## Calculus (3rd Edition)

$g$ is left-continuous at $c_1=1$ $g(1)=3$
A function has a jump discontinuity if both one-sided limits exist, but they are not equal. $\displaystyle\lim_{x\rightarrow 1^{-}} g(x)=2$ $\displaystyle\lim_{x\rightarrow 1^{+}} g(x)=3$ As $\displaystyle\lim_{x\rightarrow 1^{-}} g(x)$ and $\displaystyle\lim_{x\rightarrow 1^{+}} g(x)$, but they are not equal, the function $g$ has a discontinuity at point $x=1$. We also have: $\displaystyle\lim_{x\rightarrow 1^{-}} g(x)=g(1)$ therefore $g$ is left-continuous at $c_1=1$. In order to make $g$ right-continuous in $x=c_1=1$, we must have: $\displaystyle\lim_{x\rightarrow 1^{+}} g(x)=g(1)$ $3=g(1)$ $g(c_1)=g(1)=3$