Recall that we know that \[\lim_{x\to0} \frac{\sin(x)}{x} = 1.\] The key issue is that the thing (\(x\), in this case) inside the sine function and the thing in the denominator are the same. Any time that happens, the limit will be one.

Now in problem 3, we've got \[\lim_{\theta\to0} \frac{\sin(3\theta)}{\theta}.\] The thing inside the sine function and the thing in the denominator are not the same. However, we can multiply top and bottom by 3 to make them the same. We can then isolate the \(\sin(3\theta)\) upstairs with the \(3\theta\) downstairs, let that go to one, and see what's left.

Here's what that looks like: \[ \begin{aligned} \frac{\sin(3\theta)}{\theta} &= \frac{3\sin(3\theta)}{3\theta} \\ &= 3\times{\frac{\sin(3\theta)}{3\theta}} \\ &\to 3\times {1} = 3. \end{aligned} \]