解析:(1) f(x)=sin(x²)有y=sinu和u=x²复合而成∵ y=sinu和u=x²在(-∞,+∞)上连续∴ f(x)=sin(x²)在(-∞,+∞)上连续(2)由三角函数性质可知对于x∈R,恒有|sinx²|≤1所以,f(x)=sin(x²)在(-∞,+∞)上有界(3) f'(x)=2xcos(x²)此函数在(-∞,+∞)无界所以,f(x)=sin(x²)在(-∞,+∞)上非“一致连续”