Question

For \( \beta \in \mathbb{R} \), define

\[ f(x, y) = \begin{cases} x^2 |x|^\beta y & \text{if } x \neq 0, \\ 0 & \text{if } x = 0. \end{cases} \]

Then, at \( (0, 0) \), the function \( f \) is

• A. continuous for \( \beta = 0 \)
• B. continuous for \( \beta>0 \)
• C. not differentiable for any \( \beta \)
• D. continuous for \( \beta<0 \)

Hint:

When dealing with piecewise functions, check both continuity and differentiability at the boundary points by computing limits and applying the definition of differentiability.

Answer 1

The Correct Option is B

Step 1: Check continuity at \( (0, 0) \).
For \( x \neq 0 \), \( f(x, y) = x^2 |x|^\beta y \). At \( x = 0 \), we check the limit of \( f(x, y) \) as \( x \to 0 \). The limit does not exist for \( \beta \neq 0 \) because \( x^2 |x|^\beta \) behaves differently depending on the value of \( \beta \).
Step 2: Conclusion.
The function is not differentiable at \( (0, 0) \) for any value of \( \beta \). Thus, the correct answer is \( \boxed{(C)} \).