0

It is a well known fact that, for a real function $f$, if $f'(x)$ has a limit as $x\to x_0$, then $f$ is differentiable at $x_0$. In fact, there has been some questions on this site about proving this fact. However, consider the function $$f(x)=\begin{cases} x, x\neq 0 \\ 1, x=0 \end{cases}.$$ It seems to me that the condition is satisfied: $f'(x)=1$ everywhere except $x=0$ and so $\lim\limits_{x\to 0}f'(x)=1$. However this function is clearly not differentiable at $0$. Where did I go wrong in this reasoning, or is there perhaps an extra constraint on the function $f$ (maybe it has to be surjective)?

Theo Bendit
  • 50,900
YiFan Tey
  • 17,431
  • 4
  • 28
  • 66
  • 9
    Not so well known, if one forgets to include the hypothesis that $f$ must be continuous at $x_0$. –  Sep 06 '18 at 01:06
  • Are you sure? This does not seem to be given in https://math.stackexchange.com/q/1033952/496634 and https://math.stackexchange.com/q/2906939/496634 . Or does $f$ being differentiable everywhere except $x_0$ already imply that it is continuous at $x_0$? – YiFan Tey Sep 06 '18 at 01:09
  • 1
    Ah, we were all so blinded by the other glaring blunder in the question in the second link that we forgot to point out this other fact (my comment still sneakily holds, what a relief). If you have a look at the answers in your first link you'll see that they say what I said. –  Sep 06 '18 at 01:13
  • Ah, I see. Looks like I'll need to learn to look at answers other than the accepted one. Thanks! – YiFan Tey Sep 06 '18 at 01:15
  • Nice catch! The accepted answer mentions a condition that is immediately equivalent to the existence of a continuous function $g:\Bbb R\to\Bbb R$ such that $f(x)=g(x)$ for all $x\ne x_0$. This turns out to be incorrect, because in principle $f(x_0)$ could be $\ne g(x_0)$ like in your example. However, $g$ will be differentiable at $x_0$. –  Sep 06 '18 at 01:31
  • Anybody can err. A theorem of John von Neumann, one the the most accomplished 20th century mathematicians, was destroyed when it was found that he had accidentally changed a $-$ into a $+$. – DanielWainfleet Sep 06 '18 at 03:13
  • Here is an older question with the correct assumptions: https://math.stackexchange.com/questions/257907/prove-that-fa-lim-x-rightarrow-afx – Hans Lundmark Sep 06 '18 at 08:41

0 Answers0