Let $D$ be the unit open disk in $C$ and $S^1$ be its boundary.
Let me start by defining $z_n\in D\to 1 \in S^1$ "non-tangentially" ( abbreviated n.t. ) as $z_n\to 1 $ but all $ z_n $ lie in an angular domain $A$ at $1\in S^1$, i.e. $A$ is the set of all $z\in D : | Arg ( 1 - z ) | \le \theta $ for some $\theta > 0$. You can think of $A$ as the domain bounded in $D$ by the two straight lines that make an angle $\theta $ and $-\theta$ with the radial line segment $[0,1]$.
Let $z_n = x_n + i y_n$. It is easy to show that there is$ M > 0 $ such that the hyperbolic distance $d_n = d_D(z_n,x_n) \le M \forall n \ge 1 $. The proof is to write $e^{d_n} = \frac{1+|\frac{z_n - x_n}{1 - x_n. z_n}|}{1-|\frac{z_n - x_n}{1 - x_n. z_n} |}$. Call $a_n = |\frac{z_n - x_n}{1 - x_n. z_n}| = \frac{|y_n|}{|1- x_n.z_n|}$ .Note that $ |1 - x_n. z_n|\ge |1- z_n|$ when $x_n > 0 $, therefore $ a_n \le \frac{|y_n|}{|1- z_n|} = sin (\theta_n) \le k < 1 \forall n $,here $\theta_n$= angle between $[1,z_n]$ and $[0,1]$. Hence $e^{d_n} \le \frac{1+k}{1-k}$ and it is proved.
But my question is : if $z_n\in D\to 1 \in S^1$ tangentially, i.e. NOT lying in ANY such angular domain $A$ as mentioned before, must we have $d_n = d_D(z_n,x_n)\to \infty$ as $n\to \infty $ ? In other words, must
$ a_n = |\frac{z_n - x_n}{1 - x_n. z_n}| = \frac{|y_n|}{|1- x_n.z_n|} \to 1 $ Note that to prove the non-tangential case, we used the inequality $ |1 - x_n. z_n|\ge |1- z_n|$ in a suitable way for our favor, here we cannot do that .
If the limit is not infinity, can we have some tangential convergence for some sequence $z_n \to 1$ such that $d_n = d_D(z_n,x_n) \le M \forall n \ge 1 $ ?? Can we have explicit example or characterization of such tangential convergences ?
