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Prove/disprove: $f$ has a bounded derivative and $f'$ isn't continuous on $(a,b)$, so there's a point $y\in(a,b)$ such that $\displaystyle\lim_{x\to y}f'$ does not exist.

I think that if $f'$ isn't continuous on the interval, then maybe we could have two disjoint sub-intervals, like for example $(a,c), (d,b)$ such that $d-c=\dfrac {a+b} 3$ so there's a substantial gap in the interval $(a,b)$ where $f'$ isn't defined so it follows that it won't have a limit there, for example: on $\dfrac {c+d}2$.

GinKin
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    differentiable at some point implies continous at that point. Isn't that enough to prove the claim ? – Peter Jun 22 '14 at 18:21
  • @Peter so there can't be gaps in the domain of the derivative then ? – GinKin Jun 22 '14 at 18:24
  • OK, the limit is different from the point itself. Sorry. – Peter Jun 22 '14 at 18:44
  • This should help. It says $f'$ is continuous at $y$ if $\lim_{x\rightarrow y} f'(x)$ exists. – David Mitra Jun 22 '14 at 19:06
  • @DavidMitra so my question is the negation of the question you linked isn't it ? – GinKin Jun 22 '14 at 19:11
  • If you mean contrapositive, almost. The link shows if the limit exists for all $y\in(a,b)$, then $f'$ is continuous on $(a,b)$. This implies your statement is true. I'm not sure why the boundedness condition is in your statement. The statement is true without it. – David Mitra Jun 22 '14 at 19:32
  • @DavidMitra It's actually true ?! But here $f'$ isn't continuous so for some $y$ the limit does not exist or the limit could be different from both sides. – GinKin Jun 22 '14 at 19:36
  • Actually, at least one of the limits from the left or from the right doesn't exist (because the argument in the link would show that if a one-sided limit existed at $y$, it would have to have value $f'(y)$). – David Mitra Jun 22 '14 at 19:44

2 Answers2

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Indeed, there should be a point where the limit does not exists. The simplest way to see it is to use Darboux (intermediate value) property of the derivative. Indeed, if $f'$ is not continuous at the point $x_0$ and $\lim_{x\to x_0}f'(x)=m\ne f'(x_0)=M$ then by the intermediate value property of the derivative in the neighbourhood of $x_0$ we can find many values of $f'$ equals to say $\frac{m+M}{2}.$ This would contradict to the existence of the limit.

leshik
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The very fundamental thing one needs to observe here is that a derivative can't have jump discontinuity. If $f'(x) \to L$ as $x \to c$ then $f'(c) = L$ and thus $f'$ is continuous at $c$. Hence it is not possible for a derivative to have a limit at point and not to be continuous at that point. It follows that there will be points where limit of derivative $f'(x)$ does not exist.

Now to the proof of the statement mentioned in bold above. One proof is already given by leshik in his answer. Here is another proof based on mean value theorem. Suppose that $f(x)$ is differentiable in neighborhood of $c$. Suppose $x \to c$ and $f'(x) \to L$ as $x \to c$. Then we have $f(x) - f(c) = (x - c)f'(d)$ where $d$ lies somewhere between $c$ and $x$. Thus $$\frac{f(x) - f(c)}{x - c}=f'(d)$$ When $x\to c$ then LHS tends to $f'(c)$ and RHS tends to $L$ because $d \to c$. It follows that $f'(c) = L = \lim_{x \to c}f'(x)$. Hence $f'(x)$ is continuous at $c$.

Paramanand Singh
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