Find an example (Limits) I was asked about this problem

Find a function $f:\mathbb{R}\rightarrow\mathbb{R}$ such as the limit as $x\to0$ of $f$ doesn't exist, and $\lim\limits_{x\to0}f(x)\cdot f(2x) = 0$

I think is really interesting and I would like to know what will be the idea to came out with a solution.
 A: Consider
$$f(x)=\begin{cases}1&x\in[2^{n},2^{n+1}),\text{ for even $n$}\in\mathbb{Z}\\-1&x\in[2^{n},2^{n+1}),\text{ for odd $n$}\in\mathbb{Z}\\c&x=0\\f(-x)&x<0\end{cases}$$
where $c$ is any constant.  In any neighborhood of $0$, the function takes values $1$ and $-1$, so the limit does not exist, but $f(x)\cdot f(2x)=-1$ for all $x\ne 0$
A: Consider functions with jump discontinuities across the origin. For example, if you take $f(x)=|x|/x$
if you take $f(x)*f(2x) = |2x^2|/x^2 = 2$ (with a hole discontinuity at $x=0$)
Thus $\lim\limits_{x\rightarrow 0} f(x)f(2x) = 2$ but $\lim\limits_{x\rightarrow 0} f(x)$ does not exist
A: In order to make our function everywhere defined, we define it to be $0$ for $x\le 0$ and for $x\ge 1$.  Now we do the interesting part, $0\lt x\lt 1$.
For $1/2\le x\lt 1$, let $f(x)=1$. For $1/4 \le x\lt 1/2$, let $f(x)=0$. For $1/8\le x\lt 1/4$, let $f(x)=1$. And so on.
In general, if $n\ge 1$, with $n$ odd, and $\frac{1}{2^n}\le x\lt \frac{1}{2^{n-1}}$, let $f(x)=1$. 
If $n\ge 1$, with $n$ even, and $\frac{1}{2^n}\le x\lt \frac{1}{2^{n-1}}$, let $f(x)=0$.
Then for all $x$, we have $f(x)f(2x)=0$. So in particular $\lim_{x\to 0} f(x)f(2x)=0$.
But $\lim_{x\to 0}f(x)$ does not exist, since every open neighbourhood of $0$ contains $x$ such that $f(x)=0$, and $x$ such that $f(x)=1$.
