calculus - Determine $lim_{x to 0}{frac{x-sin{x}}{x^3}}=frac{1}{6}$, without L'Hospital or Taylor

How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$

without using L'Hospital or Taylor series?

thanks :)

Answer

Let $L = \lim_{x \to 0} \dfrac{x - \sin(x)}{x^3}$. We then have
\begin{align}
L & = \underbrace{\lim_{y \to 0} \dfrac{3y - \sin(3y)}{27y^3} = \lim_{y \to 0} \dfrac{3y - 3\sin(y) + 4 \sin^3(y)}{27y^3}}_{\sin(3y) = 3 \sin(y) - 4 \sin^3(y)}\\
& = \lim_{y \to 0} \dfrac{3y - 3\sin(y)}{27 y^3} + \dfrac4{27} \lim_{y \to 0} \dfrac{\sin^3(y)}{y^3} = \dfrac{3}{27} L + \dfrac4{27}
\end{align}
This gives us $24L = 4 \implies L = \dfrac16$