## Calculus (3rd Edition)

$g$ is discontinuous at $x=1$; it is left-continuous there. $g$ is discontinuous at $x=3 ;$ it is neither left-continuous, nor right-continuous there. $g$ is discontinuous at $x=5$; it is right-continuous there. We have a removable discontinuity at $x=3$.
From the given figure, we have $$\lim_{x\to 1^-}g(x)=g(1)=2$$ and\begin{align*} \lim_{x\to 3^-}g(x)&=4.5 \\ \lim_{x\to 3^+}g(x)&=4.5\\ g(3)&=2.5 \end{align*} and \begin{align*} \lim_{x\to 5^-}g(x)&=3.5 \\ \lim_{x\to 5^+}g(x)&=1\\ g(5)&=1 \end{align*} Thus, we see that: $g$ is discontinuous at $x=1$; it is left-continuous there. $g$ is discontinuous at $x=3 ;$ it is neither left-continuous, nor right-continuous there. $g$ is discontinuous at $x=5$; it is right-continuous there. We have a removable discontinuity at $x=3$ because if we had $g(3)=4.5$ instead of $g(3)=2.5$, the function would be continuous there.