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Chapter 14: Problem 71

Each of Exercises \(69-74\) 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 \((x, y)\) \(\sqrt{x^{2}+y^{2}}<\delta \Rightarrow|f(x, y)-f(0,0)|<\epsilon.\) $$f(x, y)=(x+y) /\left(x^{2}+1\right), \quad \epsilon=0.01$$

Short Answer

Expert verified
Choose \( \delta = 0.005 \) to ensure the condition \( |f(x, y) - 0| < 0.01 \) is satisfied for \( \sqrt{x^2 + y^2} < \delta \).

Step by step solution

01

Understand the Given Function

We are given the function \( f(x, y) = \frac{x+y}{x^2+1} \) and need to show that there exists some \( \delta > 0 \) such that for all \( (x, y) \), if \( \sqrt{x^2 + y^2} < \delta \), then \( |f(x, y) - f(0, 0)| < \epsilon = 0.01 \). First, we need to understand \( f(0, 0) \) which is the value of the function at the point \( (0,0) \).
02

Calculate f(0,0)

Substitute \( x = 0 \) and \( y = 0 \) into the function: \( f(0, 0) = \frac{0 + 0}{0^2 + 1} = 0 \). So, \( f(0, 0) = 0 \).
03

Express |f(x,y) - f(0,0)|

To find \( |f(x, y) - f(0, 0)| \), calculate \( |f(x, y) - 0| = \left| \frac{x+y}{x^2+1} \right| \). This is the expression we need to make less than \( \epsilon = 0.01 \).
04

Simplify the Expression

Our goal is to make \( \left| \frac{x+y}{x^2+1} \right| < 0.01 \). Simplify by observing that \( |x+y| \leq |x|+|y| \). This gives \( \left| \frac{x+y}{x^2+1} \right| \leq \frac{|x| + |y|}{x^2+1} \).
05

Use the Condition x虏+y虏 < 未虏

From \( \sqrt{x^2 + y^2} < \delta \), we have \( x^2 + y^2 < \delta^2 \). Further, we can say \( |x|, |y| < \delta \) because of absolute value properties. Thus, \( |x|+|y| < 2\delta \). Substitute into the previous expression: \( \frac{|x| + |y|}{x^2+1} < \frac{2\delta}{x^2+1} \).
06

Find 未 That Satisfies the Inequality

We need \( \frac{2\delta}{x^2+1} < 0.01 \) for the condition to hold true. To ensure this inequality holds, and considering \( x^2+1 \geq 1 \) always (since minimum value of \( x^2 \) is 0), choose \( 2\delta < 0.01 \). Thus, \( \delta < 0.005 \).
07

Conclusion: Verify the Condition with 未

Thus, we can conclude that if we choose \( \delta = 0.005 \), then for all \( (x, y) \) satisfying \( \sqrt{x^2 + y^2} < 0.005 \), it holds that \( |f(x, y) - 0| < 0.01 \), as desired.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limits and Continuity
In real analysis, understanding limits and continuity is essential. Limits help us understand how a function behaves as its input approaches a particular point. If a function approaches a specific value as we get closer to the input point, it's said to have a limit.
Continuity takes this concept further. A function is continuous at a point if the limit as approaches that point equals the function's value at the point itself. This means that small changes in the input cause small changes in the output, ensuring no abrupt jumps or breaks in the function.
For multivariable functions like the one in this exercise, determining limits and continuity involves considering every direction from which you can approach the point. This often requires tools like the epsilon-delta definition to precisely show stability in such complex cases.
Epsilon-Delta Definition
The epsilon-delta definition is a mathematical formalism of limits that provides a rigorous framework for defining a limit. It states that a function \( f(x, y) \) approaches a limit \( L \) as \( (x, y) \) approaches \( (a, b) \), if for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever
  • \( \sqrt{(x-a)^2 + (y-b)^2} < \delta \)
then
  • \( |f(x, y) - L| < \epsilon \)
This definition captures the idea of the function values being arbitrarily close to the limit.
Applying this to our function, we establish \( \delta \) in response to \( \epsilon = 0.01 \). Our task is to choose \( \delta \) small enough to ensure the function stays within \( \epsilon \) of its limit, effectively demonstrating the continuity at \( (0, 0) \). This involves strategically solving inequalities, focusing on how changes in \( \delta \) affect the entire expression.
Multivariable Functions
Functions involving more than one variable, like the function \( f(x, y) = \frac{x+y}{x^2+1} \), are called multivariable functions. These expand the concept of limits and continuity to multiple dimensions.
This adds complexity, as you must consider paths or approaches from every direction toward a point. A key technique is to represent conditions in terms of distances such as \( \sqrt{x^2 + y^2} \), which evaluates how far \( (x, y) \) is from the origin.
  • Here, you check how small changes in \( x \) and \( y \) influence the function's output.
  • Such functions often require careful analysis to ensure all directions lead to the same behavior.
For our given function, the goal is to confirm that changes in \( x \) and \( y \) won't lead outside a predefined tolerance (\( \epsilon \)). By managing the bounds with a calculated \( \delta \), we show that the function is stable, highlighting both its limits and continuity properties.

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Most popular questions from this chapter

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