Given a sequence of real-analytic functions $$f_n:(-\varepsilon_n,1+\varepsilon_n)\to\mathbb{R}$$ and assume that each of their derivatives converges uniformly on $[0,1]$: $$\forall k=0,1,\ldots,\qquad f_n^{(k)}(t)\to g_k(t)\text{ uniformly on }t\in[0,1]$$ It would hence imply that $g_k=g_0^{(k)}$ and in particular, $g_0$ is a smooth function.
My question is that: is $g_0$ also real-analytic on $(0,1)$?
P.S. I see that it doesn't have to be analytic at the endpoints, for example, $f_n(t)=F(t+1/n)$, where $F(u)=e^{-1/u}$. We would have $g_0=F$ with $g_0(0)=0$.