Let $G \subseteq \mathbb{C}$ be a domain (one may assume that $G$ is bounded and simply connected, if needed). Suppose $(f_n)_{n \in \mathbb{N}}$ is a sequence of bounded (i.e. $\sup_{z \in G} |f_n(z)| < \infty$ for all $n$) quasiconformal mappings $f_n: G \rightarrow \mathbb{C}$ that converges uniformly to a bounded continuous (not necessarily quasiconformal!) mapping $f \in C(G)$, i.e. $$d_{\sup}(f_n, f) \overset{n \to \infty}{\longrightarrow} 0$$ with $d_{\sup}(f_n, f) := \sup_{z \in G} |f_n(z) - f(z)|$ denoting the supremum metric. By a well-known result on sequences of quasiconformal mappings, the limit mapping $f$ will either be a quasiconformal mapping of $G$ into $\mathbb{C}$ or some degenerated function (e.g. constant), provided that the maximal dilatations $K(f_n)$ of the mappings $f_n$ are uniformly bounded, i.e. $K(f_n) \leq K$ for all $n$ and some fixed $K \in [1, \infty)$ (see for example Lehto/Virtanen, Quasiconformal Mappings in the Plane, Theorem 5.3, p. 74). Utilizing the complex dilatation $\mu_n := \frac{(f_n)_{\overline{z}}}{(f_n)_z}$ of the mappings $f_n$, this may also be written as $$ \| \mu_n \|_{L^\infty} \leq k < 1 $$ for corresponding $k := \frac{K-1}{K+1} \in [0,1)$.
My question is: Can it happen in this situation that the maximal dilatations $K(f_n)$ are unbounded (or equivalently, $\| \mu_n \|_{L^\infty} \overset{n \to \infty}{\longrightarrow} 1$)? Or does the uniform convergence of the $f_n$ somehow force the corresponding real sequence $(K(f_n))_{n \in \mathbb{N}}$ to be necessarily bounded?
Any hints, (counter)-examples, reference requests etc. on this (and related) question(s) are highly appreciated!
