How to compute the limit,$\lim_{x\rightarrow 0}\frac{3x^2-3x\sin x}{x^2+x\cos1/x}$ How to compute the limit,
$$\lim_{x\rightarrow 0}\frac{3x^2-3x\sin x}{x^2+x\cos\frac{1}{x}}$$

As noticed the limit doesn't exist indeed we can consider the sequence  as $x_n \to 0$ such that
$$\cos\frac{1}{x_n}=2x_n  \implies \frac{3x^2-3x\sin x}{x^2+x\cos\frac{1}{x}}=\frac{3x_n^2-3x_n\sin x_n}{3x_n^2}=1-\frac{\sin x_n}{x_n} \to 1-1=0$$
and  the sequence $x_n \to 0$ such that
$$\cos\frac{1}{x_n}=-x_n+x_n^3  \implies \frac{3x^2-3x\sin x}{x^2+x\cos\frac{1}{x}}=3\frac{x_n^2-x_n\sin x_n}{x_n^4}=3\frac{x_n-\sin x_n}{x_n^3}\to \frac12$$
indeed as $t \to 0$ we have that $\frac{t-\sin t}{t^3} \to \frac16$ wich can be proved by l'Hopital, Taylor or by the method shown here: Are all limits solvable without L'Hôpital Rule or Series Expansion.
For the issue already discussed here in detail by Paramanand Singh about the not satisfactory way to show that the limit doesn't exist because not defined at "infinitely many points", for a useful discussion refer also to the related


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*What is $\lim_{x \to 0}\frac{\sin(\frac 1x)}{\sin (\frac 1 x)}$ ? Does it exist?
