Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 7 / 1 / i / Solution

If $\Vert x_n^{-1}\Vert \Vert x-x_n\Vert <1$, thenis invertible by the Neumann series, a contradiction. Thus every $n$ satisfiesHere $x\ne x_n$ because $x_n$ is invertible and $x$ is not. Since $x_n\to x$, this proves the inverse norm divergence at noninvertible boundary points: