Answer
$$
f(x)=\left\{\begin{array}{ll}{x-1} & {\text { if } x<3} \\ { 2} & {\text { if } 3 \leq x \leq 5} \\ {x+3} & {\text { if } x>5}\end{array}\right.
$$
We obtain that:
(a)
$$
\lim _{x \rightarrow 3} f(x) =2
$$
(b)
$$
\lim _{x \rightarrow 5} f(x)
$$
does not exist since
$$
\lim _{x \rightarrow 5^{+}} f(x) \neq \lim _{x \rightarrow 5^{-}} f(x)
$$
Work Step by Step
$$
f(x)=\left\{\begin{array}{ll}{x-1} & {\text { if } x<3} \\ { 2} & {\text { if } 3 \leq x \leq 5} \\ {x+3} & {\text { if } x>5}\end{array}\right.
$$
A function $f(x)$ defined by two or more cases is called a piecewise function. The domain of $f(x)$ is all real numbers.
To determine the limit as $x$ approaches 0, 3, we are concerned
only with the values of $f(x)$ when $x$ is close but not equal to 3 , 5
Once again,
(a)
$$
\lim _{x \rightarrow 3^{-}} f(x)=2
$$
and
$$
\lim _{x \rightarrow 3^{+}} f(x)=2
$$
Therefore
$$
\lim _{x \rightarrow 3} f(x) =2
$$
(b)
$$
\lim _{x \rightarrow 5^{-}} f(x)=2
$$
and
$$
\lim _{x \rightarrow 5^{+}} f(x)=8
$$
Therefore
$$
\lim _{x \rightarrow 5} f(x)
$$
does not exist since
$$
\lim _{x \rightarrow 5^{+}} f(x) \neq \lim _{x \rightarrow 5^{-}} f(x)
$$
