Is there an uniformly continuous $[0,\infty) \rightarrow \mathbb R$ surjection?

Question

I think there isn't but I can't write a proof. I tried assuming that such function is (as it must be) continuous, and showing that it can't be uniformly cont., but I'm not sure what to do.

Answer 1

How about $x\mapsto \sqrt{x}\sin(\sqrt x)$?

Answer 2

The function $$f(x) = x \sin ( \log (x+1))$$ has bounded derivative (hence it is uniformly continuous) and oscillates between $+ \infty$ and $- \infty$ (hence it is surjective).