Checking the three-part continuity test at x = 0

First check condition 1: is $p(0)$ defined? By the definition of the piecewise function, $p(0) = 0$, so this condition is satisfied. Next check condition 2: does $\lim_{x\to 0} p(x)$ exist? Since $p(x) = \sin(x)/x$ for all $x \neq 0$ near $0$, and we are given that $\lim_{x\to 0} \sin(x)/x = 1$, the limit exists and equals $1$.

Finally check condition 3: does $\lim_{x\to 0} p(x) = p(0)$? We found $\lim_{x\to 0}p(x) = 1$ but $p(0) = 0$, so these values are not equal. Therefore condition 3 fails, and $p$ is not continuous at $x=0$. Since the limit exists but does not match the function value, this is a removable discontinuity — the graph has a hole at $(0,0)$ that could be patched by redefining $p(0) = 1$.