Prove that $\lim_{x \to 0} \frac{x - \sin x}{x^3} = \frac16$ using notable limits

Question

Can we prove that $\lim_{x \to 0} \frac{x - \sin x}{x^3} = \frac 16$ using just algebra, trigonometric theorems and notable limits $\left( \mathrm{i.e. }\quad \frac{\sin x}{x} \to 1 \quad \mathrm{and} \quad \frac{1 - \cos x}{x^2} \to \frac 12 \right) $ and no l'Hospital rule, no series expansion, no Taylor series?

Answer 1

$\sin x=\sin (x/3)\cos(2x/3)+\cos(x/3)\sin(2x/3)=\sin (x/3)(1-2\sin^2(2x/3))+2\cos^2(x/3)\sin(x/3)=\sin (x/3)-2\sin^3 (x/3)+2\sin (x/3)-2\sin^3 (x/3)=3\sin (x/3)-4\sin^3 (x/3)$

$\frac{x-\sin x}{x^3}=\frac{x-3\sin (x/3)-4\sin^3 (x/3)}{x^3}=\frac{x-3\sin (x/3)}{x^3}-\frac{4\sin^3 (x/3)}{x^3}=1/9\frac{x/3-\sin (x/3)}{(x/3)^3}-4/27 \frac{\sin^3 (x/3)}{(x/3)^3}$

$m=\lim_{x\rightarrow 0} \frac{x-\sin x}{x^3}= 1/9\lim_{x\rightarrow 0}\frac{x/3-\sin (x/3)}{(x/3)^3}-4/27\lim_{x\rightarrow 0} \frac{\sin^3 (x/3)}{(x/3)^3}=1/9M-4/27 \Rightarrow 8/9M=4/27\Rightarrow M=1/6.$