Problem 24 - Calculus for Biology and Medicine

Question

Show that $$f(x, y)=\left\\{\begin{array}{cl}\frac{3 r(y+x)}{x^{2}+y^{3}} & \text { for }(x, y) \neq(0,0) \\\0 & \text { for }(x, y)=(0,0)\end{array}\right.$$ is discontinuous at \((0,0)\).

Step-by-Step Solution

Answer

The function is discontinuous at \((0, 0)\) because the limit does not exist along multiple paths.

Step 1: Understanding the Function

The function is given as \( f(x, y) = \frac{3r(y+x)}{x^2 + y^3} \) for \((x, y) eq (0, 0)\) and \( f(0, 0) = 0 \). We need to show that this function is discontinuous at \((0, 0)\). This requires finding if the limit as \((x, y)\) approaches \((0, 0)\) is different from the value of the function at \((0, 0)\).

Step 2: Finding the Path-Based Limit

One common approach to check discontinuity is to evaluate the limit along different paths. Let's first choose the path where \( y = mx \), a straight line path. Substituting \( y = mx \) into \( f(x, y) \), we have: \[ f(x, mx) = \frac{3r(x + mx)}{x^2 + (mx)^3} = \frac{3r(x + mx)}{x^2 + m^3x^3} = \frac{3r(1 + m)x}{x^2(1 + m^3x)} \] As \( x \rightarrow 0 \), the function simplifies to \( \frac{3r(1 + m)}{x(1 + m^3x)} \) which diverges, indicating the limit does not exist along this path.

Step 3: Trying Another Path for Confirmation

Let's try another path where \( x = ky^2 \). Substituting \( x = ky^2 \) into \( f(x, y) \), we get: \[ f(ky^2, y) = \frac{3r(y + ky^2)}{(ky^2)^2 + y^3} = \frac{3r(y + ky^2)}{k^2y^4 + y^3} = \frac{3r(1 + ky)}{k^2y(1 + \frac{1}{y})} \] As \( y \rightarrow 0 \), the function again diverges, indicating the limit does not exist along this path either.

Step 4: Analyzing the Function at (0,0)

The value of \( f(0, 0) \) is given as 0. For the function to be continuous at this point, the limit of \( f(x, y) \) as \( (x, y) \rightarrow (0, 0) \) must equal the function value \( f(0, 0) = 0 \). However, from our path evaluations, the limits along different paths do not converge to a single value and diverge instead.

Step 5: Conclusion

Because the limit of \( f(x, y) \) as \( (x, y) \rightarrow (0, 0) \) diverges along multiple paths and does not agree with the function value at \( (0, 0) \), \( f(x, y) \) is discontinuous at \( (0, 0) \). This discrepancy confirms the discontinuity.

Key Concepts

Limit Evaluation

Limit evaluation is a crucial concept in calculus. It helps determine how a function behaves as the input approaches a certain point. For multivariable functions, this can be more complicated than in single-variable situations.

In our example, we are examining the function \( f(x, y) = \frac{3r(y+x)}{x^2 + y^3} \) as it approaches the point \((0,0)\). A limit evaluation involves finding whether the function's value approaches a specific number as \((x, y)\) gets closer to \((0,0)\).

If the limit exists and is equal to the function's value at that point, the function is continuous there. If not, the function is discontinuous. Therefore, by understanding the result of the limit evaluation at \((0,0)\) and how these results compare to \( f(0,0) \), we gain insights into the function's continuity.

Multivariable Function

A multivariable function includes more than one variable and can be expressed as \( f(x, y) \), where both \( x \) and \( y \) are independent variables. These functions represent a surface in three-dimensional space.

Understanding such functions requires considering how a change in one or both variables affects the function's output.

For the function \( f(x, y) = \frac{3r(y+x)}{x^2 + y^3} \), the output depends on both \( x \) and \( y \), and its behavior is complex.

• Multivariable functions can be analyzed using partial derivatives, helping to understand how the function changes along different dimensions.
• This type of function analysis is important when studying gradients, directional derivatives, and divergence.

In our example, the values that \( x \) and \( y \) take significantly influence the result of \( f(x, y) \) and how these variables approach the discontinuous point.

Path-Based Limit

In multivariable calculus, a path-based limit is used to determine if a limit exists as a point is approached via different paths.

To check a path-based limit, you substitute different paths for the variables and evaluate the result.

For example, with \( y = mx \) and \( x = ky^2 \) as paths to approach \((0,0)\), both calculations showed the limits diverged in our function \( f(x, y) \).

• The path \( y = mx \) drove the function to infinite values depending on \( m \).
• When \( x = ky^2 \), the same divergence occurred, confirming different behavior along different paths.

The essence is to ensure the limit is independent of the path taken, which would mean the same value is approached in all ways.

Since the function doesn’t converge to a single value from all paths towards \((0,0)\), we conclude that the limit is path-dependent, and hence, the function is discontinuous at that point.