Limit without l'Hopital or Taylor series: $\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}$ find the limit without l'Hôpital and Taylor rule :
$$\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}=?$$
My Try :
$$\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}\\=\lim\limits_{x \to 0}\frac{x\cos x \sin x- \sin x\sin x}{x^3\sin x}=\\\lim\limits_{x \to 0}\frac{x\sin 2x- \sin^2 x}{2x^3\sin x}=$$?
what now ?

If you are allowed to use the well-known limit
$$\lim_{x \to 0}\frac{\sin x}{x}=1$$
then
$$\lim_{x \to 0}\frac{\tan x}{x}=1$$
follows easily and with a bit more effort (see here), you have:
$$\color{blue}{\lim_{x \to 0}\frac{\tan x-x}{x^3}=\frac{1}{3}\tag{1}}$$
Now for your limit and using $\color{blue}{(1)}$:
$$\lim_{x \to 0}\frac{x\cos x- \sin x}{x^3}=\lim_{x \to 0}\left(\cos x\frac{x- \tan x}{x^3}\right)=-\lim_{x \to 0} \cos x \color{blue}{\lim_{x \to 0}\frac{\tan x - x}{x^3}} = -\frac{1}{3}$$
