Answer
The hole is located at $$(0,3)$$
Work Step by Step
So, the general function is $$f(x)=\frac{sin(3t)}{t}$$
With the graph, which can be seen down below, you can see and assume that there would be a hole at exactly $0,3$ since when you plug in $0$ into the original function, $0$ will be in the denominator by itself and anything divided by $0$ is undefined.
So, we can create a table of the points to between $0$, from both the left and the right of $0$, to see what is happening on both sides, which can be written as the following:
$$\begin{matrix}
t & &f(t) \\
-0.1 & \frac{sin(3(-0.1))}{(-0.1)} &2.96 \\\\
-0.01 & \frac{sin(3(-0.01))}{(-0.01)} &2.9996 \\\\
-0.001 & \frac{sin(3(-0.001))}{(-0.001)} &3 \\\\
0 & \frac{sin(3(0))}{(0)} &DNE \\\\
0.001 & \frac{sin(3(0.001))}{(0.001)} &3 \\\\
0.01 & \frac{sin(3(0.01))}{(0.01)} &2.9996 \\\\
0.1 & \frac{sin(3(0.1))}{(0.1)} & 2.96
\end{matrix}$$
Based on this table, the limit is appearing to be at $3$ when $t$ is approaching $0$ despite it does not exist at exactly $0$.
So, this graph does have an undefined point, or hole, when $(0,C(t))$ so to (properly and not assume) figure out the $y$-value you would have to find the limit of the function as $t$ is approaching $0$, which can be $t \rightarrow 0$, which can be written as the following:
$$\begin{matrix}
\lim _{t\to 0}\left(\frac{\sin \left(3t\right)}{t}\right)&=\lim _{t\to 0}\left(\frac{f'(\sin \left(3t\right))}{f'(t)}\right)\\
&=\lim _{t\to 0}\left(\frac{f'(3t)\cos \left(3t\right)}{1}\right)\\
&=\lim _{t\to 0}\left(\frac{3\cos \left(3t\right)}{1}\right)\\
&=\lim _{t\to 0}3\cos \left(3t\right)\\
&=3\cos \left(3(0)\right)\\
&=3\cos \left(0\right)\\
&=3\left(1\right)\\
C(t)&=\mathbf{3}\\
\end{matrix}$$
Since $C(t)=3$, then this means that the hole is exactly at $$(0,3)$$
