If $f,g$ real analytic and $\lim_{t \to t_0} f(t)/g(t)$ exists then $f/g$ is analytic

Question

If $f,g$ are real analytic at $t_0$ and $\lim_{t \to t_0} f(t)/g(t)$ exists then is it true that $f/g$ with the limiting value filled in at $t= t_0$ is real analytic at $t_0$? I know the complex version is true, so that if $\lim_{z \to z_0} f(z)/g(z)$ exists, then $f/g$ can be analytically continued to the point $t_0$, but I wonder if the same holds for just real analytic. I'm thinking yes, just take the power series and view them as complex functions, but I'm not entirely sure.

Answer 1

Yes, exactly. If $I \subset \mathbb{R}$ is an interval, and $t_0 \in I$, a function $f \colon I \to \mathbb{C}$ is real-analytic at $t_0$ if and only if there is an open set $U\subset \mathbb{C}$ with $t_0 \in U$ and a holomorphic $F \colon U \to \mathbb{C}$ such that $F\lvert_{U\cap I} = f\lvert_{U\cap I}$.

So if $f,g$ are real-analytic at $t_0 \in (a,b)$, there are holomorphic functions $F,G\colon U \to \mathbb{C}$, where $U$ is an open set containing $t_0$ such that $F\lvert_{U\cap (a,b)} = f\lvert_{U\cap (a,b)}$ and ditto for $G$ and $g$. By assumption $g$ doesn't vanish identically, hence neither does $G$, and so $\dfrac{F}{G}$ is meromorphic on $U$. But the existence of $\lim\limits_{t\to t_0} \dfrac{f(t)}{g(t)}$ rules out the possibility that $\dfrac{F}{G}$ has a pole at $t_0$, so $\dfrac{F}{G}$ is holomorphic on a neighbourhood $V$ of $t_0$, and thus its restriction $\dfrac{f}{g}$ is real-analytic at $t_0$.