Problem 28 - Calculus with Concepts in Calculus

Question

Explain why \(f\) is continuous. $$ f(x, y)=\left\\{\begin{array}{ll} \frac{x^{2} y^{2}}{x^{2}+y^{2}} & \text { for }(x, y) \neq(0,0) \\ 0 & \text { for }(x, y)=(0,0) \end{array}\right. $$

Step-by-Step Solution

Answer

The function \(f\) is continuous because it is defined at all points, including \( (0,0) \), and the limit as \( (x,y) \to (0,0) \) equals \(f(0,0) = 0\).

Step 1: Understand the Definition of Continuity

A function of two variables, \(f(x, y)\), is continuous at a point \((a, b)\) if it is defined at \((a, b)\), and the limit of \(f(x, y)\) as \((x, y)\) approaches \((a, b)\) equals \(f(a, b)\). For our function, we need to prove its continuity at all points in its domain, especially focusing on \((0,0)\).

Step 2: Check Continuity at (0,0)

We first look at the point \((0,0)\). The function is defined since \(f(0,0) = 0\). Next, we need to show that \(\lim_{(x, y) \to (0, 0)} f(x, y) = 0\).

Step 3: Set Up the Limit and Simplify

Consider approaching \((0, 0)\) along a general path given by \(y = mx\). Substitute \(y = mx\) into \(f(x, y)\) to get \(\frac{x^2 (mx)^2}{x^2 + (mx)^2} = \frac{x^2 m^2 x^2}{x^2 (1 + m^2)} = \frac{m^2 x^4}{x^2 + m^2 x^2} = \frac{m^2 x^4}{x^2 (1 + m^2)}\). This simplifies to \(\frac{m^2 x^4}{x^2 (1 + m^2)} = \frac{m^2 x^2}{1 + m^2}\).

Step 4: Evaluate the Limit

As \(x \to 0\), \(\frac{m^2 x^2}{1 + m^2} \to 0\). Thus, \(\lim_{x \to 0} \frac{m^2 x^2}{1 + m^2} = 0\). Regardless of the path chosen (value of \(m\)), the limit is \(0\).

Step 5: Extending to Any Path

Now consider paths like \( x = ky^2 \), \( r\cos(\theta) \), \( r\sin(\theta)\), etc. Each path also gives \(\lim_{(x, y) \to (0, 0)} f(x, y) = 0\). This means the limit is independent of the path.

Step 6: Verify Continuity at Other Points

For \((x, y) eq (0,0)\), the function is continuous since rational functions are continuous at all points where they are defined and do not have division by zero. Thus, they are continuous by basic algebraic continuity properties.

Step 7: Conclusion of Continuity

Since \(f(x, y)\) equals \(0\) at \((0,0)\), the limit \(\lim_{(x, y) \to (0, 0)} f(x, y) = 0\), and the function is continuous at all other points \((x, y) eq (0,0)\); \(f\) is continuous everywhere in its domain.

Key Concepts

Understanding Continuity

Continuity is a fundamental concept in calculus that helps us understand how functions behave close to specific points. In the case of functions of two variables, \(f(x, y)\), a function is continuous at a point \((a, b)\) if:

• It is defined at \((a, b)\).
• The limit of \(f(x, y)\) as the point \((x, y)\) approaches \((a, b)\) is equal to \(f(a, b)\).

To check the continuity of our specific function, \(f(x, y)\), we are particularly interested in the point \((0,0)\). The value of \(f(0,0)\) is given as 0, which gives us a starting point for confirming its continuity through limits.

Exploring Limits

Limits are an essential tool in multivariable calculus that help us analyze how a function behaves as its input approaches a certain point. For functions of two variables, we must consider how \(f(x, y)\) behaves along different paths approaching a specific point.
For our function, to prove continuity at \((0,0)\), we need to evaluate the limit as \((x,y)\) approaches \((0,0)\). We can start by considering different paths:

• Along the path \(y=mx\), we simplify \(rac{x^2y^2}{x^2 + y^2}\) to \(rac{m^2x^2}{1+m^2}\).
• As \((x,y)\) approaches \((0,0)\), \(x o 0\) implies \(rac{m^2x^2}{1+m^2} o 0\).

This result shows that the limit is zero regardless of the path taken, meaning that the limit is independent of the direction, an important step in verifying continuity.

Applying Classical Analysis

Classical analysis allows us to use known theorems and logical deductions to assess continuity. A pivotal part of this is evaluating the path-independence of limits by applying various paths, such as:

• Standard paths \(y = mx\).
• Curved paths like \(x = ky^2\).
• Polar coordinates \((r\cos(\theta), r\sin(\theta))\).

By ensuring the limit converges to zero along all these paths as \((x, y)\) approaches \((0,0)\), we confirm the absence of directional bias in reaching the point. Thus, classical analysis supports that \(f(x, y)\) is continuous at the origin.

Grasping Functions of Two Variables

In multivariable calculus, understanding functions of two variables is key to analyzing continuity and limits. These functions, written as \(f(x, y)\), vary their output depending on both \(x\) and \(y\).
For the given function, we derived its continuity by dissecting it:

• At \((x, y) = (0,0)\), the function has an explicitly defined value of 0.
• For \((x, y) eq (0,0)\), it is a rational function, which is known to be continuous where it does not involve division by zero.

Thus, by evaluating how the function behaves across its entire domain, we ensured that it maintains continuity everywhere. This exploration within multivariable contexts gives comprehensive insights into how such functions can be continuous at specific and all points in their domain.