How to solve this limit question without using L'Hopital's rule? 
Find the limit $$\lim_{x\rightarrow 0} \frac{\sin x-x}{\tan^3 x}$$

I found the limit which is $-\frac{1}{6}$ by using L'Hopital Rule. Is there another way to solve it without using the rule? Thanks in advance.

$$\lim_{x\rightarrow 0} \frac{\sin x-x}{\tan^3 x}= \lim_{x\rightarrow 0} \frac{\sin x-x}{x^3} \underbrace{\left(\frac {x^3}{\tan^3 x}\right)}_{=1}$$
Now, let $$\mathrm L= \lim_{x\rightarrow 0} \frac{\sin x-x}{x^3}$$
Let $x=3y$, Since $x\to 0 \implies y \to 0$.
$$ \mathrm L= \lim_{y\rightarrow 0} \frac{\sin (3y)-3y}{(3y)^3}$$
\begin{align}
\mathrm L &= \lim_{y\rightarrow 0} \frac{3\sin y-4\sin^3 y-3y}{27y^3}\\
&= \lim_{y\rightarrow 0}\frac{3}{27} \left(\frac{\sin y-y}{y^3}\right) -\frac{4}{27}  \lim_{y\rightarrow 0}\left(\frac{\sin y}{y}\right)^3\\
&=\frac{3}{27} \cdot \mathrm L-\frac{4}{27}
\end{align}
$$\implies \mathrm L=-\frac{1}{6}$$
