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Chapter 4: Problem 96

For the following exercises, determine the region in which the function is continuous. Explain your answer. $$f(x, y)=\frac{\sin \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}$$

Short Answer

Expert verified
Function is continuous for all \((x, y) \in \mathbb{R}^2\).

Step by step solution

01

Understand the Function

The given function is \( f(x, y) = \frac{\sin(x^2 + y^2)}{x^2 + y^2} \). The function has the form \( \frac{\text{numerator}}{\text{denominator}} \), which is a rational function.
02

Identify Points of Discontinuity

A function is continuous at all points where the denominator is not zero. Identify points where \( x^2 + y^2 = 0 \). This equation holds true only when \( x = 0 \) and \( y = 0 \) simultaneously.
03

Analyze Continuity Around Identified Points

At the origin \((0, 0)\), the function is indeterminate because the denominator is zero. To determine if a limit exists at this point, we need to evaluate \( \lim_{(x,y) \to (0,0)} f(x, y) \).
04

Determine the Limit at Discontinuity Points

Consider polar coordinates where \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \), thus \( x^2 + y^2 = r^2 \). The function becomes \( \frac{\sin(r^2)}{r^2} \). As \( r \to 0 \), \( \frac{\sin(r^2)}{r^2} \to 1 \) using the standard limit result \( \lim_{r \to 0} \frac{\sin r}{r} = 1 \).
05

Conclusion on Continuity

Since the limit exists and is equal to \( 1 \) at \( (0,0) \), the function approaches a defined value. Therefore, \( f(x, y) \) is continuous everywhere in \( \mathbb{R}^2 \).

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Key Concepts

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

Rational Functions
Rational functions are quotients of two functions, specifically a numerator and a denominator. In general, a rational function has the form \( \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials. In our exercise's context, the function \( f(x, y) = \frac{\sin(x^2 + y^2)}{x^2 + y^2} \) is a rational function of two variables. Here, the numerator is \( \sin(x^2 + y^2) \) and the denominator is \( x^2 + y^2 \).
  • Rational functions are only undefined at points where the denominator equals zero. For our function, the potential point of discontinuity occurs when both \( x \) and \( y \) are zero, making \( x^2 + y^2 = 0 \).
  • Identifying these points is crucial because they often indicate where the function might have holes or discontinuities.
Understanding rational functions is important to analyze their behavior, especially around points where the denominator equals zero. This sets the foundation for studying limits and continuity.
Limits and Continuity
The concepts of limits and continuity are fundamental to calculus. A function is continuous at a point if there is no interruption or undefined value at that point.
To determine the continuity of \( f(x, y) = \frac{\sin(x^2 + y^2)}{x^2 + y^2} \), examining limits is essential, particularly where the function indeterminately becomes zero over zero.
  • An important limit to understand in this context is \( \lim_{r \to 0} \frac{\sin r}{r} = 1 \). This limit helps simplify the analysis around the origin \((0, 0)\).
  • If the limit of \( f(x, y) \) as \((x, y)\) approaches \((0,0)\) exists and equals a specific value, then the potential discontinuity can be resolved, indicating continuity.
In our exercise, applying limits in polar coordinates reveals that as \( r \to 0 \), the function approaches a value of 1. This signifies that the function is continuous at the origin, effectively removing any points of discontinuity on \( \mathbb{R}^2 \).
Polar Coordinates
Polar coordinates provide a powerful method for analyzing functions with rotational symmetry or radial components.
In the exercise, we consider \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). Then \( x^2 + y^2 = r^2 \), which simplifies the function \( f(x, y) \) to \( \frac{\sin(r^2)}{r^2} \).
  • Converting to polar coordinates is helpful when evaluating limits at the origin. It provides a simpler path to approach a limit by reducing the problem to a single variable \( r \).
  • Using polar coordinates, we determined that \( \lim_{r \to 0} \frac{\sin(r^2)}{r^2} = 1 \), solving the indeterminate form present in Cartesian coordinates.
By adopting polar coordinates, we gain insights into the behavior of functions in two-dimensional space, making it easier to assess continuity and limits.

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