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 ?

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
$$\lim_{x \to 0} \frac{x\cos x-\sin x}{x^3} = -\frac{1}{3}. $$
