a multivariable function? 2. What are common techniques for proving the continuity of functions at specific points? 3. How does differentiability relate to continuity in multivariable calculus? 4. What is the significance of approaching a point along different paths? 5. How can a function be continuous but not differentiable at a point? **Tip**: Always check continuity before checking differentiability when dealing with multivariable functions.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Limits and Continuity
Piecewise Functions

Formulas

lim_{(x, y) → (0, 0)} f(x, y)
(xy(x² - y²))/(x² + y²)
Path analysis for limits

Theorems

Continuity at a point
Limit definition of continuity

Suitable Grade Level

College-level Calculus

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