Question

Let \[ f(x, y) = \begin{cases} \frac{x^2 y}{1 + x^2} \sin\left(\frac{1}{x}\right), & \text{if} \ x \neq 0 \\ 0, & \text{if} \ x = 0 \end{cases} \]
Which of the following statements is/are TRUE?

• A. \( \lim_{y \to 0} \lim_{x \to 0} f(x, y) \) exists.
• B. \( \frac{\partial f}{\partial x} \) is continuous at the point \( (0, 1) \).
• C. \( \frac{\partial f}{\partial y} \) is continuous at the point \( (0, 1) \).
• D. \( \lim_{(x, y) \to (0, 0)} \frac{f(x, y)}{\sqrt{x^2 + y^2}} \) does not exist.

Hint:

When evaluating limits of multivariable functions, consider the behavior of the function along different paths to determine if the limit exists.

Answer 1

The Correct Option is A, C

Step 1: Analyze option (A).
For \( x \neq 0 \), the function is smooth, and the limit as \( x \to 0 \) exists because the sine term is bounded. However, as \( y \to 0 \), the behavior of \( f(x, y) \) does not change. Therefore, the limit exists, and option (A) is true.

Step 2: Analyze option (B).
The derivative with respect to \( x \) at the point \( (0, 1) \) involves evaluating limits that depend on the oscillatory behavior of the sine term, which is not continuous. Therefore, option (B) is false.

Step 3: Analyze option (C).
The derivative with respect to \( y \) is continuous at \( (0, 1) \) because the function does not depend on \( y \) in a way that would introduce any discontinuities. Hence, option (C) is true.

Step 4: Analyze option (D).
The limit \( \lim_{(x, y) \to (0, 0)} \frac{f(x, y)}{\sqrt{x^2 + y^2}} \) does not exist because the function behaves differently depending on the path taken to approach \( (0, 0) \). Therefore, option (D) is true.

Step 5: Conclusion.
The correct answer is (D), as the limit in option (D) does not exist due to the behavior of the function.