Differentiability

This is a textbook JEE Advanced example showing that differentiability at a point is decided by the definition of the derivative, not by the behaviour of the derivative formula elsewhere. For continuity, |p(x)| = |x^2 \sin(1/x)| \le x^2 \to 0, so by the squeeze theorem p(x) \to 0 = p(0), confirming continuity. For differentiability, compute p'(0) = \lim_{x \to 0} \frac{x^2 \sin(1/x) - 0}{x} = \lim_{x \to 0} x \sin(1/x), and since |x \sin(1/x)| \le |x| \to 0, the derivative at 0 exists and equals 0. Thus the third statement is correct. The first option is false because the squeeze theorem secures continuity. The second is false because the difference quotient does converge. The fourth wrongly reports the value 1, which would require the difference quotient to tend to a nonzero number. The governing pattern is the squeeze theorem applied inside the limit definition of the derivative. Plausibility check: although p'(x) = 2x \sin(1/x) - \cos(1/x) oscillates and has no limit as x \to 0, the point derivative p'(0) is still well defined at 0, illustrating that derivative existence at a point need not imply derivative continuity there.