If \( f(x)=\begin{cases}x~sin(\frac{1}{x}),&x\ne0\\0,&x=0\end{cases} \) , then \( f(x) \) is $$ \text{(a) continuous for all \( x\in\mathbb{R} \) } $$ $$ \text{ (b) continuous at 0, 1 only } $$ $$ \text{(c) not continuous at 1 } $$ $$ \text{(d) not continuous at 0} $$
Step-by-step Solution:
Given:
\[ f(x)= \begin{cases} x\sin\!\left(\dfrac{1}{x}\right), & x\neq 0,\\[6pt] 0, & x=0. \end{cases} \]
For any \(x\neq 0\), the function \(x\mapsto x\) and \(x\mapsto\sin(1/x)\) are both continuous on their domains, and the product of continuous functions is continuous. Therefore \(f\) is continuous at every point \(x\neq 0\).
We check \(\displaystyle\lim_{x\to 0} x\sin\!\left(\frac{1}{x}\right)\). Using the squeeze (sandwich) theorem:
\[ -|x| \le x\sin\!\left(\frac{1}{x}\right) \le |x| \] and \(\lim_{x\to 0} -|x| = 0 = \lim_{x\to 0} |x|\).
Hence \[ \lim_{x\to 0} x\sin\!\left(\frac{1}{x}\right) = 0. \] Since \(f(0)=0\), the limit equals the function value, so \(f\) is continuous at \(x=0\).
\(f\) is continuous for every real \(x\). Therefore the correct choice is:
(a) continuous for all \(x\in\mathbb{R}\)