Let \(f: X \to Y\) be proper with \(Y\) Noetherian, \(\mcf\) coherent on \(X\), and \(y \in Y\) with maximal ideal \(\mfm_y\). Write \(X_n\) for the subscheme cut out by \(\mfm_y^n \OO_X\). Then the completion of the stalk satisfies \[ \hat{\qty{R^i f_* \mcf}_y} \cong \varprojlim_n H^i\qty{X_n, \mcf/\mfm_y^n \mcf} . \]
The theorem identifies the completed stalk \(\hat{(R^if_*\mcf)_y}\) with the inverse limit of the cohomology groups \(H^i(X_n,\mcf/\mfm_y^n\mcf)\) over all infinitesimal thickenings \(X_n\) of the fibre. Properness is load-bearing and is what makes the inverse limit converge to something algebraic; the statement is false for open immersions.
Two corollaries are the reason it is on a syllabus. Zariski’s connectedness theorem: if \(f: X \to Y\) is proper with \(f_* \OO_X = \OO_Y\), then every fibre is connected. And the version of Zariski’s main theorem for a birational proper morphism to a normal variety: the fibres are connected, so no fibre breaks into pieces, which is what “no blowing down to two points” means in practice.