real analysis - Prove that $lim_{x rightarrow 0} mathrm {sgn} sin (frac{1}{x})$ does not exist.

My progress:

Using the sequential criterion for limits, I constructed two sequences $(x_n), (y_n)$ with $\lim(x_n)=\lim(y_n)=0$, such that $\lim(f(x_n))\neq \lim(f(y_n))$, where $f(x)=\sin\frac 1 x$.

So, $\lim_{x \rightarrow 0} \sin (\frac{1}{x})$ does not exist.

I also showed separately in the same way that $\lim_{x \rightarrow 0} \mathrm{sgn} (x)$ does not exist.

I know that $$\lim_{x \rightarrow 0} f(x)=M ,\, \lim_{x \rightarrow 0} g(x)=N \Rightarrow \lim_{x \rightarrow 0} (fg)(x)=MN$$

Here, I have two functions $f,g$ which do not have limits at $x=0$. Does it follow from here that $\lim_{x \rightarrow 0} (fg)(x)$ also doesn't exist?

Is it any other way to approach the problem?

Answer

Hint do you know for which $\theta$s you have $\sin \theta = 1$? how about $\sin \theta = -1$? Can you choose a sequence of $(x_n)$ and $(y_n)$ converging to 0 such that $\sin (1/x_n) = 1$ always and $\sin (1/y_n) = -1$ always?