proof question??

Question

prove that the function x*sinx is not uniformly continuous on the real numbers

do you just negate the definition and prove it from that or use a proof by contradiction...?

Answer 1

Find the counterexample and you are done. The function may be continuous, but if the changes in f(x) grow without bound (or are undefined) at some point, it is not uniformly continuous.

To rigorous prove this, you would need to construct a delta-epsilon proof, and show that for every real number ε > 0 there exists δ > 0 such that for all x,y with d1(x,y) < δ, you have d2(f(x),f(y)) < ε.

Answer 2

Negate the definition and prove it.

That is, you need to prove: There exists epsilon > 0 such that for every delta > 0, there exists real x, y such that |x - y| < delta but |x * sin x - y * sin y| > epsilon.