alt Hacker NewsI'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?
Replies
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.
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.
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.