Suppose that a<b are extended real numbers and that f is differentiable on (a,b). If f' is bound on (a,b) prove f is uniformly continuous.
We know that (f(y)-f(x))/(y-x) = f'(a-x,y) for all x < y in the domain
where a is in (x,y), from the Mean Value Theorem.
So if we assume that |f'| is boudned by M, then
(f(y) - f(x))/(y-x) <= M for all x < y.
Or |f(x)-f(y)| <= M*|x-y|.
Now if we are given e>0, we can choose d = e/M, and if
|x-y| < d, then |f(x)-f(y)| < e, as required.