One minor nit: A function can be differentiable at a and discontinuous at a even with the standard definition of the derivative. A trivial example would be the function f(x) = (x²-1)/(x-1) which is undefined at x=1, but f'(1)=1 (in fact derivatives have exactly this sort of discontinuity in them which is why they’re defined via limits). In complex analysis, this sort of “hole” in the function is called a removable singularity¹ which is one of three types of singularities that show up in complex functions.
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1. Yes, this is mathematically the reason why black holes are referred to as singularities.
I'm not understanding what you're saying. The standard definition of the derivative of f at c is
f'(c) = lim_{h → 0} (f(c + h) - f(c))/h
The definition would not make sense if f wasn't defined at c (note the "f(c)" in the numerator). For instance, it can't be applied to your f(x) = (x² - 1)/(x - 1) at x = 1, because f(1) is not defined.
And it's a standard result (even stated in Calc 1 classes) that if a function is differentiable at a point, then it's continuous there. For example:
5.2 Theorem. Let f be defined on [a, b]. If f is differentiable at a point x ∈ [a, b], then f is continuous at x.
(Walter Rudin, "Principles of Mathematical Analysis", 3rd edition, p. 104)
Or:
Theorem 2.1 If f is differentiable at x = a, then f is continuous at x = a.
(Robert Smith and Roland Minton, "Calculus -Early Transcendentals", 4th edition, p. 140)
It's true that your f(x) = (x² - 1)/(x - 1) has a removable discontinuity at x = 1, since if we define g(x) = f(x) for x ≠ 1 and g(1) = 2, then g is continuous. Was this what you meant?
This is correct. You cannot have a discontinuity with any accepted definition of a derivative (and your definition is explicit about this: the value f(c) must exist). Just allowing the limits on both sides to be equal already has a mathematical definition which is that of a functional limit, the function in this case being (f(x) - flim(c))/ (x-c) where flim(c) is the value of a (different) functional limit of f(x): x->c (as f(c) doesn't exist).
and yes, by defining a new function with that hole explicitly filled in with a defined value to make it continuous is the typical prescription. It does not imply the derivative exists for the other function as the other post posits.
https://en.m.wikipedia.org/wiki/Classification_of_discontinu... is responsive and quite accessible. It notes that there doesn't have to be an undefined point for a function to be discontinuous (and that terminology often conflates the two), and matches what I recall of determining that if the limit of the derivative from both sides of the discontinuity exists and is equal, the derivative exists.
> ... there doesn't have to be an undefined point for a function to be discontinuous.
That's right. In the example f(x) = (x² - 1)/(x - 1) for x ≠ 1, if we further define f(1) = 0, the function is now defined at x = 1, but discontinuous there.
> ... if the limit of the derivative from both sides of the discontinuity exists and is equal, the derivative exists.
(You probably mean "both sides of the point", since if there's a discontinuity there the derivative can't exist.) Your point that, if the left and right-hand limits both exist and are equal, then the derivative exists (and equals their common value) is true for all limits.
Also, there's a difference between the use of the word "continuous" in calc courses and in topology. In calc courses where functions tend to take real numbers to real numbers, a function may be said to be "not continuous" at a point where it isn't defined. So f(x) = 1/(x - 2) is "not continuous at 2". But in topology, you only consider continuity for points in the domain of the function. So since the (natural) domain of f(x) = 1/(x - 2) is x ≠ 2, the function is continuous everywhere (that it's defined).
I was actually aiming for the situation where a function is defined on all reals but still discontinuous (e.g. the piecewise function in the wiki article for the removable discontinuity). So there's a discontinuity (x=1), however the function is defined everywhere.
The standard definition of a derivative c involves the assumption that f is defined at c.
However, you could also (probably) define the derivative as lim_{h->0} (f(c+h) - f(c-h))/2h, so without needing f(c) to be defined. But that's not standard.
> However, you could also (probably) define the derivative as lim_{h->0} (f(c+h) - f(c-h))/2h, so without needing f(c) to be defined. But that's not standard.
Although this gives the right answer whenever f is differentiable at c, it can wrongly think that a function is differentiable when it isn't, as for the absolute-value function at c = 0.
> this sort of “hole” in the function is called a removable singularity
It's called "removable" because it can be removed by a continuous extension - the original function itself is still formally discontinuous (of course, one would often "morally" treat these as the same function, but strictly speaking they're not). An important theorem in complex analysis is that any continuous extension at a single point is automatically a holomorphic (= complex differentiable) extension too.
I don't think it makes sense to allow derivatives of a function f to have a larger domain than the domain of f.
>which is why they’re defined via limits
They're defined via studying f(x+h) - f(x) with a limit h -> 0. But, your example is taking two limits, h->0 and x->1, simultaneously. This is not the same thing.
You are wrong. In order for you to make sense of what you are saying, you first must REDEFINE f(x) to be f(x) = (x^2 - 1)(x - 1) when x != 1 and define f(1) = 2. Of course, then f will be continuous at x = 1 also.
A function is continuous at x = a if it is differentiable at x = a.
You do understand the concept, but your precision in the definitions is lacking.
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1. Yes, this is mathematically the reason why black holes are referred to as singularities.