Infinite differentiability with a removable discontinuity?

Question

I'm still a beginner with calculus. But this puzzled me. Let's say you had $f(x) = \frac{x^2-1}{x+1}$. It's discontinuous at one point. If you took the derivative infinitely many times, would the resulting function be continuous at every point except at the hole?My intuition is as follows: because the limit to a function relies on two points which converge to an infinitesimal difference (which we call one point), doing this an infinite amount of times would make the difference actually significant. If this does not make sense, please allow me to explain more.

Answer 1

Note that $x^2-1=(x-1)(x+1)$ and hence the point $x=-1$ is a removable discontinuity of $f$. Therefore, if you define the derivative with, say, the central difference

$$ f^{\prime}(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x-h)}{2h}, $$

$f^\prime(-1)$ is well-defined (and so too are all subsequent derivatives). Having said that, this approach is messy, and it would be better to simply forget about $f$ in its original form and consider instead the modification

$$ \hat{f}(x)=\lim_{y\rightarrow x}f(x) $$

for which the removable discontinuity is no longer an issue.

Answer 2

The function you have is $f:\mathbb{R} \setminus \{-1\} \to \mathbb{R}$, $x \mapsto x-1$. Every $c \in \mathbb{R} \setminus \{-1\}$ can be enclosed inside an open interval where $f$ does not have a hole (and is hence locally a polynomial) and so $f'(c), f''(c), f^{(n)}(c)$ exist and are continuous for all $n$.