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

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

Consider the following three statements:

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

  • A.S2 and S3 are TRUE and S1 is FALSE
  • B.S1 and S2 are TRUE and S3 is FALSE
  • C.S1 and S3 are TRUE and S2 is FALSE
  • D.S1, S2 and S3 all are TRUE

Answer: B

The correct option is S1 and S2 are TRUE and S3 is FALSE.

S1: for fixed \(x \neq 0\), letting \(y \to 0\) gives \(f(x, 0) = 1 + x\sin(1/x^{2})\). Then as \(x \to 0\), \(|x\sin(1/x^{2})| \le |x| \to 0\), so the repeated limit is 1.

S2: for fixed \(y \neq 0\), letting \(x \to 0\) gives \(f(0, y) = -1\), so the repeated limit is -1.

S3: the second term tends to 0 in any manner, but the first term equals \(\cos 2\theta\) in polar form and takes every value between -1 and 1 near the origin, so the double limit does not exist. The two repeated limits differing (1 and -1) already shows this. The paper allows several answers, and the key gives only this one.

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