Question

Let \( f : \mathbb{R}^2 \to \mathbb{R} \) be defined by \[ f(x, y) = \begin{cases} \frac{(x^2 + \sin(xy))^2}{x^2 + y^2} & \text{if } (x, y) \neq (0, 0), \\ 0 & \text{if } (x, y) = (0, 0). \end{cases} \] Then, which of the following is/are TRUE?

• A. \( \lim_{(x, y) \to (0, 0)} f(x, y) \) exists and \( \lim_{(x, y) \to (0, 0)} f(x, y) = 1 \).
• B. \( \lim_{(x, y) \to (0, 0)} f(x, y) \) exists and \( \lim_{(x, y) \to (0, 0)} f(x, y) = 0 \).
• C. \( f \) is differentiable at \( (0, 0) \).
• D. \( f \) is NOT differentiable at \( (0, 0) \).

Hint:

When checking limits of functions in two variables, try evaluating along different paths and using polar coordinates to confirm consistency.

Answer 1

The Correct Option is B

Step 1: Check if the limit exists.
We need to check if the limit of \( f(x, y) \) as \( (x, y) \to (0, 0) \) exists. To do this, we evaluate the function along different paths: - Along the \( x \)-axis (\( y = 0 \)): \[ f(x, 0) = \frac{(x^2 + \sin x \cdot 0)^2}{x^2 + 0^2} = \frac{x^4}{x^2} = x^2. \] As \( x \to 0 \), \( f(x, 0) \to 0 \). - Along the \( y \)-axis (\( x = 0 \)): \[ f(0, y) = \frac{(0^2 + \sin 0 \cdot y)^2}{0^2 + y^2} = 0. \] So, \( f(0, y) = 0 \) for all \( y \), and as \( y \to 0 \), \( f(0, y) \to 0 \). Since the limit along both axes is 0, we can infer that the function tends to 0 along these paths. 
Step 2: Evaluate the limit using polar coordinates.
To confirm the limit in all directions, we convert to polar coordinates: \[ x = r \cos \theta, \quad y = r \sin \theta. \] In polar coordinates, the function becomes: \[ f(r, \theta) = \frac{(r^2 \cos^2 \theta + \sin(r \cos \theta) r \sin \theta)^2}{r^2}. \] As \( r \to 0 \), the numerator tends to zero and the denominator tends to \( r^2 \), so: \[ f(r, \theta) \to 0. \] Thus, the limit as \( (x, y) \to (0, 0) \) is 0. 
Step 3: Conclusion.
Since the limit exists and equals 0, we can confirm that the function is continuous at \( (0, 0) \). 
Final Answer: \[ \boxed{\lim_{(x, y) \to (0, 0)} f(x, y) = 0.} \]