Answer
$$
f(x)=\left\{\begin{array}{ll}{x^{3}+2} & {\text { if } x \neq-1} \\ {5} & {\text { if } x=-1}\end{array}\right.
$$
We obtain that:
$$
\lim _{x \rightarrow-1^{-}} f(x)=1
$$
and
$$
\lim _{x \rightarrow-1^{+}} f(x)=1
$$
Therefore
$$
\lim _{x \rightarrow-1} f(x)=1
$$
Work Step by Step
$$
\text { Let } f(x)=\left\{\begin{array}{ll}{x^{3}+2} & {\text { if } x \neq-1} \\ {5} & {\text { if } x=-1}\end{array}\right.
$$
A function $f(x)$ defined by two cases is called a piecewise function. The domain of $f(x)$ is all real numbers.
To determine the limit as $x$ approaches -1, we are concerned
only with the values of $f(x)$ when $x$ is close but not equal to -1
Once again,
$$
\lim _{x \rightarrow-1^{-}} f(x)=1
$$
and
$$
\lim _{x \rightarrow-1^{+}} f(x)=1
$$
Therefore
$$
\lim _{x \rightarrow-1} f(x)=1
$$