Suppose that f:R--> R is differentiable on R and that the derivative of f is bounded on R. Prove that f is uniformly continuous on R.
Three words: MEAN VALUE THEOREM
$\displaystyle \frac{|f(x)-f(y)|}{|x-y|}=|f'(t)|\leq M\implies |f(x)-f(y)|\leq M|x-y|$.
Let $\displaystyle \delta=\frac{\epsilon}{M}$ and you're done.
If $\displaystyle d(f(x),f(y))\leq Md(x,y)$, then $\displaystyle f$ is said to be Lipschitz continuous, with the Lipschitz constant being the smallest such $\displaystyle M$ that satisfies that inequality. Lipschitz continuity always implies uniform continuity, but not necessarily the other way around.