Let be $C_n$ a circle of radius $1/n$ and center $(1/n,0)$. Let be $X_n$ the union of the first n $C_k$ and H the Hawaiian earring, i.e. the union of all $C_k$. Is true that H is both the direct limit of $X_n$ with the inclusion as transition map and the inverse limit of $X_n$ with maps that collapse the new circles to the origin?
Thanks in advance.
Interesting, this seems to be true. Let us call $X_k := \bigvee_{k = 1}^k S^1$ to be the bouquet of $k$ circles with all of the circles wedged at a specified point $w \in S^1$, and, as you described, $r_k : X_k \to X_{k-1}$ be the map given by collapsing the $k$-th circle in the bouquet. Then $\{X_k, r_k\}_{k \in \Bbb N}$ defines an inverse system of topological spaces, so $X = \varprojlim X_k$ be the inverse limit. Fix based homeomorphisms $g_j : (C_j, (0, 0)) \to (S^1, w)$ for all $j \in \Bbb N$.
Let $H \subset \Bbb R^2$ denote the Hawaiian earring, and $f_k : H \to X_k$ denote the map obtained from collapsing $H \setminus (C_1 \cup \cdots \cup C_k)$ to a point, and then sending $C_1 \cup \cdots \cup C_k$ homeomorphically to the bouquet $X_k$ by applying $g_j$ on each $C_j$. Note that $r_k \circ f_k = f_{k-1}$ is satisfied, so these family of maps $\{f_k : H \to X_k\}_{k \in \Bbb N}$ are compatible with our inverse system. By universal property of inverse limits, there is a unique continuous map $g : H \to X$ that makes the obvious diagram commute.
For any point $x \in H$, suppose $x$ belongs to some circle $C_n \subset H$. Then $f_k(x) = w$ for all $1 \leq k < n$, and $f_k(x) = g_n(x)$ for all $k \geq n$. Therefore, by the universal property, $$g(x) = (w, w, \cdots, w, g_n(x), g_n(x), \cdots) \in X \subset \prod_j X_j$$ where the first occurrence of $g_n(x)$ is at the $n$-th entry. For two points $x, y \in H$ such that $g(x) = g(y)$, by comparing the tuples in $\prod_j X_j$ as above, we see that $x$ and $y$ both belongs to some $C_m \subset H$, and $g_m(x) = g_m(y)$ implies $x = y$ as $g_m$ is a homeomorphism, which proves $g$ is an injection. Moreover every element of $X$ must be of the form above, so $g$ is also a surjection.
Therefore $g : H \to X$ is a bijective continuous map, but the Hawaiian earring $H$ is compact, and $X$, being inverse limit Hausdorff spaces, is Hausdorff. This forces $g$ to be a homeomorphism.
EDIT: As Kevin Carlson points out in the comment below, the direct limit $\varinjlim X_k$ is not the Hawaiian earring, but the infinite wedge of circles. This is because $\bigcup_k C_k$ gets the subspace topology from $\Bbb R^2$, not the final topology from the subspaces $\bigcup_{k = 1}^n C_k$, which makes the direct limit noncompact.
As an aside, I suppose this gives an example of an inverse system of path-connected compact Hausdorff spaces with path-connected inverse limit such that $\pi_1$ does not commute with $\varprojlim$, as $\pi_1(H)$ is far more complicated than $\varprojlim F_n$ (where the connecting maps $F_n \to F_{n-1}$ are given by killing the $n$-th generator). Indeed, $\pi_1(H)$ embeds in this inverse limit, but there are words which are not contained in the image, as they'd have to be realized by paths which traverse a specific circle in $H$ infinitely often; impossible as paths are continuous images of the compact interval $[0, 1]$.