Problem 17 - The Calculus with Analytic Geometry

Question

In Exercises 8 through 17, determine the region of continuity of \(f\) and draw a sketch showing as a shaded region in \(R^{2}\) the region of continuity of \(f\). $$ f(x, y)= \begin{cases}\frac{\sin (x+y)}{x+y} & \text { if } x+y \neq 0 \\ 1 & \text { if } x+y=0\end{cases} $$

Step-by-Step Solution

Answer

The function is continuous everywhere in \(\mathbb{R}^2\), and the shaded region is the entire plane.

Step 1 - Understand the Given Function

Identify the piecewise definition of the function: \[ f(x, y) = \begin{cases}\frac{\sin (x+y)}{x+y} & \text{if } x+y eq 0 \ 1 & \text{if } x+y = 0 \end{cases} \]

Step 2 - Continuity for x+y eq 0

For the part where \(x+y eq 0\), observe that the function \(f(x,y) = \frac{\sin(x+y)}{x+y}\) represents a continuous function because both the sine function and division are continuous when \(x+y eq 0\).

Step 3 - Continuity for x+y = 0

For the part where \(x+y = 0\), the function is defined as 1. To check if the function is continuous at these points, investigate the limit \(\lim_{(x,y) \to (a,-a)} f(x,y)\), ensuring it equals the function value.

Step 4 - Check the Limit at x+y=0

Compute the limit as \((x, y)\) approaches a point on the line \(x+y=0\): \[ \lim_{(x,y) \to (a,-a)} \frac{\sin(x+y)}{x+y} = \frac{\sin(0)}{0} \] Using L'Hôpital's Rule, this limit simplifies to 1, which matches the defined value when \(x+y=0\).

Step 5 - Conclusion on Continuity

Since the limit matches the function’s value on the line \(x+y=0\) and it is continuous elsewhere, the function \(f(x,y)\) is continuous everywhere in \(\mathbb{R}^2\).

Step 6 - Sketch the Region

Draw a sketch of the region in \(\mathbb{R}^2\): 1. The entire plane represents \(\mathbb{R}^2\). 2. Shade the entire plane to indicate that the function is continuous everywhere.

Key Concepts

Piecewise Functions

Piecewise functions are mathematical functions defined by multiple sub-functions, each applying to a specific interval of the domain. In this exercise, the function is defined differently depending on whether the condition \(x + y eq 0\) or \(x + y = 0\) is met.
This means we are considering two cases:

• \(f(x, y) = \frac{\sin(x+y)}{x+y}\), when \(x + y eq 0\)
• \(f(x, y) = 1\), when \(x + y = 0\)

Note that handling piecewise functions often requires validating their continuity at points where the definition changes. For instance, at \(x + y = 0\), we need to ensure the limit equals the value defined by the piecewise function.

Multivariable Limits

Multivariable limits extend the concept of limits to functions of more than one variable. Here, to check the continuity at \(x + y = 0\), we need to evaluate the limit: \[ \lim_{(x, y) \to (a, -a)} \frac{\sin(x+y)}{x+y} \] This can be tricky because changes occur in multiple dimensions simultaneously. One must ensure the limit is approached from all directions within the x-y plane. For the given function, as \((x,y)\) approaches a point on the line \(x+y=0\), the limit simplifies thanks to L'Hôpital's Rule, leading to \[ \lim_{(x, y) \to (a, -a)} \frac{\sin(x+y)}{x+y} = 1 \] Hence, the limit matches the value specified by the piecewise definition, confirming continuity at \(x+y=0\).

L'Hôpital's Rule

L'Hôpital's Rule is a significant tool for evaluating limits that result in an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In this exercise, the limit \[ \lim_{(x, y) \to (a, -a)} \frac{\sin(x+y)}{x+y} \] initially appears to be \( \frac{0}{0} \) at \((a, -a)\). Applying L'Hôpital's Rule to \[ \lim_{(t) \to (0)} \frac{\sin t}{t} \] where \(t = x + y\), helps us resolve it to 1. This approach is critical in confirming that discontinuities do not occur at problem points and ensuring comprehensive continuity within the region.

Region of Continuity

The region of continuity of a function is the subset of the domain where the function is continuous. Here, the function is continuous everywhere in \(\mathbb{R}^2\).
For the given piecewise function:

• \(f(x,y) = \frac{\sin(x+y)}{x+y} \) is continuous when \( x+y eq 0 \)
• Checking when \( x + y = 0 \), we examine the limit as \((x, y)\) approaches the line \( x + y = 0 \).

Since the limit matches the function's defined value (as shown using the limit and L'Hôpital's Rule), the function remains continuous throughout. Therefore, the continuous region is the entire plane \( \mathbb{R}^2 \). This conclusion allows us to draw the entire plane shaded, representing the region where the function maintains continuity seamlessly.