Question

Give an example of a function \(f\) on \(\mathbb{R}^{2}\) such that \(f\) is not continuous at (0,0) , but \(f(0, y)\) is a continuous function of \(y\) on \((-\infty, \infty)\) and \(f(x, 0)\) is a continuous function of \(x\) on \((-\infty, \infty)\).

Short answer

Question: Construct a function not continuous at (0,0), but continuous on the x-axis and y-axis. Answer: A possible function is: $$ f(x, y) = \begin{cases} \frac{xy}{x^2 + y^2}, & \text{for}\ (x, y) \neq (0,0) \\ 0, & \text{for}\ (x, y) = (0,0) \\ \end{cases} $$ This function is continuous on the x-axis and y-axis but discontinuous at (0,0).

Step-by-step solution

Define the function

To find such a function, we will define the function as follows: $$ f(x, y) = \begin{cases} \frac{xy}{x^2 + y^2}, & \text{for}\ (x, y) \neq (0,0) \\ 0, & \text{for}\ (x, y) = (0,0) \\ \end{cases} $$

Check the continuity along x-axis and y-axis

We need to show that this function is continuous along both the \(x\)-axis and the \(y\)-axis. First, consider the function when \(y=0\). The function becomes: $$ f(x, 0) = \frac{x\cdot0}{x^2 + 0^2} = 0 $$ Similarly, consider the function when \(x=0\). The function becomes: $$ f(0, y) = \frac{0\cdot y}{0^2 +y^2} = 0 $$ In both cases, the function is constant at 0 and, thus, is continuous along both the x-axis and the y-axis.

Check for discontinuity at (0, 0)

To prove that the function is not continuous at \((0,0)\), we can try to find two different paths that lead to different function values when approaching \((0,0)\). Let's try the path \(y=x\): $$ \lim_{x \to 0} f(x, x) = \lim_{x \to 0} \frac{x^2}{2x^2} = \lim_{x \to 0} \frac{1}{2} = \frac{1}{2} $$ Now let's try the path \(y=0\): $$ \lim_{x \to 0} f(x, 0) = \lim_{x \to 0}0 = 0 $$ Since the limits along these two paths are different, we can conclude that the function \(f(x,y)\) is not continuous at \((0,0)\).

Key concepts

Continuity

In calculus, continuity of a function is a crucial concept. For a function to be continuous at a point, the function's limit as it approaches that point must equal the function's value at that point. This means no sudden jumps or breaks around that point.
For a function defined on multiple variables, like the one given in the exercise, continuity implies that as you approach a specific point from any direction on the plane, the function's value should steadily match its value at that point.
\[\text{At point } (x_0, y_0), \ f(x, y) \text{ is continuous if:} \]

• \( \lim_{(x, y) \to (x_0, y_0)} f(x, y) = f(x_0, y_0)\)
• \(f(x_0, y_0)\) is defined
• The limit exists

In this exercise, the function is continuous along the axes because the function's output stays constant at zero as you move along each axis. However, it becomes tricky when checking the origin, since different paths lead to different limits.

Functions of Several Variables

Functions of several variables are an extension of single-variable functions we often see. Instead of just depending on one input, these functions depend on two or more variables. The domain of such functions can be in 2D or 3D spaces, like \(\mathbb{R}^2\) or \(\mathbb{R}^3\).For example, a function \(f(x, y)\) takes two inputs, \(x\) and \(y\). These functions create surfaces in 3D space, which can be visualized as landscapes with hills and valleys.
The function given in the problem is such a function, defined across the plane except for at the origin. By setting values along specific inputs such as \(x\) or \(y\), parts of the surface can be analyzed separately, as seen when the continuity along the axes is assessed.
To visualize these functions, consider:

• The function has specific behavior along the axes, like a flat line.
• As variables change, the function creates geometry in space.
• They're useful in physics for modeling real-world scenarios.

Limits of Functions

The concept of limits is essential in understanding the behavior of functions as they approach a specific point. A limit examines the function's tendency as the variable gets infinitely close to a point.
In functions of several variables, evaluating limits can be more complex. Different paths may lead to different limits. When this happens, it indicates the function is not continuous at that point.
The given function in the exercise illustrates this beautifully. By taking different paths—such as \(y=x\) and along the \(x\)-axis itself—we observe different observed limit points. This suggests two significant things:

• If limits along different paths to a point differ, the overall limit does not exist.
• Path-dependent limits indicate a discontinuity at the point.

For the function \(f(x, y)\), as you approach the origin through \(y=x\), the limit is \(\frac{1}{2}\) whereas directly along the axis, it is zero, showing discontinuity.