Limit without Taylor expansion

Question

$\lim_{x\to0}\frac{\sin x - x}{x^3}$

I know it can be easily done by using Taylor expansion of sine function and L'Hopital. However, can we come up with a way to solve the limits using properties.

Answer 1

Let $x = 3y$

$\sin(x) = \sin(3y) = 3\sin(y) - 4\sin^3(y)$

$$L = \lim_{y\to0}\frac{3\sin(y) - 4\sin^3(y) - 3y}{(3y)^3} = \lim_{y\to0}\frac{\sin(y) + y}{9y^3}+\lim_{y\to0}\frac{- 4\sin^3(y) }{27y^3}$$

$$L = \frac{L}{9} - \frac{4}{27} $$

$$\frac{8L}{9} = - \frac{4}{27} $$

$$L = \frac{-1}{6}$$