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 $\color{blue}{(1)}$: $$\lim_{x \to 0}\frac{x\cos x- \sin x}{x^3}=\lim_{x \to 0}\left(\cos x\frac{x- \tan x}{x^3}\right)=-\lim_{x \to 0} \cos x \color{blue}{\lim_{x \to 0}\frac{\tan x - x}{x^3}} = -\frac{1}{3}$$
Write
$$\frac{x\cos x-\sin x}{x^3}=\frac{\cos x-1}{x^2}+\frac{x-\sin x}{x^3} $$
The first fraction goes to $-\frac{1}{2}$ (it follows from $\frac{\sin x}{x} \to 1$, no de l'Hopital or Taylor needed), while the second goes to $\frac{1}{6}$ (see Solving $\lim\limits_{x\to0} \frac{x - \sin(x)}{x^2}$ without L'Hospital's Rule. ). Overall
$$\lim_{x \to 0} \frac{x\cos x-\sin x}{x^3} = -\frac{1}{3}. $$
By a scaling of the variable, $$L:=\lim_{x\to 0}\frac{x\cos x-\sin x}{x^3}=\lim_{x\to 0}\frac{3x\cos3x-\sin3x}{27x^3}.$$
Then by the triple angle formulas,
$$3x\cos3x-\sin3x=3x\cos x(1-4\sin^2x)-3\sin x+4\sin^3x\\ =(3-4\sin^2x)(x\cos x-\sin x)-8x\cos x\sin^2x,$$ so that
$$L=\lim_{x\to0}(3-4\sin^2x)\cdot\frac L{27}-\lim_{x\to0}\frac{8x\cos x}{27x}\cdot\lim_{x\to0}\frac{\sin^2x}{x^2}.$$
Using the $\text{sinc}$ limit, we can conclude
$$\frac89L=-\frac8{27}.$$
