If $f(x)=\begin{cases}x\sin(\frac{1}{x}), & x\ne0 \\ 0, & x=0\end{cases}$, then $f(x)$ is

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

Step 1: Continuity at \(x=0\)

\[ \lim_{x \to 0} x \sin\!\left(\tfrac{1}{x}\right). \] Since \(|\sin(1/x)| \leq 1\), \[ -|x| \;\leq\; x \sin\!\left(\tfrac{1}{x}\right) \;\leq\; |x|. \] By the squeeze theorem, \[ \lim_{x \to 0} f(x) = 0 = f(0). \] ✅ Thus, \(f(x)\) is continuous everywhere.