Question

Let \( f(x, y) = \begin{cases} x^2 \sin \dfrac{1}{x} + y^2 \sin \dfrac{1}{y}, & xy \ne 0 \\ x^2 \sin \dfrac{1}{x}, & x \ne 0, y = 0  \\ \\  y^2 \sin \dfrac{1}{y}, & y \ne 0, x = 0 \\ 0, & x = y = 0 \end{cases} \). 

Which of the following is true at \( (0, 0)? \)
 

• A. \( f \) is not continuous
• B. \( \dfrac{\partial f}{\partial x} \) is continuous but \( \dfrac{\partial f}{\partial y} \) is not continuous
• C. \( f \) is not differentiable
• D. \( f \) is differentiable but both \( \dfrac{\partial f}{\partial x} \) and \( \dfrac{\partial f}{\partial y} \) are not continuous

Hint:

A function can be differentiable even if its partial derivatives are not continuous; continuity of partials is sufficient but not necessary for differentiability.

Answer 1

The Correct Option is D

Step 1: Check continuity at \( (0,0) \).
\[ |f(x, y)| \le x^2 + y^2. \] Thus, \( \lim_{(x, y) \to (0, 0)} f(x, y) = 0 = f(0, 0). \) Hence, \( f \) is continuous.

Step 2: Find partial derivatives at \( (0,0). \)
\[ f_x(0, 0) = \lim_{h \to 0} \dfrac{f(h, 0) - f(0, 0)}{h} = \lim_{h \to 0} \dfrac{h^2 \sin(1/h)}{h} = \lim_{h \to 0} h \sin(1/h) = 0. \] Similarly, \( f_y(0, 0) = 0. \)

Step 3: Check differentiability.
Near \( (0, 0) \): \[ |f(x, y) - f(0, 0) - 0| = |f(x, y)| \le x^2 + y^2. \] Hence, \( \dfrac{|f(x, y)|}{\sqrt{x^2 + y^2}} \to 0 \). Thus, \( f \) is differentiable at \( (0, 0). \)

Step 4: Check continuity of partial derivatives.
\[ f_x(x, 0) = 2x \sin(1/x) - \cos(1/x), \] which oscillates as \( x \to 0. \) Similarly, \( f_y(0, y) \) is also discontinuous.

Final Answer: \( f \) is differentiable at \( (0, 0) \), but both partial derivatives are not continuous there.