Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 1 / 1A / ii / Solution

Not uniformly continuous. ChooseTheir separation is $\pi /(x_n+y_n)\to 0$, while $\sin (x_n^2)=1$ and $\sin (y_n^2)=-1$. Thus the function values stay distance two apart, contradicting uniform continuity. Boundedness of the function does not prevent increasingly rapid oscillations.