Let $(A_r)_{r\geq0}$ be a one-parameter family of linear operators, with $A_r$ being the (weak) first derivative operator on $L^2(S_r)$, $S_r$ being the one-dimensional circle, with a multiplicative factor $-i$ to ensure symmetry. Each $A_r$ is known to be self-adjoint on its Hilbert space.
Intuitively, one may expect $A_r$ to "converge" to the first-derivative operator on the real line (again with a factor $-i$), say $A_\infty$, which is self-adjoint as well. However, all operators $A_r$ act on distinct Hilbert spaces, so all usual notions of operator convergence (e.g. resolvent ones), as to my knowledge, are of no use.
Is there a suitable definition of convergence of operators acting on distinct Hilbert spaces, such that $A_r\to A_\infty?$
The solution is to write down explicit expressions in coordinates. The circle of radius $r$ is parametrized as $$ r\mathbb S^1=\{(r\cos \theta, r \sin \theta)\, :\, \theta\in (-\pi, \pi)\}, $$ so the operator you refer to is $$ i\partial_\theta f := i\partial_\theta [f(r\cos \theta, r \sin \theta)], $$ where $f\in C^\infty(\mathbb R^2)$. The chain rule shows that $$i\partial_\theta= ir (-\sin \theta, \cos \theta)\cdot \nabla, $$ so, in some sense, it diverges as $r\to \infty$.
Thus, I expect that the right operator to consider is the normalized one, $\frac{i\partial_\theta}{r}.$