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An explicit disk map of a slit half disk makes the two boundary approaches visible without assuming an extension theorem for the boundary. We construct the map as a succession of bijections and then determine its limits.

First $w=z^2$ maps the upper half-disc of radius two bijectively onto $|w|<4$ slit along $[0,4)$. Indeed, its inverse chooses the square root whose argument lies in $(0,\pi )$. The removed segment on the imaginary axis becomes $[-1,0)$, so the image of $D$ is preciselyThe Möbius transformation $v=4(w+1)/(w+16)$ maps the radius-four disc bijectively onto the unit disc, taking $-1$ to zero and $4$ to one. This follows also by writing it as $(w/4+1/4)/(1+w/16)$, an automorphism of the unit disk after the scaling. It takes the removed real interval to $[0,1)$, so the image is the unit disc with this radial slit removed. Choosing $0<\arg v<2\pi$ defines a single-valued square root $u=\sqrt{v}$ and maps that slit disc bijectively ontoAll these maps have nonzero derivatives on their respective domains, since their potentially problematic zero or branch point is on the removed slit.

Next the Joukowski map $J(u)=u+1/u$ is a conformal bijection from $U$ onto the lower half-plane. For $u\in U$,If $J(u)=J(t)$, then $(u-t)(1-1/(ut))=0$, and $|ut|<1$ rules out the second factor vanishing. To prove surjectivity, take $q$ in the lower half-plane and consider $u^2-qu+1=0$. Neither root lies on the unit circle, where $u+1/u$ is real. Their product is one, so exactly one has modulus less than one. For that root the displayed imaginary-part identity forces $\mathrm{Im}u>0$. Thus it lies in $U$. Also $J^{\prime }(u)=1-u^{-2}$ cannot vanish there. Finally $C(q)=(q+i)/(q-i)$ maps the lower half-plane bijectively onto the unit disc: the inequality $|q+i|<|q-i|$ is exactly $\mathrm{Im}q<0$, and its inverse is $q=i(\zeta +1)/(\zeta -1)$. Consequentlyis a conformal bijection of $D$ onto the disc.

At the slit point $z_0=i/2$ the continuous rational expression for $v$ tends to $4/21$. In a sufficiently small neighborhood of $z_0$, points of $D$ have nonzero real part. Since $\mathrm{Im}(z^2)=2\mathrm{Re}z\mathrm{Im}z$ and the real-coefficient Möbius map $w\mapsto v$ preserves the sign of the imaginary part, right-side approach gives $\arg v\to 0$, while left-side approach gives $\arg v\to 2\pi$. HenceThe two limits of $F$ are thereforeThey have modulus one and are distinct, since $L>0$ and the Möbius map $C$ is injective on the extended real line. The two-sided cluster set at an interior slit point is a geometric consequence of opening the slit into two boundary intervals.

For the given map $f$, the composition $B=f\circ F^{-1}$ is an automorphism of the unit disk. To see its precise boundary behavior, let $a=B^{-1}(0)$ and compose $B$ with the inverse of $T_a(\zeta )=(\zeta -a)/(1-\overline{a}\zeta )$. This gives a disc automorphism fixing zero. Applying Schwarz lemma to it and its inverse shows it is a rotation. ThusIts denominator never vanishes on the closed disc, and its inverse has the same property; it therefore extends to a continuous bijection of the unit circle. Put $\alpha =B({\gamma }_{+})$ and $\beta =B({\gamma }_{-})$. These are distinct points of that circle, independent of the approaching sequence.

For any sequence approaching $z_0$, every infinite right-side subsequence has image tending to $\alpha$, and every infinite left-side subsequence has image tending to $\beta$. At least one side occurs infinitely often. Conversely, any convergent image subsequence has a further subsequence confined to one side, so its limit must be the corresponding one of these two points. The cluster set is therefore exactly $\{\alpha \}$, $\{\beta \}$ or $\{\alpha ,\beta \}$, according to which sides occur infinitely often. Both sides really occur in the domain, for example along $i/2\pm {\epsilon }_n$ with positive ${\epsilon }_n\downarrow 0$.