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We are assigned to deal with the following task.

Assume that $f(x)$ is derivable for any $x \in \mathbb{R}$. We want to research $\lim\limits_{x \to x_0}f'(x)$ where $x_0 \in \mathbb{R}$.

Notice that \begin{align*} f'(x_0)=\lim_{x \to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x \to x_0}\frac{f'(\xi)(x-x_0)}{x-x_0}=\lim_{x \to x_0}f'(\xi), \end{align*} where we applied Lagrange's Mean Value Theorem, and $ x_0 \lessgtr \xi \lessgtr x.$ Since $\xi$ is squeezed by $x_0$ and $x$, then $x \to x_0$ implies $\xi \to x_0$. Thus $$f'(x_0)=\lim_{x \to x_0}f'(\xi)=\lim_{\xi \to x_0}f'(\xi).$$

What does this say? It shows that $f'(x)$ is always continuous at any point $x=x_0$, which is an absurd conclusion, because we know safely $f'(x)$ may probably has the discontinuity point (of the second kind). But where dose the mistake occur during the reasoning above?

mengdie1982
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  • Recall that $\xi$ is actually a function of $x$. Thus, $\lim_{x\rightarrow x_0} f'(\xi(x))$ does not imply $\lim_{\xi \rightarrow x_0} f'(\xi)$ – Severin Schraven Dec 15 '18 at 09:08
  • it is just a $\ksi$ not every $x$... – dmtri Dec 15 '18 at 09:09
  • @SeverinSchraven Do you know the squeeze theorem? $x_0 \lessgtr \xi(x)\ \lessgtr x$. Let $x \to x_0$，then $\xi(x) \to x_0$. It has no fault here. – mengdie1982 Dec 15 '18 at 09:25
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    Reread my comment, I did NOT claim that $\xi(x) \rightarrow x_0$ is wrong. Your line of reasoning is the following: Let $$ f(x) = \begin{cases} 0;& x\in \mathbb{Q} \ 1,& x\in \mathbb{R}\setminus \mathbb{Q} \end{cases} $$ now take $x_0=0$ and $\xi(x)$ be a rational number between $0$ and $x$. Clearly we have $\lim_{x\rightarrow 0} f(\xi(x))=0$ and thus $\lim_{\xi \rightarrow 0} f(\xi)= \lim_{x \rightarrow 0} f(\xi(x))$. But $\lim_{\xi \rightarrow 0} f(\xi)$ does not even exist. That is, because you do not hit every number with $\xi(x)$. – Severin Schraven Dec 15 '18 at 09:32
  • @SeverinSchraven You have pointed out some essential thing！ – mengdie1982 Dec 15 '18 at 09:36
  • Yes, it is possible that $f'$ is discontinuous at a point $x_0$. But it is not so obvious that I would call the contrary assertion absurd. – GEdgar Dec 15 '18 at 15:49

3 Answers3

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Your reasoning only shows that if $\lim_{x \to x_0} f'(x)$ exists, then it's equal to $f'(x_0)$. But, as you say, it may not exist.

This is a rather classical exercise, which has been treated many times on this site, for example in this question.

Hans Lundmark
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Actually，we can make a comment for the reasoning like this:

The fact that $\lim\limits_{\xi \to x_0}f'(\xi)$ exists dose not imply $\lim\limits_{x \to x_0}f'(x)$ also exsits, because, according to Heine's Theorem, the latter one necessitates that $f'(x_n)$ converges for any sequence $x_n \to x_0$. As we can see, $\xi_n$ is only a specific sequence. Even though $f'(\xi_n)$ is convergent, this is not enough to guarantee $f'(x_n)$ converges as well.

mengdie1982
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Actually, you are right, this also called 'theorem of derivative limitation' in some textbooks. If you already assumed that $f(x)$ is derivable for any $x \in \mathbb{R}$.

With an additional condition the limitation of derivative $\lim_{x \to x_{0}}f'(x)$ exist.

You can write your conclusion:

$$\lim_{x \to x_{0}}f'(x)=f'(x_{0})$$

The condition $f(x)$ is derivable for any $x \in \mathbb{R}$ can also be weaken to $f(x)$ is continuous in $U(x_{0})$ which is the neighborhood of $x_{0}$, and derivable in $U°(x_{0})$ which is the punctured neighborhood of $x_{0}$.

The only problem is that you have not assumed the limitation of your derivative should always exist, which means the derivative can not divergence to infinity, like you say sometimes it would be the discontinuity point of the second kind.

And this theorem can be simply remembered as in other 'not so strict (please notice the comment below)' ways like:

If a continuous function has (finite) derivative, its derivative function is also continuous. (but maybe not derivable)

Nanayajitzuki
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  • It would be fair to say that you're wrong. The fact $f(x)$ is derivable does not imply $f'(x)$ is continuous. Here is a counterexample: $$f(x)=\begin{cases}x^2\sin\dfrac{1}{x},&x \neq 0,\0,&x=0.\end{cases}.$$ You may verify that $f(x)$ is derivable over $(-\infty,+\infty)$，but $f'(x)$ is not continuous at $x=0$. – mengdie1982 Dec 15 '18 at 16:16
  • But of course, if we add a condition that $\lim\limits_{x \to x_0}f'(x)$ exists, then $\lim\limits_{x \to x_0}f'(x)=f'(x_0)$. In another word, $f'(x)$ is continuous at $x=x_0$ now！ – mengdie1982 Dec 15 '18 at 16:24
  • @mengdie1982 You still remind a good example for me. Yes, I should explain that the 'not so strict' place this the limitation of derivative must be exist not only the derivative exist. Your comment point out it. – Nanayajitzuki Dec 15 '18 at 16:30