If $f$ is a differentiable function with $f'$ discontinuous at $0$, show that the set of all real $y$ 's such that $\lim_{n\rightarrow \infty}f'(x_n)=y$, where $x_n$ is a real sequence such that $\lim_{n\rightarrow \infty} x_n = 0$, is equal to the segment $[p,q]$ for some $p\not = q$ (can be infinity).

I'm wondering if this problem states the valid thing... Because if we take $f(x)=\ln x$ then, indeed, $f'$ is discontinuous at $0$ but if we take any $\{x_n\}: x_n\rightarrow 0,n\rightarrow\infty$, then $\lim_{n\rightarrow\infty}f'(x_n)$ is either $+\infty$ or $-\infty$ not a segment (even with infinity bounds as the problem allows)

  • $\begingroup$ Because the question talks about $\lim_{n \rightarrow} f'(x_n)$ not $\lim_{n \rightarrow \infty} f(x_n)$ $\endgroup$ – viru May 2 '20 at 12:40
  • $\begingroup$ @viru actually, the same argument for the limit of $f'$ $\endgroup$ – VIVID May 2 '20 at 12:47
  • $\begingroup$ $f'(x) = 1/x$ and hence your set is empty because there is no $y$ which are $\textbf{reals}$ s.t $\lim_{n \rightarrow \infty} \frac{1}{x_n} = y$ and $x_n \rightarrow 0$ $\endgroup$ – viru May 2 '20 at 12:48
  • $\begingroup$ Since $f(x)$ is differentiable, so it is continuous. But $f(x)=\ln x$ is discontinuous at $x=0$. $\endgroup$ – xpaul May 2 '20 at 13:33
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    $\begingroup$ Nice question. +1 I had never seen this property of derivatives stated in this manner. $\endgroup$ – Paramanand Singh May 2 '20 at 15:30

I guess that the problem should be states as “$f$ is differentiable at 0 with $f’$...”.

As a hint you can try to work out the standard example of a differentiable function with discontinuous derivative, namely $x^2\sin(1/x)$. Or write down the definition of differentiabilty explicitly and what that is mean for $f’$ to be discontinuous.

Hope it helps.


By Darboux theorem (or mean value theorem) it follows that derivatives can't have jump discontinuity. Thus we are left with only discontinuity which is of oscillatory kind.

Let $$p=\liminf_{x\to 0}f'(x),q=\limsup_{x\to 0}f'(x)$$ then we have $p<q$ and you can establish what is asked in question. You will need to use the fact that derivatives have intermediate value property (Darboux theorem). You need to show that for every $y\in[p, q] $ there is a sequence $x_n$ with $x_n\to 0$ and $f'(x_n) =y$ and if $y\notin [p, q] $ there is no such sequence.

Let me know if you have any difficulty writing a detailed proof based on the outline above.

  • $\begingroup$ Would the downvoter care to explain? The downvote is not related the typical case of close votes for a question. $\endgroup$ – Paramanand Singh May 4 '20 at 1:41

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