Solution (with MathJax)

Given:

\[ f(x)= \begin{cases} x\sin\!\left(\dfrac{1}{x}\right), & x\neq 0,\\[6pt] 0, & x=0. \end{cases} \]

Continuity for \(x\neq 0\)

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 \(x\neq 0\).

Continuity at \(x=0\)

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 \(x=0\).

Conclusion

\(f\) is continuous for every real \(x\). Therefore the correct choice is:

(a) continuous for all \(x\in\mathbb{R}\)