Answer
See the explanation below.
Work Step by Step
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.
