Inversion is not uniformly continuous

Idea. Uniform continuity means: for every $\eta>0$ there is $\delta>0$ with $\|A-B\|<\delta\Rightarrow\|A^{-1}-B^{-1}\|<\eta$ for *all* $A,B\in GL_n$. To refute it, exhibit pairs at distance tending to $0$ whose inverses stay far apart.

Step 1: the pairs. For $0<\varepsilon\le1$ let
$$A_\varepsilon=\operatorname{diag}(\varepsilon,1,\dots,1),\qquad B_\varepsilon=\operatorname{diag}(2\varepsilon,1,\dots,1).$$
Both are invertible.

Step 2: distances. $A_\varepsilon-B_\varepsilon=\operatorname{diag}(-\varepsilon,0,\dots,0)$, whose operator norm is $\varepsilon$. And
$$A_\varepsilon^{-1}-B_\varepsilon^{-1}=\operatorname{diag}\left(\tfrac1\varepsilon-\tfrac1{2\varepsilon},0,\dots,0\right)=\operatorname{diag}\left(\tfrac1{2\varepsilon},0,\dots,0\right),$$
with norm $\frac1{2\varepsilon}$.

Step 3: conclusion. As $\varepsilon\to0$, $\|A_\varepsilon-B_\varepsilon\|=\varepsilon\to0$ while $\|A_\varepsilon^{-1}-B_\varepsilon^{-1}\|=\frac1{2\varepsilon}\to\infty$. For $\eta=1$ no $\delta>0$ works.

Numeric check. $\varepsilon=10^{-2}$: distance $0.01$, inverse difference $50$. $\varepsilon=10^{-4}$: $10^{-4}$ and $5000$.

Remark.
- The ratio $\frac{\|A^{-1}-B^{-1}\|}{\|A-B\|}=\frac1{2\varepsilon^2}$ matches the Lipschitz constant $M^2=\varepsilon^{-2}$ of Problem 72 (up to the factor $2$).
- Inversion *is* uniformly continuous on each set $\{\sigma_{\min}\ge c\}$ with $c>0$, e.g. on the compact group $O(n)$ or on any bounded set at distance $\ge c$ from the singular matrices.
- Compare with $x\mapsto1/x$ on $(0,1)$ the case $n=1$.