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If a function is non-continuously differentiable at a point $c$, then it has a derivative at $c$, but the derivative is not continuous at $c$. This makes no sense to me. It is seemingly the conjunction of two contradictory statements:

$$\lim_{x \rightarrow c} \frac{f(x) - f(c)}{x-c} \in \Bbb R \tag1$$

$$$$

$$\lim_{x \rightarrow c}\lim_{x \rightarrow c} \frac{f(x) - f(c)}{x-c} \not\in \Bbb R \tag2$$

The second statement is true, because $f'(x)$ is discontinuous at $c$. Thus, either the left- and right-hand limits are not equal, or, the limit does not exist at all, both of which are captured by $(2)$. But composing limits is idempotent, so $(1)$ and $(2)$ are contradicting each other.

What am I missing?

user110391
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    Your statement (2) doesn't make sense. You can't take a limit with respect to the same variable twice, and how are you getting "$\notin\Bbb R$" from non-continuity? I suggest reviewing the definition of continuous. The standard example of a function with this property is $x^2\sin(\frac1x)$ at $x=0$. – Karl Nov 01 '23 at 18:59
  • You've also thrown in $|\cdot|$ signs which shouldn't be there – MPW Nov 01 '23 at 19:05
  • @Karl A derivative is always undefined at its discontinuities, since a derivative at a point is by definition the left- and right-hand limit as $x$ approaches that point. Thus, a jump discontinuity is also undefined for as long as we're dealing with a derivative. Also, why can I not take the limit with respect to the same variable twice? Is it not just an idempotent composition? – user110391 Nov 01 '23 at 19:07
  • You seem to be confusing the continuity of $f$ with the continuity of its derivative. Have you thought about the $x^2\sin(\frac1x)$ example? – Karl Nov 01 '23 at 19:43
  • @Karl See this Q&A for insight into my mistunderstanding: https://math.stackexchange.com/questions/4772434/why-is-a-derivative-undefined-at-its-discontinuities – user110391 Nov 01 '23 at 19:57

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The problem is in your second statement. Indeed, as you've written it, the first and second equations contradict each other. To calculate the derivative at a point $c$, we take a limit of a ratio involving $f$ as a dummy variable (which you write as $x$) approaches $c$. This dummy variable only exists within the context of this limit. To check the continuity of of the derivative at a point $c$, we need to calculate the limit $f'$ as a dummy variable approaches $c$--in particular this needs to be a new dummy variable. So I'd rewrite your second equation as

$$f''(c) = (f')'(c) = \lim_{x \to c} f'(x) = \lim_{x \to c} \lim_{y \to x} \frac{f(x) - f(y)}{x - y} \; \text{does not exist}.$$

It may help to see how discontinuities happen in derivatives by looking at some real examples: for that, take a look at this pdf which works through the classic example $$f(x) = \begin{cases} x^2\sin(1/x) & x \neq 0 \\ 0 & x = 0\end{cases}$$ (or check it for yourself with just the limit definition!)

George K.
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  • @User110391 [responding to a now-deleted comment] $f'(c)$ does exist; the question is, does the limit of $f'(x)$ exist as $x$ approaches $c$? The simplification you suggest is of the form (for some function $g$) $\lim_{x\to c} g(x) = g(c)$; but this is manifestly the definition of continuity. Since not all functions are continuous at all points, we can't just remove the limit--we need to prove the two sides are equal, which is not always the case. – George K. Nov 01 '23 at 19:42
  • Yes I realize my simplification was mistaken, so I deleted the comment in hopes to replace it with a better question before you replied. I guess my only remaining confusion comes from me over-generalizing the reasoning behind George Ivey's answer. Whenever $f$ has a break-point, the derivative has a jump discontinuity, AND it is undefined. But sometimes, the derivative may have a jump discontinuity, yet it is defined nonetheless. Correct? – user110391 Nov 01 '23 at 19:46
  • All good :) I'm actually not in love with George Ivy's answer. In the function you give, the problem is that the slope of the tangent looks like 1 from the right, and looks like 2 from the left, so there's no meaningful way to assign a derivative to $f$ at 1. As for jump discontinuities: There are functions whose derivatives are discontinuous and defined at the discontinuity; the paper I link above goes over an example. But these functions don't usually look like traditional "jumps." – George K. Nov 01 '23 at 19:57
  • @user110391 For what it's worth, your previous comment is not correct in two points: 1. When $f$ has a break-point (or jump discontinuity), the derivative $f'$ at that point does not exist. In your linked example, the limits of one-sided derivatives exist and are equal, but that does not imply differentiability. (!) 2. As it happens, a derivative cannot have a jump discontinuity. The fact that "derivatives satisfy the intermediate value property" is often called Darboux's theorem. – Andrew D. Hwang Nov 01 '23 at 20:22
  • "In the function you give, the problem is that the slope of the tangent looks like $1$ from the right, and looks like $2$ from the left, so there's no meaningful way to assign a derivative to $f$ at $1$." Wouldn't that be the case for any function with a discontinuous derivative, however? The slope has to be different in order for there to exist a discontinuity. It's just that sometimes this difference causes a lack of definition, and other times it does not. I am still trying to grasp what the meaningful difference between these scenarios are. – user110391 Nov 01 '23 at 20:24
  • @AndrewD.Hwang Your comment led me to realize I have conflated discontinuities with singularities (I may be misusing the term singularity here). That is, I thought any undefined point in a function was a discontinuity, but according to a Wikipedia article on classes of discontinuities, it is not. I believe this helps me with my most recent comment, taken together with no jump discontinuities existing in derivatives. – user110391 Nov 01 '23 at 20:33
  • @user110391 In George K's example, the derivative exists at $0$ (and everywhere else), but does not have a limit at $0$ from either side. That's another way a function can be discontinuous. – Andrew D. Hwang Nov 01 '23 at 21:14
  • @AndrewD.Hwang I think I see what you mean, but what's weird is that $x^2\sin(1/x)$ seems to have a dampened oscillation as it approaches zero, eventually arriving at a removable discontinuity at $x = 0$, which is filled in by the piecewise function's definition of $f(x) = 0$, though that's besides the point. The limit as $x^2\sin(1/x)$ approaches $x=0$ seems to be $0$ due to the dampening, so why isn't it? Does it have to do with the acceleration of oscillation perhaps? – user110391 Nov 01 '23 at 22:46
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    @user110391 That dampening is in $f$, not $f'$. (Separately, the fact $f'(0) = 0$ can't be deduced piecewise, but has to be established from the definition, by taking limits of difference quotients. ;) – Andrew D. Hwang Nov 02 '23 at 00:11
  • Let us continue this discussion in chat. – user110391 Nov 02 '23 at 13:18