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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.

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  • $\begingroup$ The fundamental theorem of calculus might have something to say about that. $\endgroup$ – Arthur May 10 '17 at 8:18
  • $\begingroup$ @Arthur Not really. Don't know about the integrability of $f'$. $\endgroup$ – MathematicsStudent1122 May 10 '17 at 8:19
  • $\begingroup$ @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? $\endgroup$ – Arthur May 10 '17 at 8:27
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    $\begingroup$ @Arthur No. See this $\endgroup$ – MathematicsStudent1122 May 10 '17 at 8:29
  • $\begingroup$ Cool. I think I've seen it before somewhere, but I didn't remember it. $\endgroup$ – Arthur May 10 '17 at 8:31
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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.

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  • $\begingroup$ Wow, it requires all that machinery of analysis to prove such a simple statement? $\endgroup$ – ashpool May 10 '17 at 8:46
  • $\begingroup$ @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. $\endgroup$ – MathematicsStudent1122 May 10 '17 at 8:49
  • $\begingroup$ (I'm assuming the domain is $\mathbb{R}_{>0}$, since that's what the problem suggests) $\endgroup$ – MathematicsStudent1122 May 10 '17 at 8:59

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