If $f:X\to\mathbb{R}$ is derivable in $\sigma\in X\subset \mathbb{R}$ then $\sigma$ is an accumulation point of $X$? My book says that a function $f:X\to\mathbb{R}$ is derivable at point $\sigma\in X$ if there exists a function $\varphi_{\sigma}:X\to\mathbb{R}$ which is continuous at $\sigma$ such that $f(x)-f(\sigma)=\varphi_{\sigma}(x)(x-\sigma)$. My book says that since every function is continuous at a isolated point of its domain, then it only considers the derivatives of points of accumulation of the domain of $f$. So I'm curious to know if all the points in which $f$ have derivatives are points of accumulation of the domain.
 A: The derivative at a point $x_0$ is defined to be $\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}$. The condition $x\rightarrow x_0$, itself needs $x_0$ to be an accumulation point.
A: $f$ has a derivative but rather trivial.
Let $\sigma$ be an isolated point and;
\begin{equation}
\varphi_\sigma(x) = 
\begin{cases}
\ \ \ \ \ a\quad &if \quad x=\sigma\\
\frac{f(x)-f(\sigma)}{x-\sigma} \quad &\quad\ \ \  \ o/w
\end{cases}
\end{equation}
where $a\in\mathbb{R}$. Now, $\varphi_\alpha$ is continous at $\sigma$ for any $a$ because $\sigma$  is an isolated point. Moreover, $f(x)-f(\sigma)=\varphi_\sigma(x)(x-\sigma)$ is satisfied again for any $a$.
A: A functional limit is defined on punctured neighborhoods of the variable, but there are no sequences that converges to $x_0$ in any punctured neighborhood of $x_0$, thus (by the sequential characterization of functional limits) a functional limit cannot be defined in an isolated point. Consequently any functional limit can be defined only in accumulation points.
