Question

The problem: Let \( f : \mathbb{R}^2 \to \mathbb{R} \) be a function such that \[ f(x, y) = \begin{cases} \left( 1 - \cos\left(\frac{x^2}{y^2}\right) \right) \sqrt{x^2 + y^2}, & \text{if } y \neq 0, x \in \mathbb{R}, \\ 0, & \text{otherwise.} \end{cases} \] Which of the following is/are correct?

• A. \( f \) is continuous at \( (0, 0) \), but not differentiable at \( (0, 0) \)
• B. \( f \) is differentiable at \( (0, 0) \)
• C. All the directional derivatives of \( f \) at \( (0, 0) \) exist and are equal to zero
• D. Both the partial derivatives of \( f \) at \( (0, 0) \) exist and they are equal to zero

Hint:

For piecewise functions, analyze continuity and differentiability carefully by considering limits and behavior near the point of interest.

Answer 1

The Correct Option is A

Step 1: Checking continuity at \( (0, 0) \). As \( (x, y) \to (0, 0) \), the value of \( f(x, y) \) approaches 0. Thus, \( f \) is continuous at \( (0, 0) \).
Step 2: Checking differentiability at \( (0, 0) \). The function \( f \) is not differentiable at \( (0, 0) \) because the term \( \cos\left(\frac{x^2}{y^2}\right) \) oscillates infinitely for \( y \to 0 \) and \( x \neq 0 \). 
Step 3: Partial derivatives at \( (0, 0) \). Both partial derivatives of \( f \) at \( (0, 0) \) exist and are equal to 0, as the limiting values along coordinate axes are zero. 
Step 4: Conclusion. The function is continuous but not differentiable at \( (0, 0) \), and the partial derivatives exist and are zero. The correct answers are \( {(1), (4)} \).