Suppose that f:R-->R is continuous and has the property that for every epsilon>0, there is M>0 such that if abs(x)>=M, then abs(f(x))<epsilon. Show that f is uniformly continuos.
The given function by Jose27 is not continuous.
There are a lot of threads of similar questions. Observe that in $\displaystyle [-M,M]$ a continuos function is always uniformly continuous and apply the condition for $\displaystyle \varepsilon/2$.