Answer
(a) does not exist
(b) $1$
(c) $4$
Work Step by Step
Given $$g(t)=\left\{\begin{array}{ll}{t-2,} & {t<0} \\ {t^{2},} & {0 \leq t \leq 2} \\ {2 t,} & {t>2}\end{array}\right.$$
Then
(a) Since $\lim_{t\to 0^-}g(t)=\lim_{t\to 0^-}(t-2)=-2 $
and
$\lim_{t\to 0^+}g(t)=\lim_{t\to 0^+}(t^2)=0 $, then
$\lim_{t\to 0}g(t)$ does not exist
(b) Since
\begin{align*}
\lim_{t\to 1}g(t) &=\lim_{t\to 1}t^2\\
&=1
\end{align*}
(c) Since $\lim_{t\to 2^-}g(t)=\lim_{t\to 2^-}(t^2)=4 $
and
$\lim_{t\to 2^+}g(t)=\lim_{t\to 2^+}(2t)=4 $, then
$\lim_{t\to 2}g(t)=4$
