Question

For \( x \in \mathbb{R} \), let \( f(x) = \begin{cases x^3 \sin \left( \frac{1}{x} \right), & x \neq 0
0, & x = 0 \end{cases} \) . Then which one of the following is FALSE?}

• A. \( \lim_{x \to 0} \frac{f(x)}{x} = 0 \)
• B. \( \lim_{x \to 0} f(x) = 0 \)
• C. \( f(x) \) has infinitely many maxima and minima on the interval \( (0,1) \)
• D. \( f(x) \) is continuous at \( x = 0 \) but not differentiable at \( x = 0 \)

Hint:

For functions involving oscillatory terms like \( \sin \left( \frac{1}{x} \right) \), check for limits, continuity, and differentiability carefully by analyzing the behavior near critical points like \( x = 0 \).

Answer 1

The Correct Option is D

Step 1: Analyzing the behavior of \( f(x) \).
For \( x \neq 0 \), \( f(x) = x^3 \sin \left( \frac{1}{x} \right) \), and as \( x \to 0 \), the term \( \sin \left( \frac{1}{x} \right) \) oscillates between -1 and 1, so \( f(x) \to 0 \).

Step 2: Continuity and Differentiability.
Since \( \lim_{x \to 0} f(x) = 0 \), we know that \( f(x) \) is continuous at \( x = 0 \). However, due to the oscillatory behavior of \( \sin \left( \frac{1}{x} \right) \), \( f(x) \) is not differentiable at \( x = 0 \).

Step 3: Conclusion.
Option (C) is false because the function does not have infinitely many maxima and minima on the interval \( (0, 1) \). It oscillates, but not in the manner described.