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I was reading this question.

The simplest examples of continuous functions, with discontinuous derivatives in some point, are usually of the form:

$$ f(x) = \begin{cases} x^2 \sin(1/x) &\mbox{if } x \neq 0 \\ 0 & \mbox{if } x=0. \end{cases} $$ The derivative of $f$ is $$ f'(x) = \begin{cases} 2 x \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)&\mbox{if } x \neq 0 \\ 0 & \mbox{if } x=0, \end{cases} $$

The derivative is discontinuous because the limit of $\cos \left(\frac{1}{x}\right)$ does not exist for $x\rightarrow 0$.

Is there an example where the derivative is still discontinuous but with existing limit?

Thanks

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An old man in the sea.
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    No. If the limit exists at 0, then this means that the derivative can be continuously extended to 0 making $f$ continuously differentiable at 0. Moreover, one can show that if $f$ is differentiable at 0, but $f'$ is discontinuous at 0, then $f'$ oszillates around 0. This is a consequence of the Intermediate Value Theorem for Derivatives, see https://en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis) – sranthrop Aug 31 '16 at 16:45
  • @sranthrop thanks for your comment. Would you like to write an answer for the second part of your comment? +1 ;) – An old man in the sea. Aug 31 '16 at 17:21

1 Answers1

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Suppose $f$ is differentiable in some neighborhood $(x-\delta,x+\delta)$ of $x$, and $\lim\limits_{t\rightarrow x}{f'(t)}$ exists. Define $y:(x-\delta,x+\delta)\rightarrow\mathbb{R}$ such that $y(t)$ is strictly between $x$ and $t$ and $$f(t)-f(x) = f'(y(t))(t-x)$$ for every $t\in(x-\delta,x+\delta)$. The existence of such a function is guaranteed by the Mean Value Theorem. Since $y(t)$ is between $x$ and $t$ for every $t$, this implies that $\lim\limits_{t\rightarrow x}{y(t)} = x$, and since $y(t)\ne x$ for $t\ne x$ as well, we have $\lim\limits_{t\rightarrow x}{f'(y(t))}=\lim\limits_{s\rightarrow x}{f'(s)}$ by the composition law (think of $s = y(t)$ in this substitution). This implies that $$f'(x) = \lim\limits_{t\rightarrow x}{\frac{f(t)-f(x)}{t-x}} = \lim\limits_{t\rightarrow x}{f'(y(t))} = \lim\limits_{t\rightarrow x}{f'(t)},$$ i.e. $f'$ is continuous at $x$.

Remark: Typically, the composition law is phrased as follows: if $\lim\limits_{x\rightarrow c}{g(x)} = a$ and $f$ is continuous at $a$, then $\lim\limits_{x\rightarrow c}{f(g(x))} = \lim\limits_{u\rightarrow a}{f(u)}$. In our problem, we obviously cannot assume $f'$ is continuous at $x$, since that is what we are trying to show. However, the above conclusion still holds if we merely require that $g(x)\ne a$ if $x\ne c$ in some neighborhood of $c$. A proof of this can be found here (look for "Hypothesis 2").

Joey Zou
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  • Joey, I have a doubt. I understand how y tends to x, but I cannot understand why f'(y) can have the same limit as f'(t). What are you using there? – An old man in the sea. Aug 31 '16 at 17:14
  • @Anoldmaninthesea. I am using the composition law for limits: basically, $\lim\limits_{t\rightarrow x}{f'(y(t))} = \lim\limits_{s\rightarrow\lim\limits_{t\rightarrow x}{y(t)}}{f'(s)}$. However, there is a subtlety that I need to address, and I will edit the answer shortly to address it. – Joey Zou Aug 31 '16 at 17:27
  • @Anoldmaninthesea. I have updated the answer. I hope it is more clear now. – Joey Zou Aug 31 '16 at 17:36
  • Yes, it is. I had never heard of that condition, i.e., it's the first time I'm seeing it . Is it common in more advanced subjects? – An old man in the sea. Aug 31 '16 at 17:45
  • @Anoldmaninthesea. If you mean the condition that $g(x)\ne a$ for $x\ne c$ implies the composition law, then I'm not sure. It is related to the distinction between a limit point and an accumulation point in point-set topology, where a limit point is a limit of a sequence whose terms do not include the point itself, while an accumulation point is the limit of a sequence whose terms can include the point itself. So every limit point is an accumulation point, but not vice versa. The distinction between the two is useful somewhere, although I can't exactly say where right now. – Joey Zou Aug 31 '16 at 17:56