Is it possible to solve this limit without Hopital / Taylor / derivatives: $\lim\limits_{x \to 0} \frac{x-\sin(x)}{x^3} = \frac{1}{6}$?

Question

It's simple to prove with Hopital that

$$ \lim_{x \to 0} \frac{x-\sin(x)}{x^3} = \frac{1}{6}$$

Is it possible to solve this limit without Hopital or Taylor (without derivatives)?

Answer 1

Let us assume $$\displaystyle \lim_{x\rightarrow 0}\frac{x-\sin x}{x^3} = L$$ (A finite quantity).

Now replace $x\rightarrow 3y$, then we get $$\displaystyle \lim_{y\rightarrow 0}\frac{3y-\sin 3y}{27y^3} = L$$

Now, using the formula $$\sin 3y = 3\sin y-4\sin^3 y$$

we get $$\displaystyle \lim_{y\rightarrow 0}\frac{3y-3\sin y+4\sin^3 y}{27y^3} = L$$

So $$\displaystyle \frac{1}{9}\lim_{y\rightarrow 0}\frac{y-\sin y}{y^3}+\frac{4}{27}\displaystyle \lim_{y\rightarrow 0}\left(\frac{\sin y}{y}\right)^3=L$$

So $$\frac{1}{9}L+\frac{4}{27} = L\Rightarrow \frac{8}{9}L = \frac{4}{27}\Rightarrow L=\frac{1}{6}$$