Limit at infinity of the differential implies uniform continuity Assume that $f: \mathbb{R}^{+} \to \mathbb{R}$ is a differentiable function, with $\lim_{x\to\infty}f'(x)=1$. Is $f$ then uniformly continuous on $\mathbb{R}^+$? I'm not really sure where to get started so any help is appreciated.
 A: Hint


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*Let $N>0$ s.t. $|f'(x)|\leq 2$ for all $x\geq N$ (such $N$  exist by the fact that $f'(x)\to 1$ when $x\to \infty $. Then, by mean value theorem, you can prove that $f$ is Lipschitz and thus uniformly continuous.

*The fact that $f$ is uniformly continuous on $[0,N]$ is a famous theorem.

*Check what happen when $x\leq N\leq y$ and conclude.
A: The point is that every continuous function on a closed and bounded interval is uniformly continuous by Cantor theorem. So in some sense it is enough to show that $f$ is uniformly continuous only for all $x>M$ for some $M$.
Formally, since $\lim_{x\rightarrow\infty} f'(x)=1$ there exists $M$ such that for all $x>M$ we have that $|f'(x)|<2$.
Lagrange theorem implies that $|f(a)-f(b)|<2|a-b|$ for all $a,b>M$.
So $f$ is uniformaly continuous on $[0,M]$ by cantor theorem. On the other hand for every $\varepsilon>0$ we can take $\delta=\varepsilon/2$ so if $a,b\in[M,\infty)$ such that $|a-b|<\delta$ we have by Lagrange theorem that $|f(a)-f(b)|<2\delta=\varepsilon$.
Therefore $f$ is uniformly continuous on $[0,M]$ and on $[M,\infty)$, hence uniformly continuous on $\mathbb{R}^+$. 
