Problem 19 - The Calculus with Analytic Geometry

Question

In Exercises 18 through 21, prove that the function is discontinuous at the origin. Then determine if the discontinuity is removable or essential. If the discontinuity is removable, define \(f(0,0)\) so that the discontinuity is removed. $$ f(x, y)=(x+y) \sin \frac{x}{y} $$

Step-by-Step Solution

Answer

The function is discontinuous at the origin and the discontinuity is essential.

Step 1: Identify the Function and Point of Discontinuity

Given function: \[f(x, y) = (x+y) \sin \frac{x}{y}\]The function needs to be examined at the origin, which is the point (0, 0).

Step 2: Check Continuity at the Origin

The function is potentially discontinuous at the origin because the term \( \frac{x}{y} \) is undefined when \( y = 0 \). Therefore, computation of the limit as \((x, y) \to (0, 0)\) is needed.

Step 3: Compute the Limit in Different Paths

Examine the limit of \(f(x, y)\) as \((x, y) \to (0, 0)\) along different paths:- Path 1: Let \(x = ky\). Then \[\lim_{y \to 0} f(ky, y) = \lim_{y \to 0} (ky + y) \sin \frac{ky}{y} = \lim_{y \to 0} y(k + 1) \sin k = 0.\]- Path 2: Let \(y = mx\). Then \[\lim_{x \to 0} f(x, mx) = \lim_{x \to 0} (x + mx) \sin \frac{x}{mx} = \lim_{x \to 0} x(1 + m) \sin \frac{1}{m}.\]Since path 2 depends on \(m\) and may not equal zero, the limit does not necessarily approach a single value.

Step 4: Conclude the Discontinuity and Type

From Step 3, the limit of the function \(f(x, y)\) as \((x, y) \to (0, 0)\) is not the same for different paths. Therefore, it is discontinuous at the origin. Since there is no single value that can be assigned to \(f(0, 0)\) to make the function continuous, the discontinuity is essential.

Key Concepts

Limit Paths

In multivariable functions, understanding limit paths is crucial for analyzing continuity. Limit paths are different ways you can approach a point, typically the origin \(0,0\), in a two-dimensional function. For example, you can take the path where \(x=ky\) or \(y=mx\). These represent scaling the variables by constants \(k\) and \(m\), respectively. When you approach a point along different curves or lines and find varying limits, this indicates that the limit does not exist universally for all paths. It is a helpful method to determine if a point of discontinuity exists in multivariable functions.

Removable Discontinuity

A removable discontinuity occurs when a function is not defined at a point, but its limit exists as you approach that point. In simpler terms, a hole in the graph that can be 'plugged'. This happens if you can redefine the function at the point to make it continuous. To identify a removable discontinuity, check if the limit from all paths to the point is the same. If it is, you can redefine the function at that point. For instance, \(g(x) = (x^2 - 4)/(x - 2)\) has a discontinuity at \(x = 2\). The limit as \(x \rightarrow 2\) is \(4 \), so you could redefine \(g(2) = 4\) to remove the discontinuity.

Essential Discontinuity

Essential discontinuities are more complex and occur when the limit does not exist as you approach a point from different paths. This means you can't simply redefine the function to 'fix' it. In the given problem, the function \(f(x, y) = (x + y) \sin \frac{x}{y}\) has an essential discontinuity at the origin. Different paths like \(x = ky\) and \(y = mx\) yield different limits. Thus, no single value can be assigned to \(f(0,0)\) to make the function continuous at the origin. Recognizing these kinds of discontinuities often involves detecting varying limits that depend on the path taken. This means there's an inherent 'jump' or 'gap' in the function's behavior at that point.