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I drew pictorially what each derivative looks like. For the second derivative, the derivative limits on both sides of x=0 approach 0, so we should have the third derivative equals 0 even though 1) the first derivative is undefined and 2) the second derivative is undefined.

user3180
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    First deivative at $0$ does not exist so there is no question of existence of second derivative. – Kavi Rama Murthy Feb 24 '21 at 08:45
  • Just because the second derivative does not exist at 0 (aka second derivative is not continuous at 0) does not mean the third derivative cannot exist at 0 @KaviRamaMurthy – user3180 Feb 24 '21 at 08:50
  • The existence of equal left and right limits does not mean the existence of the limit. – user Feb 24 '21 at 08:50
  • @user Why? If the limit is equal to $L$ from all possible directions, then the limit is $L$. For a scalar function, the only possible directions are from left and right. – user3180 Feb 24 '21 at 08:51
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    The definition of $f'(x)$ is $\lim_{h\to 0} \frac {f(x+h)-f(x)} h$ and not $\lim_{y\to x} f'(y)$ You have to revisit the definition of derivative. – Kavi Rama Murthy Feb 24 '21 at 09:11

2 Answers2

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The definition of the third derivative is $$ f'''(0) = \lim_{h \to 0} \frac{f''(h) - f''(0)}{h} $$ so if $f''(0)$ doesn't exist (as in this case) it's meaningless to even start talking about this limit, and therefore $f'''(0)$ doesn't exist either.

Hans Lundmark
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The limit of the third derivative $f'''(x)$ as $x$ approaches zero exists and is indeed zero. However, for the derivative to be continuous in this point (or even to exist in this point), it is necessary for this limit to be equal to the function ($f'''(x)$) evaluated in that point. In this case, $f'''(x)$ is not defined in $x=0$, as $f''(x)$ is also not defined in $x=0$ $(f^{(n+1)}(x)$ can only be defined at most in the intervals where $f^{(n)}(x)$ is defined), so $f'''(0)$ does not exist.

In summary, it is clear that the existence of the limit and the existence of the function are two different things. A function can be undefined in a point and yet have a limit at that point, as is the case here for $f'''(x)$.

Clerni
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  • Hans just said above why the first line is not true. By definition we need $f''(0)$ to exist. – user3180 Feb 24 '21 at 09:29
  • @user3180: No, the first line here is correct; the limit of $f'''(x)$, as $x \to 0$, exists and is zero. – Hans Lundmark Feb 24 '21 at 10:16
  • @user3180: The only thing I would complain slightly about is the use of the word “discontinuity” here, since I don't like applying that word to points where the function in question is undefined. (See discussion here, for example: https://math.stackexchange.com/questions/1087623/is-function-f-mathbb-c-0-rightarrow-mathbb-c-prescribed-by-z-rightarrow.) – Hans Lundmark Feb 24 '21 at 10:19
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    @HansLundmark You're right. I changed it now. – Clerni Feb 24 '21 at 11:09