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 ?

• No l'Hôpital or series, but what are you allowed to use? Certain standard/special limits? – StackTD Mar 16 '17 at 14:55
• Take out $\cos x$ and use math.stackexchange.com/questions/387333/… – lab bhattacharjee Mar 16 '17 at 15:05
• These problems and the link referenced giving special limits are all built from using Taylor expansions, then we are asked to solve without it. Seems like ostriches who put their heads in the sand to not see the Taylor expansion... – zwim Mar 16 '17 at 16:03

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

$$\frac89L=-\frac8{27}.$$