Answer
$$\lim_{x\rightarrow 0}\frac{sin(nx)}{x} = n$$
Work Step by Step
First off, based on the graphing utility, the following are the general limitations, despite that when you actually have to solve them it would be thanks to L'Hospital's Rule:
$$\begin{matrix}
\lim_{x\rightarrow 0}\frac{sin(x)}{x} = 1\\\\
\lim_{x\rightarrow 0}\frac{sin(2x)}{x} = 2\\\\
\lim_{x\rightarrow 0}\frac{sin(3x)}{x} = 3\\\\
etc.
\end{matrix}$$
Here is an example for how L'Hospital's Rule would be implied for figuring these out despite this was only based on the graphing utility:
$$\begin{matrix}
\lim_{x\to 0} \left(\frac{\sin \left(x\right)}{x}\right) & =\lim _{x\to 0}\left(\frac{f'(\sin \left(x\right))}{f'(x)}\right) \\
&=\lim _{x\to 0}\left(\frac{\cos \left(x\right)}{1}\right) \\
&=\lim _{x\to 0}\left(\cos \left(x\right)\right) \\
&=cos(0) \\
&=1
\end{matrix}$$
So, since the coefficient of term inside of $sin(x)$ determines the solution, then that would be the following: $$\lim_{x\rightarrow 0}\frac{sin(nx)}{x} = n$$
