Does the existence of $\lim_{x\to 0^+}f'(x)$ imply the existence of $\lim_{x\to0^+}f(x)$?

Let us say that $f(x)$ is differentiable on $(0,\infty)$. Does the existence of $\lim_{x\to 0^+}f'(x)$ imply the existence of $\lim_{x\to0^+}f(x)$ ? I think it should be true, but I can't seem to prove it.

• The fundamental theorem of calculus might have something to say about that. – Arthur May 10 '17 at 8:18
• @Arthur Not really. Don't know about the integrability of $f'$. – MathematicsStudent1122 May 10 '17 at 8:19
• @MathematicsStudent1122 We know the antiderivative of $f'$ exists on $(0,\infty)$ and $f'$ is bounded on, say, $(0,1)$. You're certain we can't leverage something from that? – Arthur May 10 '17 at 8:27
• @Arthur No. See this – MathematicsStudent1122 May 10 '17 at 8:29
• Cool. I think I've seen it before somewhere, but I didn't remember it. – Arthur May 10 '17 at 8:31

Note that since the limit exists, $f'$ is locally bounded near $x=0$, hence $f$ is uniformly continuous on $(0, \delta)$ for some $\delta$, hence $f$ can be continuously extended to $x=0$. This implies the claim.
• @ashpool I tried playing around with the mean value theorem. Problem is that $f$ isn't defined at $x=0$. Though, there's probably a simpler solution. – MathematicsStudent1122 May 10 '17 at 8:49
• (I'm assuming the domain is $\mathbb{R}_{>0}$, since that's what the problem suggests) – MathematicsStudent1122 May 10 '17 at 8:59