Answer
$$
g(x)=\left\{\begin{array}{ll}{0} & {\text { if } x=-2} \\ {\frac{1}{2}x^{2}-3} & {\text { if } x\neq -2}\end{array}\right.
$$
We obtain that:
$$
\lim _{x \rightarrow-2^{-}} g(x)=-1
$$
and
$$
\lim _{x \rightarrow-2^{+}} g(x)=-1
$$
Therefore
$$
\lim _{x \rightarrow-2} g(x)=-1
$$
Work Step by Step
$$
g(x)=\left\{\begin{array}{ll}{0} & {\text { if } x=-2} \\ {\frac{1}{2}x^{2}-3} & {\text { if } x\neq -2}\end{array}\right.
$$
A function $g(x)$ defined by two cases is called a piecewise function. The domain of $g(x)$ is all real numbers.
To determine the limit as $x$ approaches -2, we are concerned
only with the values of $g(x)$ when $x$ is close but not equal to -2
Once again,
$$
\lim _{x \rightarrow-2^{-}} g(x)=-1
$$
and
$$
\lim _{x \rightarrow-2^{+}} g(x)=-1
$$
Therefore
$$
\lim _{x \rightarrow-2} g(x)=-1
$$
