Problem 30 - Calculus with Concepts in Calculus

Question

Let $$ f(x, y)=\left\\{\begin{array}{ll} \frac{\sin x y}{x^{2}+y^{2}} & \text { for }(x, y) \neq(0,0) \\ 0 & \text { for }(x, y)=(0,0) \end{array}\right. $$ Show that \(f\) is not continuous at \((0,0)\).

Step-by-Step Solution

Answer

The function is not continuous at \((0, 0)\) as the limit depends on the path taken.

Step 1: Understand the Definition of Continuity

A function is continuous at a point \((a, b)\) if the limit of the function as \((x, y)\) approaches \((a, b)\) is equal to the function's value at \((a, b)\). This means \[\lim_{(x, y) \to (a, b)} f(x, y) = f(a, b).\]

Step 2: Identify the Target Point

We need to show that the function \(f(x, y)\) is not continuous at the point \((0, 0)\). According to the definition, we need to check if the limit \[\lim_{(x, y) \to (0, 0)} \frac{\sin(xy)}{x^2 + y^2}\] e 0.

Step 3: Approach Along the x-axis (y=0)

When approaching \((0, 0)\) along the x-axis where \(y = 0\), the function simplifies to\[\frac{\sin(0 \cdot x)}{x^2 + 0^2} = \frac{0}{x^2} = 0.\]Thus, as \(x\) approaches 0, the limit along the x-axis is 0.

Step 4: Approach Along the y-axis (x=0)

When approaching \((0, 0)\) along the y-axis where \(x = 0\), the function simplifies to\[\frac{\sin(0 \cdot y)}{0^2 + y^2} = \frac{0}{y^2} = 0.\]Thus, as \(y\) approaches 0, the limit along the y-axis is 0.

Step 5: Approach Along the Line y=x

Consider the path where \(y = x\). Substituting this into the function gives:\[\frac{\sin(x \cdot x)}{x^2 + x^2} = \frac{\sin(x^2)}{2x^2}.\]As \(x\) approaches 0, the limit becomes:\[\lim_{x \to 0} \frac{\sin(x^2)}{2x^2} = \frac{1}{2}.\]

Step 6: Conclude the Discontinuity

The results from Steps 3, 4, and 5 show different limits depending on the path taken to \((0, 0)\). Specifically,\(-)\) Along the x-axis, the limit is 0.\(-)\) Along the y-axis, the limit is 0.\(-)\) Along the line \(y=x\), the limit is \(\frac{1}{2}\).Since the limits are not the same, \(\lim_{(x, y) \to (0, 0)} f(x, y)\) does not exist, and thus, \(f\) is not continuous at \((0, 0)\).

Key Concepts

Continuity

Continuity of a function is an important principle in calculus and mathematical analysis. It describes a function that has no interruptions, jumps, or gaps anywhere in its domain. For a function to be continuous at a point

• the limit of the function as it approaches that point must exist,
• this limit must be equal to the value of the function at the point itself.

In mathematical terms, a function \( f(x, y) \) is continuous at \((a, b)\) if \( \lim_{(x, y) \to (a, b)} f(x, y) = f(a, b) \).
At each point in the domain, this condition must hold true for the function to be considered continuous over its entire domain.
However, if the limit does not exist or does not equal the function's value at that point, the function is said to be not continuous at that point. This discontinuity means there could be any number of different values the function might take near that point depending on the path the inputs take.

In our given problem, we evaluated the function \( f(x, y) = \frac{\sin xy}{x^2 + y^2} \) and found that it has different limits along different paths toward \((0,0)\). This inconsistency in limits along paths such as \( y=0 \), \( x=0 \), and \( y=x \) illustrates the discontinuity at the origin.

Limits

Limits are a fundamental concept in calculus, essential for defining both continuity and derivatives. They describe the value that a function approaches as the input approaches a particular point. When dealing with functions of two variables, such as \( f(x, y) \), we consider the limit of \( f \) as \((x, y)\) approaches \((a, b)\). This means that \( f(x, y) \) should get arbitrarily close to a specific number, as both \(x\) and \(y\) get close to \(a\) and \(b\), respectively.
But unlike single variable functions, two-variable limits can depend on the path taken to approach \((a, b)\). In our case, for the function \( f(x, y) = \frac{\sin xy}{x^2 + y^2} \), we evaluated its limit along different paths:

• Along the x-axis \((y = 0)\) and y-axis \((x = 0)\), the limit was 0.
• Along the line \(y = x\), the limit was \(\frac{1}{2}\).
• These differing limits mean the overall limit \(\lim_{(x, y) \to (0, 0)} f(x, y)\) does not exist.

For continuity, the same limit must be reached from every possible path, highlighting why knowing limits is crucial to determine function behavior at specific points.

Two-Variable Functions

Two-variable functions are expressions like \( f(x, y) \) that depend on two independent variables. These functions are mapped on a coordinate plane, taking inputs from a plane and providing a scalar output. The study of these functions allows us to understand contours, surfaces, and how one variable might change with respect to another.
In the exercise provided, we deal with a two-variable function, \( f(x, y) \), given by different rules depending on the input being \((x, y) eq (0,0)\) or \((x, y) = (0,0)\). Such scenarios are common in multivariable calculus, where functions have different expressions or values depending on the region of the input plane.
When analyzing properties of functions like continuity or limits with two variables, it's crucial to understand how changes in these multiple input directions affect the function’s output. Calculations in multivariable calculus open doors to practical applications, such as 3D modeling or physics problems where outcomes depend on more than one variable, providing robust techniques to predict behaviors across many fields.