= Solution
The non-inverting input is grounded, so the finite <voltage gain> gives $V_{\rm OUT}=-AV^-$, or $V^-=-V_{\rm OUT}/A$. Infinite <input impedance> makes <Kirchhoff's current law> at the inverting input
$$
\frac{V_{\rm IN}-V^-}{R_1}+\frac{V_{\rm OUT}-V^-}{R_2}=0.
$$
Substitute $V^-=-V_{\rm OUT}/A$ and multiply by $R_1R_2$:
$$
R_2V_{\rm IN}+\left(R_1+\frac{R_1+R_2}{A}\right)V_{\rm OUT}=0.
$$
Hence the <inverting operational amplifier> has
$$
\boxed{\frac{V_{\rm OUT}}{V_{\rm IN}}=-\frac{R_2}{R_1+(R_1+R_2)/A},\qquad B=\frac{R_1+R_2}{A}.}
$$
As $A\to\infty$, the <voltage gain> tends to $-R_2/R_1$. Setting $V^-=0$ before taking that limit would discard precisely the correction being calculated.
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