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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.