If $\epsilon>0$ and $f_{\epsilon}$ is the Riemann mapping from $\mathbb{D}$ (unit disk) to the union of $\mathbb{D}$ and the unit disk centered at $2-\epsilon$, made unique by specifying $f_{\epsilon}(0)=0$ and $f_{\epsilon}'(0) >0$, then is it true that $f_{\epsilon} \to $ identity as $\epsilon \to 0$ in some sense of "convergence"?
Intuitively, I want to show that $f_{\epsilon}$ is the identity on most of $\mathbb{D}$ and "blows up" at $1$. In particular this would show that the image of $\lim f_{\epsilon}$ is not the limit if the image of $f_{\epsilon}$.
I am not sure how to proceed. Any help or pointers to relevant theorems would be appreciated.
I was about to say "Montel" when I saw the comment above...
Normalize $f_\epsilon$ so $f_\epsilon(0)=0$, $f_\epsilon'(0)>0$. "Subordination" (ie the Schwarz lemma applied to a certain composition) shows that in fact $f_\epsilon'(0)>1$.
Now say $f$ is the limit of $f_\epsilon$ (or some subsequence). Since $f(\Bbb D)$ is open and connected it's clear that $f(\Bbb D)\subset \Bbb D$. And $f(0)=0$, $f'(0)\ge1$ now shows that $f(z)=z$.