If $\Omega$ is a bounded domain in $\mathbb{R}^n$, let $\lambda(\Omega)$ be an eigenvalue of the problem $$ \begin{cases} -\Delta u=\lambda u & \mbox{in }\Omega\\ u=0 & \mbox{on }\partial\Omega. \end{cases}$$
If $A_k=\{x \in \mathbb{R}^n, \frac{1}{k} < |x|< 1 + \frac{1}{k}\}$ and $B=\{x \in \mathbb{R}^n, |x|<1\}$. Is it known whether we have the limit identity $$\lim_{k\to +\infty}\lambda(A_k)=\lambda(B)?$$
In my opinion it should work at least for the first eigenvalue $\lambda_1$ when $n\ge 3$ at least if we consider $$ A_k = \{ x\in\mathbb R^n\mid \frac1k\le |x|\le 1+\frac1k \}. $$ Since it is clear that $\lambda_1(B)\le\lambda_1(A_k)$ it suffices to prove that $\lim_k \lambda_1(A_k)\le \lambda_1(B)$.
Letting $E_k(f) = \int_{A_k}|\nabla f|^2\,dx$ be the energy associated with the operator $\Delta$ on $A_k$, by the min-max principle we know that $$ \lambda_1(A_k) = \min\left\{ \frac{E_k(f)}{\|f\|_{L^2(A_k)}^2} \mid f \in H^1_0(A_k), f\neq 0 \right\}. $$ Similarly for $\lambda_1(B)$ using the energy $E(f)=\int_B |\nabla f|^2\,dx$.
Consider the eigenfunction $\phi$ of the problem on the ball $B$ and a piece-wise smooth cut-off function $\chi_k$ such that $$ 0\le\chi_k\le1, \quad \chi|_{B\setminus A_{2k}}=0, \quad \chi|_{A_{k}}=1, \quad |\nabla \chi_k|\le k \text{ a.e.}. $$ Then, $\phi_k := \chi_k \phi\in H^1_0(A_{2k})\subset H^1_0(B)$. Exploiting the cut-off function is simple to see that $$ \|\phi-\phi_k\|_{L^2(B)}^2 \lesssim \frac1{k^n}, \quad E(\phi-\phi_k)\lesssim \frac 1 {k^{n-2}}. $$ Thus, if $n\ge 3$, $\phi_k\rightarrow \phi$ in $H^1_0(B)$. Then the claim follows by $$ \lim_k\lambda_1(A_{2k})\le \lim_{k} \frac{E_{2k}(\phi_k)}{\|\phi_k\|_{L^2(A_{2k})}^2} = \lim_{k} \frac{E(\phi_k)}{\|\phi_k\|_{L^2(B)}^2} = \frac{E(\phi)}{\|\phi\|_{L^2(B)}^2} = \lambda_1(B). $$
Maybe this idea can be generalized to higher eigenvalues by exploiting the correct min-max principle.
This is basically a comment on Dario's answer. I'm going to compare the Dirichlet problem on $B$ with the one on $B_0\equiv B\setminus\{ 0\}$ (though I'm not going to justify formally that this is what the limit in the OP will give), and Dario's argument will show that these are identical. In particular, all eigenvalues agree.
Recall that we may obtain the Dirichlet Laplacian on an open set $\Omega\subset\mathbb R^n$ from its quadratic form $$ Q(f) = \int_{\Omega} |\nabla f|^2\, dx , \quad\quad f\in H_0^1(\Omega) ; $$ in other words, $H_0^1$ is the domain of $(-\Delta_D)^{1/2}$, and $\Delta_D$ denotes the self-adjoint operator on $L^2(\Omega)$ we are interested in here (the Dirichlet Laplacian). See for example here for background.
Now in the situation we are interested in and for $n\ge 3$, we have that $$ H_0^1(B)=H_0^1(B_0) \quad\quad\quad\quad (1) $$ (as subspaces of $L^2(B)$, say).
(1) is established by the argument given by Dario. Since $H_0^1(\Omega)$ may be obtained as the closure of $C_0^{\infty}(\Omega)$ under the Sobolev norm $(Q(f)+\|f\|^2)^{1/2}$, it suffices to show that any $f\in C_0^{\infty}(B)$ can be approximated by smooth functions whose support is separated from $0$. This we do as in Dario's answer by taking $f_n(x)=\chi_n(|x|) f(x)$, with $0\le \chi_n\le 1$, $\chi_n=0$ near $0$, $\chi_n(r)=1$ for $r\ge 1/n$ and $\chi'_n\lesssim n$.
It seems that limits $\mu=\lim_{k\to\infty}\lambda(A_k)$ of eigenvalues of the domains $A_k$ should be eigenvalues for the problem $$ \begin{cases} -\Delta u=\mu u & \mbox{in }B\setminus\{0\}\\ u=0 & \mbox{on }\partial B\cup\{0\}. \end{cases} $$ In other words, the limit of the sets $A_k$ in a spectral sense should be the punctured ball, not the ball.
If you take the normalized positive first eigenfunction of the Dirichlet Laplacian on $A_k$, they all vanish at the inner boundary $\partial B(0,1/k)$, so the limit function will certainly vanish at the origin. The limit function will be strictly positive elsewhere, so it will not solve the PDE at the origin. However, the first eigenfunction in the ball $B$ (normalized and chosen positive) is strictly positive and smooth at the origin. Therefore $\lambda_1(B)<\lim_{k\to\infty}\lambda_1(A_k)$, where the subscript refers to the first eigenvalue. I guess that the same inequality holds for all eigenvalues, perhaps with the exception of eigenfunctions of the ball that vanish at the origin.
