Classifying the discontinuity of v(x) at x = 0

First check the three conditions for continuity at $x=0$. We are told $v(0) = 2$, so the function value is defined. Next, $\lim_{x\to 0} v(x) = \lim_{x\to 0} \frac{\sin(x)}{x} = 1$, since for $x \neq 0$ near $0$ the function agrees with $\sin(x)/x$, so the limit exists and equals $1$.

Now compare the limit to the function value: $\lim_{x\to 0} v(x) = 1$, but $v(0) = 2$, so these are not equal. Since $f(a)$ is defined and the limit exists but they disagree, the third condition of continuity fails, and this is a removable discontinuity — the graph has a hole at $(0,1)$ that could be patched by redefining $v(0) = 1$.