Problem 18 - 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. $$ f(x, y)=\frac{\sqrt{x y}}{x+y} $$

Step-by-Step Solution

Answer

The function is discontinuous at the origin with an essential discontinuity.

Step 1: Evaluate the function at the origin

Substitute \(x = 0\) and \(y = 0\) into \(f(x, y)\). This gives \[f(0, 0) = \frac{\root{}{\root{}{0 \times 0}}}{0 + 0} = \frac{0}{0}.\] The expression \(\frac{0}{0}\) is undefined.

Step 2: Investigate the limit along different paths

Consider the limit as \( (x, y) \rightarrow (0,0) \) along the line \( y = mx. \) Substitute \( y = mx \): \[ f(x, mx) = \frac{\root{}{x \times mx}}{x + mx} = \frac{\root{}{mx^2}}{x(1+m)} = \frac{\root{}{m} \times x}{x(1+m)} = \frac{\root{}{m}}{1+m} \] Since this limit depends on the constant \( m \), the limit varies with different paths.

Step 3: Conclude the function's behavior at the origin

Since the limit depends on the path taken (different values of \( m \) give different results), the limit \( \text{lim}ewcommand{f(x,y)}_{(x, y) \rightarrow (0, 0)} f(x, y) \) does not exist. Therefore, \ f(x, y) \ is discontinuous at the origin.

Step 4: Determine the type of discontinuity

Given that the function's limit depends on the path taken, the discontinuity at \( (0,0) \) is essential. An essential discontinuity cannot be removed by redefining the function at \( (0,0) \).

Key Concepts

undefined limits

In calculus, an **undefined limit** occurs when a specific value isn't assigned to a function in a given approach. For example, consider the function in the problem: \[ f(x, y) = \frac{\sqrt{xy}}{x + y}. \]When we substitute \( x = 0 \) and \( y = 0 \), the result is \( \frac{0}{0} \), which is undefined. This is because division by zero is not valid in mathematics. Undefined limits are crucial to identify because they indicate potential discontinuities in the function, prompting further investigation.

path dependence in limits

**Path dependence** in limits means that the value of a limit can differ based on the path chosen while approaching a specific point. This is demonstrated in the given problem by testing different paths towards the origin. Consider the paths:

• Path 1: Approaching along the line \( y = mx \)

If we substitute \( y = mx \) into the function, we get: \[ f(x, mx) = \frac{\sqrt{mx^2}}{x + mx} = \frac{\sqrt{m} x}{x(1 + m)} = \frac{\sqrt{m}}{1 + m}. \]Here, the limit varies based on the constant \( m \). Therefore, the limit isn't unique or fixed and depends on the path taken. This variability means the overall limit as \( (x, y) \rightarrow (0,0) \) does not exist.

essential discontinuity

An **essential discontinuity** is a type of discontinuity that cannot be 'fixed' by redefining the function at a specific point. It arises when a function's limit does not exist as it approaches a point from multiple directions, or when the limit is path-dependent. In the problem, the function \(f(x, y)\) shows different limit values depending on the path taken towards the origin, indicating an essential discontinuity at \((0,0)\). This means the function cannot be modified to have a valid value at that point and remains discontinuous.

removable discontinuity

A **removable discontinuity** occurs when a function is undefined at a point, but by redefining the function at that point, the discontinuity can be 'removed' or 'fixed'. For example, a function may be undefined at a point due to division by zero, but if we can assign a unique limit, the discontinuity can be corrected. However, in our exercise, the function does not have a removable discontinuity because the limit does not exist universally and is path-dependent. Thus, we cannot define \( f(0,0) \) to fix the discontinuity, categorizing it as essential.