calculus - Calculate the limit : $lim_{x to 0}frac{x-sin{x}}{x^3}$ WITHOUT using L'Hopital's rule

I was given a task to find

$$\lim_{x\to0}\frac{x-\sin{x}}{x^3}$$ at my school today. I thought it was an easy problem and started differentiating denominator and numerator to calculate the limit but the teacher then said we aren't allowed to use L'Hopital's rule, but to "play around" with known limits and limit definition. I got stuck here since I can't really think of a way to do this, and according to my teacher, there are at least 4 ways. A subtle hint would be enough. Thanks.

Answer

$$L=\lim_{x\to0}\frac{x-\sin x}{x^3}\\ =\lim_{x\to0}\frac{2x-\sin2x}{8x^3}\\ 4L-L = \lim_{x\to0}\frac{x-\frac12\sin2x-x+\sin x}{x^3}$$
which simplifies to the product of three expressions of the form $\frac{\sin y}{y}$