It will suffice to prove this for each fixed x. So fix x.

Let and (I prefer to use a subscript rather than a superscript to denote the "negative part"). Then . Also, , and .

If then for all sufficiently large n, hence for all sufficiently large n, and therefore . In this case, also, and so .

Now suppose that . This implies that infinitely often. But whenever , and therefore . Thus in this case it is again true that .

I hope that argument is convincing. I found it quite hard to write it out coherently, and I'm sure that I would also have got confused if I tried to present it to a class.