How would you show that sin(x^2) is not uniformly continuous in the real numbers?

Answer 1

If it were uniformly continuous, then foe any e>0 you could find a d>o such that |sin((x+d)^2)-sin(x^2)|
But notice that (x+d)^2 -x^2+2*x*d+d^2. d^2 will be really small for small d -- but you can find SOME value of X where 2*x*d is as large as you like. Choose it so that it is near an n off multiple of pi -- and you will see that |sin((x+d)^2)-sin(x^2)| is close to two for some values of x. This is greater than e.

My argument is a little sloppy -- but you should be able to clean it up.