= Solution
Use the positive <mass accretion rate> $\dot M=-\mathcal J$, and retain $\delta=\gamma-1$, $q=2/\beta-1/2$. Define $h=1-q\delta$, which is positive under the condition in (b). Matching the <Bernoulli function> to the reservoir and using the <polytropic equation of state> gives
$$
c_*^2=\frac{c_\infty^2}{h},\qquad
r_*=\left(\frac{A\beta h}{2c_\infty^2}\right)^{1/\beta},
\qquad
\rho_*=\rho_\infty h^{-1/\delta}.
$$
The <transonic spherical accretion rate in a power-law potential> is therefore
$$
\boxed{\dot M=
4\pi\rho_\infty
\left(\frac{A\beta}{2}\right)^{2/\beta}
c_\infty^{\,1-4/\beta}
h^{\,q-1/\delta}.}
$$
The combination $A/c_\infty^2$ has dimensions of length to the power $\beta$, so this expression has dimensions of mass per time. The <sonic point> selects the flux that connects the subsonic reservoir to the inward supersonic <transonic branch>.
For the <endpoint limits of power-law spherical accretion>, hold $A,\beta,\rho_\infty,c_\infty$ fixed. As $\delta\to0$, $\ln h=-q\delta+O(\delta^2)$, so $(q-1/\delta)\ln h\to q$. Hence
$$
\boxed{\lim_{\gamma\to1}\dot M=
4\pi\rho_\infty
\left(\frac{A\beta}{2}\right)^{2/\beta}
c_\infty^{\,1-4/\beta}e^{\,2/\beta-1/2}.}
$$
This is also obtained directly from an <isothermal equation of state>: the <Bernoulli function> becomes $u^2/2+c_\infty^2\ln(\rho/\rho_\infty)-A/r^\beta=0$, so $\rho_*/\rho_\infty=e^q$. For $\beta=1$, $A=GM$, it recovers the <isothermal Bondi accretion rate>.
For $0<\beta<4$, $q>0$ and $\gamma\uparrow f(\beta)$ means $h\downarrow0$. Since $\delta=(1-h)/q$,
$$
h^{q-1/\delta}
=\exp\left[-\frac{qh\ln h}{1-h}\right]\longrightarrow1.
$$
Thus
$$
\boxed{\lim_{\gamma\uparrow f(\beta)}\dot M=
4\pi\rho_\infty
\left(\frac{A\beta}{2}\right)^{2/\beta}
c_\infty^{\,1-4/\beta}.}
$$
The finite limiting flux accompanies $r_*\to0$ and $c_*,\rho_*\to\infty$; it does not assert a finite-radius <sonic point> at the endpoint. For $\beta=1$ this is the familiar $\gamma\uparrow5/3$ limit $\pi G^2M^2\rho_\infty/c_\infty^3$.
For completeness, the printed range $\beta\geq4$ has the endpoint $f=+\infty$. At $\beta=4$, $q=0$ and the flux is independent of $\gamma$, namely $4\pi\rho_\infty(2A)^{1/2}$. For $\beta>4$, $q<0$ and $h\sim |q|\delta$, so $\dot M\to0$ as $\gamma\to\infty$. These are the corresponding extended endpoint limits.
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