<s>[INST] 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 ?
 [/INST] 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
$$\frac89L=-\frac8{27}.$$
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