Given strictly increasing sequence $x_n$ of rational numbers with $\sup x_n = x$.
In which case (sufficient condition on $x_n$) there exists real-analytic function $f:U_\epsilon(0)\to\mathbb{R}$ defined on some $\epsilon$-neighborhood of $0$ such that $f(t_k)=x_{n_k},~f(0)=x$ for some sequence $t_k\to0$ and some subsequence $x_{n_k}$?
If you are happy with a sufficient condition, you can get one from the necessary and sufficient condition for the interpolation by bounded holomorphic functions on the complex unit disc $\mathbb{D}$, also known as Nevanlinna-Pick interpolation. Basically, such an $f \in H^\infty(\mathbb{D})$, with $f(t_k) = x_k$ without the need to take subsequences, exists iff the matrices $$ \begin{pmatrix} \frac{\bar{x}_{k_i} + x_{k_j}}{1-\bar{t}_{k_i} t_{k_j}} \end{pmatrix}_{i=1,\ldots,N}^{j=1,\ldots,N} $$ are positive semi-definite as a complex Hermitian matrix (since all the values are real here, the complex conjugation can be dropped) for any finite subset $(k_i)_{i=1}^N \subset \mathbb{N}$.
On the other hand, I don't know what the necessary and sufficient conditions would be if one insists on real analytic functions (meaning one doesn't put any minimal requirements on how far they need to extend into the complex plane). The following rather recent paper proves that any sequence $(x_k)$ can be interpolated (in much more generality, in fact), as long as the points $(t_k)$ form a discrete set.
Bonet, José; Domański, Paweł; Vogt, Dietmar, Interpolation of vector-valued real analytic functions, J. Lond. Math. Soc., II. Ser. 66, No. 2, 407-420 (2002). ZBL1027.46048.
However, the example $t_k \to 0$ you seem to be interested is obviously not discrete.
