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Assume that $f:\mathbb{R}\to\mathbb{R}$ is continuous and differentiable everywhere but at $0$.

If $\displaystyle\lim_{x\to0} f'(x) = L$ exists, then does it follow that $f'(0)$ exists?

Prove or disprove.

I think it has to be true. I know that by definition $\displaystyle f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}$, but I could not able to further steps from here. could you please help me out.

Michael Hardy
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user145993
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    Analyse the function $f(x) = x^2$ for $x \ne 0$ and $f(0) = 1$ – vonbrand Oct 03 '15 at 16:07
  • Doesn't the existence of the limit imply the existence of the derivative at the point? – najayaz Oct 03 '15 at 16:07
  • @G-man, there might be a limit even if the function isn't defined at the point. – vonbrand Oct 03 '15 at 16:10
  • @vonbrand it given that the function is continuous. – najayaz Oct 03 '15 at 16:11
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    @G-man: The function is actually not continuous. "Assume that f:R↦Rf:R↦R is continuous and differentiable everywhere but at 0." – Daniel R. Collins Oct 03 '15 at 16:43
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    Comments below show there's a difference in opinion about parsing the assumption as "(continuous and differentiable) everywhere but at 0" versus "continuous and (differentiable everywhere but at 0)". Perhaps clarification is needed on that point. – Daniel R. Collins Oct 03 '15 at 16:54
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    @DanielR.Collins: If it were continuous everywhere but differentiable only everywhere but at $x = 0$, then the limit of the derivative at $x = 0$ could not exist and equal $L$. – Brian Tung Oct 03 '15 at 16:56
  • Actually "everywhere but at 0" might a bit misleading in itself, since it seems to say that the derivative mustn't exist at 0 (although it's clear from the actual question that this can't be the intended meaning). "Differentiable for $x \neq 0$" would perhaps be clearer. – Hans Lundmark Oct 03 '15 at 17:01
  • @G-man : "Doesn't the existence of the limit imply the existence of the derivative at the point?" That seems to be the question being posed here. Except that you phrase it negatively, instead of saying "Does the existence of the limit imply the existence of the derivative at the point?" Why do that? – Michael Hardy Aug 26 '17 at 23:07
  • I have taken the liberty of correcting an error in notation: The two arrows $\to$ and $\mapsto$ in standard usage mean two different things. – Michael Hardy Aug 26 '17 at 23:33
  • I took the first sentence to mean "Assume that $f:\mathbb R\to\mathbb R$ is continuous, and differentiable everywhere but at $0$." and NOT "Assume that $f:\mathbb R\to\mathbb R$ is: continuous and differentiable, everywhere but at $0$." – Michael Hardy Aug 26 '17 at 23:34
  • @MichaelHardy: Yeah, that description is pretty ambiguous. – Brian Tung Aug 26 '17 at 23:35

4 Answers4

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By the mean value theorem there is $c_h\in(0,h)$ such that

$$\frac{f(h)-f(0)}h=f'(c_h)$$ so pass to the limit $h\to0^+$ and you get $f_r'(0)=L$. Similarly you get $f'_l(0)=L$. Conclude.

user66407
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    What if $f(0)$ isn't defined, or is just different from $\lim_{x \to 0} f(x)$? – vonbrand Oct 03 '15 at 16:11
  • @vonbrand $f$ is assumed to be continuous. – Cameron Williams Oct 03 '15 at 16:13
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    If $f$ isn't defined on $0$ then there is no possibility to get that $f$ is differentiable at $0$ and the question in this case doesn't make sense@vonbrand – user66407 Oct 03 '15 at 16:15
  • @CameronWilliams ??? I'm not sure what you're saying doesn't work. The proof given in the answer seems fine to me... – David C. Ullrich Oct 03 '15 at 16:41
  • @CameronWilliams: The function is actually not continuous. "Assume that f:R↦Rf:R↦R is continuous and differentiable everywhere but at 0." – Daniel R. Collins Oct 03 '15 at 16:41
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    @DanielR.Collins The intended grouping is "Assume that $f\colon \mathbb{R}\to \mathbb{R}$ is continuous and (differentiable everywhere but at $0$)", not "… (continuous and differentiable) everywhere but at $0$". – Daniel Fischer Oct 03 '15 at 16:48
  • @Daniel Fischer: That seems at least ambiguous. – Daniel R. Collins Oct 03 '15 at 16:51
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    @DanielR.Collins: I agree with Daniel Fischer here. With the "(continuous and differentiable)" interpretation, the word "continuous" would be completely redundant... – Hans Lundmark Oct 03 '15 at 17:03
  • @DavidC.Ullrich You're right. I had slightly misremembered the conditions for the MVT. – Cameron Williams Oct 03 '15 at 17:13
  • @vonbrand : The domain of $f$ is given to be $\mathbb{R}$. Consequently, $f(0)$ is defined. A (total) function is defined on its entire domain. (If you want to talk about partial functions, see http://cstheory.stackexchange.com/ ) – Eric Towers Oct 03 '15 at 20:59
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This is an application of L'Hospital's Rule:

$$f'(0) = \lim_{t \to 0} \frac{f(t)-f(0)}{t-0} = \lim_{t \to 0} f'(t)=L$$

where the second equality is given by the rule.

