# Thread: removable discontinuity and the derivative

1. ## removable discontinuity and the derivative

I understand the proof that says a differentiable function must be continuous. I also "see" that the two sided limits of the difference quotient must be equal in order to talk about the derivative. But if the derivative is a limit why doesn't it exist about a "hole" or removable discontinuity when other limits do?

2. Originally Posted by zg12
I understand the proof that says a differentiable function must be continuous. I also "see" that the two sided limits of the difference quotient must be equal in order to talk about the derivative. But if the derivative is a limit why doesn't it exist about a "hole" or removable discontinuity when other limits do?

Because that precise limit implies the definition of the function on the point: $f'(a):=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}\Longrightarrow f(a)$ must

exist for the derivative to have any chance to exist.

Tonio

3. So, if it were otherwise you would have in the numerator f(x)-undefined/(x-a) which doesn't really make since. Is this is what you're saying? I think I can buy that.

4. Originally Posted by zg12
So, if it were otherwise you would have in the numerator f(x)-undefined/(x-a) which doesn't really make since. Is this is what you're saying? I think I can buy that.

Otherwise the definition makes no sense...not only that: the function must be defined in the point a and also in some

open neighborhood of this point for us to be capable to begin evaluating that limit...

Tonio

5. It is possible to generalize the definition of derivative to $\lim_{h\to 0}\frac{f(a+h)- f(a- h)}{2h}$ and avoid the question of the value at a. But that is, as I say, a generalization and not the standard "derivative".