Abel's theorem for the derivative of a power series Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is a function, $(a_0, a_1, \dots)$ is a sequence of real numbers and $x_*$ is a positive real number, such that the following two conditions hold: for all $x \in (-x_*, x_*)$, the power series $\sum_{n = 0}^\infty a_n x^n$ converges to $f(x)$ and $\sum_{n = 0}^\infty a_n {x_*}^n$ converges. Abel's theorem tells us that
$$
\lim_{\substack{x \rightarrow x_* \\ x < x_*}} f(x) = \sum_{n = 0}^\infty a_n {x_*}^n
$$
Suppose $f$ is continuous from the left at $x_*$.
It is a standard result that $f$ is differentiable inside $(-x_*, x_*)$ and that for every $x \in (-x^*, x^*)$, the series $\sum_{n = 1}^\infty n a_n x^{n - 1}$ converges to $f'(x)$. Suppose the last series converges at $x_*$ as well, and denote this sum by $s$.
Is it the case that $f$ is differentiable from the left at $x_*$?
Suppose $f$ happens to be differentiable from the left at $x_*$. Is it the case that $f'(x^*) = s$, where the derivative is the left-sided one?

Yes and yes.
$$
\lim_{\substack{x \rightarrow x_* \\ x < x_*}} \frac{f(x) - f(x_*)}{x - x_*} = \lim_{\substack{x \rightarrow x_* \\ x < x_*}} f'(x) = s
$$
The first equality is due to L'Hôpital's rule and the second one to Abel's theorem.
