$z,\zeta\to 0$ nontangentially is bounded uniformly in norm as $w$ varies in the unit ball of $\mathcal A$.

By Proposition \[prop:3.1\], $$\begin{aligned}
\lefteqn{\left\|\left(\Im f(\alpha+zv_1)\right)^{-\frac12}
\Delta f(\alpha+zv_1,\alpha+\zeta v_2)(w)
\left(\Im f(\alpha+\zeta v_2)\right)^{-\frac12}\right\|}\\
&\leq&\left\|(\Im zv_1)^{-\frac12}w(\Im\zeta v_2)^{-\frac12}\right\|.
\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\end{aligned}$$ Multiplying by $(\Im z\Im\zeta)^{1/2}$ we obtain $$\left\|\left[\frac{\Im f(\alpha+zv_1)}{\Im z}\right]^{-\frac12}
\Delta f(\alpha+zv_1,\alpha+\zeta v_2)(w)\left[\frac{\Im f(\alpha+\zeta v_2)}{\Im\zeta}\right]^{-\frac12}
\right\|\leq\left\|v_1^{-\frac12}wv_2^{-\frac12}\right\|.$$ Let $\varepsilon\ge0$ be fixed, and denote $f_\varepsilon(a)=
f(a)+\varepsilon a$, i.e. $f_\varepsilon=f+\varepsilon{\rm Id}$. Since $\text{Id}$ is completely positive, $f_\varepsilon$ is still a noncommutative self-map of the noncommutative upper half-plane of $\mathcal A$, so that $$\begin{aligned}
\lefteqn{\left
\|\left(\frac{\Im f(\alpha+zv_1)}{\Im z}+\varepsilon v_1\right)^{-\frac12}
(\Delta f(\alpha+zv_1,\alpha+\zeta v_2)(w)+\varepsilon w)\right.\times}\\
& & \left.\left(\frac{\Im f(\alpha+\zeta v_2)}{\Im\zeta}+\varepsilon v_2
\right)^{-\frac12}\right\|\\
& \leq & \left\|v_1^{-\frac12}wv_2^{-\frac12}\right\|.\quad\quad\quad
\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad
\quad\quad\quad\quad\quad\quad\quad\quad\end{aligned}$$ For simplicity, we denote $A_1(\Im z,\varepsilon)=
\frac{\Im f(\alpha+zv_1)}{\Im z}+\varepsilon v_1$, $A_2(\Im\zeta,\varepsilon)=
\frac{\Im f(\alpha+\zeta v_2)}{\Im\zeta}+\varepsilon v_2$, $W(z,\zeta,\varepsilon)=
\Delta f(\alpha+zv_1,\alpha+\zeta v_2)(w)+\varepsilon w$, and $K=\left\|
v_1^{-\frac12}wv_2^{-\frac12}\right\|^2$. As noted in , and following the same procedure as in the proof fo the previous proposition, the above is equivalent to $$A_2(\Im\zeta,\varepsilon)^{-\frac12}W(z,\zeta,\varepsilon)^*
A_1(\Im z,\varepsilon)^{-1}W(z,\zeta,\varepsilon)A_2(\Im\zeta,\varepsilon)^{-\frac12}
\leq K1.$$ As $A_j(\cdot,\varepsilon)\ge\varepsilon1$, we obtain by the same methods as in the proof of Proposition \[Hyper\] that $$\|W(z,\zeta,\varepsilon)\|^2\leq K\|A_1(\Im z,\varepsilon)\|\|A_2(\Im\zeta,\varepsilon)\|.$$ Let $\mathcal H$ be the Hilbert space on which $\mathcal A$ acts as a von Neumann algebra. By our hypothesis, $\lim_{y\downarrow 0}\frac{\langle\Im f(\alpha+iyv_j)\xi,\xi\rangle}{y}=
\lim_{y\downarrow 0}\left\|\left(\frac{\Im f(\alpha+iyv_j)}{y}
