Verifying continuity at x = 0 using the three-part test

First check that $b(0)$ is defined: by the piecewise definition, $b(0) = 0$, so condition 1 holds. Next, find $\lim_{x\to 0} b(x)$. Since $-1 \le \sin(1/x) \le 1$ for $x \neq 0$, multiplying by $|x|$ gives $-|x| \le x\sin(1/x) \le |x|$. As $x \to 0$, both $-|x|$ and $|x|$ approach $0$, so by the Squeeze Theorem $\lim_{x\to 0} x\sin(1/x) = 0$. Thus condition 2 holds: the limit exists and equals $0$.

Finally, compare the limit to the function value: $\lim_{x\to 0} b(x) = 0$ and $b(0) = 0$, so these are equal and condition 3 holds. Since all three conditions of the continuity definition are satisfied, $b(x)$ is continuous at $x = 0$.