# Determining if a function is uniform continuous based on information of the limit of its derivative

I'm studying real analysis more specifically derivative functions from $$\mathbb{R}$$ to $$\mathbb{R}$$. I'm struggling a bit between the relation of the derivative of a function and the function itself. For example let's say we have a function $$f:\mathbb{R}^+ \rightarrow \mathbb{R}$$ which is derivable and for which $$\lim_{x \to \infty}f'(x) = 1$$. Can we using this info determine if the function is uniformly continuous on $$\mathbb{R}$$?

I started of by noticing that for such function the $$\lim_{x \to \infty}f(x) = \infty$$ and also that if I can somehow show the derivative function is bounded that we can easily conclude that the function is indeed uniformly continuous. Now I don't now how to proceed with this. I could maybe say that if the derivative function is also continuous that is should be bounded?

Or maybe because of the fact that we don't know if the derivative function is continuous we can't really say it is bounded and thus we can't conclude if f is uniformly continuous, but in this case and in general I can't find a counter-example.

Any tips on how to furthur proceed with this problem would be greatly appreciated :))

P.S: $$\mathbb{R}^+ = [0,\infty)$$

• What is $\mathbb{R}^+$? Is it $[0,\infty)$? Or is it $(0,\infty)$? Aug 18 at 12:26
• It's [0,$\infty$) Aug 18 at 12:27
• Does this answer your question? How does the existence of a limit imply that a function is uniformly continuous Aug 18 at 12:28
• @AnotherUser No, the hypothesis in that link is different. Aug 18 at 12:30
• @AnotherUser No they assume the limit of f at infinity is a real number whereas here it is infinity Aug 18 at 12:31

There exists $$C$$ such that $$|f'(x)| <2$$ whenever $$x\geq C$$. By Mean Value Theorem $$|f(x)-f(y)| <2|x-y|$$ for $$x, y \geq C$$. This shows that $$f$$ is uniformly continuous on $$[C,\infty)$$. Since $$f$$ is also uniformly continuous on any compact interval it follows that it is uniformly continuous on$$[0,\infty)$$.