Answer
Graph
This would mean that $$_{t\rightarrow 3.5}^{\lim}\textrm{C(t)}=12.36$$
This would mean that $$_{t\rightarrow 3}^{\lim}\textrm{C(t)}=11.57$$
Therefore, $_{t\rightarrow 3}^{\lim}\textrm{C(t)}$ doesn't exist
Work Step by Step
The original function for this problem is $$C(t)=9.99-.79[[-(t-1)]]$$
If you want to know how to write it on Desmos, then you would have to write (for the greatest integer): $$y=9.99-.79((-(floor(x)-1)))$$
Anyways, to continue, the graph is down below, which is for Part A. Now, onward to Part B, which is the first of 2 plotting points with regards to the limitations. So, this would be figuring out the limitation for when "x is approaching 3.5", which means to find the limit of 3.5 (from both sides), which can be solved by the following:
$$\begin{matrix}
t & 3 & 3.1 & 3.2 & 3.3 & 3.4 & \mathbf{3.5} & 3.6 & 3.7 & 3.8 & 3.9 & 4\\
C & 11.57 & 12.36 & 12.36 & 12.36 & 12.36 & \mathbf{12.36} & 12.36 & 12.36 & 12.36 & 12.36 & 12.36
\end{matrix}$$
This would mean that $$_{t\rightarrow 3.5}^{\lim}\textrm{C(t)}=12.36$$
Now, onto the Part C, which is the same concept as Part B, but only from $2\:to\:4$ instead of from $3\:to\:4$ and finding the limit of 3 (from both sides), which can be solved by the following:
$$\begin{matrix}
t & 2 & 2.1 & 2.5 & 2.9 & \mathbf{3} & 3.1 & 3.5 & 3.9 & 4 \\
C & 10.78 & 11.57 & 11.57 & 11.57 & \mathbf{11.57} & 12.36 & 12.36 & 12.36 & 12.36
\end{matrix}$$
This would mean that $$_{t\rightarrow 3}^{\lim}\textrm{C(t)}=11.57$$
Since $$\begin{matrix}
_{t\rightarrow 3}^{\lim}\textrm{C(t)}=11.57 & and & _{t\rightarrow 3.5}^{\lim}\textrm{C(t)}=12.36
\end{matrix}$$ this means that $_{t\rightarrow 3}^{\lim}\textrm{C(t)}$ doesn't exist "thanks" to $C$'s values while it's approaching various values while $t$ is approaching $3$, which can be written as $t\rightarrow 3$, but from both sides.
