Question

Let

\[ f(x, y) = \begin{cases} \frac{|x|}{|x| + |y|} \sqrt{x^4 + y^2}, & (x, y) \neq (0, 0), \\ 0, & (x, y) = (0, 0). \end{cases} \]

Then at \( (0, 0) \),

• A. \( f \) is continuous
• B. \( \frac{\partial f}{\partial x} = 0 \) and \( \frac{\partial f}{\partial y} \) does not exist
• C. \( \frac{\partial f}{\partial x} \) does not exist and \( \frac{\partial f}{\partial y} = 0 \)
• D. \( \frac{\partial f}{\partial x} = 0 \) and \( \frac{\partial f}{\partial y} = 0 \)

Hint:

To check differentiability and continuity, compute the limits along various paths and verify the existence of partial derivatives.

Answer 1

The Correct Option is A, D

Step 1: Check the continuity at \( (0, 0) \).
To check the continuity of \( f \) at \( (0, 0) \), we compute the limit: \[ \lim_{(x, y) \to (0, 0)} f(x, y). \] Using different paths (e.g., \( y = 0 \) and \( x = 0 \)), we find that the limit is 0 along both paths. Hence, \( f(x, y) \) is continuous at \( (0, 0) \).
Step 2: Check the partial derivatives.
The partial derivative of \( f \) with respect to \( x \) at \( (0, 0) \) is: \[ \frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(h, 0) - f(0, 0)}{h} = 0. \] However, the partial derivative with respect to \( y \) does not exist because \( f(x, y) \) is not symmetric in \( y \) and the limit approaches from different directions does not give a finite value.
Step 3: Conclusion.
Thus, the correct answer is \( \boxed{(B)} \).