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A simple question which I came across recently. Just wanted to confirm if my logic on it is right....

Suppose $f(x)$ is a function and it's given that it's differentiable everywhere except possibly at $0$, but for the point "$0$", it's given that $\lim_{x\to 0} f'(x) = 0.$ We need to prove that its differentiable everywhere.

What I think is the left hand derivative will be equal to $\lim_{x\to 0^-} f'(x) = 0$ and similarly right hand derivative will be equal to $0$ too on similar lines.

Is this right? I thought of the question myself as I was working on some other question and this bumped in my head. Can we do it this way or there should be some more condition given in the question to work our proof? Thanks.

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    $\begingroup$ Yes, but you'll need to prove the one-sided derivatives (the limit of the difference quotients) actually exist and equal $0$. Also, for your statement to be true, you need to assume that $f$ is continuous at $0$. See this post and this post for ideas. $\endgroup$ – David Mitra Jun 17 '13 at 2:05
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    $\begingroup$ Consider the function that is $1$ at $0$ and $0$ everywhere else. $\endgroup$ – Javier Jun 17 '13 at 2:39
  • $\begingroup$ Thanks David for the links... got the concept! $\endgroup$ – under root Jun 17 '13 at 12:18
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Suppose $f$ is continuous at $0$. Without this assumption, user math_man's answer shows that the statement is not true.

We need to argue that $\displaystyle f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}h$ exists. But this is now immediate from L'Hôpital's rule: The assumptions that $f$ is continuous at $0$ (and therefore everywhere), and differentiable away from $0$ give us that we can apply the mean-value theorem. This allows us to conclude that for any $h$ there is a $\xi_h$ in between $0$ and $h$ such that $$ \frac{f(h)-f(0)}h=\frac{f'(\xi_h)}1=f'(\xi_h). $$ Now note that $\xi_h\to0$ as $h\to 0$. Since we are given that $\lim_{t\to0}f'(t)=0$, it follows that $\lim_{h\to0}f'(\xi_h)=0$ as well, so $f'(0)$ exists and is $0$.

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  • $\begingroup$ this cleared everything...thanks a lot! :) $\endgroup$ – under root Jun 17 '13 at 12:00
  • $\begingroup$ Why does it follow that limit is equal to $0$? $f'$ needs not be continuous, so why $\xi_h\to0$ as $h\to0$ and $f'(t)\to0$ as $t\to0$ imply $f'(\xi(h))\to0$ as $h\to0$? $\endgroup$ – sas Nov 24 '16 at 14:45
  • $\begingroup$ @sas This is immediate from the definition of limit. $\endgroup$ – Andrés E. Caicedo Nov 24 '16 at 14:48
  • $\begingroup$ Suppose $\xi(h)=0$ everywhere and $f'(t)=0$ everywhere except $f'(0)=1$ (I know that can'be true for derivative, but let it be arbitrary function). Then $\xi\to0$ as $h\to0$, $f'\to0$ as $t\to0$ but limit of $f'(\xi(h))$ is equal to $1$ $\endgroup$ – sas Nov 24 '16 at 14:52
  • $\begingroup$ @sas $\xi_h$ is not zero by definition. $\endgroup$ – Andrés E. Caicedo Nov 24 '16 at 14:54
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Well, if you imagine the function $f(x)=1 \;\;x\in [0,\infty)$ and $f(x)=-1\;\;x\in(-\infty,0)$ then it is differentiable everywhere except in $0$ and $f'(x)=0\;\forall x\neq 0$. Anda $\lim_{x\to 0}f'(x)=0$. And this functions is not continuous in $0$, and so it is not differentiable.

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  • $\begingroup$ thanks for your feedback...got the visualization now... $\endgroup$ – under root Jun 17 '13 at 12:01

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