Chapter 14: Problem 73
Gives a function \(f(x, y)\) and a positive number \(\epsilon\) In each exercise, show that there exists a \(\delta > 0\) such that for all \((x, y)\) $$ \sqrt{x^{2}+y^{2}} < \delta \Rightarrow|f(x, y)-f(0,0)| < \epsilon $$ $$f(x, y)=\frac{x y^{2}}{x^{2}+y^{2}} \text { and } f(0,0)=0, \quad \epsilon=0.04$$
Short Answer
Step by step solution
Understand the Problem
Transform the Function
Bounding the Function
Apply Delta Condition
Solve for Delta
Verify the Choice of Delta
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit of a Function
Ultimately, you want \( f(x, y) \) to approach a target value, typically \( f(0, 0) \), which is zero in this case. By focusing on the distance \( \sqrt{x^2 + y^2} \), you're effectively examining how the function should behave as \((x, y)\) approaches \( (0, 0) \) from any direction. In our exercise, we determine a certain range, a small disk of radius \( \delta \), that ensures the function remains within a specific \( \epsilon \) range from zero.
Epsilon-Delta Definition
Here's how it works in practice:
- Pick a positive number \( \epsilon \) (e.g., 0.04 from our exercise) which represents a tiny window of tolerance around the target limit value of the function.
- You seek a \( \delta \) such that whenever \( \sqrt{x^2 + y^2} < \delta \), the value of the function \( |f(x, y) - f(0, 0)| < \epsilon \).
Continuity
In the context of this exercise, we seek continuity around the point \((0, 0)\), meaning we want the function \( f(x, y) = \frac{xy^2}{x^2 + y^2} \) to not behave erratically as \( (x, y) \) approaches the origin.
To achieve this:
- The function must be defined at the point \( (0, 0) \), which it is, with \( f(0, 0) = 0 \).
- The limit of \( f(x, y) \) as \( (x, y) \) approaches \( (0, 0) \) must equal \( f(0, 0) \).
- The epsilon-delta criteria need to be satisfied.