Find limit of trigonometric function $$\lim_{x\rightarrow0} \frac{\tan^3(3x)-\sin^3(3x)}{x^5}$$
I think it should be decomposed with $$\lim_{x\to0} \frac{\sin x}{x}=1$$ but I'm always getting indefinity $\frac{0}{0}$.  

$$\begin{align}&\lim_{x \to 0} \dfrac{\tan^3 3x - \sin^3 3x}{x^5} &\\= &\lim_{x \to 0} (\sin^3 3x)\dfrac{1 - \cos^3 3x}{x^5}&\\=\,  &27 \lim_{x \to 0} \dfrac{(1 - \cos 3x)}{x^2}(1 + \cos 3x + \cos^2 3x) &\\=\, &81 \lim_{x\to 0} \dfrac{(1 - \cos 3x)}{x^2}\end{align}$$
Let $x = 2y$ 
$$\lim_{y \to 0} \dfrac{(1 - \cos 6y)}{(2y)^2} = \lim_{y \to 0} \dfrac{(2\sin^2 3y)}{(2y)^2} = \lim_{y \to 0} \dfrac{(\sin^2 3y)}{2(y)^2} = \dfrac92$$
So, $$\lim_{x \to 0} \dfrac{\tan^3 3x - \sin^3 3x}{x^5} = \dfrac{729}2$$.
