The limit of a power series as the module of the variable approaches the radius of convergence

Question

Suppose that a power series $\sum\limits_{n=0}^{\infty}a_{n}x^{n}$ converges for $|x|<r$ ($r>0$). We will call the sum of this series $f(x)$. Let $x_{0}$ be a point such that $|x_{0}|=r$. Clearly for all $\epsilon>0$ the function $$f(x_{0}-\epsilon)=\sum\limits_{n=0}^{\infty}a_{n}(x_{0}-\epsilon)^{n}$$ is defined. Here is my question: if $f(x)>0$ for all $|x|<r$, can we conclude that $\lim_{\epsilon\to0}f(x_{0}-\epsilon) $ is either a finite number, or $+\infty$?

Answer 1

Consider $$ f(x)=\cos^2\left(\frac{1}{(x-x_0)^2}\right)+1 $$ Then, for every $x_0\ne 0$, the function $f$ is expressible as a power series with radius of convergence $r=|x_0|$, i.e., $$ f(x)=\sum_{n=0}^\infty a_nx^n, \quad x\in (-r,r), $$
and also $f(x)>0$, in the same interval. But the limit $\lim_{x\to r}f(x)$ DOES NOT exist in $\mathbb R\cup\{\infty\}$.