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