Question

Let 𝑓: R² → R be the function defined by

Then

• A. 𝑓 is not continuous at (0, 0)
• B. 𝑓_𝑥 (0, 0) = 0
• C. 𝑓_𝑦 (0, 0) = −1
• D. 𝑓_𝑥 (0, 0) does not exists

Answer 1

The Correct Option is A, C, D

The given function is:

To determine the properties of the function \(𝑓(x, y)\) at the point \((0, 0)\), we need to address continuity and partial derivatives.

1. Continuity at (0, 0):

The function is continuous at \((0, 0)\) if the limit as \((x, y) \rightarrow (0, 0)\) of \(f(x, y)\) equals \(f(0, 0) = 0\). Let's evaluate this:

\[\lim_{(x, y) \to (0, 0)} \frac{x^2 - y^3}{x^2 + y^2}\]

Approaching along \(x = 0\), we have:

\[\frac{0^2 - y^3}{0^2 + y^2} = \frac{-y^3}{y^2} = -y\]

As \(y \to 0\), the limit is 0.

Approaching along \(y = 0\), we have:

\[\frac{x^2 - 0^3}{x^2 + 0^2} = 1\]

The limits are different, indicating that \(f(x, y)\) is not continuous at \((0, 0)\).

2. Partial Derivatives:

The partial derivative with respect to \(x\) at \((0, 0)\) is given by:

\[f_x(0, 0) = \lim_{h \to 0} \frac{f(h, 0) - f(0, 0)}{h}\]\[= \lim_{h \to 0} \frac{\frac{h^2}{h^2} - 0}{h} = \lim_{h \to 0} \frac{1}{h}\]

This limit does not exist; therefore, \(f_x(0, 0)\) does not exist.

The partial derivative with respect to \(y\) at \((0, 0)\) is given by:

\[f_y(0, 0) = \lim_{k \to 0} \frac{f(0, k) - f(0, 0)}{k}\]\[= \lim_{k \to 0} \frac{0 - (-k)}{k} = \lim_{k \to 0} -1 = -1\]

Thus, \(f_y(0, 0) = -1\).

Conclusion:

The correct answers are:

• \(f\) is not continuous at \((0, 0)\).
• \(f_y(0, 0) = -1\).
• \(f_x(0, 0)\) does not exist.