How to show that $\lim\limits_{x \to 0} \frac{3x-\sin 3x}{x^3}=9/2$? $\lim\limits_{x \to 0} \frac{3x-\sin 3x}{x^3}$

I need to prove that this limit equals to $\frac{9}{2}$.
Can someone give me a step by step solution?

EDIT: I am sorry. The $x$ goes to $0$, not $1$.
If you allow Taylor expansions, recall that

$$\sin(x)=x-\frac16x^3+\mathcal O(x^5)$$

Thus,

$$\sin(3x)=3x-\color{red}{\frac92}x^3+\mathcal O(x^5)$$

Thus,

$$\begin{align}\frac{3x-\sin(3x)}{x^3}&amp;=\frac{\frac92x^3+\mathcal O(x^5)}{x^3}\\&amp;=\frac92+\mathcal O(x^2)\\&amp;\to\frac92\end{align}$$
first :

theorem :
let $f(0)=0 , f'(0)$ have existed then :
$$\lim_{x\to 0} \frac{f(x)-\sin(f(x))}{x^3}= \frac{1}{6}(f'(0))^3$$

now :
$$\lim_{x\to 0} \frac{3x-\sin(3x)}{x^3}= \frac{1}{6}(3)^3=\frac{9}{2}$$