= Solution
A solution reaching arbitrarily small radii must obey $\sqrt\lambda h(\mathcal M)=1+3x/2$ with $h\ge2$. Taking $x\downarrow0$ therefore requires $2\sqrt\lambda\le1$, or $\lambda\le1/4$. This upper bound is attained: at $\lambda=1/4$ the decreasing branch $0<\mathcal M<1$ has one solution for every $x>0$ and approaches Mach one as $x\to0$. Hence
$$
\boxed{\dot M_{\max}=4\pi\left(\frac14\right)G^2M^2\rho_0c_0^{-3}
=\pi G^2M^2\rho_0v_{s0}^{-3}.}
$$
The argument establishes the maximum allowed by a globally continuing steady flow, without incorrectly claiming a finite-radius <sonic point> at $\gamma=5/3$.
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