## Thomas' Calculus 13th Edition

If we want $g$ to be discontinuous at $-1$, make $g(-1)$ undefined. The easiest way is to place $(x+1)$ in the denominator. Now, we do not want a numerator that contains $(x+1)$ as a factor, because cancelling leads to the existence of a limit at x=-1, in which case the discontinuity would be removable. We select $g(x)=\displaystyle \frac{1}{x+1}$ Approaching $x=-1$ from the left, the denominator becomes a negative number approaching zero. So, $\displaystyle \lim_{x\rightarrow-1^{-}}g(x)=-\infty$ Approaching $x=-1$ from the right, the denominator becomes a positive number approaching zero. So, $\displaystyle \lim_{x\rightarrow-1^{+}}g(x)=+\infty$ Thus, a limit at $x=-1$ does not exist, and the discontinuity is not removable.