Question

If \(\; f(x)=\begin{cases} x\sin\!\left(\tfrac{1}{x}\right), & x\neq 0 \\[6pt] 0, & x=0 \end{cases}, \; \text{then } f(x) \text{ is}\)
 

• A. continuous for all $x\in\mathbb{R}$
• B. continuous at $0$ only
• C. not continuous at $1$
• D. not continuous at $0$

Hint:

For oscillatory terms like $\sin(1/x)$ near $0$, pair them with a vanishing factor ($x$) and apply the Squeeze Theorem.

Answer 1

The Correct Option is A

Step 1: Continuity for $x\neq 0$.
For $x\neq 0$, $f(x)=x\sin(1/x)$ is a product of continuous functions, hence continuous.

Step 2: Limit at $x=0$.
Since $-1\le \sin(1/x)\le 1$, multiplying by $x$ gives $-|x|\le x\sin(1/x)\le |x|$. As $x\to 0$, both bounds $\to 0$; by the Squeeze Theorem, $\lim\limits_{x\to 0}x\sin(1/x)=0$.

Step 3: Compare with $f(0)$.
$f(0)=0$, which equals the limit, so $f$ is continuous at $0$.

Step 4: Conclusion.
$f$ is continuous for all $x\in\mathbb{R}$.