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$f$ is differentiable on $(a, b)$. (1)

Let $\alpha \in (a, b)$.

$f'(\alpha) = \lim_{h \to 0}\frac{f(\alpha + h) - f(\alpha)}h$.

(1) $\implies f$ is differentiable between $\alpha$ and $\alpha + h$, inclusive.

By the mean value theorem, $\exists c_n$ between $\alpha$, $\alpha + h$: $f'(c_n) = \frac{f(\alpha + h) - f(\alpha)}{\alpha + h - \alpha}$.

So, $f'(\alpha) = \lim_{h \to 0}f'(c_n)$.

As $h \to 0$, $c_n \to \alpha$.

So, $f'(\alpha) = \lim_{c_n \to \alpha}f'(c_h) \implies f'$ is continuous at $\alpha \forall\alpha\in(a, b)$.

$\therefore f$ is differentiable on $(a, b) \implies f'$ is continuous on $(a, b)$.

But clearly the statement must not be true as for some functions like the following, the derivative exists, but is not continuous, at zero:

$f(n) = \begin{cases} n^2 \sin(\frac{1}{n^2}) & n \in \mathbb R \\ 0 & n = 0. \end{cases}$

$f'(0) = 0$. But $\lim_{n \to 0}f'(n)$ does not exist.

José Carlos Santos
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Prajwal Samal
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3 Answers3

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Yes, $f'(c_h) \to f'(\alpha)$ as $h \to 0$ (sort of, it's not quite well defined, but you can find a sequence $h_n$ such that...), but that doesn't rule out the possibility that there is some other sequence $x_n \to \alpha$ such that $f'(x_n)$ does not tend to $f'(\alpha)$.

Hans Lundmark
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  • Why is it not well defined? We can fix $c_h$ for each $h > 0$, which makes it a function $h \mapsto c_h$ and $$\lim_{h \to 0} f'(c_h)$$ is perfectly fine. – Adayah Oct 07 '17 at 18:27
  • I just meant that $c_h$ is not uniquely defined by $h$, so there are in general many such functions $h \mapsto c_h$. (And they don't need to be continuous, of course, which is why the argument in the question fails.) – Hans Lundmark Oct 07 '17 at 20:45
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Consider the assertion$$\displaystyle f'(\alpha)=\lim_{c_h\to\alpha}f'(c_h).\tag{1}$$There are several problems here:

  1. What does it even mean? You are acting as if $c_h$ is a well-determined number depending on $\alpha$ and $h$. It is not. All we know is that, for each $\alpha$ and each $h>0$, there is some $c_h$ with a certain property.
  2. Being continuous at $\alpha$ means that $\displaystyle\lim_{\beta\to\alpha}f'(\beta)=f'(\alpha)$, which is not what $(1)$ says.

As an example, consider the function$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}\sin\left(\frac1x\right)&\text{ if }x\neq0\\0&\text{ otherwise.}\end{cases}\end{array}$$It is not continuous at $0$. Now, for each $h>0$, let $c_h=\frac1{n\pi}$ where $n$ is the smallest natural number such that $\frac1{n\pi}<h$. Then it is clear that $f(c_h)=0$. Furthermore, $f(0)=0$. However, you can't deduce from this that $f$ is continuous at $0$.

José Carlos Santos
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  • As to the first problem: we can easily make $c_h$ a well-determined number by fixing the choice of the number with the wanted property. I believe there is no problem here. Second problem: I think it would be clearer what you mean if $(1)$ was stated as: $$f'(\alpha) = \lim_{h \to 0} f'(c_h),$$ because this is really what the fallacy was: we are choosing a particular ($h$-indexed) "sequence" instead of a general one. – Adayah Oct 07 '17 at 18:26
  • @Adayah This can be done, but it is actually more subtle than one would expect, see here. – Andrés E. Caicedo Nov 26 '17 at 21:09
  • @AndrésE.Caicedo Subtlety lies in avoiding usage of the axiom of choice, not in doing what I've done. – Adayah Nov 27 '17 at 10:12
  • @Adayah Sorry, I am not sure what it is that you have done, so I am not sure I understand what you are saying. – Andrés E. Caicedo Nov 28 '17 at 04:09
  • @AndrésE.Caicedo Let $\delta > 0$ be such that $(\alpha-\delta, \alpha+\delta) \subseteq (a, b)$. I'm saying that since for each $h \in (-\delta, \delta)$ there is some $c$ between $\alpha$ and $\alpha+h$ such that $$\frac{f(\alpha+h)-f(\alpha)}{h} = f'(c),$$ by the axiom of choice there is a function $c : (-\delta, \delta)$ (we can write $c_h$ instead of usual $c(h)$) s.t. for each $h \in (-\delta, \delta)$ we have $$\frac{f(\alpha+h)-f(\alpha)}{h} = f'(c_h).$$ As you point out, finding such function $c$ without the axiom of choice needs more care, by I did mean to use the axiom of choice. – Adayah Nov 28 '17 at 18:45
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There was no problem with line 5. The mistake started at line 6 but not for the reason Jose Carlos Santos gave in their answer so their answer also incorrectly implies that there's a problem with line 5. Line 5 was fine because you defined for any α in the interval (a, b) and nonzero h such that α + h is in the interval (a, b) what cn is. Line 6 is perfectly well meaningful. The problem with line 6 is that for each α, you can only prove line 6 from line 5 under the assumption that $\lim_{h \to 0}f'(c_n)$ exists which can't be proven.

Timothy
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    There may be many numbers that satisfy the definition of $c_h$, so using such a notation is indeed bad form, as it suggests that a choice has been specified, while this is not the case. It is actually possible to specify what value to take (that is, to define a function $h\mapsto c_h$), but this requires some effort, and it is certainly not present in the sketch. Your identification of the problem, on the other hand, is incorrect, since it is indeed the case that $\lim_{h\to0} f'(c_h)$ exists. In fact, it is obvious from the assumption that $f'(\alpha)$ exists. – Andrés E. Caicedo Nov 26 '17 at 21:14
  • What is the point of making an edit on an incorrect answer to reiterate its mistake? – Andrés E. Caicedo Dec 05 '17 at 18:12