Question

Let \( D \subseteq \mathbb{R}^2 \) be defined by \[ D = \mathbb{R}^2 \setminus \{(x,0) : x \in \mathbb{R}\}. \] Consider the function \( f : D \to \mathbb{R} \) defined by \[ f(x,y) = x \sin\!\left(\frac{1}{y}\right). \] Then

• A. \(f\) is a discontinuous function on \(D.\)
• B. \(f\) is a continuous function on \(D\) and cannot be extended continuously to any point outside \(D.\)
• C. \(f\) is a continuous function on \(D\) and can be extended continuously to \(D \cup \{(0,0)\}.\)
• D. \(f\) is a continuous function on \(D\) and can be extended continuously to the whole of \(\mathbb{R}^2.\)

Hint:

When checking continuity of two-variable functions, use inequalities like \(|\sin(1/y)| \le 1\) to control oscillations near singularities.

Answer 1

The Correct Option is C

Step 1: Continuity on \(D.\)
For each \(y \ne 0\), \(f(x,y) = x\sin(1/y)\) is continuous in \(x\). Also, for fixed \(x\), \(\sin(1/y)\) is bounded and continuous for \(y \ne 0\). Hence, \(f\) is continuous on \(D\).
Step 2: Behavior near \((0,0).\)
As \((x,y)\to(0,0)\), \[ |f(x,y)| = |x\sin(1/y)| \le |x| \to 0. \] Thus, the limit exists and equals 0.
Step 3: Define extension.
Define \(f(0,0)=0\). The extended function is continuous at \((0,0)\).
Step 4: Conclusion.
Hence, (C) is correct.