Let $f$ be a continuous map defined on $[1; +\infty)$ and differentiable on $(1; +\infty)$. Let also suppose that $\lim_{x\to+\infty}f'(x)$ exists.
Is the function uniformly continuous at $[1; +\infty)$?
The correct answer is "yes". Please explain the effect of the derivative at infinity on the function itself.
0) Let $M >0$; $f$ is uniformly continuos on compactum $[0,M]$.
1) $\lim_{x \rightarrow \infty} f'(x)=: L$, real.
Let $\epsilon$ be given. There is a $M>0$ s.t.
$x >M(\epsilon)$ implies $|f'(x)-L|<\epsilon$.
For $x \in (M,\infty):$ $|f'(x)|$ is bounded,
i.e. $|f'(x)| <B(M)$.
Hence
$ |f(x)-f(y)| = |f'(t)||x-y| \le B(M)|x-y|,$
$t\in (\min(x,y),\max(x,y)).$
In $(M,\infty)$ $f$ is Lipschitz hence uniformly continuos.
2) For a $\epsilon/2$ there is a $\delta_1$ for $[0,M(\epsilon)]$, a $\delta_2$ for $(M(\epsilon),\infty)$.
Let $\delta =\min(\delta_1,\delta_2)$.
3) What happens if $x\in [0,M]$, and $y \in (M,\infty)$?