Answer
(a) 0
(b) 0
(c) 0
(d) $f$ is continuous at $x=2$
Work Step by Step
Given $$
f(x)=\left\{\begin{array}{ll}{2^{x}-4} & {\text { when } x<2} \\ {x^{2}-4} & {\text { when } x \geq 2}\end{array}\right.
$$
(a) Since
\begin{align*}
\lim _{x \rightarrow2^{-}} f(x)&=\lim _{x \rightarrow2^{-}}(2^x-4)\\
&=\lim _{x \rightarrow2^{-}}(4-4)\\
&=0
\end{align*}
(b) Since
\begin{align*}
\lim _{x \rightarrow2^{+}} f(x)&=\lim _{x \rightarrow2^{+}} (x^2-4)\\
&=\lim _{x \rightarrow2^{+}} (4-4)\\
&=0
\end{align*}
(c) $f(2)= (2^2-4)=0$
(d) Since $$\lim _{x \rightarrow2^{+}} f(x)=\lim _{x \rightarrow2^{-}} f(x)=f(2)=0$$
Then $f$ is continuous at $x=2$
