Evaluating $\lim\limits_{x \to 0} \frac{x}{\sin(1/x^2)}$ I'm trying tp calculate the limit: 
$$\lim\limits_{x \to 0} \frac{x}{\sin(1/x^2)}$$
I'm pretty sure it doesn't exist, but I can't find a formal proof. 
 A: Let consider for $n\to+\infty$
$$\frac1{x_n^2}=\frac1n+2n\pi\to+\infty\implies x_n=\frac{1}{\sqrt{\frac1n+2n\pi}}\to0\quad \sin\left(\frac1{x_n^2}\right) =\frac1n+o\left(\frac1n\right)$$
$$\frac1{y_n^2}=-\frac1n+2n\pi\to+\infty\implies y_n=\frac{1}{\sqrt{-\frac1n+2n\pi}}\to0\quad \sin\left(\frac1{y_n^2}\right)=-\frac1n+o\left(\frac1n\right)$$
thus
$$\frac{x_n}{ \sin\left(\frac1{x_n^2}\right)  }=\frac{1}{\sqrt{\frac1n+2n\pi}\left(\frac1n+o\left(\frac1n\right)\right)}=\frac{n\sqrt n}{\sqrt{1+2n^2\pi}\left(1+o\left(1\right)\right)}\to+\infty$$
$$\frac{y_n}{ \sin\left(\frac1{y_n^2}\right)}=\frac{1}{\sqrt{-\frac1n+2n\pi}\left(-\frac1n+o\left(\frac1n\right)\right)}=\frac{-n\sqrt n}{\sqrt{-1+2n^2\pi}\left(1+o\left(1\right)\right)}\to-\infty$$
then the limit does not exist.
A: Hint:$$\lim_{x\to0}\frac x{\sin\left(\frac1{x^2}\right)}=\lim_{x\to\infty}\frac{\frac1x}{\sin(x^2)}=\lim_{x\to\infty}\frac1{x\sin(x^2)}.$$
A: There are vertical asymptotes of $\frac{x}{\sin (1/x^2)}$ at $x = \pm \sqrt{1/k\pi}$ for every integer $k$. Any interval around zero contains an infinite number of such asymptotes, so the limit at $x \to 0$ doesn't exist.
