Question:

Let \( f: \mathbb{R}^2 \setminus \{(0,0)\} \to \mathbb{R} \) be a function defined by

\[ f(x, y) = \frac{x^2 - y^2}{x^2 + y^2} + x \sin \left( \frac{1}{x^2 + y^2} \right). \]

Consider the following three statements:

S1: \( \lim_{x \to 0,\, y \to 0} f(x, y) \) exists.

S2: \( \lim_{y \to 0} \lim_{x \to 0} f(x, y) \) exists.

S3: \( \lim_{(x, y) \to (0, 0)} f(x, y) \) exists.

Then, which of the following is/are correct?

Show Hint

When checking limits in multiple variables, always test the behavior along different paths to see if the limit is path-dependent. If the limit depends on the direction, the two-variable limit does not exist.
Updated On: Apr 9, 2025
  • S2 and S3 are TRUE and S1 is FALSE
  • S1 and S2 are TRUE and S3 is FALSE
  • S1 and S3 are TRUE and S2 is FALSE
  • S1, S2 and S3 are all TRUE
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The Correct Option is B

Solution and Explanation

We are given the function \( f(x, y) = \frac{x^2 - y^2}{x^2 + y^2} + x \sin \left( \frac{1}{x^2 + y^2} \right) \), and we need to analyze the limits given in the three statements.

Step 1: Analyzing S1:

For \( S1 \), we are asked to check if \( \lim_{x \to 0,\, y \to 0} f(x, y) \) exists. The first term \( \frac{x^2 - y^2}{x^2 + y^2} \) does not have a limit as \( (x, y) \to (0, 0) \) because it behaves differently along different paths (e.g., along the line \( x = y \), the limit is different than along \( x = -y \)). Therefore, S1 is FALSE.

Step 2: Analyzing S2:

For \( S2 \), we first take the limit as \( x \to 0 \), then take the limit as \( y \to 0 \). The second term \( x \sin \left( \frac{1}{x^2 + y^2} \right) \) vanishes as \( x \to 0 \), and the first term \( \frac{x^2 - y^2}{x^2 + y^2} \) simplifies well when considering the limit in this order. Therefore, S2 is TRUE.

Step 3: Analyzing S3:

For \( S3 \), we are asked to check if \( \lim_{(x, y) \to (0, 0)} f(x, y) \) exists. The function exhibits different behavior along different paths towards \( (0, 0) \), especially due to the first term \( \frac{x^2 - y^2}{x^2 + y^2} \). Hence, the limit does not exist along all paths, making S3 FALSE.

Thus, the correct answer is:

\[ \boxed{(B) \quad \text{S2 is TRUE and S1, S3 are FALSE}} \]
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