Let \(f(x) = \begin{cases} x^{2}\sin\left(\displaystyle\frac{1}{x}\right), & x \ne 0\\ 0, & x = 0\end{cases}\). Consider the following statements:
(A) f is continuous at \(x = 0\).
(B) \(f'(0) = 0\)
(C) \(f'(x) = 2x\sin\left(\displaystyle\frac{1}{x}\right) - \cos\left(\displaystyle\frac{1}{x}\right)\) for \(x \ne 0\).
(D) \(\displaystyle\lim_{x \to 0} f'(x) = 0\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (C) only
Since \(\left|\sin\frac{1}{x}\right| \le 1\), the function is squeezed:
\( \displaystyle |f(x)| \le x^{2} \to 0 = f(0) \)
At x = 0, by the definition of the derivative:
\( \displaystyle \frac{f(h) - f(0)}{h} = h\sin\frac{1}{h} \to 0 \)
For x ≠ 0, by the product and chain rules:
\( \displaystyle f'(x) = 2x\sin\frac{1}{x} + x^{2}\cos\frac{1}{x} \cdot \left(-\frac{1}{x^{2}}\right) \)
\( \displaystyle = 2x\sin\frac{1}{x} - \cos\frac{1}{x} \)
As x → 0, the first term tends to 0 but \(\cos\frac{1}{x}\) keeps taking the values 1 and −1 (at \(x = \frac{1}{2n\pi}\) and \(x = \frac{1}{(2n + 1)\pi}\)).
So f is continuous at 0 with f′(0) = 0, f′(x) has the form in (C) for x ≠ 0, and f′(x) has no limit as x → 0.
Checking the statements
(A) f is continuous at \(x = 0\). — Correct. \(|f(x)| \le x^{2}\), which tends to 0 = f(0).
(B) \(f'(0) = 0\) — Correct. The difference quotient \(h\sin\frac{1}{h}\) tends to 0.
(C) \(f'(x) = 2x\sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)\) for \(x \ne 0\). — Correct. The derivative of \(\sin\frac{1}{x}\) is \(-\frac{1}{x^{2}}\cos\frac{1}{x}\), and the \(x^{2}\) cancels.
(D) \(\lim_{x \to 0} f'(x) = 0\) — Incorrect. f′(x) is close to −1 at \(x = \frac{1}{2n\pi}\) and close to 1 at \(x = \frac{1}{(2n + 1)\pi}\), so the limit does not exist.