Michael Hardy
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Roberto Nunez
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    Doesn't L'Hospital's Rule require that both nominator and denominator are both $0$ or both $\pm\infty$? To use it you would have to assume that $\lim_{t \to 0} (f(t) - f(0)) = 0$ so $\lim_{t \to 0} f(t) = f(0)$ and therefore the function would be continuous at $f(0)$. But as function may not be continuous at $f(0)$ then there is no reason for $\lim_{t \to 0} f(t) = f(0)$ and the nominator might not tend to 0. So $\lim_{t \to 0} \frac{f(t) - f(0)}{t - 0}$ might also diverge or not even exist (Take $f(x) = \frac{1}{x}$ for $x \neq 0$ and $f(0) = 0$ or $f(x) = e^{-\left|x\right|}$). – Maja Piechotka Oct 03 '15 at 19:48
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    @ MaciejPiechotka The function is not differentiable at 0 (a priori) but it is continuos at 0. If the function is not even continuos at 0, then the statement of the problem is not true. See @DanielR.Collins comments above. – Roberto Nunez Oct 03 '15 at 23:38
  • Still $f(x) = e^{-\left|x\right|}$ is continuous at $0$ as $\lim_{x \to 0^+} e^{-\left|x\right|} = \lim_{x \to 0^-} e^{-\left|x\right|} = e^{-\left|x\right|} = 1$. Or even simpler $f(x) = |x|$. – Maja Piechotka Oct 03 '15 at 23:56
  • @ MaciejPiechotka these functions don't satisfy $\lim{x \to 0} f^{'}(x)=L$ for any L, which is one of the hypothesis. – Roberto Nunez Oct 04 '15 at 00:02
  • Sorry. Missed that one. – Maja Piechotka Oct 04 '15 at 00:14
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There is an ambiguity in your question. I assume you mean $f(x)$ is continuous everywhere but at $x=0$ and differentiable everywhere but at $x=0$. In this case what you said and the converse of what you said can be violated.

I give you some examples. Consider the following function

$$f(x) = \left\{ \matrix{ {x^2}\sin ({1 \over x})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \ne 0 \hfill \cr 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \hfill \cr} \right.$$

It's derivative is

$$f'(x) = \left\{ \matrix{ 2x\sin ({1 \over x})\, - \cos ({1 \over x})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \ne 0 \hfill \cr 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \hfill \cr} \right.$$

You can simply see that $f'(0)$ exist since

$$f'(0) = \mathop {\lim }\limits_{x \to 0} {{f(x) - f(0)} \over {x - 0}} = \mathop {\lim }\limits_{x \to 0} {{f(x)} \over x} = \mathop {\lim }\limits_{x \to 0} {{{x^2}\sin ({1 \over x})} \over x} = \mathop {\lim }\limits_{x \to 0} x\sin ({1 \over x}) = 0$$

but $\mathop {\lim f'(x)}\limits_{x \to 0} $ doesn't exist. Hence, the existence of $f'(0)$ doesn't imply the existence of $\mathop {\lim f'(x)}\limits_{x \to 0} $.

The vice versa can also happen, i.e., $\mathop {\lim f'(x)}\limits_{x \to 0} $ exists but $f'(0)$ doesn't exist. For this case you can consider the following simple function

$$f(x) = \left\{ \matrix{ \sin (x)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \ne 0 \hfill \cr 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \hfill \cr} \right.$$

which is a discontinuous function at $x=0$ and hence $f'(0)$ doesn't exist but simply you can check that $\mathop {\lim f'(x)}\limits_{x \to 0} $ exists.

Michael Hardy
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Hosein Rahnama
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    Irrelevant. The first example just shows that the converse is false. The last example doesn't apply because the OP assumed $f$ was continuous. – David C. Ullrich Oct 03 '15 at 16:38
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    About the first part you are right. I just wanted to say the converse is also false. About the second part, question says that it can be discontinous at $x=0$! :) – Hosein Rahnama Oct 03 '15 at 16:44
  • @H.R.: The only reasonable interpretation of the question is "assume $f$ is continuous (everywhere), and also assume $f$ is differentiable for $x \neq 0$". Otherwise the word "continuous" would be superfluous. – Hans Lundmark Oct 03 '15 at 16:55
  • I think we have some problems in comprehension now! – Hosein Rahnama Oct 03 '15 at 17:00
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    @ Hans Lundmark: I do not agree with you! When $f(x)$ is not differentiable at $x=0$, there are two possibilities for continuity at $x=0$. This makes an ambiguity in the question. – Hosein Rahnama Oct 03 '15 at 17:05
  • Who voted me down!? That's unfair! It's the ambiguity of question! – Hosein Rahnama Oct 03 '15 at 17:18
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    I agree with you that the question is sloppily formulated, and in principle can be interpreted in several ways. But the only interesting interpretation is the one where you assume that $f$ is continuous at 0, since if not, (1) obviously $f'(0)$ couldn't exist, and (2) continuity at the other points would be automatic, so why mention it at all? (I didn't downvote your answer, by the way.) – Hans Lundmark Oct 04 '15 at 08:58
  • That's OK. However, in each case, it's a good question to ask. :) I think the down-voter changed his mind since I am voted up now. – Hosein Rahnama Oct 04 '15 at 09:02
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If you take:

$$\begin{align} f(x) = \begin{cases} x^2 & x \ne 0 \\ 1 & x = 0 \end{cases} \end{align}$$

then $f'(x) = 2 x$ unless $x = 0$, and $\lim_{x \to 0} f'(x) = 0$. But the definition of the derivative at $x = 0$ yields:

$$ \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \frac{h^2 - 1}{h} $$

This just doesn't exist. The derivative has a "hole" (more precisely, a removable singularity) at $x = 0$.

Michael Hardy
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vonbrand
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