For the given function $f$, it is indeed true that $\underset{x\to 0}{lim}f\left(x\right)\=1\=f\left(0\right)$, so $f$ is indeed continuous at $x\=1$. This result is intuitively consistent with Figure 1.6.1(a) if the hole at $\left(0\,1\right)$ is filled in with the function value $f\left(0\right)\=1.$

The function $g\left(x\right)\=\sin \left(x\right)\/x$ has a removable discontinuity at $x\=0$ because it can be extended to a function $f$ , whose domain includes $x\=0$, in such a way as to make $f$ continuous at $x\=0$. If a function has a removable discontinuity, its graph can display the discontinuity via the Option "showremovable" in the plot command.