Limits practice question 18

The top, $1 - \cos^{2}(3x)$, is not yet in a form the standard limit $\tfrac{\sin u}{u} \to 1$ can use. Which identity changes that?

Use $\sin^{2}\theta + \cos^{2}\theta = 1$ to rewrite the numerator as $\sin^{2}(3x)$.

Write $\dfrac{\sin^{2}(3x)}{4x^{2}} = \dfrac{1}{4}\left(\dfrac{\sin(3x)}{x}\right)^{2}$, and make the denominator match the argument: $\dfrac{\sin(3x)}{x} = 3\cdot\dfrac{\sin(3x)}{3x}$.